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Holomorphic functions are real analytic and smooth in their two real coordinates

Statement

Let $U \subseteq C$ be open and let $f = u + i v : U \to C$ be holomorphic. Under the coordinate identification $C = R^{2}$ , the map $(u, v) : U \to R^{2}$ is real analytic in the sense of Real-analytic maps between open subsets of the coordinate plane and is of class $C^{k}$ for every natural $k$ , hence smooth.

Facts & Assumptions

Given: The identification of the complex plane with the real coordinate plane from $C$ is the real coordinate plane, with coordinate arithmetic, an open set $U$ , and a holomorphic function $f = u + i v$ on $U$ .

[L1]

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

[L2]

For complex $z, w$ and a natural $n$ , $(z + w)^{n} = \sum_{p \leq n} (\binom{n}{p}) z^{p} w^{n - p}$ , with each binomial coefficient regarded as a complex scalar (The binomial theorem over the complex field).

[L3]

A smooth planar map is real analytic when each component equals its total-degree Taylor series on a neighbourhood of every point (Real-analytic maps between open subsets of the coordinate plane).

[L4]

A holomorphic function has complex derivatives of every natural order locally (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

[L5]

If $f = u + i v$ is complex differentiable, then $f^{'} = u_{x} + i v_{x} = v_{y} - i u_{y}$ and the Cauchy–Riemann equations hold (Complex differentiability is equivalent to real total differentiability together with a complex-linear derivative, with $\partial_{\overset{ˉ}{z}} f = 0$ , or with the Cauchy–Riemann equations).

[L6]

A real function is $C^{k}$ when every coordinate-derivative word of length at most $k$ , including the word of length zero, exists and is continuous ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[L7]

Every complex power series converges absolutely and uniformly on closed subdiscs strictly inside its disc of convergence (A complex power series converges absolutely and uniformly on every closed subdisc strictly inside its disc of convergence).

[L8]

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

[L9]

If $p \leq n$ , then the complex scalar corresponding to $(\binom{n}{p})$ is $n! / (p! (n - p)!)$ ( $(\binom{n}{k}) k! (n - k)! = n!$ for $k \leq n$ ; hence $(\binom{n}{k}) k! = n^{\underline{k}}$ , the quotient $n! / (k! (n - k)!)$ is a natural number, and $(\binom{n}{k}) = (\binom{n}{n - k})$ ).

Proof

technique · direct

1.1L1

Fix $a \in U$ . By [L1], there is $R > 0$ such that $f (a + h) = \sum_{n \geq 0} c_{n} h^{n}$ for $∣ h ∣ < R$ , where $c_{n} = f^{(n)} (a) / n!$ .

2.1step 1.1L2L7algebra

Write $h = x + i y$ . By [L2], $h^{n} = \sum_{p \leq n} (\binom{n}{p}) x^{p} (i y)^{n - p}$ ; if $∣ x ∣ + ∣ y ∣ < R$ , [L7] and the binomial identity give absolute convergence of the resulting total-degree series because the sum of the absolute values in degree $n$ is $∣ c_{n} ∣ (∣ x ∣ + ∣ y ∣)^{n}$ .

3.1step 1.1step 2.1L4L5L9algebra

From [L5], $\partial_{x} f = f^{'}$ and $\partial_{y} f = i f^{'}$ ; induction using [L4] therefore gives $\partial_{x}^{p} \partial_{y}^{q} f (a) = i^{q} f^{(p + q)} (a)$ . Taking $n = p + q$ in step 2.1 and using [L9], the coefficient of $x^{p} y^{q}$ is $c_{p + q} (\binom{p + q}{p}) i^{q} = D^{(p, q)} f (a) / (p! q!)$ .

4.1step 2.1step 3.1

Taking real and imaginary parts in the absolutely convergent expansion of step 2.1, and using the coefficient identification of step 3.1, gives the total-degree Taylor series of $u$ and $v$ on $∣ x ∣ + ∣ y ∣ < R$ .

5.1step 3.1step 4.1L3L4L5L6L8∎

More generally, every coordinate-derivative word with $p$ occurrences of $x$ and $q$ occurrences of $y$ is the corresponding real or imaginary component of $i^{q} f^{(p + q)}$ ; [L4] makes the next complex derivative exist, [L8] makes every $f^{(p + q)}$ continuous, and the word of length zero is $f$ itself, so [L6] makes both components $C^{k}$ for every natural $k$ . Thus the map is smooth, and step 4.1 now satisfies the opening hypothesis of [L3], proving real analyticity as well.

Holomorphic functions are real analytic and smooth in their two real coordinates

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources