---
tag: BTV8962E31
kind: proof
name: Proof that holomorphic functions are infinitely differentiable
msc: [30A99]
proves: 0QEBN8DGVE
uses: [0DG0Z72CQD, 0D1C9Y1EXA, 0826SGV3R0, 0YMP521ZR5, 045EWAXY9P, 0T71SSYZ39, 05AXG3QRKR]
sources:
  - url: https://en.wikipedia.org/wiki/Holomorphic_function
    locator: Properties
---

## Proof

We use the local power series of $f$ and differentiate it term by term.

1. Fix $c \in U$. Since $U$ is open, choose $r > 0$ with $D = \set{z : \abs{z - c} < r} \subseteq U$. Since [[holomorphic functions are analytic|0826SGV3R0]], there are $a_n$ with $f(z) = \sum_{n \ge 0} a_n (z - c)^n$ for all $z \in D$.
2. Let $0 < s < r$. The series converges at $z = c + s$, so its terms tend to $0$. So $(\abs{a_n} s^n)$ is bounded. By the definition of the [[radius of convergence|045EWAXY9P]] $R$, $s \le R$. Hence $R \ge r$.
3. Let $S(z)$ be the sum of the series on its disc of convergence, which contains $D$. Since [[power series are holomorphic in the disc of convergence|0YMP521ZR5]], $S'(z) = \sum_{n \ge 1} n a_n (z - c)^{n-1}$, a power series with the same radius $R$.
4. For $z \in D$, $f$ and $S$ agree on the open set $D$ around $z$. The difference quotients at $z$ agree for small $h$, so $f'(z) = S'(z)$.
5. By step 3 and the same theorem applied to the series of $S'$, the function $S'$ is complex differentiable at each point of $D$. By step 4, $f' = S'$ on $D$, so $f'$ is [[complex differentiable|0D1C9Y1EXA]] at $c$.
6. Since $c \in U$ was arbitrary, $f'$ is [[holomorphic|0DG0Z72CQD]] on $U$. Applying this to $f', f'', \dots$ in turn, by induction each $f^{(n)}$ exists and is holomorphic on $U$.
7. Let $h = U_h + i V_h$ be holomorphic on $U$, with real and imaginary parts $U_h, V_h$. By the [[Cauchy–Riemann equations|0T71SSYZ39]], at each point the [[partial derivatives|05AXG3QRKR]] exist and $h' = (U_h)_x + i (V_h)_x$.
8. Also $(U_h)_y + i (V_h)_y = -(V_h)_x + i (U_h)_x = i h'$. So, taking partial derivatives on real and imaginary parts, $\partial h / \partial x = h'$ and $\partial h / \partial y = i h'$.
9. We show by induction on $k \ge 0$ that every partial derivative of $f$ of order $k$ exists on $U$. It equals $i^m f^{(k)}$, where $m$ counts the derivatives in $y$. For $k = 0$ the only one is $f = i^0 f^{(0)}$.
10. For the inductive step, apply step 8 to the holomorphic function $h = i^m f^{(k)}$, whose derivative is $i^m f^{(k+1)}$. We get $\partial_x h = i^m f^{(k+1)}$ and $\partial_y h = i^{m+1} f^{(k+1)}$.
11. Each $f^{(k)}$ is complex differentiable, hence continuous: $f^{(k)}(z + h) - f^{(k)}(z) = h \cdot (\text{difference quotient}) \to 0$. So all partial derivatives of $f$ of all orders exist and are continuous on $U$.

## Idea

Near each point, $f$ is a power series, and the derivative of a power series is again a power series with the same radius. So $f'$ is again holomorphic, and we can repeat forever. The Cauchy–Riemann equations turn each partial derivative into a complex derivative times a power of $i$.