Analyticity Implies Infinite Differentiability

Enjoy life. This is not a dress rehearsal, Friedrich Nietzsche

Introduction

A complex function $f(z)$ maps $z = x + iy \in \mathbb{C}$ to another complex number. For example: $f(z) = z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy, f(z) = \frac{1}{z}, f(z) = \sqrt{z^2 + 7}$.

A contour is a continuous, piecewise-smooth curve defined parametrically as: $z(t) = x(t) + iy(t), \quad a \leq t \leq b$.

Definition (Smooth Contour Integral). Let ᵞ be a smooth contour (a continuously differentiable path in the complex plane), $\gamma: [a, b] \to \mathbb{C}$. Let $f: \gamma^* \to \mathbb{C}$ be a continuous complex-valued function defined on the trace $\gamma^*$ of the contour (i.e. along the image of $\gamma$). Then, the contour integral of f along $\gamma$ is defined as $\int_{\gamma} f(z)dz := \int_{a}^{b} f(\gamma(t)) \gamma^{'}(t)dt$.

Properties

• Cauchy Integral Formula. If a function f is analytic in a simply connected domain D and γ is a simple closed contour, positively oriented in D. Then, for any point $z_0$ inside γ we have $f(z_0) = \frac{1}{2\pi i}\cdot \int_{\gamma} \frac{f(z)}{z-z_0}dz$ (Figure D).

• The Residue Theorem states that the integral of a function f(z) around a simple (doesn't cross itself), and positively oriented (counter-clockwise) contour γ is equal to 2πi times the sum of the residues of all singularities inside the contour, $\int_{\gamma} f(z)dz = 2\pi i \sum Res(f, z_k)$ where $z_k$ is a reside inside γ. f must be analytic inside the contour γ, except for a finite number of isolated singularities.

1. For a simple pole at a point $z_k$ (i.e., the denominator has a simple root) is calculated using the formula $Res(f, z_k) = \lim_{z \to z_k}(z-z_k)f(z)$.
2. For a pole of order m, $Res(f, z_k) = \frac{1}{(m-1)!}\lim_{z \to z_k} \frac{d^{m-1}}{dz^{m-1}}[(z-z_k)^mf(z)]$

• Proposition. Linearity of the contour integral, $\int_{\gamma} (af(z) + bg(z))dz = a\int_{\gamma} f(z)dz + b\int_{\gamma} g(z)dz$
• Proposition. Properties of contour integrals under path manipulation. Let γ be a contour, defined by a function $\gamma: [a, b] \to \mathbb{C}$ and let f be a continuous functions $f: \gamma^* \to \mathbb{C}$ defined along the image of ᵞ. Then,

1. Reversal of orientation. $\int_{-\gamma} f(z)dz = -\int_{\gamma} f(z)dz$ where $-\gamma:[a,b] \to \mathbb{C}$ is the reverse of the contour (traversing the same path in the opposite direction) defined by $(- \gamma)(t)=\gamma(a+b-t)$. This property states that reversing the direction of a contour changes or flips the sign of the integral. Intuitively, this is analogous to how reversing the limits of integration in real analysis changes the sign: $\int_a^b f(x)dx = -\int_b^a f(x)dx$. In complex analysis, the orientation of the contour matters because the integral depends on the direction in which we traverse the path.
2. Additivity under subdivision. Suppose a < c < b, let split $\gamma$ into two sub-contours $\gamma_1 = \gamma|_{[a, c]} \text{ and } \gamma_2 = \gamma|_{[c, b]}$, then $\int_{\gamma} f(z)dz = \int_{\gamma_1} f(z)dz + \int_{\gamma_2} f(z)dz$. This property states that integrating over the whole contour is the same as integrating over the pieces successively. This is the complex analogue of the additive property of definite integrals: $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx $ for a < c < b.
3. Invariance of Contour Integrals Under Reparameterization. Let $\tilde{\gamma}$ be a contour, defined by a function $\tilde{\gamma}: [c, d] \to \mathbb{C}$. If $\tilde{\gamma}$ is another parameterization of the same oriented path $\gamma$ meaning there exists a one-to-one, continuously differentiable map $\psi: [c,d]\to[a,b]$ with a positive derivative $\psi'(t)>0$ such that $\tilde{\gamma}(t) = \gamma(\psi(t))$, then $\int_{\gamma} f(z)dz = \int_{\tilde{\gamma}} f(z)dz$

• Deformation of Contours. If two contours $\gamma_1$ and $\gamma_2 $ are homotopic (i.e., one can be continuously deformed into the other without crossing any singularities of f) in a domain where f(z) is analytic, then: $\int_{\gamma_1} f(z)dz = \int_{\gamma_2} f(z)dz.$

• Fundamental Theorem of Calculus for Contours. Suppose $\gamma$ is a contour (piecewise smooth path) from a to b, f is defined on a domain D containing $\gamma^*$ (the image of $\gamma$) and admits a primitive (antiderivative) F on D (i.e., $F'(z) = f(z)$), then $\int_{\gamma} f(z)dz = F(\gamma(b)) - F(\gamma(a))$. In particular, if $\gamma$ is a closed contour (i.e., $\gamma(a)=\gamma(b)$), this integral evaluates to zero, $\int_{\gamma} f(z)dz = 0.$

• Estimation Theorem or the Triangle Inequality for Integrals. The triangle inequality for integrals in complex analysis states thatfor any continuous complex function $f:[a,b] \to \mathbb{C}$ on a closed real interval [a,b] (f(t) = u(t) + iv(t), t a real parameter), the following holds: $∣\int_a^b f(t)dt| \leq \int_a^b |f(t)|dt$.

• Estimation Lemma (ML Inequality) for contour integrals. For any continuous complex function $f:[a,b] \to \mathbb{C}$ on a closed real interval [a,b] (f(z) = u(x, y) + iv(x, y)) with f bounded by some constant M along the entire contour, |f(z)| ≤ M for all $z \in \gamma^*$ (the image/trace of the contour in the complex plane), the following holds: $∣\int_\gamma f(z)dz| \leq M \cdot l(\gamma)$ where l(γ) is the arc length of the contour γ given by $\int_a^b |\gamma^{'}(t)|dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}dt \text{ where } \gamma(t) = x(t) + iy(t)$.

Analyticity Implies Infinite Differentiability

In real-variable calculus, a function can be differentiable once, but fail to be differentiable a second time (for example, $f(x) = x|x|$ is differentiable everywhere, $f'(x) = 2|x|$, but $f''(0)$ does not exist). This theorem proves that complex analysis is entirely different: if a complex function is differentiable just once in an open set, it is infinitely smooth forever. This is one of the most striking rigidity phenomena in mathematics: holomorphic functions are, in a sense, “perfect.”

Theorem. Analyticity Implies Infinite Differentiability. Let $f$ be holomorphic (analytic) in an open set $G \subseteq \mathbb{C}$. Then:

1. $f'$ is also holomorphic in $G$.
2. $f$ has derivatives of all orders $f', f'', f''', \ldots$ in $G$, and each is holomorphic.
3. For any point $a \in G$ and any circle $\gamma$ centered at $a$ whose closed interior lies in $G$: $\boxed{f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{n+1}}\,dw, \qquad n = 0, 1, 2, \ldots}$

Proof.

We know that $f$ is analytic in an open set $G$, which means its first derivative $f'(z)$ exists. We want to prove that its second derivative $f''(a)$ also exists for any arbitrary point $a$ inside $G$. If we can prove this, the logic loops infinitely: if $f'$ has a derivative, then $f'$ is analytic, which means its derivative ($f''$) must also be analytic, and so on.

Fix an arbitrary point $a$ inside our open set $G$. Because $G$ is open, we can draw a perfectly safe, positively oriented circular contour $\gamma$ centered at $a$ with a radius of 2r, ensuring the entire closed disk $\overline{B(a; 2r)}$ lies completely inside $G$.

Let $ \gamma=\partial B(a;2r)$ (positively oriented circle). Because f is analytic in G, Cauchy’s formula for the first derivative holds on $\operatorname{Int}\gamma$: $f'(z) = \frac{1}{2\pi i}\oint_{\gamma} \frac{f(w)}{(w-z)^2}dw$.

We will evaluate it at $a$ and at a slightly perturbed point $a+h$. We choose $h$ strictly such that $|h| < r$, trapping $a+h$ safely inside the contour.

We are going to apply the same reasoning as the previous proof:

$$ \begin{aligned} \frac{f'(a+h)-f'(a)}{h} &\text{ [We apply the Cauchy’s Integral Formula for the First Derivative to the points a + h and a] } \\[2pt] &=\frac{1}{2\pi i h} \left[ \oint_{\gamma} \frac{f(w)}{(w-a-h)^2}dw - \oint_{\gamma} \frac{f(w)}{(w-a)^2}dw \right] \\[2pt] &\text{Linearity Property of Contour Integrals} \\[2pt] &=\frac{1}{2\pi i h} \oint_{\gamma}\left[ \frac{f(w)}{(w-a-h)^2} - \frac{f(w)}{(w-a)^2} \right]dw \\[2pt] &\text{Combine the fractions inside the brackets using the common denominator} \\[2pt] &=\frac{1}{2\pi i h} \oint_{\gamma} \frac{f(w)}{(w-a-h)^2 (w-a)^2} [(w-a)^2 - (w-a-h)^2]dw \\[2pt] &\text{It simplifies to } A^2 - B^2 = (A-B)(A+B) = h[2(w-a)-h] \\[2pt] &=\frac{1}{2\pi i h} \oint_{\gamma} \frac{f(w)}{(w-a-h)^2 (w-a)^2} [(-h)^2 +2wh - 2ah]dw \\[2pt] &\text{The factor h cancels:} \\[2pt] &=\frac{1}{2\pi i} \oint_{\gamma} \frac{f(w)}{(w-a-h)^2 (w-a)^2} [-h +2w - 2a]dw \\[2pt] &=\frac{1}{2\pi i} \oint_{\gamma} \frac{f(w)}{(w-a-h)^2 (w-a)^2} [2w - 2a -2h + h]dw \\[2pt] &=\frac{1}{2\pi i} \oint_{\gamma}\left[ \frac{f(w) \cdot 2(w - a -h)}{(w-a-h)^2 (w-a)^2} + \frac{f(w)h}{(w-a-h)^2 (w-a)^2}\right]dw \\[2pt] &=\frac{1}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2}{(w-a-h) \cdot (w-a)^2} + \frac{h}{(w-a-h)^2 (w-a)^2}\right]dw \text{ (i) } \end{aligned} $$

We strongly suspect the second derivative is going to be exactly what we get if we differentiate the Cauchy formula again $f''(a) = \frac{2}{2\pi i} \oint_{\gamma} \frac{f(w)}{(w-a)^3}dw$. To prove this, we subtract our expected formula from our actual difference quotient (i) and try to show that the “Error” difference shrinks to 0.

$$ \begin{aligned} |Error| &=|\frac{f'(a+h)-f'(a)}{h} -\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw| \\[2pt] &=\text{(i)} \\[2pt] &=|\frac{1}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2}{(w-a-h) \cdot (w-a)^2} + \frac{h}{(w-a-h)^2 (w-a)^2} -\frac{2}{(w-a)^3} \right]dw| \\[2pt] &\text{Combine the fractions inside the brackets using the common denominator} \\[2pt] &=|\frac{1}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2(w-a) -2(w-a-h)}{(w-a-h) \cdot (w-a)^3} + \frac{h}{(w-a-h)^2 (w-a)^2} \right]dw| \\[2pt] &=|\frac{1}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2h}{(w-a-h) \cdot (w-a)^3} + \frac{h}{(w-a-h)^2 (w-a)^2} \right]dw| \\[2pt] &\text{Combine the fractions inside the brackets using the common denominator and move h outside the integral} \\[2pt] &=|\frac{h}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2(w-a-h)+(w-a)}{(w-a-h)^2 \cdot (w-a)^3}\right]dw| \end{aligned} $$

Bounding the Error with ML-Estimation.

We must prove that as $h \to 0$, the absolute value of this Error integral is forced to zero. We apply the ML-Inequality ($| \oint f(z) dz | \le \text{Max}(|f|) \cdot \text{Length}$).

1. Contour Length L. The length of the contour $\gamma$ is $L = 2\pi (2r) = 4\pi r$.
2. Bound for |f(w)|: Since f is analytic, it is continuous on the compact (closed and bounded) contour $\gamma$. By the Extreme Value Theorem (Weierstrass Theorem), f is bounded: $|f(w)| \le M$ for some M, $M = \max_{w \in \gamma} |f(w)|$.
3. Denominator Bounds. Bound for |w - a|: For w on the circle $\gamma, w \in \gamma^*$, the distance is exactly |w - a| = 2r. Bound for |w -a -h|: We chose |h| < r. By the Reverse Triangle Inequality: $|w-a-h| \ge ||w-a| - |h|| = |2r - |h||$. Since |h| < r, then $-|h| > -r \leadsto 2r - |h| > 2r - r = r > 0, |w-a-h| \ge |2r - |h|| = 2r - |h| \gt r$.
4. Numerator Bound: $|2(w-a-h) + (w-a)| = |3(w-a) - 2h| \le |3(w-a)| + |-2h| = 3|w-a| + 2|h| < 3(2r) + 2(r) = \mathbf{8r}$

$$ \begin{aligned} |\frac{f'(a+h)-f'(a)}{h} -\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw| &=|\frac{h}{2\pi i} \oint_{\gamma} f(w)\left[ \frac{2(w-a-h)+(w-a)}{(w-a-h)^2 \cdot (w-a)^3}\right]dw| \\[2pt] &\text{By the ML-Estimation Lemma} \\[2pt] &\le \frac{|h|}{2\pi} M \frac{8r}{r^2\cdot (2r)^3} 4\pi r\\[2pt] &\text{Simplify the factors:} \\[2pt] &=\frac{|h| 32r^2}{16r^5} = \frac{|h|2M}{r^3} \end{aligned} $$

Plugging all of our maximum numerators and minimum denominators into the Error formula gives the absolute upper bound:

$0 \le |\frac{f'(a+h)-f'(a)}{h} -\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw| \le \frac{2M}{r^3}|h|$

Since $M$ and $r$ are fixed constants, as $h \to 0$, the upper bound $\frac{2M}{r^3}|h|$ approaches or collapses to $0$. $\lim_{h \to 0} \frac{2M}{r^3}|h| = \lim_{h \to 0} 0 = 0 \leadsto[\text{By the Squeeze Theorem, the Error itself must equal 0}]$

$\lim_{h \to 0} |\frac{f'(a+h)-f'(a)}{h} -\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw| = 0 \leadsto \lim_{h \to 0} \frac{f'(a+h)-f'(a)}{h} -\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw = 0$.

Since $\frac{2}{2\pi i} \oint_\gamma \frac{f(w)}{(w-a)^3}dw$ is constant related to h, the second derivative exists, and its value is $f''(a) = \frac{2}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^3}dw$

Since $f''(a)$ exists for an arbitrary point $a \in G$, $f'(z)$ is differentiable on $G$, meaning $f'$ is analytic in $G$. (This proves part (i)).

We normally use the following notation: $f''(z) = \frac{2!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-z)^3}dw, \forall z \in \mathcal{Int}(\gamma)$

(ii) We proved that if f is analytic in G then f’ is analytic in G by (i). Since f’ is analytic in G, we can replace f by f’ in the previous statement, so f’’ is analytic in G, and we can continue this process indefinitely (infinite induction), so f is infinitely differentiable in G.

Furthermore, the same Cauchy integral argument at each stage yields the general formula: $\boxed{f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{n+1}}\,dw}$ for any $n \geq 0$ and any circle $\gamma$ centered at $a$ with $\overline{B(a; R)} \subset G$. $\quad\square$

Consequences

• Cauchy’s Inequality

From the formula $f^{(n)}(a) = \frac{n!}{2\pi i}\oint_\gamma \frac{f(w)}{(w-a)^{n+1}}\,dw$ with $\gamma$ the circle of radius $R$, applying the ML inequality:

$$|f^{(n)}(a)| \leq \frac{n!}{2\pi}\cdot\frac{M}{R^{n+1}}\cdot 2\pi R = \frac{n!\,M}{R^n},$$

where $M = \max_{|w-a|=R}|f(w)|$. This is Cauchy’s inequality, which bounds derivatives of holomorphic functions and leads directly to Liouville’s theorem.

The ML-inequality states: $\left|\oint_\gamma g(w)\,dw\right| \leq M \cdot L$ where $M = \max_{w \in \gamma}|g(w)|$ and $L = \text{length}(\gamma)$. $\left|f^{(n)}(a)\right|\leq \frac{n!}{2\pi} \cdot \underbrace{\frac{M}{R^{n+1}}}_{\text{bound on } |g(w)|} \cdot \underbrace{2\pi R}_{\text{length}}$

Liouville’s theorem. A bounded entire function is constant.

Suppose $|f(z)| \leq M$ for all $z \in \mathbb{C}$. Apply Cauchy’s inequality with $n=1$ at any point $a$: $|f'(a)| \leq \frac{M}{R}$.

This holds for any $R > 0$. Let $R \to \infty$: $|f'(a)| \leq \lim_{R\to\infty}\frac{M}{R} = 0$

So $f'(a) = 0$ for all $a \in \mathbb{C}$, hence $f$ is constant. $\square$

• Taylor Series Representation.

Since all derivatives exist, $f$ has a Taylor series at every point $a \in G$: $\boxed{f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n,}$ and this series converges to $f(z)$ on the largest open disk centered at $a$ contained in $G$. In complex analysis, “holomorphic” and “analytic” (representable by a convergent power series) are equivalent — a fact that fails spectacularly in real analysis.

$$\frac{1}{w-z} = \frac{1}{w-a}\cdot\frac{1}{1-\frac{z-a}{w-a}} = \frac{1}{w-a}\sum_{n=0}^{\infty}\left(\frac{z-a}{w-a}\right)^n$$

This converges uniformly for $|z-a| < |w-a| = R$. Substitute into: $f(z) = \frac{1}{2\pi i}\oint_{C_R}\frac{f(w)}{w-z}\,dw = \sum_{n=0}^{\infty}\left(\frac{1}{2\pi i}\oint_{C_R}\frac{f(w)}{(w-a)^{n+1}}\,dw\right)(z-a)^n$

The coefficient is exactly $\frac{f^{(n)}(a)}{n!}$ by the generalized Cauchy Integral Formula.

• Holomorphic Functions Are Real-Analytic.

The real and imaginary parts $u = \operatorname{Re}(f)$, $v = \operatorname{Im}(f)$ are not just $C^\infty$ — they are real-analytic (expressible as convergent real power series in $x$ and $y$). They are also harmonic: $\nabla^2 u = \nabla^2 v = 0$.

Examples

• $f(z) = e^z$ is entire. Its derivatives are $f^{(n)}(z) = e^z$ for all $n$ — an infinite sequence of holomorphic functions, all equal.

• $f(z) = \frac{1}{1-z}$ is holomorphic on $\mathbb{C}\setminus\{1\}$. Its derivatives are $f^{(n)}(z) = \frac{n!}{(1-z)^{n+1}}$, all holomorphic on the same domain.

  $f'(z)=\frac{1}{(1-z)^2}. f''(z)=\frac{2}{(1-z)^3}. f^{(n)}(z)=\frac{n!}{(1-z)^{n+1}}$

• Real counterexample. The function $f(x) = e^{-1/x^2}$ (with $f(0) = 0$) is $C^\infty$ on $\mathbb{R}$ and all its derivatives at $x = 0$ are zero: $f^{(n)}(0) = 0$ for all $n$. However, $f$ is not identically zero near $0$, so its Taylor series (which is identically $0$) does not converge to $f$. This function is $C^\infty$ but not real-analytic. No such pathology exists in complex analysis: a holomorphic function always equals its Taylor series.

Morera’s theorem application

A sequence of holomorphic functions $\{ f_n \}$ on a domain $G \subset \mathbb{C}$ such that $f_n \to f$ converges uniformly on compact subsets of G (i.e., $\forall K \subset G$, K compact and every $\varepsilon \gt 0, \exist N, \text{ s.t. } n \ge N \implies |f_n(z) -f(z)| \lt \varepsilon, \forall z \in K$), then $f$ is holomorphic (by Morera’s theorem). By the theorem above, $f$ is then automatically $C^\infty$. Moreover, $f_n^{(k)} \to f^{(k)}$ uniformly on compact subsets for every $k$ — uniform convergence of holomorphic functions propagates to all derivatives.

Proof.

Morera’s Theorem: If $f$ is continuous on a domain $G$ and $\oint_\gamma f(z)\, dz = 0$ for every closed triangle $\gamma \subset G$, then $f$ is holomorphic on $G$.

1. $f$ is continuous. Each $f_n$ is continuous. Uniform convergence on compact sets implies the limit f is continuous on every compact subset, hence on all of G.
2. Integrals over triangles vanish. Take any closed triangle $\gamma \subset G$. Since each $f_n$ is holomorphic: $\oint_{\gamma }f_n(z)\, dz=0.$ Because $\gamma$ is compact and $f_n\rightarrow f$ uniformly on $\gamma$, we can pass the limit inside the integral: $\oint _{\gamma }f(z)\, dz=\lim _{n\rightarrow \infty }\oint _{\gamma }f_n(z)\, dz=0.$
3. Apply Morera’s theorem. If a continuous function has zero integral over every triangle, it is holomorphic. Thus, f is holomorphic on G.

Holomorphic functions are infinitely differentiable and even real‑analytic. The moment we know f is holomorphic, the entire Cauchy‑theoretic machinery becomes available.

For any closed curve $\gamma$ around z: $\boxed{f^{(k)}(z)=\frac{k!}{2\pi i}\oint _{\gamma }\frac{f(w)}{(w-z)^{k+1}}\, dw.}$ This important result gives:

• Existence of all derivatives.
• Continuity of all derivatives
• Analyticity (power series expansion).

Furthermore, uniform convergence propagates to derivatives. If $f_n\rightarrow f$ uniformly on compact sets and each $f_n$ is holomorphic, then $f_n^{(k)}\rightarrow f^{(k)}$ uniformly on compact sets for every $k\geq 0$. 🌟 Holomorphicity is stable under uniform convergence on compact sets.