Morera's Theorem, Gauss’s Mean Value Theorem & Cauchy Estimates

It ain’t about how hard you can hit. Its about how hard you can get hit, and how much you can take, and keep moving forward, Rocky Balboa

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 $\int_a^b |\gamma^{'}(t)|dt = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}dt \text{ where } \gamma(t) = x(t) + iy(t)$.

Morera’s theorem

Morera’s theorem. Suppose that f is a continuous function on an open set $G \subseteq \mathbb{C}$. If the contour integral of f is zero for every closed contour $\gamma$ in G, i.e., $\oint_\gamma f(z)dz = 0$. Then, f is analytic (holomorphic) in G.

Morera’s Theorem is the converse of Cauchy’s (or Cauchy-Goursat) Theorem. Cauchy’s Theorem states that if f is analytic, any closed contour integral having f as an integrand is zero. In other words, analyticity implies vanishing closed contour integrals. Morera’s Theorem flips this logic. It states that if all closed contour integrals vanish and the function is continuous, then the function must be analytic. Together, they provide a complete characterization: a continuous function on an open set is analytic if and only if all its closed contour integrals vanish.

It provides a way to prove analyticity without checking the Cauchy–Riemann equations or computing derivatives directly.

Proof.

1. Construct an antiderivative.
   Let $a \in G$ be an arbitrary element in G. Since G is open, there exists a radius r such that $\overline{B(a; r)} \subseteq G$.
   By the antiderivative theorem, since $\oint_\gamma f(z)dz = 0$ for every closed contour $\gamma$ in G, f has an antiderivative function, i.e., there exists a function F defined in G, F: $G \to \mathbb{C}$ such that F’(z) = f(z).
   Explicitly, fixing a point $a \in G$, and defining: $F(z) = \int_a^z f(w)dw$ (along any path in G from a to z).

   The Antiderivative Theorem (Path Independence) states that a continuous function f in a domain G has an antiderivative F in G (meaning F’(z) = f(z)) if and only if the integral of f is path-independent in G (which is equivalent to $\oint_{\gamma} f(z)dz = 0$ for every closed contour $\gamma$ in G).

2. F is analytic. Since F is complex differentiable at every point of G (its derivative F′(z) = f(z) exists everywhere in G), F is holomorphic on G.
3. f is analytic. Since f is the derivative of an analytic function F, by Analyticity Implies Infinite Differentiability, (i) [Suppose that f is analytic (or holomorphic) in an open set G. Then, (i) f' is also analytic in G; (ii) f has derivatives of all orders ($f', f'', f''', \dots$) in G] f is also analytic (the roles of f and F are reversed compared to typical applications, F is analytic in G ⇒ the derivative of F, which is F’(z) = f(z), must also be analytic in G).

Triangles suffice. In practice, it is enough to verify that $\oint_T f(z)dz = 0$ for every triangle T in G. The triangle version implies the general version because any closed contour integral can be approximated by sums of integrals over triangles.

Continuity is essential. Without the continuity hypothesis, the conclusion can fail. There exist discontinuous functions for which all contour integrals vanish.

Application: Limits of analytic functions. A classical use of Morera’s theorem: if $\{f_n\}$ is a sequence of analytic functions converging uniformly to $f$, $f_n\rightarrow f$ on compact subsets of $G$, then $f$ is analytic. Proof:

1. Uniform convergence on compact sets implies continuity of the limit f (analytic functions are continuous, and uniform limits of continuous functions are continuous).
2. For any closed contour $\gamma \subset G: \oint_{\gamma }f(z)\, dz = \lim_{n\rightarrow \infty } \oint_{\gamma }f_n(z)\, dz$. Uniform convergence allows interchanging limits and integrals.
3. Each $f_n$ is analytic, so by Cauchy’s theorem: $\oint_\gamma f\,dz = \lim \oint_\gamma f_n\,dz = 0$.
4. Therefore, $\oint_{\gamma }f(z)\, dz=0.$
5. Finally, apply Morera’s theorem: Since f is continuous and all contour integrals vanish, f is analytic.

Gauss’s mean value theorem

Gauss’s mean value theorem. If f(z) is analytic on and inside a circle $\gamma$ with center $z_0$ and radius R, $\gamma(z_0; R)$, then$\boxed{f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0 + Re^{i\theta})\,d\theta.}$

This theorem states that the value of an analytic function at the center of a circle $f(z_0)$ is precisely the average value of $f(z)$ on the circle $\gamma$. This is a purely complex-analytic phenomenon — it reflects the rigidity of holomorphic functions and has no analogue for general smooth real functions.

Proof

We begin with the Cauchy Integral Formula, which is the source and core of this theorem. It tells us the value of an analytic function at a point $z_0$ based on its values on a contour $\gamma$ enclosing it, equals:

$$\begin{aligned} f(z_0) &=\frac{1}{2\pi i}\int_{\gamma(z_0; R)} \frac{f(w)}{w-z_0}dw \\[2pt] &\text{Parametrize the Contour} \gamma(z_0; R): w = z_0 + Re^{i\theta}, 0 \le \theta \le 2\pi \leadsto dw = iRe^{i\theta}d\theta \\[2pt] &= \frac{1}{2\pi i}\int_0^{2\pi} \frac{f(z_0 + Re^{i\theta})}{Re^{i\theta}}iRe^{i\theta}d\theta \\[2pt] &\text{This looks complex, but it simplifies beautifully.} \\[2pt] &=\frac{1}{2\pi}\int_0^{2\pi}f(z_0 + Re^{i\theta})d\theta \end{aligned} $$

Connection to the Mean Value Property of Harmonic Functions

Let $f = u + iv$ be analytic in a domain D containing the disk $|z - z_0| \le R$, then $u$ and $v$ are harmonic (satisfy Laplace’s equation: $\nabla^2 u = \nabla^2 v = 0$).

For $f$ analytic on and inside the circle $C_R(z_0) = \{z : |z - z_0| = R\}$: $f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0 + Re^{i\theta})\, d\theta$. This states that the value at the center equals the average of values on the surrounding circle.

Decompose into Real and Imaginary Parts, $u(z_0) + iv(z_0) = \frac{1}{2\pi}\int_0^{2\pi} \left[u(z_0 + Re^{i\theta}) + iv(z_0 + Re^{i\theta})\right] d\theta$

By linearity of integration and uniqueness of complex decomposition: $\boxed{u(z_0) = \frac{1}{2\pi}\int_0^{2\pi} u(z_0 + Re^{i\theta})\, d\theta}$, $\boxed{v(z_0) = \frac{1}{2\pi}\int_0^{2\pi} v(z_0 + Re^{i\theta})\, d\theta}$

This is the mean value property for harmonic functions, which we encountered earlier. Gauss’s theorem for analytic functions is the complex-valued version that implies it.

Application: Bounding Derivatives on the Unit Disk

• Let f(z) be analytic on the unit disk and suppose that $|f(z)| \le \frac{1}{1-|z|}$ for each z ∈ D. Show that $|f^{(n)}(0)| \le (n+1)!e$

Solution.

1. Choose a contour.
   The core idea is to find a contour, so that the n-th derivative of our function at zero can be bounded appropriately.
   Consider the contour $\gamma$ is a circle centered at the origin with radius $1 - \frac{1}{k}$ (a radius that is less than 1), $\forall z \in \gamma, |z| = 1 - \frac{1}{k} \leadsto |f(z)| \le \frac{1}{1-|z|} = \frac{1}{1-(1-\frac{1}{k})} = k$.
2. By the Cauchy Integral Formula for Derivatives (the order is n, the singularity is zero). It connects the n-th derivative at a point “a” (a = 0) to an integral around that point $f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z-a)^{n+1}}dz, f^{(n)}(0) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{f(z)}{z^{n+1}}dz$
3. By the ML-Inequality (Estimation Lemma), which bounds the magnitude of an integral: $|\oint_{\gamma} g(z) dz| \le M \cdot L$, where L is the path length and M is the max value of $|g(z)|$ on the path, $|f^{(n)}(0)| \le \frac{n!}{2\pi} \oint_{\gamma} \frac{|f(z)|}{|z^{n+1}|}|dz| \le \frac{n!}{2\pi} \oint_{\gamma} \frac{k}{(1-\frac{1}{k})^{n+1}}|dz|$
4. Parametrize $\gamma, z = re^{i\theta}, r = 1-\frac{1}{k}, 0 \le \theta \le 2\pi$. Then, $dz = rie^{i\theta}d\theta, |dz| = r|d\theta| = (1-\frac{1}{k})d\theta$
   Therefore, $|f^{(n)}(0)| \le \frac{n!}{2\pi}\int_0^{2\pi} \frac{k}{(1-\frac{1}{k})^{n+1|}}(1-\frac{1}{k})d\theta = \frac{n!}{2\pi}\int_0^{2\pi} \frac{k}{(1-\frac{1}{k})^n}d\theta = \frac{n!}{2\pi}\int_0^{2\pi} \frac{k^{n+1}}{(k-1)^n}d\theta = \frac{n!}{2\pi}\int_0^{2\pi} (\frac{k}{k-1})^n k d\theta$
5. Optimize the choice of $k$.
   We have an inequality that depends on k. To get the tightest possible bound, we need to choose the value of k that minimizes and simplifies this expression. Let’s make the choice: k = n + 1.
   $|f^{(n)}(0)| \le \frac{(n+1)!}{2\pi}\int_0^{2\pi} (\frac{n + 1}{n})^n d\theta = \frac{(n+1)!}{2\pi}(1+\frac{1}{n})^n \int_0^{2\pi}d\theta = \frac{(n+1)!}{2\pi}(1+\frac{1}{n})^n 2\pi = (n+1)!(1+\frac{1}{n})^n$
6. Use the classical limit.
   It is a well-know result that $(1+\frac{1}{n})^n$ is an increasing sequence of real number whose limit as $n \to \infty$ is e. Because the sequence increases towards $e$, for all $n \geq 1$: $\left(1 + \frac{1}{n}\right)^n \leq e.$ Substituting this into our inequality gives the final answer: $\boxed{|f^{(n)}(0)| \leq (n+1)!\,e.} \quad\square$

Remark. The key technique — choosing the radius of the contour to optimize the ML bound — is a standard and powerful method in complex analysis.