Morera's Theorem

Morera's theorem provides a powerful converse to the Cauchy-Goursat theorem: if the integral of a continuous function over every closed curve vanishes, then the function must be holomorphic.

Statement

Theorem3.11Morera's theorem

Let $f$ be continuous on a domain $D \subseteq \mathbb{C}$ . If

$\oint_T f(z)\,dz = 0$

for every triangle $T$ in $D$ (or equivalently, for every closed contour in $D$ ), then $f$ is holomorphic on $D$ .

RemarkStrength of the conclusion

The assumption is merely continuity plus vanishing integrals, yet the conclusion gives holomorphicity — and hence infinite differentiability and analyticity. This reflects the rigidity of holomorphic functions: the integral condition alone forces a continuous function to be complex-analytic.

Proof

Fix $z_0 \in D$ and let $r > 0$ be such that $D(z_0, r) \subset D$ .

Step 1. Define $F(z) = \int_{z_0}^z f(w)\,dw$ where the integral is taken along the line segment from $z_0$ to $z$ (valid for $z \in D(z_0, r)$ ). The hypothesis that $\oint_T f\,dz = 0$ for every triangle ensures this definition is path-independent.

Step 2. We show $F'(z) = f(z)$ . For small $|h|$ :

$\frac{F(z+h) - F(z)}{h} - f(z) = \frac{1}{h}\int_z^{z+h} [f(w) - f(z)]\,dw.$

$\left|\frac{F(z+h) - F(z)}{h} - f(z)\right| \leq \frac{1}{|h|} \cdot \varepsilon \cdot |h| = \varepsilon.$

Therefore $F'(z) = f(z)$ .

Step 3. Since $F$ is holomorphic (it has a complex derivative $f$ ), and $f = F'$ is the derivative of a holomorphic function, $f$ is also holomorphic on $D(z_0, r)$ . Since $z_0$ was arbitrary, $f$ is holomorphic on $D$ . $\blacksquare$

■

Applications

ExampleUniform limits of holomorphic functions

Morera's theorem implies that a uniform limit of holomorphic functions is holomorphic. If $f_n \to f$ uniformly on compact subsets of $D$ and each $f_n$ is holomorphic, then for any triangle $T$ in $D$ :

$\oint_T f\,dz = \lim_{n \to \infty} \oint_T f_n\,dz = 0.$

Since $f$ is continuous (uniform limit of continuous functions), Morera's theorem gives that $f$ is holomorphic.

ExampleIntegrals depending on a parameter

Consider $F(z) = \int_0^1 \frac{g(t)}{t - z}\,dt$ for $z \notin [0,1]$ , where $g$ is continuous. Then $F$ is continuous, and for any closed triangle $T$ not meeting $[0,1]$ :

$\oint_T F(z)\,dz = \int_0^1 g(t) \left(\oint_T \frac{dz}{t-z}\right) dt = 0$

by Fubini's theorem and the fact that $1/(t-z)$ is holomorphic in $z$ for $z \notin [0,1]$ . By Morera's theorem, $F$ is holomorphic on $\mathbb{C} \setminus [0,1]$ .

Variants and Strengthenings

RemarkWeakened hypotheses

Several strengthenings of Morera's theorem exist:

It suffices to check $\oint_T f\,dz = 0$ for triangles with sides parallel to the coordinate axes.
Painleve's theorem: If $f$ is continuous on $D$ , holomorphic on $D \setminus \gamma$ for a smooth curve $\gamma$ , then $f$ is holomorphic on all of $D$ .
If $f$ is continuous on $D$ and holomorphic on $D \setminus E$ where $E$ has zero area, then $f$ is holomorphic on $D$ .

These removability results are fundamental in the theory of analytic continuation.

RemarkComparison with real analysis

In real analysis, there is no analogue of Morera's theorem. A continuous real function whose integral over every interval is zero must be zero, but this does not force differentiability. The complex-analytic setting is far more rigid.