Morera’s Theorem

Fix $z_0\in \Omega$ and attempt to define a primitive $F(z) \coloneqq\int_{z_0}^z f(\xi) \,d\xi$, integrating along any path connecting $z_0$ to $z$. This will be well-defined since integrating along 2 different paths $\gamma, \mu$ will yield $\int_\gamma f + \int_\mu f = \int_{\gamma \cdot \mu} f = 0$ by assumption since it bounds a closed region. Then just show $F' = f$.

Commute limit with integral and apply Morera’s theorem.

Taken from Gamelin: the proof is based on Morera’s theorem. Idea: break region into nested cubes:

Let $R$ be a closed rectangle in $D$. We subdivide $R$ into four equal subrectangles. Since the integral of $f(z)$ around $\partial R$ is the sum of the integrals of $f(z)$ around the four subrectangles, there is at least one of the subrectangles, call it $R_{1}$, for which $$\begin{align*} \left|\int_{\partial R_{1}} f(z) d z\right| \geq \frac{1}{4}\left|\int_{\partial R} f(z) d z\right| \end{align*}$$ Now subdivide $R_{1}$ into four equal subrectangles and repeat the procedure. This yields a nested sequence of rectangles $\left\{R_{n}\right\}$ such that $$\begin{align*} \left|\int_{\partial R_{n}} f(z) d z\right| \geq \frac{1}{4}\left|\int_{\partial R_{n-1}} f(z) d z\right| \geq \cdots \geq \frac{1}{4^{n}}\left|\int_{\partial R} f(z) d z\right| . \end{align*}$$ Since the $R_{n}$ ’s are decreasing and have diameters tending to 0 , the $R_{n}$ ’s converge to some point $z_{0} \in D$. Since $f(z)$ is differentiable at $z_{0}$, we have an estimate of the form $$\begin{align*} \left|\frac{f(z)-f\left(z_{0}\right)}{z-z_{0}}-f^{\prime}\left(z_{0}\right)\right| \leq \varepsilon_{n}, \quad z \in R_{n}, \end{align*}$$ where $\varepsilon_{n} \rightarrow 0$ as $n \rightarrow \infty$. Let $L$ be the length of $\partial R$. Then the length of $\partial R_{n}$ is $L / 2^{n}$. For $z$ belonging to $R_{n}$ we have the estimate $$\begin{align*} \left|f(z)-f\left(z_{0}\right)-f^{\prime}\left(z_{0}\right)\left(z-z_{0}\right)\right| \leq \varepsilon_{n}\left|z-z_{0}\right| \leq 2 \varepsilon_{n} L / 2^{n} . \end{align*}$$ From the $M L$-estimate and Cauchy’s theorem, we obtain $$\begin{align*} \begin{aligned} \left|\int_{\partial R_{n}} f(z) d z\right| &=\left|\int_{\partial R_{n}}\left[f(z)-f\left(z_{0}\right)-f^{\prime}\left(z_{0}\right)\left(z-z_{0}\right)\right] d z\right| \\ & \leq\left(2 \varepsilon_{n} L / 2^{n}\right) \cdot\left(L / 2^{n}\right)=2 L^{2} \varepsilon_{n} / 4^{n} \end{aligned} \end{align*}$$ Hence $$\begin{align*} \left|\int_{\partial R} f(z) d z\right| \leq 4^{n}\left|\int_{\partial R_{n}} f(z) d z\right| \leq 2 L^{2} \varepsilon_{n} \end{align*}$$ Since $\varepsilon_{n} \rightarrow 0$ as $n \rightarrow \infty$, we must have $$\begin{align*} \int_{\partial R} f(z) d z=0 . \end{align*}$$ By Morera’s theorem, $f(z)$ is analytic.