# Morera’s Theorem

If $$f$$ is continuous on a domain $$\Omega$$ and $$\int_T f = 0$$ for every triangle $$T\subset \Omega$$, then $$f$$ is holomorphic.

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 $$F' = f$$.

If every integral along a triangle vanishes, implies holomorphic. Equivalently, $$f(z)\,dz$$ is a closed differential form iff $$f$$ is holomorphic.

Sometimes stated for rectangles with sides parallel to axes. The power of this theorem is that virtually no assumptions on $$f$$ are made, e.g. $$f$$ is not even assumed smooth or even differentiable.

If $$\left\{{ f_n }\right\}_{n\in {\mathbb{N}}}$$ is a holomorphic sequence on a region $$\Omega$$ which uniformly converges to $$f$$ on every compact subset $$K \subseteq \Omega$$, then $$f$$ is holomorphic, and $$f_n' \to f'$$ uniformly on every such compact subset $$K$$.

Commute limit with integral and apply Morera’s theorem.

This can be applied to series of the form $$\sum_k f_k(z)$$.

Holomorphic implies analytic.

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.

## Exercises

For $$\Omega\subseteq{\mathbf{C}}$$, show that $$A({\mathbf{C}})\coloneqq\left\{{f: \Omega \to {\mathbf{C}}{~\mathrel{\Big\vert}~}f\text{ is holomorphic, bounded}}\right\}$$ is a Banach space.

Hint: Apply Morera’s Theorem and Cauchy’s Theorem

#complex/exercise/work