# Morera's Theorem

## Uniform limit theorem #complex/exercise/completed

Suppose $$\left\{{f_n}\right\}_{n\in {\mathbb{N}}}$$ is a sequence of analytic functions on $${\mathbb{D}}\coloneqq\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert} < 1}\right\}$$.

Show that if $$f_n\to g$$ for some $$g: {\mathbb{D}}\to {\mathbf{C}}$$ uniformly on every compact $$K\subset {\mathbb{D}}$$, then $$g$$ is analytic on $${\mathbb{D}}$$.

By Morera’s theorem, it suffices to show $$\int_T f = 0$$ for all triangles $$T \subseteq {\mathbb{D}}$$. Noting that $$T$$ is closed and bounded and thus compact, $$f_n\to g$$ uniformly on $$T$$. Since the $$f_n$$ are holomorphic, $$\int_T f_n = 0$$ for all $$n$$, and thus \begin{align*} \int_T g = \int_T \lim f_n = \lim_n \int_T f_n = \lim_n 0 = 0 ,\end{align*} where moving the limit through the integral is justified by uniform convergence.

## Fourier transforms are entire #complex/exercise/completed

Suppose that $$f: {\mathbf{R}}\to{\mathbf{R}}$$ is a continuous function that vanishes outside of some finite interval. For each $$z\in {\mathbf{C}}$$, define \begin{align*} g(z) = \int_{-\infty}^\infty f(t) e^{-izt} \,dt .\end{align*}

Show that $$g$$ is entire.

By Fubini: \begin{align*} \oint_T g(z)\,dz &\coloneqq\oint_T \int_{\mathbf{R}}f(t)e^{-izt} \,dt\,dz\\ &\coloneqq\int_{\mathbf{R}}\oint_T f(t)e^{-izt} \,dz\,dt\\ &\coloneqq\int_{\mathbf{R}}f(t) \qty{ \oint_T e^{-izt} \,dz} \,dt\\ &\coloneqq\int_{\mathbf{R}}f(t) \cdot 0 \,dt\\ &= 0 ,\end{align*} where the inner integral vanishes because $$z\mapsto e^{-izt}$$ is entire by Goursat’s theorem. So $$g$$ is entire by Morera.

## Tie’s Extra Questions: Fall 2009, Fall 2011 #complex/exercise/completed

Let $$f(z)$$ be analytic in an open set $$\Omega$$ except possibly at a point $$z_0$$ inside $$\Omega$$. Show that if $$f(z)$$ is bounded in near $$z_0$$, then $$\displaystyle \int_\Delta f(z) dz = 0$$ for all triangles $$\Delta$$ in $$\Omega$$.

Write $${\mathbb{D}}_{\varepsilon}(z_0)$$ for a disc in which $$f$$ is bounded, say by $${\left\lvert {f} \right\rvert}\leq M$$ here. Note that if $$z_0$$ is not in the region enclosed by $$\Delta$$, then $$\int_\Delta f = 0$$ since $$f$$ is holomorphic throughout $$\Delta$$, so suppose $$z_0$$ is in this region.

Since $$f$$ is holomorphic in $$\Omega\setminus\left\{{ z_0 }\right\}$$, $$\Delta$$ can be deformed to $${{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)$$ without changing the integral. So \begin{align*} {\left\lvert { \oint_\Delta f } \right\rvert} &= {\left\lvert {\oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} f } \right\rvert} \\ &\leq \oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} {\left\lvert {f} \right\rvert} \\ &\leq \oint_{{{\partial}}{\mathbb{D}}_{\varepsilon}(z_0)} M \\ &= M\cdot 2\pi {\varepsilon}\\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 ,\end{align*} noting that the bound $$M$$ is constant and remains an upper bound on smaller discs by the maximum modulus principle.

## Fall 2021.2 #complex/qual/completed

Let $$\gamma(t)$$ be a piecewise smooth curve in $$\mathbb{C}, t \in[0,1]$$. Let $$F(w)$$ be a continuous function on $$\gamma$$. Show that $$f(z)$$ defined by \begin{align*} f(z):=\int_{\gamma} \frac{F(w)}{w-z} d w \end{align*} is analytic on the complement of the curve $$\gamma$$.

By Morera’s theorem, it suffices to show $$\int_\Delta f(z) \,dz= 0$$ for all triangles $$\Delta \subseteq \gamma^c$$. Claim: \begin{align*} \int_\Delta f(z) \,dz &= \int_\Delta \int_\gamma {F(w) \over w-z} \,dw\,dz\\ &= \int_\gamma \int_\Delta {F(w) \over w-z} \,dz\,dw\\ &= \int_\gamma F(w) \qty{ \int_\Delta {1 \over w-z} \,dz} \,dw\\ &= \int_\gamma F(w) \cdot 0 \,dw\\ &= 0 .\end{align*}

That the inner integral is zero follows from the fact that the function $$z\mapsto {1\over w-z}$$ is holomorphic on $$\gamma^c$$, since it has only a simple pole at $$w$$ where $$w\in \gamma$$ is fixed.

That the interchange of integrals is justified follows from Fubini’s theorem: these are continuous functions on compact sets, which are uniformly bounded and thus Lebesgue measurable and integrable.

The claim is that $$f$$ is complex differentiable, thus smooth, thus holomorphic and equal to its Taylor series expansion. The quick justification: \begin{align*} {\frac{\partial }{\partial z}\,} f(z) &= {\frac{\partial }{\partial z}\,} \int_\gamma {F(w) \over w-z}\,dw\\ &= \int_\gamma {\frac{\partial }{\partial z}\,} {F(w) \over w-z} \,dw\\ &= \int_\gamma {F(w) \over (w-z)^2} \,dw ,\end{align*} where differentiating through the integral is justified since the integrand is a continuous function of $$z$$ on $$\gamma$$ since $$w\neq z$$ on $$\gamma$$, and $$\gamma$$ is a compact set.

Slightly more rigorously, one can equivalently pass a limit through the integral to show that the defining limit exists: \begin{align*} f(z+h) - f(z) &= \int_\gamma {F(w) \over w+h-z} \,dw- \int_\gamma {F(w) \over w-z}\,dw\\ &= \int_\gamma {(w-z)F(w) - (w+h-z)F(w) \over (w+h-z)(w-z) } \,dw\\ &= \int_\gamma F(w) {h \over (w+h-z)(w-z)} \,dw\\ &\overset{h\to 0}\longrightarrow \int_\gamma {F(w) \over (w-z)^2}\,dw ,\end{align*} since the term involving $$h$$ goes to 1.

## Spring 2020 HW 2, SS 2.6.6 #complex/exercise/completed

Suppose that $$f$$ is holomorphic on a punctured open set $$\Omega\setminus\left\{{w_0}\right\}$$ and let $$T\subset \Omega$$ be a triangle containing $$w_0$$. Prove that if $$f$$ is bounded near $$w_0$$, then $$\int_T f(z) ~dz = 0$$.

Without loss of generality assume $$w_0 = 0$$. If $${\left\lvert {f(z)} \right\rvert} \leq M$$ for $${\left\lvert {z} \right\rvert} < {\varepsilon}$$, pick $$T$$ contained in $${\mathbb{D}}_{\varepsilon}$$, then \begin{align*} {\left\lvert {\oint_T f(z)\,dz} \right\rvert} \leq \oint_T {\left\lvert {f(z)} \right\rvert}\,dz\leq \oint_T M \,dz= M \mu(T) ,\end{align*} and by homotopy invariance, this integral doesn’t change as $$T$$ is perturbed. Shrinking $$T$$, we can make $$\mu(T)\to 0$$.

#complex/exercise/completed #complex/qual/completed