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