Morera's Theorem

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.

By Fubini: \begin{align*} \oint_T g(z)\,dz &\coloneqq\oint_T \int_{\mathbb{R}}f(t)e^{-izt} \,dt\,dz\\ &\coloneqq\int_{\mathbb{R}}\oint_T f(t)e^{-izt} \,dz\,dt\\ &\coloneqq\int_{\mathbb{R}}f(t) \qty{ \oint_T e^{-izt} \,dz} \,dt\\ &\coloneqq\int_{\mathbb{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.

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.

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.

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 \(\mu(T)\to 0\).