Show that if \(f\) has a primitive \(F\) on \(\Omega\) then \(\displaystyle\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
Let \(F\) be a primitive of \(f\), so \({\frac{\partial }{\partial z}\,}F = f\). Then \begin{align*} \int_\gamma f(z) \,dz= F(\gamma(1)) - F(\gamma(0)) = F(p) - F(p) = 0 .\end{align*} More explicitly, let \(z(t): [a, b]\to {\mathbf{C}}\) be any parameterization of \(\gamma\), then \begin{align*} \int_\gamma f(z) \,dz &= \int_a^b f(z(t)) z'(t)\,dt\\ &= \int_a^b F'(z(t))z'(t) \,dt\\ &= \int_a^b \tilde F'(t)\,dt&& \text{ where } \tilde F(t) \coloneqq F(z(t)) \text{ by the chain rule} \\ &= F(z(b)) - F(z(a)) && \text{ by FTC} \\ &= 0 ,\end{align*} since \(z(b) = z(a)\) for a closed curve.
Show that if \(f_n\to f\) locally uniformly and each \(f_n\) is holomorphic then \(f\) is holomorphic.
This is S&S Theorem 5.2. Statement: if \(f_n\to f\) uniformly locally uniformly on \(\Omega\) then \(f\) is holomorphic on \(\Omega\).
- Let \(D \subset \Omega\) with \(\overline{{\mathbb{D}}}\subset \Omega\) and \(\Delta \subset D\) be a triangle.
- Apply Cauchy-Goursat: \begin{align*} \int_\Delta f_n = 0 .\end{align*}
- \(f_n\to f\) uniformly on \(\Delta\) since it is closed and bounded and thus compact by Heine-Borel, so \(f\) is continuous and \begin{align*} \lim_n 0 = \lim_n \int_\Delta f_n = \int_\Delta \lim_n f_n \coloneqq\int_\Delta f .\end{align*}
- Apply Morera’s theorem: \(\displaystyle\int_\Delta f\) vanishes on every triangle in \(\Omega\), so \(f\) is holomorphic on \(\Omega\).
Prove that if \(f_n\to f\) locally uniformly with \(f_n\) holomorphic, then \(f_n'\to f'\) locally uniformly and \(f'\) is holomorphic.
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Simplifying step: for some reason, it suffices to assume \(f_n\to f\) uniformly on all of \(\Omega\)?
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Take \(\Omega_R\) to be \(\Omega\) with a buffer of \(R\), so \(d(z, {{\partial}}\Omega) > R\) for every \(z \in \overline{\Omega_R}\).
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It suffices to show the following bound for \(F\) any holomorphic function on \(\Omega\): \begin{align*} \sup_{z\in \Omega_R} {\left\lvert {F'(z)} \right\rvert} \leq {1\over R} \sup_{\zeta \in \Omega} {\left\lvert {F(\zeta)} \right\rvert} && \forall R ,\end{align*} where on the right we take the sup over all \(\Omega\).
- Then take \(F \coloneqq f_n-f\) and \(R\to 0\) to conclude, since the right-hand side is a constant not depending on \(\Omega_R\).
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For any \(z\in \Omega_R\), we have \(\overline{D_R(z)} \subseteq \Omega_R\), so Cauchy’s integral formula can be applied:
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\begin{align*} {\left\lvert {F'(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \int_{{{\partial}}D_R(z)} {F(\xi) \over (\xi-z)^2 } \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi} \int_{{{\partial}}D_R(z)} { { {\left\lvert {F(\xi)} \right\rvert} \over {\left\lvert {\xi-z} \right\rvert}^2 }} \,d\xi\\ &\leq {1\over 2\pi} \int_{{{\partial}}D_R(z)} { { \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} \over {\left\lvert {\xi-z} \right\rvert}^2 }} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} \int_{{{\partial}}D_R(z)} { { 1 \over R^2 }} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2} \int_{{{\partial}}D_R(z)} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2} 2\pi R \\ &\leq {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2}\qty{ 2\pi R} \\ &= {1\over R} \sup_{\zeta \in \Omega}{\left\lvert {F(\zeta)} \right\rvert} .\end{align*}
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Now \begin{align*} {\left\lVert {f_n' - f'} \right\rVert}_{\infty, \Omega_R} \leq {1\over R} {\left\lVert {f_n - f} \right\rVert}_{\infty, \Omega} ,\end{align*} where if \(R\) is fixed then by uniform convergence of \(f_n\to f\), for \(n\) large enough \({\left\lVert {f_n - f} \right\rVert} < {\varepsilon}/R\).
Suppose that \(f\) is entire and \(f\) has sublinear growth in the following sense: \begin{align*} {\left\lvert {f(z)\over z} \right\rvert}\to 0 \text{ as } {\left\lvert {z} \right\rvert}\to \infty .\end{align*} Show that \(f\) must be constant.
Claim: \(f'(z_0) = 0\) for every \(z_0\in {\mathbf{C}}\), so \(f'\equiv 0\), making \(f\) constant. Fix \(z_0\), then define \begin{align*} g(z) \coloneqq \begin{cases} {f(z) - f(0) \over z-0} & z\neq 0 \\ f'(0) & z=0. \end{cases} .\end{align*} Note that for \(z\neq 0\), \begin{align*} {\left\lvert {g(z)} \right\rvert} \coloneqq{\left\lvert {f(z) - f(0)\over z} \right\rvert} \leq {\left\lvert {f(z) \over z} \right\rvert} + {\left\lvert {f(0)\over z} \right\rvert} \overset{{\left\lvert {z} \right\rvert} \to \infty }\longrightarrow 0 ,\end{align*} where we’ve used the assumption in the last step. So for \({\left\lvert {z} \right\rvert}\geq R_{\varepsilon}\) large enough, \({\left\lvert {g(z)} \right\rvert} < {\varepsilon}\). In particular, \({\left\lvert {g(z)} \right\rvert}<{\varepsilon}\) on the circle \({\left\lvert {z} \right\rvert} = R_{\varepsilon}\), and by the MMP \({\left\lvert {g(z)} \right\rvert} < {\varepsilon}\) in the disc \({\left\lvert {z} \right\rvert}\leq R_{\varepsilon}\). Taking \({\varepsilon}\to 0\) yields \(g(z) = 0\) for all \(z\in {\mathbf{C}}\), so \(f(z) = f(0)\) is a constant for all \(z\).
Claim: \(f'(z) \equiv 0\). Choose \(R = R({\varepsilon}) \gg 1\) so that \({\left\lvert {f(z)} \right\rvert} \leq {\varepsilon}{\left\lvert {z} \right\rvert}\) for \({\left\lvert {z} \right\rvert} \geq R\), and apply Cauchy’s formula: \begin{align*} {\left\lvert {f'(z)} \right\rvert} &= {\left\lvert {{1\over 2\pi i } \int_{{\left\lvert {\xi } \right\rvert}= R} { f(\xi) \over (\xi - z)^2 }\,d\xi} \right\rvert} \\ &\leq {1\over 2\pi} \int_{{\left\lvert {\xi } \right\rvert}= R} { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^2 } \,d\xi\\ &\leq {1\over 2\pi} \int_{{\left\lvert {\xi } \right\rvert}= R} { {\varepsilon}{\left\lvert {\xi} \right\rvert} \over \qty{R - {\left\lvert {z} \right\rvert}^2 } } \,d\xi\\ &= {1\over 2\pi} \qty{{\varepsilon}R\over \qty{R-{\left\lvert {z} \right\rvert}}^2 } \cdot 2\pi R \\ &= { \mathsf{O}}\qty{{\varepsilon}R^2\over R^2} = { \mathsf{O}}({\varepsilon}) \\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 .\end{align*}
Suppose that \(f\) is entire and has polynomial growth in the following sense: \begin{align*} {\left\lvert {f(z)\over z^n} \right\rvert} \leq M \text{ for }{\left\lvert {z} \right\rvert} \geq R ,\end{align*} for some constants \(k\) and \(R\). Show that \(f\) is a polynomial of degree at most \(n\).
Since \(f\) is entire, it equals its Laurent expansion about \(z_0 = 0\), so \begin{align*} f(z) = \sum_{k\geq 0} c_k z^k, && c_k = {f^{(k)}(0)\over k! } = {1\over 2\pi i}\int_{{\left\lvert {\xi} \right\rvert} = R} {f(\xi) \over \xi^{k+1}}\,d\xi .\end{align*} A direct estimate yields \begin{align*} {\left\lvert {c_k} \right\rvert} &\leq {1\over 2\pi} \int_{{\left\lvert {\xi} \right\rvert} = R} {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &\leq {1\over 2\pi} \int_{{\left\lvert {\xi} \right\rvert} = R} {M {\left\lvert {\xi} \right\rvert}^n \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &\leq {M\over 2\pi} \int_{{\left\lvert {\xi} \right\rvert} = R} {\left\lvert {\xi} \right\rvert}^{n-(k+1)} \,d\xi\\ &\leq {M\over 2\pi} \int_{{\left\lvert {\xi} \right\rvert} = R} R^{n-(k+1)} \,d\xi\\ &= {M\over 2\pi} R^{n-k-1} \cdot 2\pi R \\ &= MR^{n-k} ,\end{align*} which converges to \(0\) as \(R\to \infty\) provided \(n-k<0\), i.e. \(k>n\).