Exercises

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 $z(t): [a, b]\to {\mathbb{C}}$ 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.

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 $\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{D}}\mkern-1.5mu}\mkern 1.5mu\subset \Omega$ 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 $\Omega$.

• Simplifying step: for some reason, it suffices to assume $f_n\to f$ uniformly on all of $\Omega$?

• Take $\Omega_R$ to be $\Omega$ with a buffer of $R$, so $d(z, {{\partial}}\Omega) > R$ for every $z \in \mkern 1.5mu\overline{\mkern-1.5mu\Omega_R\mkern-1.5mu}\mkern 1.5mu$.

• 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 $\Omega_R$.

• For any $z\in \Omega_R$, we have $\mkern 1.5mu\overline{\mkern-1.5muD_R(z)\mkern-1.5mu}\mkern 1.5mu \subseteq \Omega_R$, so Cauchy’s integral formula can be applied:

• $$\begin{align*} {\left\lvert {F'(z)} \right\rvert} &= {\left\lvert { {1\over 2\pi i} \int_{{{\partial}}D_R(z)} {F(\xi) \over (\xi-z)^2 } \,d\xi} \right\rvert} \\ &\leq {1\over 2\pi} \int_{{{\partial}}D_R(z)} { { {\left\lvert {F(\xi)} \right\rvert} \over {\left\lvert {\xi-z} \right\rvert}^2 }} \,d\xi\\ &\leq {1\over 2\pi} \int_{{{\partial}}D_R(z)} { { \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} \over {\left\lvert {\xi-z} \right\rvert}^2 }} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} \int_{{{\partial}}D_R(z)} { { 1 \over R^2 }} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2} \int_{{{\partial}}D_R(z)} \,d\xi\\ &= {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2} 2\pi R \\ &\leq {1\over 2\pi} \sup_{\zeta\in \Omega} {\left\lvert {F(\zeta)} \right\rvert} {1\over R^2}\qty{ 2\pi R} \\ &= {1\over R} \sup_{\zeta \in \Omega}{\left\lvert {F(\zeta)} \right\rvert} .\end{align*}$$

• 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 ${\left\lVert {f_n - f} \right\rVert} < {\varepsilon}/R$.

Claim: $f'(z_0) = 0$ for every $z_0\in {\mathbb{C}}$, 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 $z\in {\mathbb{C}}$, 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: $$\begin{align*} {\left\lvert {f'(z)} \right\rvert} &= {\left\lvert {{1\over 2\pi i } \int_{{\left\lvert {\xi } \right\rvert}= R} { f(\xi) \over (\xi - z)^2 }\,d\xi} \right\rvert} \\ &\leq {1\over 2\pi} \int_{{\left\lvert {\xi } \right\rvert}= R} { {\left\lvert { f(\xi) } \right\rvert} \over {\left\lvert {\xi - z} \right\rvert}^2 } \,d\xi\\ &\leq {1\over 2\pi} \int_{{\left\lvert {\xi } \right\rvert}= R} { {\varepsilon}{\left\lvert {\xi} \right\rvert} \over \qty{R - {\left\lvert {z} \right\rvert}^2 } } \,d\xi\\ &= {1\over 2\pi} \qty{{\varepsilon}R\over \qty{R-{\left\lvert {z} \right\rvert}}^2 } \cdot 2\pi R \\ &= { \mathsf{O}} \qty{{\varepsilon}R^2\over R^2} = { \mathsf{O}} ({\varepsilon}) \\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 .\end{align*}$$

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. $k>n$.