Exercises

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.

• 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 $$z \in \overline{\Omega_R}$$.

• 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 $$\overline{D_R(z)} \subseteq \Omega_R$$, 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 $${\left\lVert {f_n - f} \right\rVert} < {\varepsilon}/R$$.

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. $$k>n$$.

#complex/exercise/completed #complex/exercise/work