Fall 2018 - Problem 8

Solution.

1. Let $\set{f_n}_n$ be Cauchy in $\mathcal{D}$. Notice by the mean value property, $\set{f_n'}_n$ is uniformly bounded on compact sets, i.e., $\set{f_n'}_n$ is a normal family. Hence, $f_n' \to g$ locally uniformly in $\D$ for some holomorphic $g$ on $\D$.

   Let $f\p{z} = \int_0^z g\p{w} \,\diff{w}$, where integration is done on the segment $\br{0, z}$. Then $f\p{0} = 0$ by construction and

   $$\abs{f_n\p{z} - f\p{z}} \leq \int_0^z \abs{f_n'\p{w} - f\p{w}} \,\diff{w},$$

   so $f_n \to f$ locally uniformly as well. Thus, we see that $f_n\p{0} \to f\p{0}$ and $f_n'$ converges locally uniformly to $g$, so $f_n$ converges locally uniformly to $f$ with $f' = g$. By Fatou's lemma, we also see that $\norm{f} \leq \sup_n \norm{f_n} < \infty$, since $\set{f_n}_n$ is Cauchy, so $\mathcal{D}$ is complete.

2. First, suppose that $f$ is holomorphic in $\D$ with $f\p{0} = 0$. We may then write $f\p{z} = \sum_{n=1}^\infty a_n z^n$. Then because its Taylor series converges locally uniformly,

   $$\begin{aligned} \norm{f}^2 &= \int_\D \abs{f'\p{z}}^2 \,\diff{x} \,\diff{y} \\ &= \int_\D \p{\sum_{n=1}^\infty na_n z^{n-1}} \conj{\p{\sum_{m=1}^\infty ma_m z^{m-1}}} \,\diff{x} \,\diff{y} \\ &= \int_0^1 r \int_0^{2\pi} \p{\sum_{n=1}^\infty na_n r^{n-1} e^{i\p{n-1}\theta}} \conj{\p{\sum_{m=1}^\infty ma_m r^{m-1} e^{i\p{m-1}\theta}}} \,\diff{\theta} \,\diff{r} \\ &= \sum_{n=0}^\infty 2\pi \int_0^1 n^2\abs{a_n}^2 r^{2n-1} \,\diff{r} \\ &= 2\pi \sum_{n=0}^\infty n\abs{a_n}^2. \end{aligned}$$

   Thus, $f \in \mathcal{D}$ if and only if $a_0 = 0$ and $\sum_{n=0}^\infty n\abs{a_n}^2 < \infty$.