Spring 2009 - Problem 10

Solution.

$L^2\p{\D, \mu}$ is itself complete, so it suffices to show that $H$ is closed in $L^2\p{\D, \mu}$.

Let $\set{f_n}_n \subseteq H$ be a sequence converging to $f \in L^2\p{\D, \mu}$. Let $z \in \D$ and $0 < r < R$ small enough so that $\cl{B\p{z, R}} \subseteq \D$. Then by the mean value property,

Multiplying both sides by $2r$ and integrating over $\br{0, R}$, we get

Now we may apply the triangle inequality along with Cauchy-Schwarz:

We will show that $\set{f_n}_n$ is locally uniformly Cauchy: Let $0 < R < 1$ and let $0 < \delta < \operatorname{dist}\p{\cl{B\p{0, R}}, \D^\comp}$. This is positive since it is the distance between two closed, disjoint sets. Thus, for any $z \in B\p{0, R}$ we have that $B\p{z, \delta} \subseteq \cl{B\p{0, R + \delta}} \subseteq \D$, and so

and notice that $C_R$ is independent of $z$. Thus, since $\set{f_n}_n$ is Cauchy in $L^2\p{\D, \mu}$, it follows that $\set{f_n}_n$ is uniformly Cauchy on $\cl{B\p{0, R}}$. Since $R$ was arbitrary, it follows that $\set{f_n}_n$ is uniformly Cauchy on any compact set, so by completeness, there exists $\func{g}{\D}{\C}$ such that $f_n \to f$ uniformly on compact sets.

By assumption, $f_n \to f$ in $L^2\p{\D, \mu}$, so there exists a subsequence which converges pointwise to $f$ almost everywhere. Since $f_n \to g$ pointwise, it follows that $g = f$ almost everywhere, so the pointwise limit is the same thing as the $L^2\p{\D, \mu}$ limit.

Finally, since $f_n$ is holomorphic, it is continuous, which means that $f$ is continuous as well. Hence, we may apply Morera's theorem. By uniform convergence on compact sets, we get for any square $\partial R$ in $\D$ that

so $f \in H$. Consequently, $H$ is closed, hence complete.