Showing a subset of the Bergman space is closed The following is problem 1.10 in chapter 1 of Conway's A Course in Functional Analysis.

Let $G$ be an open subset of $\mathbb C$ and show that if $a\in G$,
  then $\{f\in L^2_a(G): f(a)=0\}$ is closed in $L^2_a(G)$.

Conway defines the space $L^2_a(G)$ to be the set of all analytic functions on $G$ with 
$$\int\int_G |f(x+iy)|^2 \ dx \ dy<\infty.$$
The expression above is the norm, and the inner product is given in the obvious way:
$$\langle f,g\rangle = \int\int_G f\overline g \ dx \ dy.$$
This space is known as the Bergman space, especially when $G$ is the open unit disk. 
 A: Let $\left\{ f_n \right\}$ be a Cauchy sequence in the given set. We need to show that it converges to another member of the set. 
Conway shows that $L^2_a(G)$ is complete, so we know the sequence converges to an analytic function $g$. (This is slightly nontrivial, so see the text for details.) We must show that $g(a)=0$. 
Because $G$ is open, there exists some closed disk centered at $a$ contained in $G$. Let this disk be called $B$, and call its radius $r$. Recall Corollary 1.12 in Conway: if $r$ is as before, then for $f$ in this space, we have 
$$|f(a)|\le \frac{1}{r \sqrt{\pi}} \|f\|_{L^2_a(G)}.$$
(For full details of the proof, refer to Conway. There are two essential steps. First, one shows that $\frac{1}{\pi r^2}\int\int_B f(z)=f(a)$. Then one can apply Cauchy-Swartz to this integral to get a bound on $f(a)$, splitting $f$ as $f\cdot 1$.)
Applying this estimate yields 
$$|f_n(a) -g(a)|= |g(a)|\le \frac{1}{r\sqrt \pi} \|f_n-g\|_{L^2_a(G)}.$$
(We use the assumption that $f_n(a)=0$.)
Take the limit as $n$ goes to infinity. The right side of the inequality converges to zero by hypothesis, since $g$ is the limit of the $f_n$ in the Hilbert space. This forces $g(a)=0$. 
