Bergman Space is a Hilbert Space Conway book

bergman-spacescomplex-analysisfunctional-analysislp-spacestopological-vector-spaces

I'm trying to understand a proof in Conway's book A Course in Functional Analysis about the completeness of the Bergman Space.

He defines the Bergman space as follows.

Definition:
Let $G$ be an open set of $\mathbb{C}$ then $L_a^2(\mathbb{C})$ denotes the collection of all analytic functions $f: G \to \mathbb{C}$ such that:
$$
\int \int_G|f(x+iy)|^2dxdy < \infty
$$

$L_a^2(\mathbb{C})$ is called the Bergman space for $G$

then he notes that $L_a^2(\mathbb{C}) \subset L^2(\mu)$ where $\mu = Area|G$.

He proves the following two results.

Lemma:
if $f$ is analytic in a neighborhood of $\overline{B(a;r)}$ then
$$
f(a) = \frac{1}{\pi r^2}\int\int_{B(a;r)} f
$$

Corollary: if $f\in L_a^2(G)$ $a\in G$ and $0<r<\text{dist}(a, \partial G)$ then
$$
|f(a)|\leq \frac{1}{r\sqrt{\pi}}||f||_2
$$

Then he claims the Bergman space is a Hilbert space and for proving completeness is enough to prove that is a closed subspace of $L^2(\mu)$. Here is the problem, the proof goes as follows:

We take a sequence in $L_a^2(G)$ $\{ f_n\}$ such that $f_n \to f $, $f\in L^2(G)$. We want to show that $f \in L_a^2(G)$.

Suppose $\overline{B(a;r)} \subset G$ and $0<\rho<\text{dist}(B(a; r), \partial G)$. By the preceding corollary there is a constant C such that $|f_n(z) – f_m(z)|\leq C||f_n – f_m||_2$ for all $m, n$ and for $|z – a|\leq \rho$. Thus $\{f_n\}$ is a uniformly cauchy sequence on any closed disk in $G$

This is the last part I don't understand, how $0<\rho<\text{dist}(B(a; r), \partial G)$ implies that we can apply the last corollary for any $z$ such that $|z-a|\leq \rho$.

I tried proving that if $z$ is such that $|z-a|\leq \rho \Rightarrow 0< \rho < \text{dist}(z, \partial G)$ so I can apply the corollary to $z$ but I haven't found the way to do it.

Any help either for clarifying Conway's proof or providing another one is appreciated.

Best Answer

Note that $$\tag1 \operatorname{dist}(a,\partial G) \geq\operatorname{dist}(B(a;r),\partial G)>\rho>0. $$ From the triangle inequality, if $|z-a|\leq\rho$, $$\tag2 \operatorname{dist}(a,\partial G) \leq\operatorname{dist}(z,\partial G)+\operatorname{dist}(a,z)\leq\operatorname{dist}(z,\partial G)+\rho. $$ Now apply the Corollary to $f_n-f_m$ and $z$ with $r=\frac12\,\operatorname{dist}(z,\partial G)$, to get \begin{align} |f_n(z)-f_m(z)|&\leq\frac1{\sqrt\pi\,\frac12\,\operatorname{dist}(z,\partial G)}\,\|f_n-f_m\|_2\\[0.3cm] &\leq\frac2{\sqrt\pi\,(\operatorname{dist}(a,\partial G)-\rho)}\,\|f_n-f_m\|_2. \end{align} Now the constant does not depend on $z$.

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