How can I prove, that the 2nd Bergman space is a Hilbert space?

Question

We consider the analytic functions on an open set $U\subset\mathbb{C}$, which are also in $L^2(U)$. I've found several posts in this topic, but all of them references to classical complex analysis results. It's obvious, that the Bergman space ($L_a^2$) is a linear manifold, so it's efficient to prove the closeness of $L_a^2(U)$ in $L^2(U)$. According to Wikipedia:

This is a consequence of the estimate, valid on compact subsets K of D, that $$\sup\limits_{z\in K}\vert f(z)\vert \le C_K \Vert f \Vert_2,$$ which in turn follows from Cauchy's integral formula. Thus convergence of a sequence of holomorphic functions in $L^2(D)$ implies also compact convergence, and so the limit function is also holomorphic.

I can't proove neither the inequality, nor the compact convergence $\Rightarrow$ closeness of $L_a^2(U)$ implication. Please help!!!

Answer 1

The value at a point of a holomorphic function is the average value of the values in a ball around it. This shows that the supremum norm os bounded by the $ L^1$ norm and, a fortiori, by the $ L^2$ norm. (You need to fill gaps, using that you only need the bound on compact subsets)

For the second question, all you need to show is that the $ L^2$ limit of holomorphic functions is holomorphic again. This holds because Cauchy's formula is sufficient for holomorphy, and you can easily show that the Cauchy formula holds for the $ L^\infty$ limit. (Note from before that you have supremum bounds even for $ L^2$ convergent functions)

Answer 2

I am writing how the inequality came. But instead of $\mathbb{D}$ I will take arbitrary open set $U\subset \mathbb{R}^n$. Let $K\subset U$ compact set and $f$ be a holomorphic function of $U$. Now, $\{B(x,r)\subset U:x\in U\}$ is an open cover of $K$. By compactness we will get finite subcover say $\{B(x_i,r_i):i=1(1)n\}$. Now, for each $B(x,r)$ by Cauchy’s integral formulae,$$|f(a)|\le \frac{1}{2\pi}\int_{|z-x|=r}\frac{|f(z)|}{|z-a|}dz\le C_r\|f\|_2$$ where $C_r=\frac1{\sqrt{2\pi r}}$ for all $a\in B(x,r)$. Hence, for all $a\in K$, $$|f(a)|\le (C_{r_1}+C_{r_2}+\dots+C_{r_n}) \|f\|_2.$$