# Bergman space of holomorphic functions on unit disc is closed in L^2 (unit disc)

• Feb 8th 2011, 10:54 AM
Mimi89
Bergman space of holomorphic functions on unit disc is closed in L^2 (unit disc)
Hello!

Let $\displaystyle A^2 (\mathbb{D} )$ be the Bergman space of all holomorphic functions on the unit disc $\displaystyle \mathbb{D}$ which also belong to $\displaystyle L^2 ( \mathbb{D})$. Let $\displaystyle f \in A^2(\mathbb{D}), \ 0<s<1$, and $\displaystyle |z| < s$. Cauchy's Integral Formula gives:
$\displaystyle rf(z) = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(z + re^{i \theta}) r d \theta \ (0<r<1-s).$
By integrating this formula for $\displaystyle 0<r<1-s$, we are supposed to show that
$\displaystyle f(z) = \frac{1}{\pi (1-s)^2} \langle f, \chi_{\mathbb{D}(z,1-s)} \rangle_{L^2 (\mathbb{D})}$,
where $\displaystyle \mathbb{D} (z, 1-s)$ is the disc of radius $\displaystyle 1-s$ with centre $\displaystyle z$.
From this, we should deduce that $\displaystyle |f(z)| \leq \frac{\| f \| _L^2 (\mathbb{D})}{(1-s)\surd{ \pi} }$
From this I know how to deduce that $\displaystyle A^2 (\mathbb{D})$ is closed in $\displaystyle L^2 (\mathbb{D})$; it's just the above calculations which confuse me.