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

Hello!

Let $A^2 (\mathbb{D} )$ be the Bergman space of all holomorphic functions on the unit disc $\mathbb{D}$ which also belong to $L^2 ( \mathbb{D})$. Let $f \in A^2(\mathbb{D}), \ 0, and $|z| < s$. Cauchy's Integral Formula gives:

$
rf(z) = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(z + re^{i \theta}) r d \theta \ (0$

By integrating this formula for $0, we are supposed to show that
$f(z) = \frac{1}{\pi (1-s)^2} \langle f, \chi_{\mathbb{D}(z,1-s)} \rangle_{L^2 (\mathbb{D})}
$
,

where $\mathbb{D} (z, 1-s)$ is the disc of radius $1-s$ with centre $z$.

From this, we should deduce that $|f(z)| \leq \frac{\| f \| _L^2 (\mathbb{D})}{(1-s)\surd{ \pi} }$

From this I know how to deduce that $A^2 (\mathbb{D})$ is closed in $L^2 (\mathbb{D})$; it's just the above calculations which confuse me.