Hello!

Could you please help with the following question:

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<s<1 , and |z| < s. Cauchy's Integral Formula gives:

<br />
rf(z) = \frac{1}{2 \pi} \int_{0}^{2 \pi} f(z + re^{i \theta}) r d \theta \ (0<r<1-s).<br />

By integrating this formula for  0<r<1-s , 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})}<br />
,

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.