The space $L_{2}(D_{1})$ of all analytic functions on $D_{1}$ let $D_{\rho}$ denote the closed disk of $\mathbb{C}$, $|Z-Z_{0}|\leq \rho$ ,$\rho >0$ , $z_{0} \in \mathbb{C}$ are fixed .
Assume that $f$ is analytic on $D_{\rho}$ with taylor series :
$$f(z)=\sum_{n=0}^{\infty} a_{n} (z-z_{0})^n$$
Prove that the space $L_{2}(D_{1})$ of all analytic functions f on $D_{1}$ for which 
$$||f(z) ||_{1}^{2}=\iint_{D_{\rho}}^{} |f(z)|^2 dx \ dy <\infty$$
is a Hilbert space by finding an appropriate inner product ,and find an orthonormal basis .
 A: I will prove the result for any open bounded subset of $\mathbb C$.
The inner product will be:
$$\langle f,g\rangle =\int_\Omega f\bar g.$$
Let $\Omega$ be an open bounded subset of $\mathbb C$. The space $A^2(\Omega)=\mathcal H(\Omega)\cap L^2(\Omega)$ is a Hilbert.

Lemma. Let $K\Subset \Omega$ and $f\in A^2(\Omega)$, then
$$\max_K\vert f\vert\leqslant \frac1{\sqrt \pi d(K,\partial \Omega)}\vert f_2\vert.$$

Proof.
Let $a$ be in $K$, $r$ s.t. $D(a,r)\subset \Omega$, $\rho\le r$.
We use Cauchy's formula on $C(a,\rho)$ to find $f(a)$, and with some manipulations we conclude the proof. (I will add details if you want.)

Let $(f_n)\in A^2(\Omega)^{\mathbb N}$ a Cauchy sequence and $K\Subset \Omega$.
We apply the lemma to $f_n-f_m$.
With Weierstrass' theorem, we know that the limit $f$ of this sequence is holomorphic.
Since $L^2(\Omega)$ is complete, the limit is in $L^2(\Omega)$.
Which conclude the proof that it is a Hilbert.

Hint
You can prove that 
$$(e_n)_{n\geqslant 0}$$
with
$$e_n(z)=\sqrt{\frac{n+1}\pi}z^n$$
is the orthonormal basis you are looking for.

Edit
For the details you can look here. It is in french, but since it is mostly calculations I think it is readable.
