I attended a presentation and I do not understand why we have the following assertion. Can someone explain to me why we have this assertion or a reference on this subject:
The space of holomorphic polynomials $\mathcal P(\mathbb C)$ is dense on the weighted Bergman space $\mathcal A^2_\alpha(\mathbb D)$ on the unit disc of the complex plane $\mathbb C$, i.e., the space of all holomorphic functions $f$ such that
$${\displaystyle \|f\|_{A_{\alpha }^{2}}:=\left((\alpha +1)\int _{\mathbb {D} }|f(z)|^{2}\,(1-|z|^{2})^{\alpha }dA(z)\right)^{1/2}<\infty}.$$
Furthermore, we have the following decomposition
$$\mathcal A^2_\alpha(\mathbb D)=\oplus_{d=0}^{\infty}\mathcal P^d(\mathbb C)$$
into a direct sum of Hilbert spaces, where $\mathcal P^d(\mathbb C)$ denotes the subspace
of homogeneous holomorphic polynomials of degree $d$.
Best Answer
This is simpler (actually "softer") than I thought.
Say $||\cdot||_\alpha$ is the weighted $L^2$ norm, and let $$w_n=||z^n||_\alpha^2.$$Say $f\in\mathcal A_\alpha^2$. We have $$f(z)=\sum_0^\infty c_nz^n,$$with uniform convergence on compact subsets of the open disk. It's enough to show the series converges in norm, since $f_n\to f$ in $L^2$ and $f_n\to g$ almost everywhere imply $f=g$.
Since the $z^n$ are orthogonal, it's enough to show that $$\sum_0^N|c_n|^2w_n\le c||f||_\alpha^2.$$But that follows since $s_N$ is the orthogonal projection of $f$ onto $\mathcal P_n$ (which in turn follows from the fact that $f-s_n\perp\mathcal P_n$.)