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[EDIT by YC: the original question's title asked about a basis for the Hardy space on the disk. It is clear from the actual question that what was meant was the Bergman space.]

In arXiv:0310.5297, Yuval Peres and Balin Virag study the roots of random power series, $f(z) = \sum a_n z^n$ (where the $a_n$ are Gaussian with mean 0 and variance 1) and show that correlations of the roots are determined by the Bergman kernel $$ \rho(z_1, \dots, z_n) = \frac{1}{\pi^n} \det\left[ \frac{1}{1 - z_i \overline{z_j}} \right]_{i,j}$$ According to Wikipedia, the Bergman kernel is the projection operator from $L^2(\mathbb{D})$ to the Bergman space $A^2(\mathbb{D})$ of $L^2$ holomorphic functions on the unit disk. What is the basis for such a space of functions?

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    $\begingroup$ I don't think you mean to say Hardy space. Hardy space is the space of holomorphic functions whose boundary values are square-integrable on the circle. What you are considering, I think is the Bergman space of the disk. By integrating in polar co-ordinates it is easy to see that the monomials are pairwise orthogonal (where the inner product is that of $L^2(D)$) and so one now just needs to normalize appropriately. $\endgroup$
    – Yemon Choi
    Commented Jun 19, 2011 at 1:45
  • $\begingroup$ By the way, what do you mean by "countable basis"? I assume that you are after an orthonormal basis for the Bergman space, but you should probably take note that there is a more general notion of (Schauder) basis for Banach spaces. Note also that the set of reproducing kernel functions for the Bergman space does not form a basis, uncountable or otherwise, in this sense. $\endgroup$
    – Yemon Choi
    Commented Jun 19, 2011 at 4:55
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    $\begingroup$ John Mangual: it seems you are extremely confused about the basic definitions of the Hardy space, Bergman space, and Bergman kernel. The Bergman space, not the Hardy space, is what is relevant for the Bergman kernel; the $\rho$ you've written down is not the Bergman kernel; and the Bergman kernel is a function, not a projection operator. An orthogonal basis for the Hardy space $H^2$ or the Bergman space $L^2_a$ is very easy: just take $\{1, z, z^2, \ldots\}$ and normalise if you wish. $\endgroup$
    – Zen Harper
    Commented Jun 20, 2011 at 1:04
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    $\begingroup$ I suggest you read the following books for a good introduction: Kehe Zhu, Operator Theory on Function Spaces (both Hardy and Bergman spaces); and P. Koosis, Introduction to $H_p$ spaces; P. Duren, Theory of $H^p$ spaces for the Hardy spaces, etc. There is a more recent book (a yellow Springer grad. text) by Hedenmalm, Korenblum, Zhu, Theory of Bergman spaces, which I haven't read but is probably very good. $\endgroup$
    – Zen Harper
    Commented Jun 20, 2011 at 1:09
  • $\begingroup$ you are right the monomials are orthogonal. why don't you post an answer for credit! $\endgroup$
    – john mangual
    Commented Aug 22, 2011 at 4:33

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Assuming that we normalize so that the area of the unit disc is $\pi$, take $f_n(z) = \sqrt{\frac{n+1}{\pi}} z^n$ for $n=0,1,2,\dots$ to get an orthonormal basis.

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