# A representation for the reproducing kernel of a weighted Bergman space

Erwin Miña-Díaz Corresponding author; Email<EMAIL_ADDRESS>The
University of Mississippi, Department of Mathematics, Hume Hall 305,P.O. Box
1848, University, MS 38677-1848, USA.

###### Abstract

For a weight function in the unit disk which is the modulus of a finite
product of powers of Blaschke factors, we give a canonical representation for
the reproducing kernel of the corresponding weighted Bergman space in terms of
the values of the kernel and its derivatives at the origin. This yields a
formula for the contractive zero divisor of a Bergman space corresponding to a
finite zero set.

Keywords: Bergman spaces, reproducing kernel, zero divisor.

##  1 Statement of results

Let $\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ be the unit disk and let $\sigma$
denote the normalized area measure $d\sigma=\pi^{-1}dA$, so that
$\sigma(\mathbb{D})=1$. Let $h\geq 0$ be an integrable function in
$\mathbb{D}$, and let $A^{2}_{h}$ denote the space of functions $f$ analytic
in $\mathbb{D}$ such that

$\displaystyle\int_{\mathbb{D}}|f(\zeta)|^{2}h(\zeta)d\sigma<\infty.$

We write $A^{2}$ for $h\equiv 1$.

If $h$ is such that point evaluation functionals $z\mapsto f(z)$ acting on
$A^{2}_{h}$ are bounded (for instance, if for some $\rho<1$,
$\inf_{\rho<|z|<1}h>0$), then there exists a unique function $K(z,\zeta)$ in
$\mathbb{D}\times\mathbb{D}$, analytic in $z$ and in $\overline{\zeta}$, such
that

$\displaystyle
f(z)=\int_{\mathbb{D}}f(\zeta)K(z,\zeta)h(\zeta)d\sigma(\zeta),\quad f\in
A^{2}_{h},\ z\in\mathbb{D}.$ (1.1)

The function $K(z,\zeta)$ is called the reproducing kernel of the weighted
Bergman space $A^{2}_{h}$, and has the symmetry property

$\displaystyle K(z,\zeta)=\overline{K(\zeta,z)}\,.$ (1.2)

In this paper we give a canonical formula for the reproducing kernel
corresponding to a weight $h$ of the form

$\displaystyle
h(z)=\prod_{k=1}^{s}\left|\frac{z-a_{k}}{1-\overline{a}_{k}z}\right|^{p_{k}}\,,$
(1.3)

where $s\geq 1$, $p_{k}>-2$, and $a_{k}\in\mathbb{D}$ for all $1\leq k\leq s$.
The formula is explicit in terms of the values of $K(z,\zeta)$ and its
derivatives at the origin, which can be numerically computed by integration.

The subclass of (1.3) consisting of weights of the form

$\displaystyle
h(z)=\left|\prod_{k=1}^{s}\left(\frac{z-a_{k}}{1-\overline{a}_{k}z}\right)^{m_{k}}\right|^{p},\qquad
p>0,\ m_{k}\in\mathbb{N},$ (1.4)

is of particular importance because for these weights there is a connection
between the reproducing kernel and the contractive zero divisors of the
Bergman space $A^{p}$ (see Section 1.1 below).

The following notation will be used throughout the paper. For all non-negative
integers $k,j$, we set

$\displaystyle K^{(k,j)}(z,\zeta):=\frac{\partial^{k}\partial^{j}}{\partial
z^{k}\partial\overline{\zeta}^{j}}K(z,\zeta).$

Note that the equality

$K^{(k,j)}(z,\zeta)=\overline{K^{(j,k)}(\zeta,z)}$

follows from the symmetry property (1.2), and that differentiating (1.1)
yields

$\displaystyle
f^{(j)}(z)=\int_{\mathbb{D}}f(\zeta)K^{(j,0)}(z,\zeta)h(\zeta)d\sigma(\zeta),\quad
f\in A^{2}_{h},\ z\in\mathbb{D}.$ (1.5)

We write

$q(z):=\prod_{k=1}^{s}(z-a_{k}),\quad
q^{*}(z):=z^{s}\overline{q(1/\overline{z})}=\prod_{k=1}^{s}(1-\overline{a}_{k}z),$

and for each $1\leq k\leq s$, set

$q_{k}(z):=\underset{j\not=k}{\prod_{1\leq j\leq s}}(z-a_{j}),\quad
q^{*}_{k}(z):=\underset{j\not=k}{\prod_{1\leq j\leq
s}}(1-\overline{a}_{j}z)\,.$

Let $c_{0},\ldots,c_{s}$ denote the coefficients of $q$, and for each $1\leq
k\leq s$, let $c_{k,0},\ldots,c_{k,s-1}$ denote the coefficients of $q_{k}$,
that is,

$q(z)=\sum_{\ell=0}^{s}c_{\ell}z^{\ell},\qquad
q_{k}(z)=\sum_{\ell=0}^{s-1}c_{k,\ell}z^{\ell},\qquad c_{s}=c_{k,s-1}=1.$

From these coefficients we construct the polynomials

$\displaystyle\begin{split}T_{k}(z):={}&z^{k}\sum_{\ell=0}^{s-2-k}\overline{c_{s-\ell}}z^{\ell},\qquad
0\leq k\leq s-2,\end{split}$ (1.6)

and

$\displaystyle L_{k}(z,\zeta):={}$
$\displaystyle\sum_{r=0}^{s-2}(z\overline{\zeta})^{r}\left(\sum_{\ell=0}^{s-2-r}\overline{c_{k,s-1-\ell}}z^{\ell}\right)\left(\sum_{j=0}^{s-2-r}c_{k,s-1-j}\overline{\zeta}^{j}\right),$
(1.7)

$1\leq k\leq s$.

Our main result is the following formula.

######  Theorem 1.1.

The kernel $K(z,\zeta)$ corresponding to the weight (1.3) can be written as

$\displaystyle K(z,\zeta)=$
$\displaystyle\frac{1}{(1-z\overline{\zeta})^{2}}+\sum_{k=1}^{s}\frac{\frac{p_{k}}{2}(1-|a_{k}|^{2})}{(1-z\overline{\zeta})(1-z\overline{a}_{k})(1-a_{k}\overline{\zeta})}+\frac{J(z,\zeta)}{\overline{q^{*}(\zeta)}q^{*}(z)},$
(1.8)

where

$\displaystyle\begin{split}J(z,\zeta)={}&\sum_{k=0}^{s-2}\sum_{j=0}^{s-2}\frac{K^{(k,j)}(0,0)}{k!\,j!}T_{k}(z)\overline{T_{j}(\zeta)}-\sum_{k=0}^{s-2}(k+1)T_{k}(z)\overline{T_{k}(\zeta)}\\\
&-\sum_{k=1}^{s}\frac{p_{k}}{2}(1-|a_{k}|^{2})L_{k}(z,\zeta).\end{split}$
(1.9)

For $s=1$, $J(z,\zeta)\equiv 0$ and (1.8) reduces to

$\displaystyle K(z,\zeta)=$
$\displaystyle\frac{1}{(1-z\overline{\zeta})^{2}}+\frac{\frac{p_{1}}{2}(1-|a_{1}|^{2})}{(1-z\overline{\zeta})(1-z\overline{a}_{1})(1-a_{1}\overline{\zeta})}.$

For $s=2$, (1.8) yields

$\displaystyle\begin{split}K(z,\zeta)={}&\frac{1}{(1-z\overline{\zeta})^{2}}+\sum_{k=1}^{2}\frac{\frac{p_{k}}{2}(1-|a_{k}|^{2})}{(1-z\overline{\zeta})(1-z\overline{a}_{k})(1-a_{k}\overline{\zeta})}\\\
&+\frac{K(0,0)-1-\frac{p_{1}}{2}(1-|a_{1}|^{2})-\frac{p_{2}}{2}(1-|a_{2}|^{2})}{(1-\overline{a}_{1}z)(1-\overline{a}_{2}z)(1-a_{1}\overline{\zeta})(1-a_{2}\overline{\zeta})}.\end{split}$
(1.10)

Formula (1.10) was obtained by Hansbo in [2, Thm. 2.7] by first using an
iterative procedure to establish the formula for integer values of $p_{1}$ and
$p_{2}$, and then passing from integer to arbitrary values of $p_{1}$ and
$p_{2}$ by means of a function theoretical argument involving Carlson’s
theorem. Hansbo’s method seems to become too intricate when higher values of
$s$ are considered, but for $s=2$ it yields the following elegant
representation for $K(0,0)$:

$K(0,0)=1+\sum_{k=1}^{2}\frac{p_{k}}{2}(1-|a_{k}|^{2})+|a_{1}-a_{2}|^{2}\frac{\sum_{j=1}^{\infty}j\binom{p_{1}/2}{j}\binom{p_{2}/2}{j}\lambda^{j}}{\sum_{j=0}^{\infty}\binom{p_{1}/2}{j}\binom{p_{2}/2}{j}\lambda^{j}},$

with $\lambda=(1-|a_{1}|^{2})(1-|a_{2}|^{2})/|1-a_{1}\overline{a}_{2}|^{2}$.
We do not know whether this representation generalizes in some satisfactory
way to all the derivatives of the kernel in (1.9), but at least in a
structural sense, formula (1.8) extends Hansbo’s (1.10) to an arbitrary $s$.

Besides Hansbo’s work, the paper by MacGregor and Stessin [3] established a
couple of structural formulas for the kernel in the context of Bergman spaces
$A^{p}$, for weights $h$ of the form (1.4) with simple zeros (i.e., $m_{k}=1$,
$1\leq k\leq s$). These formulas equally extend to the more general weight
(1.3) with little change in their statements. First, the canonical
representation (13) of [3] can now be obtained as a corollary of Theorem 1.1
as follows. Since the polynomials $q^{*}_{1},\ldots,q^{*}_{s}$ are linearly
independent, $J(z,\zeta)$ can be expanded as a sum

$J(z,\zeta)=\sum_{1\leq k,j\leq
s}d_{k,j}q^{*}_{k}(z)\overline{q^{*}_{j}(\zeta)}\,,$

where the coefficients $d_{k,j}$ are given by

$d_{k,j}=\frac{J(1/\overline{a}_{k},1/\overline{a}_{j})}{q^{*}_{k}(1/\overline{a}_{k})\overline{q^{*}_{j}(1/\overline{a}_{j})}}$

(if some $a_{k}$ equals zero, the formula for $d_{k,j}$ has to be adjusted).
This gives

$\displaystyle\begin{split}K(z,\zeta)={}&\frac{1}{(1-z\overline{\zeta})^{2}}+\sum_{k=1}^{s}\frac{\frac{p_{k}}{2}(1-|a_{k}|^{2})}{(1-z\overline{\zeta})(1-z\overline{a}_{k})(1-a_{k}\overline{\zeta})}\\\
&+\sum_{1\leq k,j\leq
s}\frac{d_{k,j}}{(1-z\overline{a}_{k})(1-a_{j}\overline{\zeta})}\,.\end{split}$
(1.11)

Taking $p_{k}=p$ in (1.11) yields an equivalent form of [3, Eq. (13)].

The second representation [3, Eq. (7)] that MacGregor and Stessin provide is
in terms of the Blaschke products

$\displaystyle B(z):=\frac{q(z)}{q^{*}(z)},\quad
B_{k}(z):=\frac{q_{k}(z)}{q^{*}_{k}(z)},$ (1.12)

and the values of the kernel at the points $a_{k}$. It results by taking
$p_{k}=p$ in the following formula, more generally valid for a weight $h$ as
in (1.3):

$\displaystyle\begin{split}K(z,\zeta)={}&\frac{B(z)\overline{B(\zeta)}}{(1-\overline{\zeta}z)^{2}}+\sum_{k=1}^{s}\frac{(\frac{p_{k}}{2}+1)B_{k}(z)\overline{B_{k}(\zeta)}(\overline{\zeta}z-|a_{k}|^{2})}{(1-\overline{\zeta}z)(1-\overline{a}_{k}z)(1-a_{k}\overline{\zeta})}\\\
&+\sum_{k=1}^{s}\sum_{j=1}^{s}\frac{(1-|a_{k}|^{2})(1-|a_{j}|^{2})B_{k}(z)\overline{B_{j}(\zeta)}}{(1-\overline{a}_{k}z)(1-a_{j}\overline{\zeta})B_{k}(a_{k})\overline{B_{j}(a_{j})}}K(a_{k},a_{j})\,.\end{split}$
(1.13)

We prove (1.13) by following the original argument in [3], just noting that
that argument works equally well for a weight $h$ as in (1.3). The first part
of the proof of Theorem 1.1 (i.e., up to (2.11)) is also based on the same
idea, but applied in a way in which values at the origin are involved instead
of values at the points $a_{k}$. The second part of the proof of Theorem 1.1
is more demanding and consists of performing the algebraic manipulations
needed to get the kernel written in the reduced and symmetrical form (1.8).

###  1.1 Contractive zero divisors

For a fixed $0<p<\infty$, let $A^{p}$ denote the Bergman space of functions
$f$ analytic in $\mathbb{D}$ and such that

$\|f\|_{p}:=\left(\int_{\mathbb{D}}|f(\zeta)|^{p}d\sigma(\zeta)\right)^{1/p}<\infty.$

Any zero set of a function in $A^{p}$ (each zero counted as many times as its
multiplicity) is called an $A^{p}$ zero set. For each $A^{p}$ zero set $Z$,
there exists a unique (up to a unimodular constant factor) function $G\in
A^{p}$ of norm $\|G\|_{p}=1$ such that for every $f\in A^{p}$ that vanishes on
$Z$, the function $f/G\in A^{p}$ and $\|f/G\|_{p}\leq\|f\|_{p}$. This function
$G$ is called the contractive zero divisor (or canonical divisor) of the zero
set $Z$. We refer the reader to [1, Chap. 5] for the relevant theory and
history.

For a finite zero set, say consisting of the points $a_{1},\ldots,a_{s}$, each
$a_{k}$ of multiplicity $m_{k}$, the canonical divisor is given by [1, Chap.
5, Thm. 4]

$\displaystyle
G(z)=K(0,0)^{-1/p}K(z,0)^{2/p}\prod_{k=1}^{s}\left(\frac{z-a_{k}}{1-\overline{a}_{k}z}\right)^{m_{k}}\,,$
(1.14)

where $K(z,\zeta)$ is the reproducing kernel for the weight (1.4). If we
therefore evaluate (1.8) at $\zeta=0$, we get a structural formula for the
canonical divisor, the evaluation resulting in

$\displaystyle\begin{split}K(z,0)={}&\frac{\overline{q(0)}z^{s}}{q^{*}(z)}+\frac{z^{s-1}}{q^{*}(z)}\sum_{k=1}^{s}\overline{q_{k}(0)}\left(1+\frac{p_{k}}{2}(1-|a_{k}|^{2})\right)\\\
&+\frac{1}{q^{*}(z)}\sum_{k=0}^{s-2}\frac{K^{(k,0)}(0,0)}{k!}T_{k}(z),\end{split}$
(1.15)

where (in this context) $p_{k}=m_{k}p$, $1\leq k\leq s$.

We remark that, indeed, to get (1.15) one does not need the reduced symmetric
representation (1.8). Evaluating at $\zeta=0$ in the less refined version
(2.10) provides a shorter proof.

Since the representation (1.14) already involves values of the kernel at $z=0$
and $\zeta=0$, it is particularly fitting that (1.15) is equally expressed in
terms of derivatives of the kernel at the origin.

###  1.2 Computing the values $K^{(k,j)}(0,0)$

Let us now see how the unknown parameters that occur in (1.9) can be
determined by solving a linear system of equations.

Observe that the polynomial $T_{k}$ defined in (1.6) can be characterized as
the unique polynomial of degree $\leq s-2$ such that

$\displaystyle\left[\frac{T_{k}}{q^{*}}\right]^{(\ell)}(0)=\delta_{k,\ell}k!,\qquad
0\leq\ell\leq s-2.$

From (2.11) and (2.9) we find

$\displaystyle\sum_{j=0}^{s-2}\frac{K^{(k,j)}(0,0)}{k!j!}\frac{\overline{T_{j}(\zeta)}}{\overline{q^{*}(\zeta)}}={}\frac{K^{(k,0)}(0,\zeta)}{k!}-\frac{\overline{P_{k}(\zeta)}}{\overline{q^{*}(\zeta)}},\qquad
0\leq k\leq s-2,$ (1.16)

where

$\displaystyle\overline{P_{n}(\zeta)}={}$
$\displaystyle\frac{\overline{\zeta}^{s-1}}{n!}\left.\frac{\partial^{n}}{\partial
z^{n}}\left(\frac{\partial}{\partial
z}\left(\frac{q(z)}{1-z\overline{\zeta}}\right)+\frac{1}{1-z\overline{\zeta}}\sum_{k=1}^{s}\frac{\frac{p_{k}}{2}(1-|a_{k}|^{2})q_{k}(z)}{1-\overline{a}_{k}z}\right)\right|_{z=0}\,.$

After computing the derivatives, we can alternatively write

$\overline{P_{n}(\zeta)}=(n+1)q(0)\overline{\zeta}^{s+n}+\overline{\zeta}^{s-1}\sum_{j=0}^{n}b_{n,j}\overline{\zeta}^{j}\,,$

with

$\displaystyle b_{n,j}={}$
$\displaystyle(n+1)c_{n+1-j}+\sum_{k=1}^{s}\frac{p_{k}}{2}(1-|a_{k}|^{2})\sum_{r=0}^{n-j}(\overline{a}_{k})^{r}c_{k,n-j-r}$

By (1.5), we have that for all $0\leq n\leq s-2$,

$\displaystyle\int_{\mathbb{D}}\frac{T_{n}(\zeta)}{q^{*}(\zeta)}K^{(k,0)}(0,\zeta)h(\zeta)d\sigma(\zeta)={}$
$\displaystyle\left[\frac{T_{n}}{q^{*}}\right]^{(k)}(0)=\delta_{n,k}n!\,,$

so that after multiplying (1.16) by $T_{n}(\zeta)/q^{*}(\zeta)$ and
integrating with respect to $hd\sigma$ we obtain

$\displaystyle\sum_{j=0}^{s-2}\frac{K^{(k,j)}(0,0)}{k!j!}\langle
T_{n},T_{j}\rangle=\delta_{n,k}-\langle T_{n},P_{k}\rangle,\qquad 0\leq n\leq
s-2,$

with $\langle f,g\rangle:=\int_{\mathbb{D}}f\overline{g}|q^{*}|^{-2}hd\sigma$.
Since the polynomials $T_{k}$ are linearly independent, the Gram matrix
$\left(\langle T_{n},T_{j}\rangle\right)_{n,j=0}^{s-2}$ has a non-zero
determinant, and so for each $0\leq k\leq s-2$, the vector
$(K^{(k,j)}(0,0))_{j=0}^{s-2}$ is the solution of a non-singular linear system
of equations whose coefficients and constants can be numerically computed by
integration.

##  2 Proofs

Both Theorem 1.1 and formula (1.13) will be proved with the help of Green’s
formula

$\frac{1}{2\pi
i}\int_{\partial\Omega}g(\zeta)d\zeta=\int_{\Omega}\frac{\partial
g}{\partial\overline{\zeta}}\,d\sigma(\zeta)\,,$

valid for $\Omega$ a finitely connected domain bounded by analytic Jordan
curves and $g\in C^{1}(\overline{\Omega})$.

###  2.1 Proof of Theorem 1.1

Let $f$ be a function analytic in the closed unit disk
$\overline{\mathbb{D}}$. By an application of the residue theorem we get that
for all $z\in\mathbb{D}$,

$\displaystyle\begin{split}f(z)={}&\frac{z^{s-1}}{q^{*}(z)}\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{f(\zeta)q^{*}(\zeta)}{\zeta^{s-1}(\zeta-z)}d\zeta+\frac{1}{q^{*}(z)}\sum_{j=0}^{s-2}\frac{f^{(j)}(0)}{j!}T_{j}(z)\,,\end{split}$
(2.1)

where

$\displaystyle
T_{k}(z)=\frac{k!\,z^{s-1}}{(s-2)!}\binom{s-2}{k}\left[\frac{\partial^{s-2-k}}{\partial
t^{s-2-k}}\left(\frac{q^{*}(t)}{z-t}\right)\right]_{t=0}\,.$ (2.2)

In terms of the coefficients of $q$, we can write $q^{*}$ as

$q^{*}(t)=t^{s}\overline{q(1/\overline{t})}=\sum_{\ell=0}^{s}\overline{c_{s-\ell}}\,t^{\ell},$

so that

$\displaystyle\frac{{q^{*}}^{(\ell)}(0)}{\ell!}=\overline{c_{s-\ell}},\quad
0\leq\ell\leq s.$ (2.3)

Hence

$\displaystyle\begin{split}T_{k}(z)={}&\frac{k!z^{s-1}}{(s-2)!}\binom{s-2}{k}\sum_{\ell=0}^{s-2-k}\binom{s-2-k}{\ell}{q^{*}}^{(\ell)}(0)\frac{(s-2-k-\ell)!}{z^{s-1-k-\ell}}\\\
={}&z^{k}\sum_{\ell=0}^{s-2-k}\overline{c_{s-\ell}}z^{\ell}.\end{split}$

Note that this is the same $T_{k}$ that was previously defined in (1.6).

We now seek to express the integral in (2.1) as an integral over $\mathbb{D}$
with respect to the weight $h$ given by (1.3). On the unit circle $|\zeta|=1$,
$h$ is constant $1$, and also $\overline{\zeta}=1/\zeta$. Then, by an
application of Green’s formula we obtain that for all $z\in\mathbb{D}$,

$\displaystyle\begin{split}\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{f(\zeta)q^{*}(\zeta)}{\zeta^{s-1}(\zeta-z)}d\zeta={}&\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{f(\zeta)\prod_{k=1}^{s}(1-\overline{a}_{k}\zeta)}{\zeta^{s-1}(\zeta-z)}d\zeta\\\
={}&\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{1}{(1-z\overline{\zeta})}\overline{q(\zeta)}h(\zeta)f(\zeta)d\zeta\\\
={}&\int_{\mathbb{D}}\frac{\partial}{\partial\overline{\zeta}}\left(\frac{1}{(1-z\overline{\zeta})}\overline{q(\zeta)}h(\zeta)\right)f(\zeta)d\sigma(\zeta)\,.\end{split}$
(2.4)

We clarify that to get (2.4) we apply Green’s formula in a subregion of
$\mathbb{D}$ that lies exterior to small circles around each $a_{k}$ for which
$p_{k}<0$, and then take limits as the radii of these circles go to $0$.

The function

$\displaystyle
w(z):=\prod_{k=1}^{s}\left(\frac{z-a_{k}}{1-\overline{a}_{k}z}\right)^{p_{k}/2}$
(2.5)

admits a single-valued branch around any point of $\mathbb{D}$ that is not an
$a_{k}$. Since $h(z)=|w(z)|^{2}$, we have

$\displaystyle\begin{split}\frac{\partial}{\partial\overline{\zeta}}\left(\frac{1}{(1-z\overline{\zeta})}\overline{q(\zeta)}h(\zeta)\right)={}&\frac{z}{(1-z\overline{\zeta})^{2}}\overline{q(\zeta)}h(\zeta)+\frac{w(\zeta)}{(1-z\overline{\zeta})}\overline{\frac{\partial}{\partial\zeta}\left(q(\zeta)w(\zeta)\right)}\,.\end{split}$
(2.6)

By direct computation of the derivative using the product rule, we find

$\displaystyle\begin{split}\frac{\partial}{\partial\zeta}\left(q(\zeta)w(\zeta)\right)={}&w(\zeta)\sum_{k=1}^{s}q_{k}(\zeta)\left(1+\frac{p_{k}}{2}\frac{1-|a_{k}|^{2}}{(1-\overline{a}_{k}\zeta)}\right).\end{split}$
(2.7)

From this point onward we will use the abbreviation

$A_{k}:=\frac{p_{k}}{2}(1-|a_{k}|^{2}),\qquad 1\leq k\leq s.$

The actual values of these constants play no role in what remains of the
proof.

The equalities (2.6) and (2.7) then combine into

$\displaystyle\begin{split}\frac{\partial}{\partial\overline{\zeta}}\left(\frac{1}{(1-z\overline{\zeta})}\overline{q(\zeta)}h(\zeta)\right)={}&h(\zeta)Q(z,\zeta)\end{split}$
(2.8)

with

$\displaystyle
Q(z,\zeta):=\frac{z\overline{q(\zeta)}}{(1-z\overline{\zeta})^{2}}+\frac{1}{(1-z\overline{\zeta})}\sum_{k=1}^{s}\overline{q_{k}(\zeta)}\left(1+\frac{A_{k}}{(1-a_{k}\overline{\zeta})}\right).$
(2.9)

If we now write the derivatives in (2.1) in the integral form

$f^{(j)}(0)=\int_{\mathbb{D}}f(\zeta)K^{(j,0)}(0,\zeta)h(\zeta)d\sigma(\zeta),$

then (2.1) can be written with the help of (2.4) and (2.8) as

$\displaystyle\begin{split}f(z)={}&\int_{\mathbb{D}}\frac{f(\zeta)}{q^{*}(z)}\left(z^{s-1}Q(z,\zeta)+\sum_{j=0}^{s-2}\frac{T_{j}(z)}{j!}K^{(j,0)}(0,\zeta)\right)h(\zeta)d\sigma(\zeta)\,.\end{split}$

By a density argument, this reproducing property also holds for every $f\in
A^{2}_{h}$, and so we must have

$\displaystyle
K(z,\zeta)=\frac{z^{s-1}Q(z,\zeta)}{q^{*}(z)}+\frac{1}{q^{*}(z)}\sum_{k=0}^{s-2}\frac{T_{k}(z)K^{(k,0)}(0,\zeta)}{k!}\,.$
(2.10)

By symmetry,

$K(z,\zeta)=\frac{\overline{\zeta^{s-1}Q(\zeta,z)}}{\overline{q^{*}(\zeta)}}+\frac{1}{\overline{q^{*}(\zeta)}}\sum_{j=0}^{s-2}\frac{\overline{T_{j}(\zeta)}K^{(0,j)}(z,0)}{j!}\,.$

Taking derivatives with respect to $z$ and evaluating at $z=0$ yields

$\displaystyle K^{(k,0)}(0,\zeta)={}$
$\displaystyle\frac{\overline{\zeta}^{s-1}}{\overline{q^{*}(\zeta)}}\left[\frac{\partial^{k}}{\partial
z^{k}}\overline{Q(\zeta,z)}\right]_{z=0}+\frac{1}{\overline{q^{*}(\zeta)}}\sum_{j=0}^{s-2}\frac{\overline{T_{j}(\zeta)}K^{(k,j)}(0,0)}{j!}\,.$
(2.11)

Substituting (2.11) into (2.10) we get

$\displaystyle\begin{split}K(z,\zeta)={}&\frac{z^{s-1}Q(z,\zeta)}{q^{*}(z)}+\frac{\overline{\zeta}^{s-1}}{\overline{q^{*}(\zeta)}q^{*}(z)}\sum_{k=0}^{s-2}\frac{T_{k}(z)}{k!}\left[\frac{\partial^{k}}{\partial
z^{k}}\overline{Q(\zeta,z)}\right]_{z=0}\\\
&+\frac{1}{\overline{q^{*}(\zeta)}q^{*}(z)}\sum_{k=0}^{s-2}\sum_{j=0}^{s-2}\frac{K^{(k,j)}(0,0)}{k!j!}T_{k}(z)\overline{T_{j}(\zeta)}\,.\end{split}$
(2.12)

This concludes the first part of the proof. The second part amounts to
performing the algebraic manipulations needed to reduce (2.12) to (1.8). We
begin by writing (2.12) as

$\displaystyle K(z,\zeta)={}$
$\displaystyle\frac{N(z,\zeta)}{(1-z\overline{\zeta})^{2}q^{*}(z)\overline{q^{*}(\zeta)}}+\frac{1}{q^{*}(z)\overline{q^{*}(\zeta)}}\sum_{k=0}^{s-2}\sum_{j=0}^{s-2}\frac{K^{(k,j)}(0,0)}{k!j!}T_{k}(z)\overline{T_{j}(\zeta)}\,,$
(2.13)

with

$\displaystyle\begin{split}N(z,\zeta):={}&(1-z\overline{\zeta})^{2}z^{s-1}\overline{q^{*}(\zeta)}Q(z,\zeta)\\\
&+(1-z\overline{\zeta})^{2}\overline{\zeta}^{s-1}\sum_{k=0}^{s-2}\frac{T_{k}(z)}{k!}\left[\frac{\partial^{k}}{\partial
z^{k}}\overline{Q(\zeta,z)}\right]_{z=0}\,.\end{split}$ (2.14)

By (2.9), and since

$q^{\prime}(t)=\sum_{k=1}^{s}q_{k}(t)\,,$

we have

$\displaystyle\begin{split}(1-z\overline{\zeta})^{2}z^{s-1}\overline{q^{*}(\zeta)}Q(z,\zeta)={}&z^{s}\overline{(q^{*}q)(\zeta)}+(1-z\overline{\zeta})z^{s-1}\overline{(q^{*}q^{\prime})(\zeta)}\\\
&+(1-z\overline{\zeta})z^{s-1}\sum_{k=1}^{s}A_{k}\overline{(q_{k}^{*}q_{k})(\zeta)}\,.\end{split}$
(2.15)

Using the expression (2.2) for $T_{k}$, we can write

$\displaystyle\sum_{k=0}^{s-2}\frac{T_{k}(z)}{k!}\left[\frac{\partial^{k}}{\partial
z^{k}}\overline{Q(\zeta,z)}\right]_{z=0}={}$
$\displaystyle\frac{z^{s-1}}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{q^{*}(t)}{z-t}\overline{Q(\zeta,t)}\right)\right]_{t=0},$
(2.16)

while (2.9) yields an expression for $\overline{Q(\zeta,z)}$ from which we get

$\displaystyle\frac{q^{*}(t)}{z-t}\overline{Q(\zeta,t)}={}$
$\displaystyle\frac{\overline{\zeta}(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})^{2}}+\frac{(q^{*}q^{\prime})(t)}{(z-t)(1-t\overline{\zeta})}+\frac{\sum_{k=1}^{s}A_{k}(q^{*}_{k}q_{k})(t)}{(z-t)(1-t\overline{\zeta})}.$
(2.17)

The relations (2.17), (2.16), (2.15) and (2.14) then allow us to write
$N(z,\zeta)$ in the form

$\displaystyle\begin{split}N(z,\zeta)={}&z^{s}\overline{(q^{*}q)(\zeta)}+(1-z\overline{\zeta})z^{s-1}\overline{(q^{*}q^{\prime})(\zeta)}+(1-z\overline{\zeta})(z\overline{\zeta})^{s-1}D(z,\zeta)\\\
&+(1-z\overline{\zeta})z^{s-1}\sum_{k=1}^{s}A_{k}\overline{(q_{k}^{*}q_{k})(\zeta)}\\\
&+\frac{(1-z\overline{\zeta})^{2}z^{s-1}\overline{\zeta}^{s-1}}{(s-2)!}\sum_{k=1}^{s}A_{k}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(q^{*}_{k}q_{k})(t)}{(z-t)(1-t\overline{\zeta})}\right)\right]_{t=0}\,,\end{split}$
(2.18)

where

$D(z,\zeta):=\frac{1}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(1-z\overline{\zeta})\overline{\zeta}(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})^{2}}+\frac{(1-z\overline{\zeta})(q^{*}q^{\prime})(t)}{(z-t)(1-t\overline{\zeta})}\right)\right]_{t=0}\,.$

We now make use of the identity

$\displaystyle\begin{split}\frac{\overline{\zeta}(1-z\overline{\zeta})(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})^{2}}+\frac{(1-z\overline{\zeta})(q^{*}q^{\prime})(t)}{(z-t)(1-t\overline{\zeta})}={}&\frac{\overline{\zeta}(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})}-\frac{\partial}{\partial
t}\left(\frac{\overline{\zeta}(q^{*}q)(t)}{(1-t\overline{\zeta})}\right)\\\
&+\frac{(q^{*}q^{\prime})(t)}{(z-t)}+\frac{\overline{\zeta}({q^{*}}^{\prime}q)(t)}{(1-t\overline{\zeta})}\,,\end{split}$
(2.19)

and the fact that derivatives at the origin of any function $f$ satisfy

$f^{(m)}(0)=\frac{1}{m+1}\left.\left(tf(t)\right)^{(m+1)}\right|_{t=0}\,,$

to get $D(z,\zeta)$ written in the form

$\displaystyle\begin{split}&\frac{1}{(s-1)!}\left[\frac{\partial^{s-1}}{\partial
t^{s-1}}\left(\frac{t\overline{\zeta}(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})}-(s-1)\frac{\overline{\zeta}(q^{*}q)(t)}{1-t\overline{\zeta}}+\frac{t\overline{\zeta}({q^{*}}^{\prime}q)(t)}{1-t\overline{\zeta}}\right)\right]_{t=0}\\\
&+\frac{1}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(q^{*}q^{\prime})(t)}{z-t}\right)\right]_{t=0}\,.\end{split}$
(2.20)

If we use the identity

$\displaystyle
t{q^{*}}^{\prime}(t)=sq^{*}(t)-{q^{\prime}}^{*}(t),\quad\quad{q^{\prime}}^{*}(t):=t^{s-1}\overline{q^{\prime}(1/\overline{t})}$
(2.21)

while rearranging terms in (2.20) we obtain

$\displaystyle D(z,\zeta)={}$
$\displaystyle\frac{z\overline{\zeta}}{(s-1)!}\left[\frac{\partial^{s-1}}{\partial
t^{s-1}}\left(\frac{(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})}\right)\right]_{t=0}$
$\displaystyle-\frac{1}{(s-1)!}\left[\frac{\partial^{s-1}}{\partial
t^{s-1}}\left(\frac{\overline{\zeta}({q^{\prime}}^{*}q)(t)}{1-t\overline{\zeta}}\right)\right]_{t=0}$
$\displaystyle+\frac{1}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(q^{*}q^{\prime})(t)}{z-t}\right)\right]_{t=0}\,,$

and after plugging this expression for $D(z,\zeta)$ in (2.18) we see that we
can represent $N(z,\zeta)$ as a sum

$\displaystyle N(z,\zeta)={}$ $\displaystyle
N_{1}(z,\zeta)+(1-z\overline{\zeta})N_{2}(z,\zeta)+(1-z\overline{\zeta})\sum_{k=1}^{s}A_{k}M_{k}(z,\zeta)\,,$
(2.22)

with

$\displaystyle N_{1}(z,\zeta):={}$ $\displaystyle
z^{s}\overline{(q^{*}q)(\zeta)}+\frac{(z\overline{\zeta})^{s}}{(s-1)!}\left[\frac{\partial^{s-1}}{\partial
t^{s-1}}\left(\frac{(1-z\overline{\zeta})(q^{*}q)(t)}{(z-t)(1-t\overline{\zeta})}\right)\right]_{t=0}\,,$
(2.23)
$\displaystyle\begin{split}N_{2}(z,\zeta):={}&z^{s-1}\overline{(q^{*}q^{\prime})(\zeta)}+\frac{(z\overline{\zeta})^{s-1}}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(q^{*}q^{\prime})(t)}{z-t}\right)\right]_{t=0}\\\
&-\frac{(z\overline{\zeta})^{s-1}}{(s-1)!}\left[\frac{\partial^{s-1}}{\partial
t^{s-1}}\left(\frac{\overline{\zeta}({q^{\prime}}^{*}q)(t)}{1-t\overline{\zeta}}\right)\right]_{t=0}\,,\end{split}$
(2.24)

and

$\displaystyle\begin{split}M_{k}(z,\zeta):={}&z^{s-1}\overline{(q_{k}^{*}q_{k})(\zeta)}+\frac{(z\overline{\zeta})^{s-1}}{(s-2)!}\left[\frac{\partial^{s-2}}{\partial
t^{s-2}}\left(\frac{(1-z\overline{\zeta})(q^{*}_{k}q_{k})(t)}{(z-t)(1-t\overline{\zeta})}\right)\right]_{t=0}\,.\end{split}$

We next show that

$\displaystyle N_{1}(z,\zeta)={}$ $\displaystyle
q^{*}(z)\overline{q^{*}(\zeta)}-(1-z\overline{\zeta})L(z,\zeta)\,,$ (2.25)

with

$\displaystyle L(z,\zeta):={}$
$\displaystyle\sum_{r=0}^{s-1}(z\overline{\zeta})^{r}\left(\sum_{j=0}^{s-1-r}c_{s-j}\overline{\zeta}^{j}\right)\left(\sum_{\ell=0}^{s-1-r}\overline{c_{s-\ell}}z^{\ell}\right)\,.$
(2.26)

Since $N_{1}(z,\zeta)$ and $M_{k}(z,\zeta)$ have the same structure, this will
simultaneously show that

$\displaystyle
M_{k}(z,\zeta)=q_{k}^{*}(z)\overline{q_{k}^{*}(\zeta)}-(1-z\overline{\zeta})L_{k}(z,\zeta),\quad
1\leq k\leq s,$ (2.27)

with $L_{k}$ given by (1.7).

Using the identity

$\frac{1-z\overline{\zeta}}{(z-t)(1-t\overline{\zeta})}=\frac{1}{z-t}-\frac{\overline{\zeta}}{1-t\overline{\zeta}}$

we find that the second term in the right-hand side of (2.23) equals

$\displaystyle\overline{\zeta}^{s}\sum_{\ell=0}^{s-1}\frac{(q^{*}q)^{(\ell)}(0)}{\ell!}z^{\ell}-z^{s}\sum_{\ell=0}^{s-1}\frac{(q^{*}q)^{(\ell)}(0)}{\ell!}\overline{\zeta}^{2s-\ell}.$
(2.28)

Since $t^{2s}\overline{(q^{*}q)(1/\overline{t})}=(q^{*}q)(t)$, we have

$\frac{(q^{*}q)^{(\ell)}(0)}{\ell!}=\frac{\overline{(q^{*}q)^{(2s-\ell)}(0)}}{(2s-\ell)!},\qquad
0\leq\ell\leq 2s,$

so that

$\displaystyle z^{s}\overline{(q^{*}q)(\zeta)}={}$ $\displaystyle
z^{s}\sum_{\ell=0}^{2s}\frac{(q^{*}q)^{(\ell)}(0)}{\ell!}\overline{\zeta}^{2s-\ell}\,.$
(2.29)

To simplify the upcoming presentation, we write

$C_{\ell}:=\frac{(q^{*}q)^{(\ell)}(0)}{\ell!},\qquad 0\leq\ell\leq 2s\,,$

so that by (2.3)

$C_{\ell}=\sum_{k=0}^{\ell}c_{k}\overline{c_{s-\ell+k}}=\sum_{k=0}^{\ell}\overline{c_{s-k}}c_{\ell-k}\,.$

Note that $C_{s}=\sum_{j=0}^{s}|c_{j}|^{2}\geq 1$.

Since the right-hand side of (2.23) is equal to the sum of (2.29) and (2.28),
we get

$\displaystyle N_{1}(z,\zeta)$
$\displaystyle={}C_{s}(z\overline{\zeta})^{s}+\overline{\zeta}^{s}\sum_{\ell=0}^{s-1}C_{\ell}z^{\ell}+z^{s}\sum_{\ell=0}^{s}\overline{C_{\ell}}\,\overline{\zeta}^{\ell}\,.$
(2.30)

Let us now define, for every integer $n\geq 0$,

$V_{n}(t):=\frac{t^{n+1}-1}{t-1}=\sum_{j=0}^{n}t^{j},$

and let us also agree that $V_{-1}\equiv 0$. With the help of these
polynomials we can then write (2.30) as

$\displaystyle N_{1}(z,\zeta)={}$
$\displaystyle(z\overline{\zeta}-1)\left(C_{s}V_{s-1}(z\overline{\zeta})+\sum_{\ell=0}^{s-1}C_{\ell}\overline{\zeta}^{s-\ell}V_{\ell-1}(z\overline{\zeta})+\sum_{\ell=0}^{s-1}\overline{C_{\ell}}z^{s-\ell}V_{\ell-1}(z\overline{\zeta})\right)$
$\displaystyle+C_{s}+\sum_{\ell=0}^{s-1}C_{\ell}\overline{\zeta}^{s-\ell}+\sum_{\ell=0}^{s-1}\overline{C_{\ell}}z^{s-\ell}\,.$
(2.31)

Letting $\ell=j-k-1$ be the new counter, one verifies the equalities

$\displaystyle\sum_{\ell=0}^{s-1}C_{\ell}\overline{\zeta}^{s-\ell}=\sum_{0\leq
k<j\leq s}\overline{c_{s-k}}c_{s-j}\overline{\zeta}^{j-k}\,,$ (2.32)

and

$\displaystyle\sum_{\ell=0}^{s-1}C_{\ell}\overline{\zeta}^{s-\ell}V_{\ell-1}(z\overline{\zeta})=\sum_{0\leq
k<j\leq
s}\overline{c_{s-k}}c_{s-j}\overline{\zeta}^{j-k}V_{s-1-(j-k)}(z\overline{\zeta})\,.$
(2.33)

After expanding the product

$q^{*}(z)\overline{q^{*}(\zeta)}=\left(\sum_{u=0}^{s}\overline{c_{s-u}}z^{u}\right)\left(\sum_{j=0}^{s}c_{s-j}\overline{\zeta}^{j}\right),$

we can express it by means of the polynomials $V_{n}$ as

$\displaystyle\begin{split}q^{*}(z)\overline{q^{*}(\zeta)}={}&(z\overline{\zeta}-1)\sum_{j=0}^{s}|c_{s-j}|^{2}V_{j-1}(z\overline{\zeta})\\\
&+(z\overline{\zeta}-1)\sum_{0\leq k<j\leq
s}\left(c_{s-k}\overline{c_{s-j}}z^{j-k}+\overline{c_{s-k}}c_{s-j}\overline{\zeta}^{j-k}\right)V_{k-1}(z\overline{\zeta})\\\
&+C_{s}+\sum_{0\leq k<j\leq
s}\overline{c_{s-k}}c_{s-j}\overline{\zeta}^{j-k}+\sum_{0\leq k<j\leq
s}c_{s-k}\overline{c_{s-j}}z^{j-k}\,.\end{split}$ (2.34)

Four terms in (2.1) can be rewritten using (2.32) and (2.33), so that
subtracting (2.34) from (2.1) yields

$\displaystyle N_{1}(z,\zeta)={}$ $\displaystyle
q^{*}(z)\overline{q^{*}(\zeta)}+(z\overline{\zeta}-1)\sum_{j=0}^{s-1}|c_{s-j}|^{2}(z\overline{\zeta})^{j}V_{s-1-j}(z\overline{\zeta})$
$\displaystyle+(z\overline{\zeta}-1)\sum_{0\leq k<j\leq
s-1}\overline{c_{s-k}}c_{s-j}z^{k}\overline{\zeta}^{j}V_{s-1-j}(z\overline{\zeta})$
$\displaystyle+(z\overline{\zeta}-1)\sum_{0\leq k<j\leq
s-1}c_{s-k}\overline{c_{s-j}}z^{j}\overline{\zeta}^{k}V_{s-1-j}(z\overline{\zeta})\,.$

If we now write these sums in the form

$\displaystyle\sum_{j=0}^{s-1}|c_{s-j}|^{2}(z\overline{\zeta})^{j}V_{s-1-j}(z\overline{\zeta})={}$
$\displaystyle\sum_{j=0}^{s-1}\left(\sum_{\ell=s-j}^{s}|c_{\ell}|^{2}\right)(z\overline{\zeta})^{j}\,,$

and

$\displaystyle\sum_{0\leq k<j\leq
s-1}\overline{c_{s-k}}c_{s-j}z^{k}\overline{\zeta}^{j}V_{s-1-j}(z\overline{\zeta})={}$
$\displaystyle\sum_{0\leq k<j\leq
s-1}z^{k}\overline{\zeta}^{j}\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\,,$

then we see that $N_{1}(z,\zeta)$ can be written as in (2.25) with

$\displaystyle\begin{split}L(z,\zeta)={}&\sum_{j=0}^{s-1}\left(\sum_{\ell=s-j}^{s}|c_{\ell}|^{2}\right)(z\overline{\zeta})^{j}+\sum_{0\leq
k<j\leq
s-1}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)z^{k}\overline{\zeta}^{j}\\\
&+\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}c_{s-k+\ell}\overline{c_{s-j+\ell}}\right)z^{j}\overline{\zeta}^{k}\,.\end{split}$
(2.35)

To verify that (2.35) coincides with (2.26), we group the terms corresponding
to the lowest values of $\ell$ to get

$\displaystyle\begin{split}L(z,\zeta)={}&\left(\sum_{j=0}^{s-1}c_{s-j}\overline{\zeta}^{j}\right)\left(\sum_{\ell=0}^{s-1}\overline{c_{s-\ell}}z^{\ell}\right)+(z\overline{\zeta})\sum_{j=0}^{s-2}\left(\sum_{\ell=s-j}^{s}|c_{\ell}|^{2}\right)(z\overline{\zeta})^{j}\\\
&+(z\overline{\zeta})\sum_{0\leq k<j\leq
s-2}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)z^{k}\overline{\zeta}^{j}\\\
&+(z\overline{\zeta})\sum_{0\leq k<j\leq
s-2}\left(\sum_{\ell=0}^{k}c_{s-k+\ell}\overline{c_{s-j+\ell}}\right)z^{j}\overline{\zeta}^{k}\,.\end{split}$
(2.36)

The sum of the last three terms equals the product of $(z\overline{\zeta})$
and a second factor which is just the sum in (2.35) with the terms
corresponding to $j=s-1$ removed. This decomposition can then be iterated to
produce (2.26).

We will also need a further decomposition of $L(z,\zeta)$, namely, writing
(2.35) as

$\displaystyle\begin{split}L(z,\zeta)={}&(z\overline{\zeta}-1)\sum_{j=0}^{s-1}\left(\sum_{\ell=s-j}^{s}|c_{\ell}|^{2}\right)V_{j-1}(z\overline{\zeta})\\\
&+(z\overline{\zeta}-1)\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)\overline{\zeta}^{j-k}V_{k-1}(z\overline{\zeta})\\\
&+(z\overline{\zeta}-1)\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}c_{s-k+\ell}\overline{c_{s-j+\ell}}\right)z^{j-k}V_{k-1}(z\overline{\zeta})\\\
&+\sum_{j=0}^{s-1}\left(\sum_{\ell=s-j}^{s}|c_{\ell}|^{2}\right)+\sum_{0\leq
k<j\leq
s-1}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)\overline{\zeta}^{j-k}\\\
&+\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}c_{s-k+\ell}\overline{c_{s-j+\ell}}\right)z^{j-k}\,.\end{split}$
(2.37)

We now focus on the quantity $N_{2}(z,\zeta)$ defined in (2.24). With
${q^{\prime}}^{*}$ as defined in (2.21), we have

$\frac{({q^{\prime}}^{*}q)^{(\ell}(0)}{\ell!}=\frac{\overline{(q^{*}q^{\prime})^{(2s-1-\ell)}(0)}}{(2s-1-\ell)!},\qquad
0\leq\ell\leq 2s-1,$

and

$\displaystyle
z^{s-1}\overline{(q^{*}q^{\prime})(\zeta)}={}z^{s-1}\sum_{\ell=0}^{2s-1}\frac{\overline{(q^{*}q^{\prime})^{(\ell)}(0)}}{\ell!}\overline{\zeta}^{\ell}.$

With the aid of these two relations, we can then expand the derivatives in
(2.24) to get

$\displaystyle N_{2}(z,\zeta)={}$
$\displaystyle\overline{\zeta}^{s-1}\sum_{\ell=0}^{s-2}\frac{(q^{*}q^{\prime})^{(\ell)}(0)}{\ell!}z^{\ell}+z^{s-1}\sum_{\ell=0}^{s-1}\frac{\overline{(q^{*}q^{\prime})^{(\ell)}(0)}}{\ell!}\overline{\zeta}^{\ell}\,.$
(2.38)

Using that

$\displaystyle\frac{(q^{*}q^{\prime})^{(\ell)}(0)}{\ell!}={}$
$\displaystyle\sum_{j=0}^{\ell}\frac{q^{(j+1)}(0)}{j!}\frac{{q^{*}}^{(\ell-j)}(0)}{(\ell-j)!}=\sum_{j=1}^{\ell+1}jc_{j}\overline{c_{s-1-\ell+j}}\,,$

we can then write (2.38) as

$\displaystyle\begin{split}N_{2}(z,\zeta)={}&(z\overline{\zeta}-1)\sum_{\ell=1}^{s}\ell|c_{\ell}|^{2}V_{s-2}(z\overline{\zeta})+\sum_{\ell=1}^{s}\ell|c_{\ell}|^{2}\\\
&+(z\overline{\zeta}-1)\sum_{\ell=0}^{s-2}z^{s-1-\ell}V_{\ell-1}(z\overline{\zeta})\sum_{j=1}^{\ell+1}j\overline{c_{j}}c_{s-1-\ell+j}\\\
&+(z\overline{\zeta}-1)\sum_{\ell=0}^{s-2}\overline{\zeta}^{s-1-\ell}V_{\ell-1}(z\overline{\zeta})\sum_{j=1}^{\ell+1}jc_{j}\overline{c_{s-1-\ell+j}}\\\
&+\sum_{\ell=0}^{s-2}\overline{\zeta}^{s-1-\ell}\sum_{j=1}^{\ell+1}jc_{j}\overline{c_{s-1-\ell+j}}+\sum_{\ell=0}^{s-2}z^{s-1-\ell}\sum_{j=1}^{\ell+1}j\overline{c_{j}}c_{s-1-\ell+j}\,.\end{split}$
(2.39)

If we now rewrite the last five summands of (2.39) in the form

$\sum_{\ell=1}^{s}\ell|c_{\ell}|^{2}=\sum_{k=0}^{s-1}\sum_{\ell=s-k}^{s}|c_{\ell}|^{2},$
$\displaystyle\sum_{u=0}^{s-2}\overline{\zeta}^{s-1-u}\sum_{j=1}^{u+1}jc_{j}\overline{c_{s-1-u+j}}={}$
$\displaystyle\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)\overline{\zeta}^{j-k},$

and

$\displaystyle\sum_{u=0}^{s-2}\overline{\zeta}^{s-1-u}V_{u-1}(z\overline{\zeta})\sum_{j=1}^{u+1}jc_{j}\overline{c_{s-1-u+j}}$
$\displaystyle\quad=\sum_{0\leq k<j\leq
s-1}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)V_{s-2-j+k}(z\overline{\zeta})\overline{\zeta}^{j-k},$

and then subtract (2.37) from (2.39) we obtain

$\displaystyle\begin{split}\frac{N_{2}(z,\zeta)-L(z,\zeta)}{(z\overline{\zeta}-1)}={}&\sum_{k=0}^{s-2}(z\overline{\zeta})^{k}V_{s-2-k}(z\overline{\zeta})\left(\sum_{\ell=s-k}^{s}|c_{\ell}|^{2}\right)\\\
&+\sum_{0\leq k<j\leq
s-2}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)V_{s-2-j}(z\overline{\zeta})z^{k}\overline{\zeta}^{j}\\\
&+\sum_{0\leq k<j\leq
s-2}\left(\sum_{\ell=0}^{k}c_{s-k+\ell}\overline{c_{s-j+\ell}}\right)V_{s-2-j}(z\overline{\zeta})z^{j}\overline{\zeta}^{k}\,.\end{split}$
(2.40)

We then replace the terms in the right-hand side of (2.40) with the equivalent
expressions

$\displaystyle\sum_{k=0}^{s-2}(z\overline{\zeta})^{k}V_{s-2-k}(z\overline{\zeta})\left(\sum_{\ell=s-k}^{s}|c_{\ell}|^{2}\right)={}$
$\displaystyle\sum_{j=0}^{s-2}(z\overline{\zeta})^{j}\sum_{\ell=0}^{j}(\ell+1)|c_{s-j+\ell}|^{2}\,,$
$\displaystyle\sum_{0\leq k<j\leq
s-2}\left(\sum_{\ell=0}^{k}\overline{c_{s-k+\ell}}c_{s-j+\ell}\right)V_{s-2-j}(z\overline{\zeta})z^{k}\overline{\zeta}^{j}$
$\displaystyle\quad={}\sum_{0\leq k<j\leq
s-2}z^{k}\overline{\zeta}^{j}\sum_{\ell=0}^{k}(\ell+1)\overline{c_{s-k+\ell}}c_{s-j+\ell},$

and after doing that, we group the terms corresponding to $\ell=0$ to obtain

$\displaystyle\begin{split}\frac{N_{2}(z,\zeta)-L(z,\zeta)}{(z\overline{\zeta}-1)}={}&\left(\sum_{j=0}^{s-2}c_{s-j}\overline{\zeta}^{j}\right)\left(\sum_{k=0}^{s-2}\overline{c_{s-k}}z^{k}\right)\\\
&+(z\overline{\zeta})\sum_{j=0}^{s-3}(z\overline{\zeta})^{j}\sum_{\ell=0}^{j}(\ell+2)|c_{s-j+\ell}|^{2}\\\
&+(z\overline{\zeta})\sum_{0\leq k<j\leq
s-3}z^{k}\overline{\zeta}^{j}\sum_{\ell=0}^{k}(\ell+2)\overline{c_{s-k+\ell}}c_{s-j+\ell}\\\
&+(z\overline{\zeta})\sum_{0\leq k<j\leq
s-3}z^{j-1}\overline{\zeta}^{k-1}\sum_{\ell=0}^{k}(\ell+2)c_{s-k+\ell}\overline{c_{s-j+\ell}}\,.\end{split}$
(2.41)

By the same argument we used to derive (2.26) from (2.35), we can iterate
(2.41) to finally get

$\displaystyle N_{2}(z,\zeta)-L(z,\zeta)={}$
$\displaystyle(z\overline{\zeta}-1)\sum_{r=0}^{s-2}(r+1)(z\overline{\zeta})^{r}\left(\sum_{j=0}^{s-2-r}c_{s-j}\overline{\zeta}^{j}\right)\left(\sum_{k=0}^{s-2-r}\overline{c_{s-k}}z^{k}\right).$

Inserting (2.25), (2.27), and the latter equality into (2.22) (recall
definition (1.6)) gives

$\displaystyle N(z,\zeta)={}$ $\displaystyle
q^{*}(z)\overline{q^{*}(\zeta)}-(1-z\overline{\zeta})^{2}\sum_{r=0}^{s-2}(r+1)T_{r}(z)\overline{T_{r}(\zeta)}$
$\displaystyle+(1-z\overline{\zeta})\sum_{k=1}^{s}A_{k}q_{k}^{*}(z)\overline{q_{k}^{*}(\zeta)}-(1-z\overline{\zeta})^{2}\sum_{k=1}^{s}A_{k}L_{k}(z,\zeta).$

In view of (2.13), this establishes Theorem 1.1.

###  2.2 Proof of formula (1.13)

The proof is basically the same given in [3], the argument simply works
equally well for a product of different powers of Blaschke factors. For every
$f$ in the disk algebra, a residue calculation gives (this is (2) in [3])

$\displaystyle\begin{split}\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{B(z)}{B(\zeta)}\frac{f(\zeta)}{\zeta-z}d\zeta={}&f(z)-B(z)\sum_{k=1}^{s}\frac{1-|a_{k}|^{2}}{B_{k}(a_{k})}\frac{f(a_{k})}{z-a_{k}}.\end{split}$
(2.42)

Letting $w$ be as in (2.5), and since $h\equiv 1$ on the unit circle, we can
apply Green’s formula to obtain

$\displaystyle\begin{split}&\frac{1}{2\pi
i}\int_{|\zeta|=1}\frac{B(z)}{B(\zeta)}\frac{f(\zeta)}{\zeta-z}d\zeta=\frac{B(z)}{2\pi
i}\int_{|\zeta|=1}\frac{\overline{\zeta}}{1-\overline{\zeta}z}\overline{B(\zeta)w(\zeta)}f(\zeta)w(\zeta)d\zeta\\\
&=\int_{\mathbb{D}}\left(\frac{B(z)\overline{B(\zeta)}}{(1-\overline{\zeta}z)^{2}}+\frac{B(z)\overline{\zeta}}{1-\overline{\zeta}z}\sum_{k=1}^{s}\frac{(\frac{p_{k}}{2}+1)(1-|a_{k}|^{2})\overline{B_{k}(\zeta)}}{(1-a_{k}\overline{\zeta})^{2}}\right)f(\zeta)h(\zeta)d\sigma(\zeta)\end{split}$
(2.43)

Writing each $f(a_{k})$ in (2.42) as the integral of $f(\zeta)$ against
$K(a_{k},\zeta)$ and replacing the left-hand side of (2.42) by the last
integral in (2.43), we obtain a representation for $f(z)$ as an integral over
$\mathbb{D}$ with respect to $h(\zeta)d\sigma(\zeta)$, whose integrand is the
product of $f(\zeta)$ and the function

$\displaystyle\begin{split}&\frac{B(z)\overline{B(\zeta)}}{(1-\overline{\zeta}z)^{2}}+\frac{B(z)\overline{\zeta}}{1-\overline{\zeta}z}\sum_{k=1}^{s}\frac{(\frac{p_{k}}{2}+1)(1-|a_{k}|^{2})\overline{B_{k}(\zeta)}}{(1-a_{k}\overline{\zeta})^{2}}\\\
&+\sum_{k=1}^{s}\frac{(1-|a_{k}|^{2})B_{k}(z)}{(1-\overline{a}_{k}z)B_{k}(a_{k})}K(a_{k},\zeta).\end{split}$

Since this function is analytic in $z$ and anti-analytic in $\zeta$, it must
coincide with $K(z,\zeta)$, and from this expression for the kernel we can
continue as in [3] to derive (1.13).

## References

  * [1] P. Duren, A. Schuster, Bergman Spaces, Mathematical Surveys and Monographs 100, American Mathematical Society, Providence, Rhode Island, 2004.
  * [2] J. Hansbo, Reproducing kernels and contractive divisors in Bergman spaces, J. Math. Sci. 92 (1998), 3657-3674.
  * [3] T. H. MacGregor, M. I. Stessin: Weighted reproducing kernels in Bergman spaces, Michigan Math. J. 41 (1994), 523-533.