Planar Beurling Transform and Bergman Type Spaces

Abstract

In this paper we describe the actions of the operator $S_\mathbb{D }$ or its adjoint $S_\mathbb{D }^*$ on the poly-Bergman spaces of the unit disk $\mathbb{D }.$ Let $k$ and $j$ be positive integers. We prove that $(S_\mathbb{D })^{j}$ is an isometric isomorphism between the true poly-Bergman subspace $\mathcal{A }_{(k)}^2(\mathbb{D })\ominus N_{(k),j}$ onto the true poly-Bergman space $\mathcal{A }_{(j+k)}^2(\mathbb{D }),$ where the linear space $N_{(k),j}$ have finite dimension $j.$ The action of $(S_\mathbb{D })^{j-1}$ on the canonical Hilbert base for the Bergman subspace $\mathcal{A }^2(\mathbb{D })\ominus \mathcal{P }_{j-1},$ gives a Hilbert base $\{ \phi _{ j , k } \}_{ k }$ for $\mathcal{A }_{(j)}^2(\mathbb{D }).$ It is shown that $\{ \phi _{ j , k } \}_{ j, k }$ is a Hilbert base for $L^2(\mathbb{D },d A)$ such that whenever $j$ and $k$ remain constant we obtain a Hilbert base for the true poly-Bergman space $\mathcal{A }_{(j)}^2(\mathbb{D })$ and $\mathcal{A }_{(-k)}^2(\mathbb{D }),$ respectively. The functions $\phi _{ j , k }$ are polynomials in $z$ and $\overline{z}$ and are explicitly given in terms of the $(2,1)$-hypergeometric polynomials. We prove explicit representations for the true poly-Bergman kernels and the Koshelev representation for the poly-Bergman kernels of $\mathbb{D }.$ The action of $S_\Pi$ on the true poly-Bergman spaces of the upper half-plane $\Pi$ allows one to introduce Hilbert bases for the true poly-Bergman spaces, and to give explicit representations of the true poly-Bergman and poly-Bergman kernels.

1 Introduction

Let $\Omega$ be a non-empty, open, connected (domain) subset of $\mathbb{C }$ equipped with the Lebesgue area measure $dA(z)=dxdy$. Define $\mathbb{Z }_\pm :=\mathbb{Z }\backslash \{0\}$ and denote the set of positive and negative integers by $\mathbb{Z }_+$ and $\mathbb{Z }_-,$ respectively. If $j\in \mathbb{Z }_+$ and $j\in \mathbb{Z }_-,$ then a smooth complex function $f$ defined on $\Omega$ and satisfying

$$\begin{aligned} \partial _{\overline{z}}^{j} f=0 ,\,\,\,\,\,\,j\in \mathbb{Z }_{+} \quad \text{ or}\quad \partial _{z}^{-j} f=0 ,\,\,\,\,\,\,j\in \mathbb{Z }_{-}, \end{aligned}$$

(1.1)

respectively, is said to be a $j$-analytic function on $\Omega$. The poly-Bergman space $\mathcal{A }^2_j(\Omega )$ consists of $j$-analytic functions on $\Omega$ which also belong to the Lebesgue space $L^2(\Omega ,dA).$ It has been shown [9, Proposition 2.3] (also see [3, 7]) that point-evaluations within the unit disk $\mathbb{D }$ are uniformly bounded on $\mathcal{A }^2_j(\mathbb{D })$, from which it follows that $\mathcal{A }^2_j(\Omega )$ is a Hilbert space equipped with the inner product induced by that in $L^2(\Omega ,dA)$ (see, e.g. [2, 3, 7, 9]). As a consequence of the continuity of point-evaluation in $\Omega ,$ the poly-Bergman space is a reproducing kernel Hilbert space (RKHS) of functions on $\Omega .$ In particular, there exists a unique reproducing kernel function $K_{ \Omega , j }(z,w),$ the so-called poly-Bergman kernel. According to [7, Proposition 2.1], for every positive integer $j,$ the poly-Bergman kernel of $\mathbb{D }$ is explicitly represented by

$$\begin{aligned} K_{\mathbb{D }, j }(z,w)=\frac{j}{\pi } \frac{\sum _{k=1}^j (-1)^{k-1} {j \atopwithdelims ()k} {j+k-1 \atopwithdelims ()j} | z - w |^{2(k-1)} | 1 - \overline{w} z |^{2(j-k)} }{(1-\overline{w} z)^{2j}}. \end{aligned}$$

(1.2)

The space $\mathcal{A }^2_1(\Omega )$ is the usual Bergman space of the domain $\Omega$, which is also denoted by $\mathcal{A }^2(\Omega ).$

The planar Beurling transform is the two-dimensional singular integral operator given by

$$\begin{aligned} S f(z):=-\frac{1}{\pi }\int _{\mathbb{C }} \frac{f(w)}{(w-z)^{2}} d A(w) ,\,\,\,\,\,\,z\in \mathbb{C }. \end{aligned}$$

(1.3)

It is well known (see, e.g. [8, 12]) that the Beurling transform is a unitary operator acting on $L^2(\mathbb{C }, dA).$ Hence, for every domain $\Omega \subset \mathbb{C },$ the compression of the Beurling transform to $L^2(\Omega ,d A)$ defines a bounded operator acting in the space $L^2(\Omega ,d A),$ i.e. the operator given by

$$\begin{aligned} S_{\Omega }:=\chi _\Omega S \chi _\Omega , \end{aligned}$$

(1.4)

where $\chi _\Omega$ denotes the characteristic function of $\Omega ,$ is a bounded linear operator acting on $L^2(\Omega ,dA).$ The singular integral operator $S_\Omega$ is related with the poly-Bergman projection $B_{\Omega ,j},$ which is defined to be the orthogonal projection of $L^2(\Omega ,dA )$ onto the poly-Bergman space $\mathcal{A }^2_{j}(\Omega ).$ For instance, in the unit disk case it is known (see [6, Theorem 2.3]) that

$$\begin{aligned} B_{\mathbb{D },j}\!=\! I \!-\! (S_{\mathbb{D }})^j (S_\mathbb{D }^*)^j,\,\,\,\,\,\,j\in \mathbb{Z }_+ \quad \mathrm{;}\quad B_{\mathbb{D },j}\!=\! I - (S_{\mathbb{D }}^*)^j (S_\mathbb{D })^j,\,\,\,\,\,\,j\in \mathbb{Z }_-.\quad \quad \end{aligned}$$

(1.5)

Considering (1.5) we will describe the compression of $S_\mathbb{D }$ to the space $\mathcal{A }^2_j(\mathbb{D })$. The assumption that the poly Bergman projection over $\Omega$ lies in the $*$-algebra generated by $S_\Omega ,$ which is satisfied e.g. if $\Omega$ is the unit disk, the complement of a disk, a half-plane or a bounded multi connected domain with a smooth boundary (see [3, 9]), is an important step in [3, 5, 11] and [14]. Accordingly to [6] and [9], the equalities in (1.5) are said to be the Dzhuraev formulas for $\mathbb{D }$.

In this paper we show that the operator $(S_\mathbb{D })^j$ is an isometric isomorphism of $\mathcal{A }_{(k)}^2 (\mathbb{D }) \ominus N_{(k),j}$ onto $\mathcal{A }_{(k+j)}^2 (\mathbb{D }),$ where $N_{(k),j}$ is a $j$-dimensional space, $\mathcal{A }_{(k)}^2 (\mathbb{D })$ is the true poly-Bergman space consisting of functions in $\mathcal{A }_{k}^2 (\mathbb{D })$ which are orthogonal to $\mathcal{A }_{k-1}^2 (\mathbb{D })$ [see (2.5)] and $j$ and $k$ are positive integers. Whenever $k=1$, we obtain, for every $j=1,2,\ldots ,$ that the following operator

$$\begin{aligned} (S_\mathbb{D })^{j}: \mathcal{A }^2 (\mathbb{D }) \ominus \mathcal{P }_{j} \rightarrow \mathcal{A }_{(j+1)}^2 (\mathbb{D }) \end{aligned}$$

is an isometric isomorphism, where $\mathcal{P }_{j}$ denotes the linear space of polynomials in the complex variable $z$ with degree not greater than $j-1.$ The action of $S_\mathbb{D }$ on the poly-Bergman spaces of the unit disk, is different from those in the cases of the upper half-plane (see [5, 14]) and of the complement of a disk (see [9]). Considering the orthogonal decomposition of the Lebesgue space $L^2 (\mathbb{D }, dA)$ on the true poly-Bergman spaces (see [6, Corollary 4.4]), we obtain a Hilbert base $\{ \phi _{ j , k } \}_{ j, k }$ for $L^2(\mathbb{D },d A)$ such that whenever $j$ and $k$ remain constant we achieve a Hilbert base for the true poly-Bergman space $\mathcal{A }_{(j)}^2(\mathbb{D })$ and $\mathcal{A }_{(-k)}^2(\mathbb{D }),$ respectively. The functions $\phi _{j,k}$ will be explicitly represented in terms of the $(2,1)$-hypergeometric polynomials. In Proposition 4.1 we prove that the latter Hilbert base for the poly-Bergman space coincide with that one introduced in [7, Theorem 1]. In Proposition 7.1 we give a more transparent proof of the representation (1.2) for $K_{\mathbb{D },j}(z,w),$ that can be found in [7, Theorem 2]. Explicit representations for the true poly-Bergman kernels of the unit disk are given in Theorem 8.2. The action of $(S_\Pi )^j$ on poly-Bergman spaces of the upper half-plane and the same kind of techniques, allow us to prove representations for the true poly-Bergman kernel of upper half-plane $\Pi ,$ different and more transparent than those found in [15, Theorem 3.4.1] (see also [13]). We also give explicit Hilbert bases for the poly Bergman spaces of the upper half-plane.

The paper is organised as follows. In Sect. 2 we define the spaces $N_{k,j}$ and $N_{(k),j}$ as the intersection of poly-Bergman and true poly-Bergman spaces of order $k,$ respectively, with the poly-Bergman space of order $-j.$ Then we prove that $(S_\mathbb{D })^j$ acting on the true poly-Bergman space $\mathcal{A }^2_{(k)}(\mathbb{D })$ has null space given by the $j$-dimensional space $N_{(k),j}$, and also that $(S_\mathbb{D })^j$ is a unitary operator from $\mathcal{A }_{(k)}^2 (\mathbb{D }) \ominus N_{(k),j}$ onto $\mathcal{A }_{(k+j)}^2.$ In Sect. 3, considering the results of the previous section, we introduce a Hilbert base $\{\phi _{j,k}\}_{j,k}$ for $L^2(\mathbb{D },d A)$ (see Proposition 3.3) and we prove that it contains Hilbert bases for the linear spaces $N_{k,j}, N_{(k),j}, N_{k,(j)}$ and $N_{(k),(j)},$ likewise for $\mathcal{A }_{j}^2 (\mathbb{D })$ and $\mathcal{A }_{(j)}^2 (\mathbb{D })$, obtained just by natural restrictions on its indices $j$ and $k.$ In Sect. 4 the functions $\phi _{j,k}$ are explicitly written in terms of $(2,1)$-hypergeometric polynomials. In Sect. 5 we compute the one dimensional projection $B_{\mathbb{D }, (j) } B_{\mathbb{D }, (-k)},$ and, as a consequence, the functions $\phi _{j,k}$ are represented in terms of the derivatives of the true poly-Bergman and poly-Bergman kernel functions. In Sect. 6 we introduce unitary operators acting on the poly-Bergman spaces $\mathcal{A }_j^2(\mathbb{D })$ and induced by the conformal automorphisms of $\mathbb{D }.$ The latter mentioned unitary operators appear in Sect. 7 to prove (1.2). Section 8 is devoted to obtain an explicit formula for the true poly-Bergman kernel of $\mathbb{D }$ [see Theorem 8.2 and (8.3)]. The poly-Bergman spaces over the upper half-plane are considered in Sect. 9. Making use of the technique on this paper, we give Hilbert bases for the true poly-Bergman space (see Proposition 9.1), from which it follows new and more transparent representations for the true poly-Bergman kernel (see Theorem 9.3). We define unitary operators acting between the poly-Bergman spaces of $\mathbb{D }$ and of $\Pi ,$ to give a new proof for the representation of the poly-Bergman kernel of the upper half-plane, found in [10, Corollary 2.5].

2 The Beurling Transform and Poly-Bergman Spaces

Let $j$ and $k$ be positive integers and consider the following linear space

$$\begin{aligned} N_{j,k}:=\mathcal{A }^2_{j}(\mathbb{D })\cap \mathcal{A }^2_{-k}(\mathbb{D }). \end{aligned}$$

(2.1)

If $\Omega$ is a bounded domain then we know that (see [9, Proposition 3.6])

$$\begin{aligned} \mathcal{A }^2_{j}(\Omega )\cap \mathcal{A }^2_{-k}(\Omega ) = \mathrm{span}\{ z^l \overline{z}^n : l=0,1,\ldots ,k-1;\, n= 0,\ldots , j-1 \}. \end{aligned}$$

(2.2)

Thus, from (2.1) and (2.2) it is straightforward to conclude that

$$\begin{aligned} \mathcal{P }_{j} = N_{1,j} \quad \text{ and}\ N_{ j , k} = \overline{N}_{k,j}. \end{aligned}$$

(2.3)

From [6, Lemma 3.1] we know that $B_{\mathbb{D },j}B_{\mathbb{D },-k}$ is the orthogonal projection of $L^2(\mathbb{D },d A)$ onto the $jk$ dimensional space $N_{j,k}.$ Therefore, the projections $B_{\mathbb{D },j}$ and $B_{\mathbb{D },-k}$ commute. By analogy with (2.1), we define

$$\begin{aligned} N_{(j),k}:= \mathcal{A }^2_{(j)}(\mathbb{D })\cap \mathcal{A }^2_{-k}(\mathbb{D }) \quad \text{ and}\quad N_{j,(k)}:= \mathcal{A }^2_{j}(\mathbb{D })\cap \mathcal{A }^2_{(-k)}(\mathbb{D }), \end{aligned}$$

(2.4)

where, for every non-zero integer $j,$ the true poly-Bergman space $\mathcal{A }_{(j)}^{2}(\mathbb{D })$ is given by

$$\begin{aligned} \mathcal{A }_{(\pm 1)}^{2}(\mathbb{D }):=\mathcal{A }_{\pm 1}^{2}(\mathbb{D }) \quad \text{ and}\quad \mathcal{A }_{(j)}^{2}(\mathbb{D }):=\mathcal{A }_{j}^{2}(\mathbb{D })\ominus \mathcal{A }_{j-\mathrm{sgn}j}^{2}(\mathbb{D }) ,\,\,\,\,\,\,|j|>1.\qquad \end{aligned}$$

(2.5)

The true poly-Bergman projection $B_{\mathbb{D }, (j)}$ is the orthogonal projection of $L^2(\mathbb{D })$ onto the space $\mathcal{A }^2_{(j)}(\mathbb{D }).$ It is clear that $B_{\mathbb{D }, (\pm 1)} = B_{\mathbb{D }, \pm 1}$ and

$$\begin{aligned} B_{\mathbb{D }, (j)} = B_{\mathbb{D }, j} - B_{\mathbb{D }, j-1} ,\,\,\,\,\,\,j>1 \quad \text{ and}\quad B_{\mathbb{D }, (j)} = B_{\mathbb{D }, j} - B_{\mathbb{D }, j+1} ,\,\,\,\,\,\,j<-1\nonumber \\ \end{aligned}$$

(2.6)

From (2.6) it follows that $B_{\mathbb{D }, (n)}$ commute with $B_{\mathbb{D }, m},$ for every non-zero integers $n$ and $m.$ Hence, the operators $B_{\mathbb{D }, (j)}B_{\mathbb{D }, -k}$ and $B_{\mathbb{D }, j}B_{\mathbb{D }, (-k)}$ are the orthogonal projections of $L^2(\mathbb{D },d A)$ onto $N_{(j),k}$ and $N_{j,(k)},$ respectively. The following operator

$$\begin{aligned} W: L^2(\mathbb{D })\rightarrow L^2(\mathbb{D }), \quad W f(z)= f(\overline{z}) \end{aligned}$$

(2.7)

is a unitary operator acting from $\mathcal{A }^2_{j} (\mathbb{D })$ onto $\mathcal{A }^2_{-j} (\mathbb{D })$. Therefore

$$\begin{aligned} W B_{\mathbb{D },j} W = B_{\mathbb{D },-j} \quad \text{ and}\quad W B_{\mathbb{D }, (j)} W = B_{\mathbb{D },(-j)}. \end{aligned}$$

Theorem 2.1

Let $j\in \mathbb{Z }_+$ and $k\in \mathbb{Z }_{\pm }.$ The operators

$$\begin{aligned} \begin{aligned} (S_{\mathbb{D }})^{j} : \mathcal{A }_{(k)}^2 (\mathbb{D }) \ominus N_{(k),j}&\rightarrow \mathcal{A }_{(k+j)}^2 (\mathbb{D }) , \quad k>0 \\ (S_{\mathbb{D }})^{j} : \mathcal{A }_{(k)}^2 (\mathbb{D })&\rightarrow \mathcal{A }_{(k+j)}^2 (\mathbb{D }) , \quad 0< j<-k \end{aligned} \end{aligned}$$

(2.8)

as well as the following ones

$$\begin{aligned} \begin{aligned} (S_{\mathbb{D }}^*)^{j} : \mathcal{A }_{(k)}^2(\mathbb{D }) \ominus N_{j,(-k)}&\rightarrow \mathcal{A }_{(k-j)}^2 (\mathbb{D }) ,\quad k<0 \\ (S_{\mathbb{D }}^*)^{j} : \mathcal{A }_{(k)}^2 (\mathbb{D })&\rightarrow \mathcal{A }_{(k-j)}^2 (\mathbb{D }) , \quad 0<j<k \end{aligned} \end{aligned}$$

(2.9)

are isometric isomorphisms. Furthermore

$$\begin{aligned} \mathrm{Ker}(S_{\mathbb{D }}^*)^{j} = \mathcal{A }_j^2(\mathbb{D }) \supset \mathcal{A }^2(\mathbb{D }) \quad \mathrm{and} \quad \mathrm{Ker}(S_{\mathbb{D }})^{j} = \mathcal{A }_{-j}^2 (\mathbb{D })\supset \mathcal{A }^2_{-1}(\mathbb{D }). \end{aligned}$$

(2.10)

Proof

It is easily seen that if $\mathcal{H }$ is a Hilbert space, $S$ is a bounded operator acting on $\mathcal{H }$ and the operator $D:=I-SS^*$ is such that $\mathrm{Im}D \subset N$, then $D$ is the orthogonal projection onto the closed subspace $N$ if and only if $S^*(N)=0.$ Thus, (2.10) follows from (1.5). Next, we suppose that $j$ is a positive integer. If $W$ is the unitary operator defined in (2.7), then straightforward substitutions and change of variable in the singular integral shows that

$$\begin{aligned} W (S_{\mathbb{D }})^j W = (S_{\mathbb{D }}^*)^j. \end{aligned}$$

Since $W$ transforms the poly-Bergman space $\mathcal{A }^2_j(\mathbb{D })$ onto $\mathcal{A }^2_{-j}(\mathbb{D })$, then

$$\begin{aligned} W (\mathcal{A }^2_{(j)}(\mathbb{D }))= \mathcal{A }^2_{(-j)}(\mathbb{D }) ,\,\,\,\,\,\,j\in \mathbb{Z }_\pm \quad \text{ and}\quad W (N_{j, (k)}) = N_{(k),j} ,\,\,\,\,\,\,j,k \in \mathbb{Z }_+. \end{aligned}$$

It is now straightforward to check that (2.9) follows from (2.8). Let us prove (2.8). Recall that if $\mathcal{H }_1$ and $\mathcal{H }_2$ are Hilbert spaces, then a bounded operator $P:\mathcal{H }_1\rightarrow \mathcal{H }_2$ is a partial isometry with initial space $N$ and final space $M$ if and only if $P^*P$ and $PP^*$ are orthogonal projections of $\mathcal{H }_1$ and $\mathcal{H }_2$ onto $N$ and $M,$ respectively. In particular, from (1.5) it follows that the operator $(S_ \mathbb{D })^j$ is a partial isometry with initial space $[\mathcal{A }^2_{-j}(\mathbb{D })]^\bot$ and with final space $[\mathcal{A }^2_{j}(\mathbb{D })]^\bot .$ Therefore, $(S_ \mathbb{D })^j$ is a unitary operator acting on the space $\mathcal{A }^2_{(k)}(\mathbb{D }) \ominus \mathcal{A }^2_{-j}(\mathbb{D }),$ which is the image of the operator $B_{\mathbb{D }, (k)}(I-B_{\mathbb{D }, -j}).$ If $k$ is a negative integer and $j<-k,$ then it is clear that

$$\begin{aligned} \mathcal{A }^2_{(k)}(\mathbb{D }) \ominus \mathcal{A }^2_{-j}(\mathbb{D }) = \mathcal{A }^2_{(k)}(\mathbb{D }). \end{aligned}$$

If $k$ is a positive integer and $P_{(k),j}$ denotes the orthogonal projection of $L^2$ onto $N_{(k),j},$ then

$$\begin{aligned} B_{\mathbb{D }, (k)}(I-B_{\mathbb{D }, -j})= B_{\mathbb{D }, (k)}-B_{\mathbb{D }, (k)} B_{\mathbb{D }, -j } = B_{\mathbb{D }, (k)}-P_{(k),j}. \end{aligned}$$

Thus, it is easy to check that

$$\begin{aligned} \mathcal{A }^2_{(k)}(\mathbb{D }) \ominus \mathcal{A }^2_{-j}(\mathbb{D }) =\mathcal{A }^2_{(k)}(\mathbb{D }) \ominus N_{(k),j} ;\,\,\,\,\,\,k\in \mathbb{Z }_+. \end{aligned}$$

For the remaining it is sufficient to prove that $(S_{\mathbb{D }})^{j} ( \mathcal{A }^2_{(k)}(\mathbb{D }))$ coincides with the true poly-Bergman space $\mathcal{A }^2_{(k+j)}(\mathbb{D })$. Since the final space of the partial isometry $(S_\mathbb{D }^*)^j$ equals $[\mathcal{A }^2_{-j}(\mathbb{D })]^\bot$, then, from

$$\begin{aligned} B_{\mathbb{D }, (k+j)}&= \displaystyle (S_{\mathbb{D }})^j B_{\mathbb{D }, (k)} (S_\mathbb{D }^*)^j = (S_{\mathbb{D }})^j B_{\mathbb{D }, (k)} \left( I - B_{\mathbb{D }, -j} \right)(S_\mathbb{D }^*)^j\\&= \displaystyle (S_{\mathbb{D }})^j B_{\mathbb{D }, (k)} \left( I - P_{(k),j} \right)(S_\mathbb{D }^*)^j \end{aligned}$$

it follows that

$$\begin{aligned} \mathcal{A }^2_{(k+j)}(\mathbb{D }) = (S_{\mathbb{D }})^{j} \left( \mathcal{A }^2_{(k)}(\mathbb{D }) \ominus N_{(k),j}\right). \end{aligned}$$

$\square$

Corollary 2.2

Let $j\in \mathbb{Z }_+$ and $k\in \mathbb{Z }_{\pm }.$ The operators

$$\begin{aligned} \begin{aligned} (S_{\mathbb{D }})^{j} : \mathcal{A }_{k}^2 (\mathbb{D }) \ominus N_{k,j}&\rightarrow \mathcal{A }_{k+j}^2 (\mathbb{D }) \ominus \mathcal{A }_{j}^2 (\mathbb{D }) , \quad k>0 \\ (S_{\mathbb{D }})^{j} : \mathcal{A }_{k}^2 (\mathbb{D })&\rightarrow \mathcal{A }_{k+j}^2 (\mathbb{D }) \ominus \mathcal{A }_{j}^2 (\mathbb{D }) ; \quad 0<j<-k \end{aligned} \end{aligned}$$

(2.11)

as well as the following ones

$$\begin{aligned} \begin{aligned} (S_{\mathbb{D }}^*)^{j} : \mathcal{A }_{k}^2(\mathbb{D }) \ominus N_{j,-k}&\rightarrow \mathcal{A }_{k-j}^2 (\mathbb{D }) \ominus \mathcal{A }_{-j}^2 (\mathbb{D }) , \quad k<0 \\ (S_{\mathbb{D }}^*)^{j} : \mathcal{A }_{k}^2 (\mathbb{D })&\rightarrow \mathcal{A }_{k-j}^2 (\mathbb{D }) \ominus \mathcal{A }_{-j}^2 (\mathbb{D }) ;\quad 0< j<k \end{aligned} \end{aligned}$$

(2.12)

are isometric isomorphisms.

Proof

As in the proof of Theorem 2.1, one can easily check that (2.12) follows from (2.11). Furthermore, in light of the definition of true poly-Bergman space, it is clear that

$$\begin{aligned} \mathcal{A }^2_j (\mathbb{D }) = \overset{\,j}{\underset{\,n=1}{\begin{array}{c}\bigoplus \end{array}}} \mathcal{A }^2_{(n)} (\mathbb{D }) ,\,\,\,\,\,\,j\in \mathbb{Z }_+. \end{aligned}$$

(2.13)

Then, to prove (2.11) it is sufficient to consider (2.8) jointly with the following

$$\begin{aligned} \overset{\,k}{\underset{\,n=1}{\begin{array}{c}\bigoplus \end{array}}} N_{(n),j} = \overset{\,k}{\underset{\,n=1}{\begin{array}{c}\bigoplus \end{array}}} \mathcal{A }^2_{-j}(\mathbb{D })\cap \mathcal{A }^2_{(n)}(\mathbb{D }) = \mathcal{A }^2_{-j}(\mathbb{D })\cap \mathcal{A }^2_{k}(\mathbb{D }) = N_{k,j}. \end{aligned}$$

(2.14)

$\square$

3 Hilbert Base for Bergman Type Spaces

For every positive integer $j,$ let

$$\begin{aligned} \phi _j(z) = \sqrt{\frac{j}{\pi }}\; z^{j-1} ,\,\,\,\,\,\,z\in \mathbb{D }. \end{aligned}$$

(3.1)

It is well known that $\{\phi _j\}_j$ is a Hilbert base for the Bergman space $\mathcal{A }^2(\mathbb{D }).$ Define the functions

$$\begin{aligned} \phi _{j,k}:= (S_{\mathbb{D }})^{j-1} \phi _{k+j-1} \quad \text{ where}\quad j, k =1,2,\ldots \end{aligned}$$

(3.2)

If $\eta _{n,m}$ denotes the function $\overline{z}^n z^m,$ then from [6, Lemma 2.2] we know that

$$\begin{aligned} \big (S_\mathbb{D }\,\eta _{n,m}\big )(z)=\frac{m}{n+1}\,{z}^{m-1}\overline{z}^{n+1} +\frac{\min \{0,n-m+1\}}{n+1}\,{z}^{m-n-2}. \end{aligned}$$

(3.3)

Therefore, $\phi _{j,k}$ lies in the set of polynomials $\mathbb{P }[z,\overline{z}]$ in the variables $z$ and $\overline{z}.$

Proposition 3.1

For every positive integer $j,$ the following sets

$$\begin{aligned} \mathcal{O }_{(j)} := \left\{ \phi _{j,m} \right\} _{m} \quad \text{ and}\quad \mathcal{O }_{j} := \displaystyle \left\{ \phi _{n,m} \right\} _{n\le j,\, m} \end{aligned}$$

are Hilbert bases for the spaces $\mathcal{A }^2_{(j)}(\mathbb{D })$ and $\mathcal{A }^2_{j}(\mathbb{D }),$ respectively.

Proof

Let $j$ be a positive integer. In light of (2.3) we know that $\mathcal{P }_{j} = N_{1,j} = N_{(1), j}.$ Since $\{\phi _n\}_n$ and $\{\phi _n\}_{n\le j - 1}$ are Hilbert bases for $\mathcal{A }^2(\mathbb{D })$ and $\mathcal{P }_{j-1},$ respectively, one obtains that the set of functions $\phi _n,$ where $n = j,\ldots ,$ is a Hilbert base for the space $\mathcal{A }^2(\mathbb{D })\ominus \mathcal{P }_{j-1}.$ From Theorem 2.1 we know that the following operator

$$\begin{aligned} (S_{\mathbb{D }})^{j-1} : \mathcal{A }^2 (\mathbb{D }) \ominus \mathcal{P }_{j-1} \rightarrow \mathcal{A }_{(j)}^2 (\mathbb{D }) \end{aligned}$$

(3.4)

is a unitary operator. Therefore, the set $\{ (S_\mathbb{D })^{j-1} \phi _n\}_{n\ge j }$ is a Hilbert base for $\mathcal{A }^2_{(j)}(\mathbb{D }),$ i.e. we conclude that $\mathcal{O }_{(j)}$ is a Hilbert base for $\mathcal{A }^2_{(j)}(\mathbb{D })$. From the decomposition (2.13) and from

$$\begin{aligned} \mathcal{O }_{j}= \overset{\,j}{\underset{\,k=1}{\begin{array}{c}\bigcup \end{array}}} \mathcal{O }_{(k)} \end{aligned}$$

it follows that $\mathcal{O }_{j}$ is a Hilbert base for the poly-Bergman space $\mathcal{A }^2_{j}(\mathbb{D })$. $\square$

Let $j$ be a positive integer and define the following sets of functions

$$\begin{aligned} \mathcal{O }_{(-j)} := \overline{\mathcal{O }}_{(j)} \quad \text{ and}\quad \mathcal{O }_{-j} := \overline{\mathcal{O }}_{j}. \end{aligned}$$

(3.5)

If $Cf = \overline{f}$ denotes the anti-linear isomorphism of complex conjugation, acting on $L^2(\mathbb{D }),$ then $C(\mathcal{A }_j(\mathbb{D }))=\mathcal{A }_{-j}(\mathbb{D })$ and $C(\mathcal{A }_{(j)}(\mathbb{D }))=\mathcal{A }_{(-j)}(\mathbb{D }).$ Therefore, the sets $\mathcal{O }_{(-j)}$ and $\mathcal{O }_{-j}$ are Hilbert bases for $\mathcal{A }^2_{(-j)}(\mathbb{D })$ and for $\mathcal{A }^2_{-j}(\mathbb{D }),$ respectively. Since $C S_\mathbb{D }C = S_\mathbb{D }^*,$ then

$$\begin{aligned} \overline{\phi }_{n,m} = (S_{\mathbb{D }}^*)^{n-1} \overline{\phi }_{m+n-1} ;\,\,\,\,\,\,n, m =1,2,\ldots . \end{aligned}$$

Corollary 3.2

Let $j$ be a positive integer. Then

$$\begin{aligned} \mathcal{O }_{(-j)} = \left\{ \phi _{m, j } \right\} _m \quad \text{ and}\quad \mathcal{O }_{-j} = \left\{ \phi _{m, n } \right\} _{n \le j, \, m}. \end{aligned}$$

Furthermore, $\mathcal{O }_{(-j)}$ and $\mathcal{O }_{-j}$ are Hilbert bases for the true poly-Bergman space $\mathcal{A }^2_{(-j)}(\mathbb{D })$ and for the Bergman space $\mathcal{A }^2_{-j}(\mathbb{D }),$ respectively.

Proof

In (4.10) we will independently prove that $\overline{\phi }_{j,m} = \phi _{m,j}$. Therefore, the result easily follows from the definition (3.5) and from the considerations below it. $\square$

Corollary 3.3

The set $\{\phi _{n,m}\}_{n,m}$ is a Hilbert base for $L^2(\mathbb{D },d A).$

Proof

From [6, Theorem 4.3] we know that the Lebesgue space $L^2(\mathbb{D },dA)$ admits the following orthogonal decompositions in terms of true poly-Bergman spaces

$$\begin{aligned} L^2(\mathbb{D })=\overset{\,\infty }{\underset{\,k=1}{\begin{array}{c}\bigoplus \end{array}}}\mathcal{A }_{(k)}^{2}(\mathbb{D }) =\overset{\,\infty }{\underset{\,k=1}{\begin{array}{c}\bigoplus \end{array}}} \mathcal{A }_{(-k)}^{2}(\mathbb{D }). \end{aligned}$$

Therefore, from Proposition 3.1 one completes the proof. $\square$

By analogy with (2.1) and (2.4) we introduce the following linear spaces

$$\begin{aligned} N_{(j),(k)}:= \mathcal{A }^2_{(j)}(\mathbb{D })\cap \mathcal{A }^2_{(-k)}(\mathbb{D }) ,\,\,\,\,\,\,j,k=1,2,\ldots . \end{aligned}$$

Proposition 3.4

Let $j$ and $k$ be positive integers. The following sets

$$\begin{aligned} \begin{array}{rllrl} \mathcal{O }_{(j),k}&:= \{\phi _{j,m}\}_{m\le k} ,&\quad \mathcal{O }_{j,(k)}&:= \{\phi _{n,k} \}_{n\le j } \\ \mathcal{O }_{j,k}&:= \{ \phi _{n,m} \}_{n\le j, m\le k} ,&\quad \mathcal{O }_{(j),(k)}&:= \{ \phi _{j,k} \} \end{array} \end{aligned}$$

are Hilbert bases for the spaces $N_{(j),k}, N_{j,(k)}, N_{j,k}$ and $N_{(j),(k)},$ respectively.

Proof

It is clear that

$$\begin{aligned} \mathcal{O }_{(j),k} = \overset{\,k}{\underset{\,m=1}{\begin{array}{c}\bigcup \end{array}}} \mathcal{O }_{(j),(m)},\,\,\, \mathcal{O }_{j, (k)} = \overset{\,j}{\underset{\,n=1}{\begin{array}{c}\bigcup \end{array}}} \mathcal{O }_{(n),(k)} \,\,\,\mathrm{and} \,\,\, \mathcal{O }_{j, k} = \overset{\,j}{\underset{\,n=1}{\begin{array}{c}\bigcup \end{array}}} \overset{\,k}{\underset{\,m=1}{\begin{array}{c}\bigcup \end{array}}} \mathcal{O }_{(m),(n)}. \end{aligned}$$

Furthermore, like in (2.14), one easily obtains

$$\begin{aligned} N_{(j),k} = \overset{\,k}{\underset{\,m=1}{\begin{array}{c}\bigoplus \end{array}}} N_{(j),(m)},\,\,\, N_{j,(k)} = \overset{\,j}{\underset{\,n=1}{\begin{array}{c}\bigoplus \end{array}}} N_{(n),(k)} \,\,\,\mathrm{and} \,\,\, N_{k,j}= \overset{\,k}{\underset{\,m=1}{\begin{array}{c}\bigoplus \end{array}}} \overset{\,j}{\underset{\,n=1}{\begin{array}{c}\bigoplus \end{array}}} N_{(n),(m)}.\qquad \end{aligned}$$

(3.6)

Hence, it is sufficient to prove the statement about $\mathcal{O }_{(j),(k)}.$ Let $m=1,2,\ldots .$ From the definition of poly-Bergman space, we know that $\phi _{m}$ belongs to $\mathcal{A }^2_{-m}(\mathbb{D }).$ It is clear that $\phi _{1}\in \mathcal{A }^2_{(-1)}(\mathbb{D }).$ Suppose $m\ge 2.$ From [6, Lemma 2.1] we know that the linear span of $\overline{z}^l z^n,$ for $l=0,\ldots$ and $n=0,\ldots , m-2,$ is dense in the space $\mathcal{A }^2_{1-m}(\mathbb{D }).$ Therefore, considering that

$$\begin{aligned} \langle \phi _m\,,\,\overline{z}^l z^k \rangle = \sqrt{\frac{m-1}{\pi }} \langle z^{m+l-1} \,,\, z^k \rangle = 0 ;\,\,\,\,\,\,k=0,\ldots , m-2,\, l= 0,\ldots , \end{aligned}$$

one concludes that $\phi _{m}$ belongs to the poly-Bergman space $\mathcal{A }^2_{(-m)}(\mathbb{D }).$ Hence, from Theorem 2.1, the function $(S_{\mathbb{D }})^{j-1} \phi _{m}$ lies in the true poly-Bergman space of order $j-m-1,$ whenever $j-m-1$ is a negative integer. If $m=j+k-1,$ then one has that $\phi _{j, k}$ lies in $\mathcal{A }_{(-k)}(\mathbb{D }).$ Since $\phi _m\in \mathcal{A }^2(\mathbb{D }),$ we know from Theorem 2.1 that $\phi _{j, k}\in \mathcal{A }_{(j)}(\mathbb{D }).$ In summary, it can be concluded that $\phi _{j, k}\in N_{(j),(k)}.$ Thus, $\dim N_{(k), (j)} \ge 1.$ On the other hand, from (2.2) and (3.6) we have

$$\begin{aligned} kj= \dim N_{k, j} = \sum _{j,k} \dim N_{(k), (j)}, \end{aligned}$$

which allows us to conclude that

$$\begin{aligned} \dim N_{(k), (j)} = 1 \quad \text{ and}\quad N_{(j), (k)}=\{\lambda \phi _{j, k} : \lambda \in \mathbb{C }\}. \end{aligned}$$

$\square$

4 Explicit Representations for the Function $\phi _{j,k}$

Let $j$ be a positive integer. From [6, Corollary 2.4] we know that

$$\begin{aligned} S_{ \mathbb{D }, j } = ( S_{\mathbb{D }}^*)^j \quad \mathrm{and} \quad S_{ \mathbb{D }, -j } = ( S_{ \mathbb{D }} )^j, \end{aligned}$$

(4.1)

where $S_{\Omega ,j}$ denotes the singular integral operator acting on $L^2(\Omega ,dA)$

$$\begin{aligned} S_{\Omega ,j} u (z):=\frac{(-1)^{j}|j|}{\pi }\int _{\Omega } \frac{(w-z)^{j-1}}{(\overline{w}-\overline{z})^{j+1}} u (w)d A(w) ,\,\,\,\,\,\,j\in \mathbb{Z }_{\pm }. \end{aligned}$$

(4.2)

Considering that for every $u$ in the Lebesgue space $L^2(\mathbb{D },dA)$ and continuous at $z,$ one has

$$\begin{aligned} \lim _{\epsilon \rightarrow 0^+}\int \limits _{|z-w|=\epsilon } \frac{(\overline{w}-\overline{z})^{j-1}}{(w-z)^{j}} u(w) d w = 0, \end{aligned}$$

we apply the following Green’s formula

$$\begin{aligned} \int _\mathbb{D }\partial _{\overline{w}} u (w) d A(w) = \frac{1}{2i} \int _{\partial \mathbb{D }} u(w) d w , \quad u\in C^1(\mathrm{cl}{\mathbb{D }}) \end{aligned}$$

(4.3)

to obtain, for every function $u\in \mathcal{A }^2(\mathbb{D }) \cap C^1(\mathrm{cl}\mathbb{D })$ that

$$\begin{aligned} (S_\mathbb{D })^{j-1} u (z) = \frac{(-1)^{j-1}}{2\pi i} \int _{\partial \mathbb{D }} \frac{(\overline{w}-\overline{z})^{j-1}}{(w-z)^{j}} u (w) d w. \end{aligned}$$

(4.4)

Let $k$ be a positive integer. From (4.1) we know that (3.2) can be rewritten as $\phi _{j,k} = S_{\mathbb{D },1-j} \phi _{j+k-1}.$ Therefore,

$$\begin{aligned} \phi _{j,k}(z) = \sqrt{ \frac{j+k-1}{\pi } } I_{j,k}(z), \end{aligned}$$

(4.5)

where $I_{j,k}(z)$ is the line integral given by

$$\begin{aligned} I_{j,k}(z) :=\displaystyle \frac{1}{2\pi i }\int _{\partial \mathbb{D }} \frac{(\overline{z}-\overline{w})^{j-1}}{(w-z)^{j}} w^{j+k-2} d w. \end{aligned}$$

(4.6)

In [7] the functions $\phi _{j,k}$ were defined by (4.7) below. There one can look for suggestions towards a different proof of the assertion that $\mathcal{O }_{j}$ is a Hilbert base for the poly-Bergman space $\mathcal{A }^2_j(\mathbb{D }),$ which is a consequence of Proposition 3.1.

Proposition 4.1

Let $j$ and $k$ be positive integers. Then

$$\begin{aligned} \phi _{j,k}(z) = \frac{ \sqrt{j+k-1}}{\sqrt{\pi } (j+k-2)!} \partial _{{z}}^{j-1} \partial _{\overline{z}}^{k-1} (\overline{z} z-1)^{j+k-2}. \end{aligned}$$

(4.7)

Proof

We consider (4.5), (4.6) and the Cauchy integral formula to obtain

$$\begin{aligned} I_{j,k}(z)&= \displaystyle \displaystyle \frac{1}{2\pi i}\int _{\partial \mathbb{D }} \frac{(\overline{z} w -1 )^{j-1}}{(w-z)^{j}} w^{k-1} d w \\&= \displaystyle \frac{\partial _{{z}}^{j-1} \left[ (\overline{z} z - 1)^{j-1} z^{k-1} \right] }{(j-1)!}= \displaystyle \frac{\partial _{{z}}^{j-1} \partial _{\overline{z}}^{k -1} (\overline{z} z - 1)^{j + k - 2}}{( j + k - 2 )!}. \end{aligned}$$

$\square$

Let $j$ and $k$ be positive integers. Next we show that the functions $\phi _{j,k}$ are related with the $(2,1)$-hypergeometric polynomials. Indeed, from Proposition 4.1 we obtain that

$$\begin{aligned} \displaystyle \frac{\sqrt{\pi } \, \phi _{j,k}(z) }{ \sqrt{j+k-1} }&= \displaystyle \frac{\partial _{\overline{z}}^{k-1} \left[ \overline{z}^{j-1} (\overline{z} z-1)^{k-1} \right] }{(k-1)!}= \frac{\partial _{\overline{z}}^{k-1} \left[ (z \overline{z})^{j-1} (\overline{z} z-1)^{k-1} \right] }{ z^{j-1} (k-1)!}\nonumber \\&= \displaystyle \frac{z^{1-j}}{ (k-1)!} \partial _{\overline{z}}^{k-1} \sum _{n=0}^{j-1} {j-1\atopwithdelims ()n } (\overline{z} z-1)^{n+k-1} \nonumber \\&= \displaystyle z^{k-j} \frac{(j-1)!}{(k-1)!} \sum _{n=0}^{j-1} (-1)^n \frac{(n+k-1)!}{(j-n-1)! (n!)^2} (1-\overline{z} z)^{n}. \end{aligned}$$

(4.8)

Define $(a)_0 :=1$ and $(a)_n :=a(a+1)\cdots (a+n-1),$ for $n=1,2,\ldots$ and $a\in \mathbb{C }$. Since

$$\begin{aligned} (-1)^n \frac{( j - 1 )! }{( k - 1 )!} \frac{(n+k-1)!}{(j-n-1)! n!} = \frac{(1-j)_n (k)_n}{(1)_n } \,;\,\, n=0,\ldots ,j-1 \end{aligned}$$

it follows from (4.8) that

$$\begin{aligned} \phi _{j,k}(z) = \frac{\sqrt{j+k-1}}{\sqrt{\pi }}z^{k-j} F(1-j,k;1; 1-\overline{z} z), \end{aligned}$$

(4.9)

where $F(-n,b;c;z)$ is the $(2,1)$-hypergeometric polynomial given for complex numbers $b$ and $z$, $c\in \mathbb{C }{\setminus }\{0,-1,-2,\ldots ,-k+1\}$ and $n=0,1,\ldots$ by

$$\begin{aligned} F(-n,b;c;z)=\sum _{k=0}^n \frac{(-n)_k (b)_k }{(c)_{k} k! }z^k. \end{aligned}$$

It is straightforward from the Proposition 4.1 to conclude that

$$\begin{aligned} \phi _{j,k}(z) = \overline{\phi }_{k,j }(z) ;\,\,\,\,\,\,j,k=1,2\ldots . \end{aligned}$$

(4.10)

Therefore, (4.9) can be rewritten in the following form

$$\begin{aligned} \phi _{j,k}(z) = \frac{\sqrt{j+k-1}}{\sqrt{\pi }}\overline{z}^{j-k} F(1-k,j;1; 1-\overline{z} z). \end{aligned}$$

(4.11)

For $n=0,1,\ldots$ the $(2,1)$-hypergeometric polynomials $F(-n,b;c;z)$ are related with the so-called Jacobi polynomials $P_n^{(\alpha ,\beta )}$. Indeed, from [1, Definition 2.5.1] we know that

$$\begin{aligned} P_n^{(\alpha ,\beta )}(x) := \frac{(\alpha + 1)_n}{n!}F(-n, n + \alpha + \beta + 1; \alpha +1; (1-x)/2). \end{aligned}$$

(4.12)

Proposition 4.2

Let $j$ and $k$ be positive integers. Then

$$\begin{aligned} \phi _{j,k}(z)&= \displaystyle \displaystyle (-1)^{j-1}\frac{\sqrt{j+k-1}}{\sqrt{\pi }} {k-1\atopwithdelims ()j-1} z^{k-j} F(1-j,k;k-j+1; \overline{z} z)\nonumber \\&= \displaystyle \frac{ \sqrt{j+k-1}}{ \sqrt{\pi }} z^{k-j} P_{j-1}^{(0,k-j)}(2\overline{z} z-1) ,\,\,\,\,\,\,j \le k. \end{aligned}$$

(4.13)

Furthermore

$$\begin{aligned} \phi _{ j , k }(z)&= \displaystyle (-1)^{k-1} \frac{\sqrt{j+k-1}}{\sqrt{\pi }} {j-1\atopwithdelims ()k-1} \overline{z}^{j-k} F(1-k,j;j-k+1; \overline{z} z)\nonumber \\&= \displaystyle \frac{ \sqrt{j+k-1}}{ \sqrt{\pi }} \overline{z}^{j-k} P_{k-1}^{(0,j-k)}(2\overline{z} z-1) ,\,\,\,\,\,\,k\le j. \end{aligned}$$

(4.14)

Proof

From (4.10) it is clear that (4.14) follows from (4.13). To prove (4.13), we combine (4.9) with the following well known Pfaff’s formula (see [1, Corollary 2.3.3])

$$\begin{aligned} F(-n,b;c; x) = \frac{(c-b)_n}{(c)_n} F(-n,b;b+1-n-c; 1-x). \end{aligned}$$

$\square$

Let $P_n(x)$ denote the Legendre polynomials $P_n(x):=P_n^{(0,0)}(x).$ From (4.12) it is clear that

$$\begin{aligned} P_n (x) = F(-n ,n+1;1; (1 - x)/2) ,\,\,\,\,\,\,n=0,1,\ldots . \end{aligned}$$

(4.15)

Corollary 4.3

Let $j\in \mathbb{Z }_+$ and let $P_j(x)$ be the Legendre polynomial. Then

$$\begin{aligned} \phi _{j,j}(z) = \sqrt{\frac{2j-1}{\pi } } P_{j-1} (2 \overline{z} z - 1). \end{aligned}$$

Note that from (3.2) and (4.10) and for every positive integer $j$, it is clear that

$$\begin{aligned} \phi _{1,j}(z) =\phi _{j}(z) = \sqrt{\frac{j}{\pi }}\; z^{j-1} \quad \text{ and}\quad \phi _{j,1}(z) = \sqrt{\frac{j}{ \pi }} \; \overline{z} ^{j-1} =\overline{\phi }_{j}(z). \end{aligned}$$

5 The Projection $B_{\mathbb{D }, (j)} B_{\mathbb{D },(-k)}$ and Representations for $\phi _{j,k}$

From [16, p. 61] and from [6, section 4] we know that the linear space $C^{\infty }(\Omega )\cap L^2(\Omega )$ is invariant for the operators $S_{\Omega }$ and $S_{\Omega }^*.$ Furthermore, if $f \in C^{\infty }(\Omega )\cap L^2(\Omega ),$ then

$$\begin{aligned} \partial _{\overline{z}}S_{\Omega }f=\partial _z f, \quad \partial _zS_{\Omega }^*f=\partial _{\overline{z}}f. \end{aligned}$$

(5.1)

The derivation formulas (5.1) will be used in the proof of the following result.

Proposition 5.1

Let $j$ and $k$ be positive integers. Then

$$\begin{aligned} B_{\mathbb{D }, (j) } B_{\mathbb{D }, (-k)} g = \langle g\,,\,\phi _{j,k} \rangle \phi _{j,k} ,\,\,\,\,\,\,g\in L^2(\mathbb{D }). \end{aligned}$$

(5.2)

Furthermore, if $g$ lies in the poly-Bergman space $\mathcal{A }^2_j(\mathbb{D }),$ then

$$\begin{aligned} \langle g\,,\,\phi _{j,k} \rangle = \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \left( \partial _z^{k-1} \partial _{\overline{z}}^{j-1} g\right) (0). \end{aligned}$$

Proof

Since the projections $B_{\mathbb{D }, (j)}$ and $B_{\mathbb{D }, (-k)}$ commute, then the product $B_{\mathbb{D }, (j)} B_{\mathbb{D }, (-k)}$ is the orthogonal projection onto the subspace $N_{(j),(k)}.$ Thus, (5.2) follows straightforward from Proposition 3.1. Furthermore, it is well known that the Bergman kernel of $\mathbb{D }$ is given by

$$\begin{aligned} K_\mathbb{D }(z,w)= \frac{1}{\pi } \frac{1}{(1- \overline{w} z )^2 } ;\,\,\,\,\,\,z,w\in \mathbb{D }. \end{aligned}$$

(5.3)

Apply the differential operator $\partial _{\overline{w}}^n$ to both sides of (5.3) to obtain that

$$\begin{aligned} \partial _{\overline{w}}^n K_\mathbb{D }( z ,w)_{|w=0} = \frac{(n+1)!}{\pi } z^n. \end{aligned}$$

Therefore, for every $f$ lying in the Bergman space $\mathcal{A }^2(\mathbb{D })$ we conclude

$$\begin{aligned} \displaystyle \frac{(n+1)!}{\pi } \langle f\!\!\,,\,z^n \rangle&= \displaystyle \langle f\!\!\,,\,\partial _{\overline{w} }^n K_\mathbb{D }( \cdot ,w) \rangle _{|w=0} = \partial _w^n \langle f\!\!\,,\, K_\mathbb{D }( w ,\cdot ) \rangle _{|w=0} \nonumber \\&= \displaystyle f^{(n)}(0). \end{aligned}$$

(5.4)

Since $(S_{\mathbb{D }}^*)^{j-1}g\in \mathcal{A }^2(\mathbb{D })$ whenever $g\in \mathcal{A }^2_j(\mathbb{D }),$ it follows from (3.2), (5.1) and (5.4) that

$$\begin{aligned} \langle g\,,\,\phi _{j,k} \rangle&= \displaystyle \langle (S_{\mathbb{D }}^*)^{j-1}g\,,\,\phi _{j+k-1} \rangle = { \frac{\sqrt{j+k-1}}{\sqrt{\pi }} } \langle (S_{\mathbb{D }}^*)^{j-1}g\,,\,z^{j+k-2} \rangle \\&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \left( \partial _z^{k-1} \partial _z^{j-1} (S_{\mathbb{D }}^*)^{j-1}g\right) (0)\\&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \left( \partial _z^{k-1} \partial _{\overline{z}}^{j-1} g\right) (0). \end{aligned}$$

$\square$

The case $j=1$ in Proposition 5.1 can be found in [6, Lemma 3.3] and [6, Lemma 3.4].

Corollary 5.2

Let $j$ and $k$ be positive integers. Then

$$\begin{aligned} \phi _{j,k}(z)&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \partial _w^{j-1} \partial _{\overline{w}}^{k-1} K_{\mathbb{D }, (j)} (z,w)_{|w=0}\\&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \partial _w^{j-1} \partial _{\overline{w}}^{k-1} K_{\mathbb{D }, j } (z,w)_{|w=0}. \end{aligned}$$

Proof

Since $\phi _{j,k}$ lies in $\mathcal{A }^2_{(j)}(\mathbb{D })$ and in $\mathcal{A }^2_{j}(\mathbb{D }),$ one obtains from Proposition 5.1 that

$$\begin{aligned} \phi _{j,k}(z)&= \displaystyle \langle \phi _{j,k} \,,\, K_{\mathbb{D }, (j) } ( \cdot , z) \rangle = \langle \phi _{j,k} \,,\, K_{\mathbb{D }, j } ( \cdot , z) \rangle \\&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \partial _w^{j-1} \partial _{\overline{w}}^{k-1} K_{\mathbb{D }, (j)} (z,w)_{|w=0}\\&= \displaystyle \frac{\sqrt{\pi (j+k-1)}}{(j+k-1)!} \partial _w^{j-1} \partial _{\overline{w}}^{k-1} K_{\mathbb{D }, j} (z,w)_{|w=0}. \end{aligned}$$

$\square$

6 Unitary Operators on the Poly-Bergman Spaces

Let $w\in \mathbb{D }$ and consider the following conformal mapping of the unit disk

$$\begin{aligned} \varphi _{w}:\mathbb{D }\mapsto \mathbb{D },\quad \varphi _{w}(z) = \frac{w-z}{1-\overline{w} z}. \end{aligned}$$

The following basic properties can be checked easily

$$\begin{aligned} \varphi _{w}(0) = w,\quad \varphi _w \circ \varphi _w (z) = z \quad \text{ and}\quad \varphi _{w}^{\prime }(z) =\frac{|w|^2-1}{(1-\overline{w} z)^2}. \end{aligned}$$

The conformal mapping of the unit disk $\varphi _w$ induces the unitary shift operator

$$\begin{aligned} V_w: L^2(\mathbb{D })\rightarrow L^2(\mathbb{D }), \quad V_w f(z)= \varphi _w^{\prime }(z) f(\varphi _w (z)). \end{aligned}$$

(6.1)

Next, we assume that $j$ is a positive integer. If $f$ belongs to the poly-Bergman space $\mathcal{A }^2_j(\mathbb{D }),$ then from [2, p. 11] it follows that there exists unique analytical functions $f_{n}$ satisfying

$$\begin{aligned} f (z) = \sum _{n=0}^{j-1} \overline{z}^n f_n(z),\,\,\,\,\,\,z\in \mathbb{D }. \end{aligned}$$

Therefore,

$$\begin{aligned} V_w f (z) = \sum _{n=0}^{j-1}\overline{\varphi }_w^n (z) g_n(z),\quad g_n:= V_w f_n. \end{aligned}$$

(6.2)

For every $w$ in the unit disk, denote by $\nu _w$ the following unitary function

$$\begin{aligned} \nu _w(z):=\frac{1-w\overline{z}}{1-\overline{w} z} ,\,\,\,\,\,\,z\in \mathbb{D }. \end{aligned}$$

Hence, considering (6.2) it is straightforward to check that

$$\begin{aligned} \nu _w^ {j-1} V_w f (z) = \sum _{n=0}^{j-1} q_{n} (z)\frac{g_n(z)}{(1-\overline{w} z)^{j-1}},\,\,\,\,\,\,z\in \mathbb{D }\end{aligned}$$

(6.3)

where the $j$-analytical polynomials $q_{n}(z)$ are given by

$$\begin{aligned} q_{n}(z):= (\overline{w} - \overline{z})^n (1-w\overline{z})^{j-n-1} ;\,\,\,\,\,\,n=0,\ldots ,j-1. \end{aligned}$$

It is clear that the Bergman space is invariant for the unitary operator $V_w.$ For the poly-Bergman case we define the following unitary operator

$$\begin{aligned} U_{w,j}: L^2(\mathbb{D })\rightarrow L^2(\mathbb{D }),\quad U_{w,j} :=\nu _w^ {j-1} V_w \end{aligned}$$

(6.4)

to obtain from (6.3), that $\mathcal{A }^2_j(\mathbb{D })$ is invariant for $U_{w,j}.$ From the following equality

$$\begin{aligned} 1-w \overline{\varphi }_w (z) = \frac{1-|w|^2}{1-w\overline{z}} ,\,\,\,\,\,\,(z,w)\in \mathbb{D }\times \mathbb{D }\end{aligned}$$

it is straightforward that

$$\begin{aligned} \nu _w(z) \nu _w (\varphi _w (z)) = 1 ,\,\,\,\,\,\,z\in \mathbb{D }. \end{aligned}$$

(6.5)

Since $\varphi _w \circ \varphi _w (z) = z,$ then $V_w^* = V_w$ and considering (6.5) we obtain

$$\begin{aligned} U_{w,j}^* = V_w \nu _w^{1-j} = (\nu _w^{1-j} \circ \varphi _w) V_w = \nu _w^{j-1} V_w = U_{w,j}. \end{aligned}$$

Therefore, for every $w\in \mathbb{D }$ and $j\in \mathbb{Z }_+$ the operator $U_{w,j}$ is unitary, self-adjoint and satisfies

$$\begin{aligned} U_{w,j}(\mathcal{A }^2_j(\mathbb{D })) = \mathcal{A }^2_j(\mathbb{D }) ,\,\,\,\,\,\,w\in \mathbb{D }. \end{aligned}$$

We consider the operator $U_{w,j}$ defined by (6.4), whenever $j$ is a non-zero integer.

Proposition 6.1

Let $w\in \mathbb{D }$ and let $j$ be a non-zero integer $j.$ The operator $U_{w,j}$ is a unitary self-adjoint operator acting from $\mathcal{A }^2_j(\mathbb{D })$ onto $\mathcal{A }^2_j(\mathbb{D }).$

Proof

It remains to consider the case when $j$ is a negative integer. Let $Cf=\overline{f}$ be the anti-linear isomorphism of complex conjugation. We know that $C U_{w,-j} C$ is a unitary operator that transforms the poly-Bergman space $\mathcal{A }^2_{j}(\mathbb{D })$ onto itself. It is straightforward to check that

$$\begin{aligned} C \nu _{w}^{-j-1} I C = \nu _{w} ^{j+1}I \quad \text{ and}\quad C V_w C =\nu _w^{-2} V_{ w}. \end{aligned}$$

Therefore, the operator $\nu _{ w} ^{ j - 1} V_{ w},$ which coincides with $U_{w , j},$ transforms $\mathcal{A }_j^2(\mathbb{D })$ onto itself. To prove that $\nu _{ w } ^{j - 1} V_{ w }$ is self-adjoint one can proceed as in the case when $j$ is a positive integer. $\square$

From Proposition 6.1 one obtains that

$$\begin{aligned} U_{w,j} B_{\mathbb{D },j} U_{w,j} = B_{\mathbb{D },j}. \end{aligned}$$

If $f\in L^2(\mathbb{D },dA),$ then straightforward substitutions and changes of variable shows that $U_{w,j} B_{\mathbb{D },j} U_{w,j} f (z),$ for $z\in \mathbb{D },$ coincides with the following integral

$$\begin{aligned} \int _{\mathbb{D }} \nu _{w}^{j-1}(z) \varphi _w^{\prime }(z)K_{\mathbb{D },j} (\varphi _w(z), \varphi _w(\xi ) ) \overline{\varphi }_w^{\prime }(\xi ) \nu _{w}^{1-j}(\xi ) f (\xi ) d A (\xi ). \end{aligned}$$

From the uniqueness of the reproducing kernel function and for every $w\in \mathbb{D }$, it follows that

$$\begin{aligned} K_{\mathbb{D },j} (z,\xi ) = \nu _{w}^{j-1}(z) \varphi _w^{\prime }(z)K_{\mathbb{D },j} (\varphi _w(z), \varphi _w(\xi ) ) \overline{\varphi }_w^{\prime }(\xi ) \overline{\nu }_{w}^{j-1}(\xi ). \end{aligned}$$

(6.6)

Let $\xi =w$ in (6.6) to conclude that

$$\begin{aligned} K_{\mathbb{D },j} (z, w) = \frac{(1-w \overline{z})^{j-1}}{(1-\overline{w} z )^{j+1}} K_{\mathbb{D },j} (\varphi _w(z), 0 ) ,\,\,\,\,\,\,j\in \mathbb{Z }_{\pm }. \end{aligned}$$

(6.7)

7 The Poly-Bergman Kernel Functions

Next, we suppose that $j\in \mathbb{Z }_+.$ Since $\mathcal{O }_{(j)}$ is a Hilbert base for $\mathcal{A }^2_{(j)}(U),$ then

$$\begin{aligned} K_{\mathbb{D },(j)} (z,w) = \sum _{k=1}^{+\infty } \phi _{j,k}(z)\overline{\phi }_{j,k}(w), \end{aligned}$$

(7.1)

where the series converges in $L^2.$ It easily follows from Proposition 4.2 that

$$\begin{aligned} \phi _{j,k}(0)= (-1)^{j-1}\sqrt{ \frac{2j-1}{\pi }}\; \delta _{j,k} ;\,\,\,\,\,\,j,k=1,2, \dots , \end{aligned}$$

where $\delta _{j,k}=1,$ if $j=k$ and $\delta _{j,k}=0,$ if $j\ne k.$ From the equality $P_n(x)=(-1)^nP_n(-x)$ together with Corollary 4.3 and (7.1), it easily follows that

$$\begin{aligned} K_{\mathbb{D },(j)} (z,0) = \displaystyle \overline{\phi }_{j,j}(0) \phi _{j,j}(z) = \displaystyle \frac{2j-1}{\pi } P_{j-1} (\xi ) ,\,\,\,\,\,\,\xi = 1- 2\overline{z} z. \end{aligned}$$

(7.2)

According to [4, p. 47 (4.3.5)], we know that

$$\begin{aligned} (2j -1) P_{j-1} (x) = P_{j}^{\prime } (x) - P_{j-2}^{\prime } (x) ,\,\,\,\,\,\,j=2,\ldots , \end{aligned}$$

(7.3)

and it is easy to see that (2.6) implies that

$$\begin{aligned} K_{\mathbb{D }, (j)} (z,w) = \displaystyle K_{\mathbb{D }, j} (z,w) - K_{\mathbb{D }, j-1} (z,w) \,\,\,\mathrm and \,\,\, K_{\mathbb{D }, j} (z,w) = \displaystyle \sum _{n=1}^j K_{\mathbb{D }, (n)} (z,w). \end{aligned}$$

Therefore, combining (7.2) and (7.3), we obtain

$$\begin{aligned} K_{\mathbb{D }, j } (z,0) = \displaystyle \sum _{n=1}^j \overline{\phi }_{ n , n }(0) \phi _{ n , n } (z) =\frac{1}{\pi }\left(P_{j}^{\prime } (\xi ) + P_{j-1}^{\prime } (\xi )\right) ,\,\,\,\,\,\,\xi = 1-2\overline{z} z.\qquad \end{aligned}$$

(7.4)

Considering the well known [4, e.g. see (9.2.2)] equality

$$\begin{aligned} F^{\prime }(\alpha ,\beta ; \gamma ; x) = \frac{\alpha \beta }{\gamma }F(\alpha +1 ,\beta +1 ; \gamma + 1; x), \end{aligned}$$

(7.5)

it follows from (4.15), (7.4) and from (7.5) that

$$\begin{aligned} K_{\mathbb{D }, j } (z,0) = \frac{ j }{2\pi } \sum _{k=0} \frac{ c_{k,j}}{ (2)_{k} k!}(\overline{z} z)^k, \end{aligned}$$

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

$$\begin{aligned} c_{k,j} := (j+1) (1-j)_ k (j+2)_k + (j-1) (2-j)_ k (j+1)_k=2j (1-j)_ k (j +1)_k. \end{aligned}$$

Thus, it is now straightforward to check that

$$\begin{aligned} K_{\mathbb{D }, j } (z,0) = \frac{j^2 }{\pi } \sum _{k=0}^{j-1} \frac{ (1-j)_ k (j+1)_k}{ (2)_{k} k!}(\overline{z} z)^k = \frac{j^2 }{\pi } F(1-j, j+1; 2; \overline{z} z).\quad \quad \end{aligned}$$

(7.6)

As we shall state in the proof of the following result, the Koshelev representation for the poly-Bergman kernel function (see [7, Proposition 2.1]) follows from (6.7) and (7.6).

Proposition 7.1

If $j\in \mathbb{Z }_{\pm },$ then $K_{\mathbb{D },j}(z,w) = K_{\mathbb{D }, -j}(w,z).$ Furthermore, if $j$ is a positive integer, then

$$\begin{aligned} K_{\mathbb{D }, j }(z,w)=\frac{j}{\pi } \frac{\sum _{k=1}^j (-1)^{k-1} {j \atopwithdelims ()k} {j+k-1 \atopwithdelims ()j} | z - w |^{2(k-1)} | 1 - \overline{w} z |^{2(j-k)} }{(1-\overline{w} z)^{2j}}. \end{aligned}$$

Proof

From the reproducing property of the poly-Bergman kernel functions it follows that $K_{\mathbb{D },j}(z,w) = K_{\mathbb{D }, -j}(w,z).$ Next we assume that $j$ is a positive integer. Simple manipulations shows that

$$\begin{aligned} j \frac{ (1-j)_ k (j+1)_k}{ (2)_{k} k!} = (-1)^k { j \atopwithdelims ()k+1} { j + k\atopwithdelims ()n} ,\,\,\,\,\,\,k=0,\ldots , j-1. \end{aligned}$$

Therefore, we can now rephrase (7.6) as follows

$$\begin{aligned} K_{\mathbb{D }, j } (z,0) = \frac{ j }{\pi } \sum _{k=1}^{j} (-1)^{k-1} {j \atopwithdelims ()k} {j+k - 1 \atopwithdelims ()j}(\overline{z} z)^{k-1}. \end{aligned}$$

(7.7)

Combining (6.7) and (7.7) we conclude that

$$\begin{aligned} K_{\mathbb{D },j} (z, w) = \displaystyle \frac{ j }{\pi } \frac{(1-w \overline{z})^{j-1}}{(1-\overline{w} z )^{j+1}} \sum _{k=1}^{j} (-1)^{k-1} {j \atopwithdelims ()k} {j+k - 1 \atopwithdelims ()j}\frac{|w-z|^{2(k-1)}}{|1-\overline{w} z|^{2(k-1)}}. \end{aligned}$$

The proof is now easily completed. $\square$

8 The True Poly-Bergman Kernel Functions

The set $\mathcal{H }$ is said to be a reproducing kernel Hilbert space (RKHS) of functions on $\mathbb{D }$ whenever $\mathcal{H }$ is a Hilbert space of complex functions on $\mathbb{D },$ the inner product $(.,.)_\mathcal{H }$ of which is such that the point-evaluations in $\mathbb{D }$ are bounded on $\mathcal{H }.$ The poly-Bergman spaces are reproducing kernel Hilbert space of functions on $\mathbb{D }$ and the space $\mathcal{A }^2_{(j)}(\mathbb{D })$ is a closed subspace of $\mathcal{A }^2_{j}(\mathbb{D }).$ Therefore, the true poly-Bergman space also is a RKHS.

Proposition 8.1

Suppose that $\mathcal{A }_1$ and $\mathcal{A }_2$ are reproducing kernel Hilbert spaces of functions on $\mathbb{D },$ having reproducing kernel functions $K_1(z,w)$ and $K_2(z,w)$, respectively. Then

$$\begin{aligned} K_2(z,w) = U ( \tau _w ) (z) \quad \text{ where}\quad \tau _w (z) = \sum _n \overline{(U \phi _n) }(w) \phi _n(z), \end{aligned}$$

where $U : \mathcal{A }_1\rightarrow \mathcal{A }_2$ is a unitary operator and $\{ \phi _n \}_n$ is a Hilbert base for $\mathcal{A }_1.$

Proof

Since $U$ is unitary, we know that $\{ U \phi _n \}_n$ is a Hilbert base for $\mathcal{A }_2.$ It follows that

$$\begin{aligned} \sum _n | U \phi _n (w) |^2 = K_2(w,w)<\infty ,\,\,\,\,\,\,w\in \mathbb{D }. \end{aligned}$$

Thus, the series defining $\tau _w$ converges in the norm of the space $\mathcal{A }_1.$ Hence

$$\begin{aligned} ( U \tau _w ) (z) = \sum _n \overline{(U \phi _n) }(w) ( U \phi _n ) (z) = K_2 (z , w ) ,\,\,\,\,\,\,(z,w)\in \mathbb{D }^2. \end{aligned}$$

$\square$

We will apply Proposition 8.1. Considering the unitary operator defined in (3.4) and that the set $\{\phi _{j+k-1} \}_{k\ge 1}$ is a Hilbert base for the space $\mathcal{A }^2 (\mathbb{D }) \ominus \mathcal{P }_{j-1}$ we obtain from (3.2) that

$$\begin{aligned} \tau _w (z) = \sum _{k=1}^\infty \overline{\phi }_{j,k} (w) \phi _{j+k-1}(z) ;\,\,\,\,\,\,z,w\in \mathbb{D }. \end{aligned}$$

(8.1)

Just insert (4.9) in (8.1) to obtain that

$$\begin{aligned} \tau _w (z) = \frac{1}{\pi } \left(\frac{z}{ \overline{w} }\right)^{j-1} \sum _{n=0}^{j-1} \frac{(1-j)_n}{(n!)^2 } (1 - \overline{w} w)^n \sum _{k=1}^\infty (j+k-1) (k)_{n} (z\overline{w})^{k-1}. \end{aligned}$$

It is straightforward to check that

$$\begin{aligned} s_{j,n} (\lambda )&:= \displaystyle \sum _{k=0}^\infty (j+k)(k+1)_n \lambda ^k = \left( j \frac{\mathrm{d}^{n}}{\mathrm{d} \lambda ^n }+ \lambda \frac{\mathrm{d}^{n+1}}{\mathrm{d} \lambda ^{n+1} } \right) \sum _{k=0}^\infty \lambda ^{k+n}\\&= \displaystyle \left( j \frac{\mathrm{d}^{n}}{\mathrm{d} \lambda ^n }+ \lambda \frac{\mathrm{d}^{n+1}}{\mathrm{d} \lambda ^{n+1} } \right) \frac{\lambda ^n}{1-\lambda } = j\ \frac{n!}{(1-\lambda )^{n+1} } + \lambda \ \frac{(n+1)!}{(1-\lambda )^{n+2} }. \end{aligned}$$

Therefore,

$$\begin{aligned} \tau _w (z) = \frac{1}{\pi } \left(\frac{z}{ \overline{w} }\right)^{j-1} \sum _{n=0}^{j-1} {j-1 \atopwithdelims ()n } \left(\frac{\overline{w} w -1}{1-\overline{w} z}\right)^n \left( \frac{j}{1-\overline{w} z} + \overline{w} z \frac{n+1}{(1-\overline{w} z)^2} \right). \end{aligned}$$

Considering the binomial sum and the following computations

$$\begin{aligned} \sum _{n=0}^{j-1} {j-1\atopwithdelims ()n} (n\!+\!1) \lambda ^n \!=\! \frac{\mathrm{d}}{\mathrm{d} \lambda } \sum _{n=0}^{j-1} {j-1\atopwithdelims ()n} \lambda ^{n+1} = j (1+\lambda )^{j-1} - (j-1)(1+\lambda )^{j-2} \end{aligned}$$

we obtain that

$$\begin{aligned} \tau _w (z)&= \displaystyle \frac{ z^{j-1} }{\pi } \left[ \frac{ (w -z)^{j-2} }{ (1-\overline{w} z)^j} ( j w -(2j-1) z) + j \frac{ z \overline{w} (w -z)^{j-1}}{(1-\overline{w} z)^{j+1}} \right]\nonumber \\&= \displaystyle \frac{z^{j-1}}{\pi } \left[ j \frac{ (w -z)^{j-1} }{ (1-\overline{w} z)^{j+1} } - (j-1) z \frac{ (w -z)^{j-2} }{ (1-\overline{w} z)^{j} } \right]\nonumber \\&= \displaystyle \frac{z^{j-1}}{\pi } \left[ \left(\partial _{{w}} \frac{ (w -z)^{j} }{ (1-\overline{w} z)^{j+1} }\right) - \left(\partial _{\overline{w}} \frac{ (w -z)^{j-2} }{ (1-\overline{w} z)^{j-1} } \right) \right]. \end{aligned}$$

(8.2)

Theorem 8.2

If $j\in \mathbb{Z }_{\pm },$ then $K_{\mathbb{D }, (j)}(z,w) = K_{\mathbb{D }, (-j)}(w,z).$ Moreover, if $j$ is a positive integer, then

$$\begin{aligned} K_{\mathbb{D }, (j)} (z,w) = \frac{\partial _{{z}}^{j-1}(\overline{z} z - 1)^{j-1}}{ \pi (j-1)! } \left[ \left(\partial _{{w}} \frac{ (w -z)^{j} }{ (1-\overline{w} z)^{j+1} }\right) - \left(\partial _{\overline{w}} \frac{ (w -z)^{j-2} }{ (1-\overline{w} z)^{j-1} } \right) \right]. \end{aligned}$$

Proof

The first part of the result follows as we have proved the first part of the Proposition 7.1. Next, suppose that $j$ is a positive integer. From Proposition 8.1 and from (4.4) it follows that

$$\begin{aligned} K_{\mathbb{D }, (j)} (z,w) = \displaystyle \frac{1}{2\pi i} \int _{\partial \mathbb{D }} \frac{(\overline{z} \xi - 1)^{j-1}}{( \xi -z)^{j}} \frac{\tau _w ( \xi ) }{ \xi ^{j-1} } d \xi = \frac{\partial _{{z}}^{j-1}}{(j-1)!} \left[\frac{(\overline{z} z - 1)^{j-1} \tau _w ( z ) }{ z^{j-1}} \right], \end{aligned}$$

where $\tau _w ( z )$ is defined in (8.1). From (8.2) one easily completes the proof. $\square$

Next we give an example of how to obtain different representations for the true poly-Bergman kernels. From Proposition 4.1 together with (7.1) it easily follows that

$$\begin{aligned} K_{\mathbb{D }, (j)} (z,w) = \displaystyle \frac{ \partial _{{z}}^{j-1} \partial _{{\overline{w}}}^{j-1}}{\pi (j-1)!^2} \left[ (\overline{z} z - 1)^{j-1} (\overline{w} w - 1)^{j-1} \displaystyle \sum _{k=1}^{+\infty } (j+k-1) (z \overline{w} )^{k-1} \right]. \end{aligned}$$

Therefore, we can check at once that

$$\begin{aligned} K_{\mathbb{D }, (j)} (z,w) = \frac{\partial _{{z}}^{j-1}\partial _{{\overline{w}}}^{j-1}}{ \pi (j-1)!^2 } \left[ \frac{ (\overline{z} z - 1)^{j-1} (\overline{w} w - 1)^{j-1} (j +(1-j)\overline{w} z) }{ (1-\overline{w} z)^{2} } \right].\quad \quad \end{aligned}$$

(8.3)

9 Hilbert Base and True Poly-Bergman Kernels for Half-Planes

The upper half-plane $\Pi := \{ z : \mathrm{Im}z >0\}$ and $\mathbb{D }$ are related by the conformal mapping

$$\begin{aligned} \varphi : \Pi \mapsto \mathbb{D },\quad \varphi (z) = \frac{z-i}{ z + i}. \end{aligned}$$

(9.1)

The map $\varphi$ induces the following unitary shift operator between Lebesgue spaces

$$\begin{aligned} V_\varphi : L^2(\mathbb{D })\rightarrow L^2(\Pi ),\quad V_\varphi f(z)= \varphi ^{\prime }(z) f(\varphi (z)). \end{aligned}$$

(9.2)

We consider the function $\psi _j (z) := V_\varphi \phi _j (z)$ $(j=1,\ldots )$ to obtain that $\{ \psi _j \}_j$ is a Hilbert base for the Bergman space, where $\phi _j$ is defined in (3.1). It has been shown [5, Theorem 2.4] (see also [15, Theorem 3.5.5]), for every positive integer $j,$ that the following operator

$$\begin{aligned} (S_{\Pi })^{j-1} : \mathcal{A }^2 (\Pi ) \rightarrow \mathcal{A }_{(j)}^2 (\Pi ) \end{aligned}$$

is a unitary operator. Suppose that $j$ and $k$ are positive integers and define $\psi _{j,k}:= (S_\Pi )^{j-1} \psi _{k}.$ Hence, $\{ \psi _{j,k} \}_k$ is a Hilbert base for the true poly-Bergman space $\mathcal{A }^2_{(j)}(\Pi ).$ Whenever $j$ is a negative integer we consider $\psi _{ j , k } (z) := \overline{\psi }_{-j,k} (z).$ It follows that for every non-zero integer $j,$ the set $\{ \psi _{j,k} \}_k$ is a Hilbert base for $\mathcal{A }^2_{(j)}(\Pi ).$ Next we show how to obtain explicit representations for $\psi _{ j , k }.$ Taking into account the invariance of the operators $S_{\mathbb{D }, j }$ $(j\in \mathbb{Z }_{\pm })$ under dilatations and translations, one obtains from (4.1) that $( S_{D(ir,r)} )^j= S_{D(ir,r) , - j},$ where $D(z,r)$ denotes the disk centred at the complex number $z$ and with radius $r>0.$ A passage to the limit $r\rightarrow +\infty$ in the later equality and the Lebesgue dominated convergence theorem shows that $(S_\Pi )^j= S_{\Pi ,- j}.$ Therefore, for every positive integers $j$ and $k$, as in the proof of (4.4), a Green formula similar to (4.3) and the order at infinity of the functions $\psi _k$ implies that

$$\begin{aligned} \psi _{j,k} (z)&= \displaystyle (S_\Pi )^{j-1} \psi _k (z) = \frac{1}{2\pi i} \int _{\mathbb{R }} \frac{( \overline{z} - t )^{j-1}}{(t-z)^{j}} \psi _{k} (t) d t \nonumber \\&= \displaystyle \sqrt{\frac{k}{\pi ^3} } \int _{\mathbb{R }} \frac{( \overline{z} - t )^{j-1} (t-i)^{k-1} }{(t-z)^{j} (t+i)^{k +1} } d t \nonumber \\&= \displaystyle \frac{ 2i \sqrt{ k } }{ \sqrt{ \pi } (j-1)! } \partial _{{z}}^{j-1} \left[ \frac{( \overline{z} - z )^{j-1}(z-i)^{k-1}}{ (z+i)^{k +1} } \right]. \end{aligned}$$

(9.3)

Proposition 9.1

Let $j\in \mathbb{Z }_{\pm }$ and let $k\in \mathbb{Z }_+.$ Then

$$\begin{aligned} \psi _{j,k} (z) = \frac{ 2 i \sqrt{ k } }{ \sqrt{ \pi } (j-1)! } \partial _{{z}}^{j-1} \left[ \frac{( \overline{z} - z )^{j-1}(z-i)^{k-1}}{ (z+i)^{k +1} } \right] ;\,\,\,\,\,\,j,k \in \mathbb{Z }_+. \end{aligned}$$

Furthermore, the set $\{ \psi _{j,k} \}_k$ is a Hilbert base for $\mathcal{A }^2_{(j)}(\Pi ).$

Next we study the poly-Bergman and the true poly-Bergman kernels of the upper half-plane. The variation of the domain technique allows us to give an explicit representation for the poly-Bergman kernel function of $\Pi$. Indeed, it has been shown [10, Corollary 2.4] that

$$\begin{aligned} K_{\Pi ,j}(z,w)=\lim _{r\rightarrow +\infty } K_{D(i r,r),j}(z,w). \end{aligned}$$

Hence, for every $j\in \mathbb{Z }_+,$ it is easy to check (see [10, Corollary 2.5]) that

$$\begin{aligned} K_{\Pi ,j}(z,w) =-\frac{j}{\pi }\frac{\sum _{k=1}^j (-1)^{j-k} {j \atopwithdelims ()k} {j+k-1 \atopwithdelims ()j } \left|z-\overline{w}\right|^{2(j-k)} \left|z-w\right|^{2(k-1)}}{(z-\overline{w})^{2j}}.\quad \quad \end{aligned}$$

(9.4)

The later formula can also be obtained by other methods. Indeed, by the same method as in the proof of Proposition 6.1 and of (6.6), one obtains the following result.

Theorem 9.2

Let $j$ be a non-zero integer and let $\varphi$ be the Möbius map given by $\varphi (z) := (az+b)/(cz+d).$ If $\Omega _1$ and $\Omega _2$ are domains such that $\varphi : \Omega _1 \rightarrow \Omega _2$ is an analytic bijection, then the following operator is a unitary operator between poly-Bergman spaces

$$\begin{aligned} U_{ \varphi ,j}: \mathcal{A }_j^2( \Omega _2 ) \rightarrow \mathcal{A }_j^2( \Omega _1),\quad U_{ \varphi ,j} :=\nu _\varphi ^ {j-1} V_\varphi \end{aligned}$$

where $V_\varphi$ is defined by (9.2) and $\nu _\varphi$ is the function $\nu _\varphi (z) := \overline{(cz+d)}/(cz+d).$ Furthermore, the poly-Bergman kernel function of $\Omega _1$ and of $\Omega _2$ are related by the following equality

$$\begin{aligned} K_{ \Omega _1 , j } (z, w ) = \nu _\varphi ^{j-1} ( z ) \varphi ^{\prime } (z) K_{ \Omega _2 , j } (\varphi (z) , \varphi (w) ) \overline{\varphi }^{\prime } (w) \overline{\nu }_\varphi ^{j-1} ( w ). \end{aligned}$$

Considering the conformal mapping defined in (9.1), one obtains from Theorem 9.2 that

$$\begin{aligned} K_{\Pi , j } (z,w) = 4 \ \frac{(\overline{z} - i)^{j-1} }{ ( z + i)^{j+1} } K_{ \mathbb{D }, j } (\varphi (z) , \varphi (w) ) \frac{(w + i)^{j-1} }{ ( \overline{w} - i)^{j+1} }. \end{aligned}$$

(9.5)

Hence, (9.4) follows from (9.5) and from Proposition 7.1. Assume $j\in \mathbb{Z }_+.$ In [15, Theorem 3.4.1] it is proved an explicit representation for the true poly-Bergman kernels of $\Pi .$ Here we give a different and more transparent representation. It follows from Proposition 9.1 that

$$\begin{aligned} K_{\Pi , (j) } ( z , w )&= \displaystyle \displaystyle \sum _{k=1}^\infty \psi _{j,k} (z) \overline{\psi }_{j,k} (w)\\&= \displaystyle \frac{ 4 \,\partial _{{z}}^{j-1} \partial _{\overline{w }}^{j-1} \left[ ( \overline{z} - z )^{j-1} ( w - \overline{w} )^{j-1} \sum _{k=1}^\infty k \lambda ^{k-1} / \mu ^{k +1} \right]}{ \pi [ (j-1)! ]^2 } , \end{aligned}$$

where, for $z$ and $w$ lying in $\Pi ,$ the complex variables $\lambda$ and $\mu$ are defined by

$$\begin{aligned} \lambda = (z-i) (\overline{w} + i) \quad \text{ and}\quad \mu = (z+i) (\overline{w} - i ). \end{aligned}$$

One can easily check that

$$\begin{aligned} \sum _{k=1}^\infty k \frac{\lambda ^{k-1} }{ \mu ^{k +1}} = \frac{1}{(\mu - \lambda )^2} = -\frac{1}{4(z-\overline{w})^2} ;\,\,\,\,\,\,z,w \in \Pi . \end{aligned}$$

Therefore, the next result follows from straightforward substitutions.

Theorem 9.3

If $j\in \mathbb{Z }_{\pm },$ then $K_{\Pi , (j) } (z,w) = K_{\Pi , (-j) } (w,z).$ Moreover, if $j\in \mathbb{Z }_+,$ then

$$\begin{aligned} K_{\Pi , (j) } ( z , w ) = -\frac{\partial _{{z}}^{j-1} \partial _{\overline{w }}^{j-1} }{ \pi (j-1)!^2 } \left[ \frac{ ( \overline{z} - z )^{j-1} ( w - \overline{w} )^{j-1} }{ (z-\overline{w})^2 } \right]. \end{aligned}$$