††thanks: Supported in part by a grant from Prince of Songkla
University††thanks: Supported in part by NSF Grant DMS-0555862

# Toeplitz operators on generalized
Bergman spaces

Kamthorn Chailuek  Department of Mathematics
Prince of Songkla University
Hatyai, Songkhla, Thailand 90112<EMAIL_ADDRESS>Brian C. Hall
Department of Mathematics
University of Notre Dame
255 Hurley Building
Notre Dame IN 46556-4618 USA<EMAIL_ADDRESS>
(Date: July 10, 2009)

###### Abstract.

We consider the weighted Bergman spaces
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ where we set
$d\mu_{\lambda}(z)=c_{\lambda}(1-\left|z\right|^{2})^{\lambda}~{}d\tau(z),$
with $\tau$ being the hyperbolic volume measure. These spaces are nonzero if
and only if $\lambda>d.$ For $0<\lambda\leq d,$ spaces with the same formula
for the reproducing kernel can be defined using a Sobolev-type norm. We define
Toeplitz operators on these generalized Bergman spaces and investigate their
properties. Specifically, we describe classes of symbols for which the
corresponding Toeplitz operators can be defined as bounded operators or as a
Hilbert–Schmidt operators on the generalized Bergman spaces.

###### Key words and phrases:

Bergman space; Toeplitz operator; quantization; holomorphic Sobolev space;
Berezin transform

###### 1991 Mathematics Subject Classification:

Primary 47B35; Secondary 32A36, 81S10

##  1\. Introduction

###  1.1. Generalized Bergman spaces

Let $\mathbb{B}^{d}$ denote the (open) unit ball in $\mathbb{C}^{d}$ and let
$\tau$ denote the hyperbolic volume measure on $\mathbb{B}^{d},$ given by

$d\tau(z)=(1-|z|^{2})^{-(d+1)}~{}dz,$ (1.1)

where $dz$ denotes the $2d$-dimensional Lebesgue measure. The measure $\tau$
is natural because it is invariant under all of the automorphisms
(biholomorphic mappings) of $\mathbb{B}^{d}.$ For $\lambda>0,$ let
$\mu_{\lambda}$ denote the measure

$d\mu_{\lambda}(z)=c_{\lambda}(1-|z|^{2})^{\lambda}~{}d\tau(z),$

where $c_{\lambda}$ is a positive constant whose value will be specified
shortly. Finally, let $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ denote
the (weighted) Bergman space, consisting of those holomorphic functions on
$\mathbb{B}^{d}$ that are square-integrable with respect to $\mu_{\lambda}.$
(Often these are defined using the Lebesgue measure as the reference measure,
but all the formulas look nicer if we use the hyperbolic volume measure
instead.) These spaces carry a projective unitary representation of the group
$SU(d,1).$

If $\lambda>d,$ then the measure $\mu_{\lambda}$ is finite, so that all
bounded holomorphic functions are square-integrable. For $\lambda>d,$ we
choose $c_{\lambda}$ so that $\mu_{\lambda}$ is a probability measure.
Calculation shows that

$c_{\lambda}=\frac{\Gamma(\lambda)}{\pi^{d}\Gamma(\lambda-d)},\quad\lambda>d.$
(1.2)

(This differs from the value in Zhu’s book [Z2] by a factor of $\pi^{d}/d!,$
because Zhu uses normalized Lebesgue whereas we use un-normalized Lebesgue
measure in (1.1).) On the other hand, if $\lambda\leq d,$ then $\mu_{\lambda}$
is an infinite measure. In this case, it is not hard to show that there are no
nonzero holomorphic functions that are square-integrable with respect to
$\mu_{\lambda}$ (no matter which nonzero value for $c_{\lambda}$ we choose).

Although the holomorphic $L^{2}$ space with respect to $\mu_{\lambda}$ is
trivial (zero dimensional) when $\lambda\leq d,$ there are indications that
life does not end at $\lambda=d.$ First, the reproducing kernel for
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ is given by

$K_{\lambda}(z,w)=\frac{1}{(1-z\cdot\bar{w})^{\lambda}}$

for $\lambda>d$. The reproducing kernel is defined by the property that it is
anti-holomorphic in $w$ and satisfies

$\int_{\mathbb{B}^{d}}K_{\lambda}(z,w)f(w)~{}d\mu_{\lambda}(w)=f(z)$

for all $f\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$ Nothing unusual
happens to $K_{\lambda}$ as $\lambda$ approaches $d.$ In fact,
$K_{\lambda}(z,w):=(1-z\cdot\bar{w})^{-\lambda}$ is a “positive definite
reproducing kernel” for all $\lambda>0.$ Thus, it is possible to define a
reproducing kernel Hilbert space for all $\lambda>0$ that agrees with
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ for $\lambda>d.$

Second, in representation theory, one is sometimes led to consider spaces like
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ but with $\lambda<d.$
Consider, for example, the much-studied metaplectic representation of the
connected double cover of $SU(1,1)\cong Sp(1,\mathbb{R}).$ This representation
is a direct sum of two irreducible representations, one of which can be
realized in the Bergman space $\mathcal{H}L^{2}(\mathbb{B}^{1},\mu_{3/2})$ and
the other of which can be realized in (a suitably defined version of) the
Bergman space $\mathcal{H}L^{2}(\mathbb{B}^{1},\mu_{1/2}).$ To be precise, we
can say that the second summand of the metaplectic representation is realized
in a Hilbert space of holomorphic functions having $K_{\lambda},$
$\lambda=1/2,$ as its reproducing kernel. See [14, Sect. 4.6].

Last, one often wants to consider the infinite-dimensional
($d\rightarrow\infty$) limit of the spaces
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$ (See, for example, [25] and
[23].) To do this, one wishes to embed each space
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ isometrically into a space of
functions on $\mathbb{B}^{d+1},$ as functions that are independent of
$z_{n+1}.$ It turns out that if one uses (as we do) hyperbolic volume measure
as the reference measure, then the desired isometric embedding is achieved by
embedding $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ into
$\mathcal{H}L^{2}(\mathbb{B}^{d+1},\mu_{\lambda}).$ That is, if we use the
same value of $\lambda$ on $\mathbb{B}^{d+1}$ as on $\mathbb{B}^{d},$ then the
norm of a function $f(z_{1},\ldots,z_{d})$ is the same whether we view it as a
function on $\mathbb{B}^{d}$ or as a function on $\mathbb{B}^{d+1}$ that is
independent of $z_{d+1}.$ (See, for example, Theorem 4, where the inner
product of $z^{m}$ with $z^{n}$ is independent of $d$.) If, however, we keep
$\lambda$ constant as $d$ tends to infinity, then we will eventually violate
the condition $\lambda>d.$

Although it is possible to describe the Bergman spaces for $\lambda\leq d$ as
reproducing kernel Hilbert spaces, this is not the most convenient description
for calculation. Instead, drawing on several inter-related results in the
literature, we describe these spaces as “holomorphic Sobolev spaces,” also
called Besov spaces. The inner product on these spaces, which we denote as
$H(\mathbb{B}^{d},\lambda),$ is an $L^{2}$ inner product involving both the
functions and derivatives of the functions. For $\lambda>d,$
$H(\mathbb{B}^{d},\lambda)$ is identical to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ (the same space of functions
with the same inner product), but $H(\mathbb{B}^{d},\lambda)$ is defined for
all $\lambda>0.$

It is worth mentioning that in the borderline case $\lambda=d,$ the space
$H(\mathbb{B}^{d},\lambda)$ can be identified with the Hardy space of
holomorphic functions that are square-integrable over the boundary. To see
this, note that the normalization constant $c_{\lambda}$ tends to zero as
$\lambda$ approaches $d$ from above. Thus, the measure of any compact subset
of $\mathbb{B}^{d}$ tends to zero as $\lambda\rightarrow d^{+},$ meaning that
most of the mass of $\mu_{\lambda}$ is concentrated near the boundary. As
$\lambda\rightarrow d^{+},$ $\mu_{\lambda}$ converges, in the weak-$\ast$
topology on $\overline{\mathbb{B}^{d}},$ to the unique rotationally invariant
probability measure on the boundary. Alternatively, we may observe that the
formula for the inner product of monomials in $H(\mathbb{B}^{d},d)$ (Theorem 4
with $\lambda=d$) is the same as in the Hardy space.

###  1.2. Toeplitz operators

One important aspect of Bergman spaces is the theory of Toeplitz operators on
them. If $\phi$ is a bounded measurable function, the we can define the
Toeplitz operator $T_{\phi}$ on
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ by
$T_{\phi}f=P_{\lambda}(\phi f),$ where $P_{\lambda}$ is the orthogonal
projection from $L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ onto the holomorphic
subspace. That is, $T_{\phi}$ consists of multiplying a holomorphic function
by $\phi,$ followed by projection back into the holomorphic subspace. Of
course, $T_{\phi}$ depends on $\lambda,$ but we suppress this dependence in
the notation. The function $\phi$ is called the (Toeplitz) symbol of the
operator $T_{\phi}.$ The map sending $\phi$ to $T_{\phi}$ is known as the
Berezin–Toeplitz quantization map and it (and various generalizations) have
been much studied. See, for example, the early work of Berezin [5, 6], which
was put into a general framework in [26, 27], along with [22, 8, 7, 10], to
mention just a few works. The Berezin–Toeplitz quantization may be thought of
as a generalization of the anti-Wick-ordered quantization on $\mathbb{C}^{d}$
(see [15]).

When $\lambda<d,$ the inner product on $H(\mathbb{B}^{d},\lambda)$ is not an
$L^{2}$ inner product, and so the “multiply and project” definition of
$T_{\phi}$ no longer makes sense. Our strategy is to find alternative formulas
for computing $T_{\phi}$ in the case $\lambda>d,$ with the hope that these
formulas will continue to make sense (for certain classes of symbols $\phi$)
for $\lambda\leq d.$ Specifically, we will identify classes of symbols $\phi$
for which $T_{\phi}$ can be defined as:

  * •

A bounded operator on $H(\mathbb{B}^{d},\lambda)$ (Section 4)

  * •

A Hilbert–Schmidt operator on $H(\mathbb{B}^{d},\lambda)$ (Section 5).

We also consider in Section 3 Toeplitz operators whose symbols are polynomials
in $z$ and $\bar{z}$ and observe some unusual properties of such operators in
the case $\lambda<d.$

###  1.3. Acknowledgments

The authors thank M. Engliš for pointing out to them several useful references
and B. Driver for useful suggestions regarding the results in Section 4. This
article is an expansion of the Ph.D. thesis of the first author, written under
the supervision of the second author. We also thank the referee for helpful
comments and corrections.

##  2\. $H(\mathbb{B}^{d},\lambda)$ as a holomorphic Sobolev space

In this section, we construct a Hilbert space of holomorphic functions on
$\mathbb{B}^{d}$ with reproducing kernel $(1-z\cdot\bar{w})^{-\lambda},$ for
an arbitrary $\lambda>0.$ We denote this space as $H(\mathbb{B}^{d},\lambda).$
The inner product on this space is an $L^{2}$ inner product with respect to
the measure $\mu_{\lambda+2n},$ where $n$ is chosen so that $\lambda+2n>d.$
The inner product, however, involves not only the holomorphic functions but
also their derivatives. That is, $H(\mathbb{B}^{d},\lambda)$ is a sort of
holomorphic Sobolev space (or Besov space) with respect to the measure
$\mu_{\lambda+2n}.$ When $\lambda>d,$ our space is identical to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$—not just the same space of
functions, but also the same inner product. When $\lambda\leq d,$ the Hilbert
space $H(\mathbb{B}^{d},\lambda),$ with the associated projective unitary
action of $SU(d,1),$ is sometimes referred to as the analytic continuation
(with respect to $\lambda$) of the holomorphic discrete series.

Results in the same spirit as—and in some cases almost identical to—the
results of this section have appeared in several earlier works, some of which
treat arbitrary bounded symmetric domains and not just the ball in
$\mathbb{C}^{d}.$ For example, in the case of the unit ball in
$\mathbb{C}^{d},$ Theorem 3.13 of [30] would presumably reduce to almost the
same expression as in our Theorem 4, except that Yan has all the derivatives
on one side, in which case the inner product has to be interpreted as a limit
of integrals over a ball of radius $1-\varepsilon.$ (Compare the formula for
$\mathcal{D}_{\lambda}^{k}$ on p. 13 of [30] to the formula for $A$ and $B$ in
Theorem 4.) See also [2, 4, 21, 31, 32]. Note, however, a number of these
references give a construction that yields, for $\lambda>d,$ the same space of
functions as $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ with a different
but equivalent norm. Such an approach is not sufficient for our needs; we
require the same inner product as well as the same space of functions.

Although our results in this section are not really new, we include proofs to
make the paper self-contained and to get the precise form of the results that
we want. The integration-by-parts argument we use also serves to prepare for
our definition of Toeplitz operators on $H(\mathbb{B}^{d},\lambda)$ in Section
4. We ourselves were introduced to this sort of reasoning by the treatment in
Folland’s book [14] of the disk model for the metaplectic representation. The
paper [16] obtains results in the same spirit as those of this section, but in
the context of a complex semisimple Lie group.

We begin by showing that for $\lambda>d,$ the space
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ can be expressed as a
subspace of $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$, with a
Sobolev-type norm, for any positive integer $n.$ Let $N$ denote the “number
operator,” defined by

$N=\sum_{j=1}^{d}z_{j}\frac{\partial}{\partial z_{j}}.$

This operator satisfies $Nz^{m}=|m|z^{m}$ for all multi-indices $m.$ If $f$ is
holomorphic, then $Nf$ coincides with the “radial derivative”
$\left.df(rz)/dr\right|_{r=1}.$ We use also the operator
$\bar{N}=\sum_{j=1}^{d}\bar{z}_{j}\partial/\partial\bar{z}_{j}.$

A simple computation shows that

$(1-|z|^{2})^{\alpha}=\left(I-\frac{N}{\alpha+1}\right)(1-|z|^{2})^{\alpha+1}=\left(I-\frac{\bar{N}}{\alpha+1}\right)(1-|z|^{2})^{\alpha+1}.$
(2.1)

We will use (2.1) and the following integration by parts result, which will
also be used in Section 4.

######  Lemma 1.

If $\lambda>d$ and $\psi$ is a continuously differentiable function for which
$\psi$ and $N\psi$ are bounded, then

$\displaystyle
c_{\lambda}\int_{\mathbb{B}^{d}}\psi(z)(1-|z|^{2})^{\lambda-d-1}dz$
$\displaystyle=c_{\lambda+1}\int_{\mathbb{B}^{d}}\left[\left(I+\frac{N}{\lambda}\right)\psi\right](z)(1-|z|^{2})^{\lambda-d}~{}dz$
$\displaystyle=c_{\lambda+1}\int_{\mathbb{B}^{d}}\left[\left(I+\frac{\bar{N}}{\lambda}\right)\psi\right](z)(1-|z|^{2})^{\lambda-d}~{}dz.$

Here $dz$ is the $2d$-dimensional Lebesgue measure on $\mathbb{B}^{d}.$

###### Proof.

We start by applying (2.1) and then think of the integral over
$\mathbb{B}^{d}$ as the limit as $r$ approaches 1 of the integral over a ball
of radius $r<1.$ On the ball of radius $r,$ we write out $\partial/\partial
z_{j}$ in terms of $\partial/\partial x_{j}$ and $\partial/\partial y_{j}.$
For, say, the $\partial/\partial x_{j}$ term we express the integral as a one-
dimensional integral with respect to $x_{j}$ (with limits of integration
depending on the other variables) followed by an integral with respect to the
other variables. We then use ordinary integration by parts in the $x_{j}$
integral, and similarly for the $\partial/\partial y_{j}$ term.

The integration by parts will yield a boundary term involving
$z_{j}\psi(z)(1-|z|^{2})^{\lambda-d}$; this boundary term will vanish as $r$
tends to 1, because we assume $\lambda>d.$ In the nonboundary term, the
operator $N$ applied to $(1-|z|^{2})^{\lambda-d}$ will turn into the operator
$-\sum_{j=1}^{d}\partial/\partial z_{j}\circ z_{j}=-(dI+N)$ applied to $\psi.$
Computing from (1.2) that $c_{\lambda}/c_{\lambda+1}=(\lambda-d)/\lambda$, we
may simplify and let $r$ tend to 1 to obtain the desired result involving $N.$
The same reasoning gives the result involving $\bar{N}$ as well. ∎

We now state the key result, obtained from (2.1) and Lemma 1, relating the
inner product in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ to the inner
product in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+1})$ (compare [14, p.
215] in the case $d=1$).

######  Proposition 2.

Suppose that $\lambda>d$ and $f$ and $g$ are holomorphic functions on
$\mathbb{B}^{d}$ for which $f,$ $g,$ $Nf,$ and $Ng$ are all bounded. Then

$\left\langle
f,g\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=\left\langle
f,\left(I+\frac{N}{\lambda}\right)g\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+1})}=\left\langle\left(I+\frac{N}{\lambda}\right)f,g\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+1})}.$
(2.2)

###### Proof.

Recalling the formula (1.1) for the measure $\tau,$ we apply Lemma 1 with
$\psi(z)=\overline{f(z)}g(z)$ with $f$ and $g$ holomorphic. Observing that
$N(\bar{f}g)=\bar{f}Ng$ gives the first equality and observing that
$\bar{N}(\bar{f}g)=\overline{(Nf)}g$ gives the second equality. ∎

Now, a general function in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ is
not bounded. Indeed, the pointwise bounds on elements of
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ coming from the reproducing
kernel, are not sufficient to give a direct proof of the vanishing of the
boundary terms in the integration by parts in Proposition 2. Nevertheless,
(2.2) does hold for all $f$ and $g$ in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ provided that one interprets
the inner product as the limit as $r$ approaches 1 of integration over a ball
of radius $r.$ (See [14, p. 215] or [30, Thm. 3.13].) We are going to iterate
(2.2) to obtain an expression for the inner product on
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ involving equal numbers of
derivatives on $f$ and $g.$ This leads to the following result.

######  Theorem 3.

Fix $\lambda>d$ and a non-negative integer $n.$ Then a holomorphic function
$f$ on $\mathbb{B}^{d}$ belongs to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ if and only if $N^{l}f$
belongs to $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ for $0\leq
l\leq n.$ Furthermore,

$\left\langle
f,g\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=\left\langle
Af,Bg\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}$ (2.3)

for all $f,g\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ where

$\displaystyle A$
$\displaystyle=\left(I+\frac{N}{\lambda+n}\right)\left(I+\frac{N}{\lambda+n+1}\right)\cdots\left(I+\frac{N}{\lambda+2n-1}\right)$
$\displaystyle B$
$\displaystyle=\left(I+\frac{N}{\lambda}\right)\left(I+\frac{N}{\lambda+1}\right)\cdots\left(I+\frac{N}{\lambda+n-1}\right).$

Let us make a few remarks about this result before turning to the proof. Let
$\sigma=\lambda+2n.$ It is not hard to see that $N^{k}f$ belongs to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma})$ for $0\leq k\leq n$ if and
only if all the partial derivatives of $f$ up to order $n$ belong to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\mu}),$ so we may describe this
condition as “$f$ has $n$ derivatives in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma}).$” This condition then implies
that $f$ belongs to $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma-2n}),$ which
in turn means that $f(z)/(1-\left|z\right|^{2})^{n}$ belongs to
$L^{2}(\mathbb{B}^{d},\mu_{\sigma}).$ Since $1/(1-\left|z\right|^{2})^{n}$
blows up at the boundary of $\mathbb{B}^{d},$ saying that
$f(z)/(1-\left|z\right|^{2})^{n}$ belongs to
$L^{2}(\mathbb{B}^{d},\mu_{\sigma})$ says that $f(z)$ has better behavior at
the boundary than a typical element of
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma}).$ We may summarize this
discussion by saying that each derivative that
$f\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma})$ has in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma})$ results, roughly speaking, in
an improvement by a factor of $(1-\left|z\right|^{2})$ in the behavior of $f$
near the boundary.

This improvement is also reflected in the pointwise bounds on $f$ coming from
the reproducing kernel. If $f$ has $n$ derivatives in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma}),$ then $f$ belongs to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma-2n}),$ which means that $f$
satisfies the pointwise bounds

$\displaystyle\left|f(z)\right|$
$\displaystyle\leq\left\|f\right\|_{L^{2}(\mathbb{B}^{d},\mu_{\sigma-2n})}\left(K_{\sigma-2n}(z,z)\right)^{1/2}$
$\displaystyle=\left\|f\right\|_{L^{2}(\mathbb{B}^{d},\mu_{\sigma-2n})}\left(\frac{1}{1-\left|z\right|^{2}}\right)^{\frac{\sigma}{2}-n}.$
(2.4)

These bounds are better by a factor of $(1-\left|z\right|^{2})^{n}$ than the
bounds on a typical element of
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\sigma}).$ See also [16] for another
setting in which the existence of derivatives in a holomorphic $L^{2}$ space
can be related in a precise way to improved pointwise behavior of the
functions.

The results of the two previous paragraphs were derived under the assumption
that $\lambda=\sigma-2n>d.$ However, Theorem 4 will show that (2.4) still
holds under the assumption $\lambda=\sigma-2n>0.$

###### Proof.

If $f$ and $g$ are polynomials, then (2.3) follows from iteration of
Proposition 2. Note that $N$ is a non-negative operator on polynomials,
because the monomials form an orthogonal basis of eigenvectors with non-
negative eigenvalues. It is well known and easily verified that for any $f$ in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ the partial sums of the
Taylor series of $f$ converge to $f$ in norm. We can therefore choose
polynomials $f_{j}$ converging in norm to $f$. If we apply (2.3) with
$f=g=(f_{j}-f_{k})$ and expand out the expressions for $A$ and $B,$ then the
positivity of $N$ will force each of the terms on the right-hand side to tend
to zero. In particular, $N^{l}f_{j}$ is a Cauchy sequences in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}),$ for all $0\leq l\leq n.$
It is easily seen that the limit of this sequence is $N^{l}f$; for holomorphic
functions, $L^{2}$ convergence implies locally uniform convergence of the
derivatives to the corresponding derivatives of the limit function. This shows
that $N^{l}f$ is in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}).$ For
any $f,g\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ choose sequences
$f_{j}$ and $g_{k}$ of polynomials converging to $f,g.$ Since $N^{l}f_{j}$ and
$N^{l}g_{j}$ converge to $N^{l}f$ and $N^{l}g,$ respectively, plugging $f_{j}$
and $g_{j}$ into (2.3) and taking a limit gives (2.3) in general.

In the other direction, suppose that $N^{l}f$ belongs to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ for all $0\leq l\leq n.$
Let $f_{j}$ denote the $j$th partial sum of the Taylor series of $f$. Then
since $Nz^{m}=|m|z^{m}$ for all multi-indices $m,$ the functions $N^{l}f_{j}$
form the partial sums of a Taylor series converging to $N^{l}f_{j},$ and so
these must be the partial sums of the Taylor series of $N^{l}f.$ Thus, for
each $l,$ we have that $N^{l}f_{j}$ converges to $N^{l}f$ in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}).$ If we then apply (2.3)
with $f=g=f_{j}-f_{k},$ convergence of each $N^{l}f_{j}$ implies that all the
terms on the right-hand side tend to zero. We conclude that $f_{j}$ is a
Cauchy sequence in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ which
converges to some $\hat{f}.$ But $L^{2}$ convergence of holomorphic functions
implies pointwise convergence, so the limit in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ (i.e., $\hat{f}$) coincides
with the limit in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ (i.e.,
$f$). This shows that $f$ is in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$ ∎

Now, when $\lambda\leq d,$ Proposition 2.2 no longer holds. This is because
the boundary terms, which involve $(1-|z|^{2})^{\lambda-d},$ no longer vanish.
This failure of equality is actually a good thing, because if we take $f=g$,
then

$c_{\lambda}\int_{\mathbb{B}^{d}}\left|f\left(z\right)\right|^{2}(1-|z|^{2})^{\lambda}~{}d\tau(z)=+\infty$

for all nonzero holomorphic functions, no matter what positive value we assign
to $c_{\lambda}.$ (Recall that when $\lambda>d,$ $c_{\lambda}$ is chosen to
make $\mu_{\lambda}$ a probability measure, but this prescription does not
make sense for $\lambda\leq d.$) Although the left-hand side of (2.2) is
infinite when $f=g$ and $\lambda\leq d,$ the right-hand side is finite if
$\lambda+1>d$ and, say, $f$ is a polynomial.

More generally, for any $\lambda\leq d,$ we can choose $n$ big enough that
$\lambda+2n>d.$ We then take the right-hand side of (2.3) as a definition.

######  Theorem 4.

For all $\lambda>0,$ choose a non-negative integer $n$ so that $\lambda+2n>d$
and define

$H(\mathbb{B}^{d},\lambda)=\left\\{f\in\mathcal{H}(\mathbb{B}^{d})\left|N^{k}f\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}),~{}0\leq
k\leq n\right.\right\\}.$

Then the formula

$\left\langle f,g\right\rangle_{\lambda}=\left\langle
Af,Bg\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}$

where

$\displaystyle A$
$\displaystyle=\left(I+\frac{N}{\lambda+n}\right)\left(I+\frac{N}{\lambda+n+1}\right)\cdots\left(I+\frac{N}{\lambda+2n-1}\right)$
$\displaystyle B$
$\displaystyle=\left(I+\frac{N}{\lambda}\right)\left(I+\frac{N}{\lambda+1}\right)\cdots\left(I+\frac{N}{\lambda+n-1}\right)$

defines an inner product on $H(\mathbb{B}^{d},\lambda)$ and
$H(\mathbb{B}^{d},\lambda)$ is complete with respect to this inner product.

The monomials $z^{m}$ form an orthogonal basis for $H(\mathbb{B}^{d},\lambda)$
and for all multi-indices $l$ and $m$ we have

$\left\langle
z^{l},z^{m}\right\rangle_{\lambda}=\delta_{l,m}\frac{m!\Gamma(\lambda)}{\Gamma(\lambda+|m|)}.$

Furthermore, $H(\mathbb{B}^{d},\lambda)$ has a reproducing kernel given by

$K_{\lambda}(z,w)=\frac{1}{(1-z\cdot\bar{w})^{\lambda}}.$

Using power series, it is easily seen that for any holomorphic function $f,$
if $N^{n}f$ belongs to $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}),$
then $N^{k}f$ belongs to $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$
for $0\leq k\,<n.$

Note that the reproducing kernel and the inner product of the monomials are
independent of $n.$ Thus, we obtain the same space of functions with the same
inner product, no matter which $n$ we use, so long as $\lambda+2n>d.$

From the reproducing kernel we obtain the pointwise bounds given by
$\left|f(z)\right|^{2}\leq\left\|f\right\|_{\lambda}^{2}(1-\left|z\right|^{2})^{-\lambda}.$

###### Proof.

Using a power series argument, it is easily seen that if $f$ and $N^{k}f$
belong to $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$, then
$\left\langle
f,N^{k}f\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}\geq 0.$ From
this, we obtain positivity of the inner product
$\left\langle\cdot,\cdot\right\rangle_{\lambda}.$ If $f_{j}$ is a Cauchy
sequence in $H(\mathbb{B}^{d},\lambda),$ then positivity of the coefficients
in the expressions for $A$ and $B$ imply that for $0\leq k\leq n,$
$N^{k}f_{j}$ is a Cauchy sequence in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}),$ which converges (as in
the proof of Theorem 3) to $N^{k}f.$ This shows that $N^{k}f$ is in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ for each $0\leq k\leq n,$
and so $f\in H(\mathbb{B}^{d},\lambda).$ Further, convergence of each
$N^{k}f_{j}$ to $N^{k}f$ implies that $f_{j}$ converges to $f$ in
$H(\mathbb{B}^{d},\lambda).$

To compute the inner product of two monomials in $H(\mathbb{B}^{d},\lambda),$
we apply the definition. Since $Nz^{m}=|m|z^{m},$ we obtain

$\displaystyle\left\langle z^{l},z^{m}\right\rangle_{\lambda}$
$\displaystyle=\delta_{l,m}\left(\frac{\lambda+|m|}{\lambda}\right)\left(\frac{\lambda+1+|m|}{\lambda+1}\right)\cdots\left(\frac{\lambda+2n-1+|m|}{\lambda+2n-1}\right)\frac{m!\Gamma(\lambda+2n)}{\Gamma(\lambda+2n+|m|)}$
$\displaystyle=\delta_{l,m}\frac{m!\Gamma(\lambda)}{\Gamma(\lambda+|m|)},$

where we have used the known formula for the inner product of monomials in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ (e.g., [Z2]).

Completeness of the monomials holds in $H(\mathbb{B}^{d},\lambda)$ for
essentially the same reason it holds in the ordinary Bergman spaces. For
$f\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$, expand $f$ in a Taylor
series and then consider $\left\langle z^{m},f\right\rangle_{\lambda}$. Each
term in the inner product is an integral over $\mathbb{B}^{d}$ with respect to
$\mu_{\lambda+2n}$, and each of these integrals can be computed as the limit
as $r$ tends to 1 of integrals over a ball of radius $r<1.$ On the ball of
radius $r,$ we may interchange the integral with the sum in the Taylor series.
But distinct monomials are orthogonal not just over $\mathbb{B}^{d}$ but also
over the ball of radius $r,$ as is easily verified. The upshot of all of this
is that $\left\langle z^{m},f\right\rangle_{\lambda}$ is a nonzero multiple of
the $m$th Taylor coefficient of $f.$ Thus if $\left\langle
z^{m},f\right\rangle_{\lambda}=0$ for all $m,$ $f$ is identically zero.

Finally, we address the reproducing kernel. Although one can use essentially
the same argument as in the case $\lambda>d,$ using the orthogonal basis of
monomials and a binomial expansion (see the proof of Theorem 12), it is more
enlightening to relate the reproducing kernel in $H(\mathbb{B}^{d},\lambda)$
to that in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}).$ We require
some elementary properties of the operators $A$ and $B$; since the monomials
form an orthogonal basis of eigenvectors for these operators, these properties
are easily obtained. We need that $A$ is self-adjoint on its natural domain
and that $A$ and $B$ have bounded inverses.

Let $\chi_{z}^{\lambda+2n}$ be the unique element of
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ for which

$\left\langle\chi_{z}^{\lambda+2n},f\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}=f(z)$

for all $f$ in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n}).$
Explicitly, $\chi_{z}^{\lambda+2n}(w)=(1-\bar{z}\cdot w)^{-(\lambda+2n)}.$
(This is Theorem 2.2 of [Z2] with our $\lambda$ corresponding to $n+\alpha+1$
in [Z2].) Now, a simple calculation shows that

$(I+N/a)(1-\bar{z}\cdot w)^{-a}=(1-\bar{z}\cdot w)^{-(a+1)},$ (2.5)

where $N$ acts on the $w$ variable with $z$ fixed. From this, we see that
$N^{k}\chi_{z}^{\lambda+2n}$ is a bounded function for each fixed
$z\in\mathbb{B}^{d}$ and $k\in\mathbb{N},$ so that $\chi_{z}^{\lambda+2n}$ is
in $H(\mathbb{B}^{d},\lambda).$

For any $f\in H(\mathbb{B}^{d},\lambda)$ we compute that

$\displaystyle\left\langle
f,(AB)^{-1}\chi_{z}^{\lambda+2n}\right\rangle_{\lambda}$
$\displaystyle=\left\langle
Af,B(AB)^{-1}\chi_{z}^{\lambda+2n}\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n)}}$
$\displaystyle=\left\langle
f,\chi_{z}^{\lambda+2n}\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n)}}=f(z).$

This shows that the reproducing kernel for $H(\mathbb{B}^{d},\lambda)$ is
given by $K_{\lambda}(z,w)=\overline{[(AB)^{-1}\chi_{z}^{\lambda+2n}](w)}.$
Using (2.5) repeatedly gives the desired result. ∎

We conclude this section with a simple lemma that will be useful in Section 5.

######  Lemma 5.

For all $\lambda_{1},\lambda_{2}>0,$ if $f$ is in
$H(\mathbb{B}^{d},\lambda_{1})$ and $g$ is in $H(\mathbb{B}^{d},\lambda_{2})$
then $fg$ is in $H(\mathbb{B}^{d},\lambda_{1}+\lambda_{2}).$

###### Proof.

If, say, $\lambda_{1}>d,$ then we have the following simple argument:

$\displaystyle\left\|fg\right\|_{\lambda_{1}+\lambda_{2}}^{2}$
$\displaystyle=c_{\lambda_{1}+\lambda_{2}}\int_{\mathbb{B}^{d}}\left|f(z)\right|^{2}\left|g(z)\right|^{2}(1-\left|z\right|^{2})^{\lambda_{1}+\lambda_{2}}~{}d\tau(z)$
$\displaystyle\leq
c_{\lambda_{1}+\lambda_{2}}\left\|g\right\|_{\lambda_{2}}^{2}\int_{\mathbb{B}^{d}}\left|f(z)\right|^{2}(1-\left|z\right|^{2})^{-\lambda_{2}}(1-\left|z\right|^{2})^{\lambda_{1}+\lambda_{2}}~{}d\tau(z)$
$\displaystyle=\frac{c_{\lambda_{1}+\lambda_{2}}}{c_{\lambda_{1}}}\left\|f\right\|_{\lambda_{1}}^{2}\left\|g\right\|_{\lambda_{2}}^{2}.$

Unfortunately, $c_{\lambda_{1}+\lambda_{2}}/c_{\lambda_{1}}$ tends to infinity
as $\lambda_{1}$ approaches $d$ from above, so we cannot expect this simple
inequality to hold for $\lambda_{1}<d.$

For any $\lambda_{1},\lambda_{2}>0,$ choose $n$ so that $\lambda_{1}+n>d$ and
$\lambda_{2}+n>d.$ Then $fg$ belongs to
$H(\mathbb{B}^{d},\lambda_{1}+\lambda_{2})$ provided that $N^{n}(fg)$ belongs
to $\mathcal{H}L^{2}(\mathbb{B}^{d},\lambda_{1}+\lambda_{2}+2n).$ But

$N^{n}(fg)=\sum_{k=0}^{n}\binom{n}{k}N^{k}f~{}N^{n-k}g.$ (2.6)

Using Theorem 4, it is easy to see that if $f$ belongs to
$H(\mathbb{B}^{d},\lambda_{1})$ then $N^{k}f$ belongs to
$H(\mathbb{B}^{d},\lambda_{1}+2k)$. Thus,

$\left|N^{k}f(z)\right|^{2}\leq
a_{k}(1-\left|z\right|^{2})^{-(\lambda_{1}+2k)}.$

Now, for each term in (2.6) with $k\leq n/2$, we then obtain the following
norm estimate:

$\displaystyle
c_{\lambda_{1}+\lambda_{2}+2n}\int_{\mathbb{B}^{d}}\left|N^{k}f(z)N^{n-k}g(z)\right|^{2}~{}(1-\left|z\right|^{2})^{\lambda_{1}+\lambda_{2}+2n}~{}d\tau(z)$
$\displaystyle\leq
c_{\lambda_{1}+\lambda_{2}+2n}a_{k}\int_{\mathbb{B}^{d}}\left|N^{n-k}g(z)\right|^{2}(1-\left|z\right|^{2})^{\lambda_{2}+2n-2k}~{}d\tau(z).$
(2.7)

Since $k\leq n/2,$ we have $\lambda_{2}+2n-2k\geq\lambda_{2}+n>d.$ We are
assuming that $g$ is in $H(\mathbb{B}^{d},\lambda_{2}),$ so that $N^{n-k}g$ is
in $H(\mathbb{B}^{d},\lambda_{2}+2n-2k),$ which coincides with
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda_{2}+2n-2k}).$ Thus, under our
assumptions on $f$ and $g,$ each term in (2.6) with $k\leq n/2$ belongs to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\lambda_{1}+\lambda_{2}+2n).$ A similar
argument with the roles of $f$ and $g$ reversed takes care of the terms with
$k\geq n/2.$ ∎

##  3\. Toeplitz operators with polynomial symbols

In this section, we will consider our first examples of Toeplitz operators on
generalized Bergman spaces, those whose symbols are (not necessarily
holomorphic) polynomials. Such examples are sufficient to see some interesting
new phenomena, that is, properties of ordinary Toeplitz operator that fail
when extended to these generalized Bergman spaces. The definition of Toeplitz
operators for the case of polynomial symbols is consistent with the definition
we use in Section 4 for a larger class of symbols.

For $\lambda>d,$ we define the Toeplitz operator $T_{\phi}$ by

$T_{\phi}f=P_{\lambda}(\phi f)$

for all $f$ in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ and all
bounded measurable functions $\phi.$ Recall that $P_{\lambda}$ is the
orthogonal projection from $L^{2}(\mathbb{B}^{d},\tau)$ onto the holomorphic
subspace. Because $P_{\lambda}$ is a self-adjoint operator on
$L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ the matrix entries of $T_{\phi}$ may be
calculated as

$\left\langle
f_{1},T_{\phi}f_{2}\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=\left\langle
f_{1},\phi
f_{2}\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda})},\quad\lambda>d,$
(3.1)

for all $f_{1},f_{2}\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$ From
this formula, it is easy to see that $T_{\bar{\phi}}=(T_{\phi})^{\ast}.$

If $\psi$ is a bounded holomorphic function and $\phi$ is any bounded
measurable function, then it is easy to see that
$T_{\phi\psi}=T_{\phi}M_{\psi}.$ Thus, for any two multi-indices $m$ and $n,$
we have

$T_{\bar{z}^{m}z^{n}}=(M_{z^{m}})^{\ast}(M_{z^{n}}).$ (3.2)

We will take (3.2) as a definition for $0<\lambda\leq d.$ Our first task,
then, is to show that $M_{z^{n}}$ is a bounded operator on
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ for all $\lambda>0.$

######  Proposition 6.

For all $\lambda>0$ and all multi-indices $n,$ the multiplication operator
$M_{z^{n}}$ is a bounded operator on $H(\mathbb{B}^{d},\lambda).$ Thus, for
any polynomial $\phi,$ the Toeplitz operator $T_{\phi}$ defined in (3.2) is a
bounded operator on $H(\mathbb{B}^{d},\lambda).$

###### Proof.

The result is a is a special case of a result of Arazy and Zhang [3] and also
of the results of Section 4, but it is easy to give a direct proof. It
suffices to show that $M_{z_{j}}$ is bounded for each $j.$ Since $M_{z_{j}}$
preserves the orthogonality of the monomials, we obtain

$\left\|M_{z_{j}}\right\|=\sup_{m}\frac{\left\|z_{j}z^{m}\right\|_{\lambda}}{\left\|z^{m}\right\|_{\lambda}}=\sup_{m}\frac{m_{j}+1}{|m|+\lambda}.$

Note that $m_{j}\leq\left|m\right|$ with equality when $m_{k}=0$ for $k\neq
j.$ Thus the supremum is finite and is easily seen to have the value of 1 if
$\lambda\geq 1$ and $1/\lambda$ if $\lambda<1.$ ∎

We now record some standard properties of Toeplitz operators on (ordinary)
Bergman spaces. These properties hold for Toeplitz operators (defined by the
“multiply and project” recipe) on any holomorphic $L^{2}$ space. We will show
that these properties do not hold for Toeplitz operators with polynomial
symbols on the generalized Bergman spaces $H(\mathbb{B}^{d},\lambda),$
$\lambda<d.$

######  Proposition 7.

For $\lambda>d$ and $\phi(z)$ bounded, the Toeplitz operator $T_{\phi}$ on the
space $\mathcal{H}L^{2}(\mathbb{B}^{d},d\mu_{\lambda}),$ which is defined by
$T_{\phi}f=P_{\lambda}(\phi f),$ has the following properties.

  1. (1)

$\|T_{\phi}\|\leq\sup_{z}|\phi(z)|$

  2. (2)

If $\phi(z)\geq 0$ for all $z,$ then $T_{\phi}$ is a positive operator.

Both of these properties fail when $\lambda<d.$ In fact, for $\lambda<d,$
there is no constant $C$ such that $\|T_{\phi}\|\leq C\sup_{z}|\phi(z)|$ for
all polynomials $\phi.$

As we remarked in the introduction, when $\lambda=d,$ the space
$H(\mathbb{B}^{d},\lambda)$ may be identified with the Hardy space. Thus
Properties 1 and 2 in the proposition still hold when $\lambda=d,$ if, say,
$\phi$ is continuous up to the boundary of $\mathbb{B}^{d}$ (or otherwise has
a reasonable extension to the closure of $\mathbb{B}^{d}$).

###### Proof.

When $\lambda>d,$ the projection operator $P_{\lambda}$ has norm 1 and the
multiplication operator $M_{\phi}$ has norm equal to $\sup_{z}|\phi(z)|$ as an
operator on $L^{2}(\mathbb{B}^{d},\mu_{\lambda})$. Thus, the restriction to
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ of $P_{\lambda}M_{\phi}$ has
norm at most $\sup_{z}|\phi(z)|.$ Meanwhile, if $\phi$ is non-negative, then
from (3.1) we see that $\left\langle f,T_{\phi}f\right\rangle\geq 0$ for all
$f\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$

Let us now assume that $0<\lambda<d.$ Computing on the orthogonal basis in
Theorem 4, it is a simple exercise to show that

$T_{\bar{z}_{j}z_{j}}(z^{m})=\frac{\Gamma(\lambda+|m|)}{m!}\frac{(m+e_{j})!}{\Gamma(\lambda+|m|+1)}z^{m}=\frac{1+m_{j}}{\lambda+|m|}z^{m}.$
(3.3)

If we take $\phi(z)=|z|^{2},$ then summing (3.3) on $j$ gives

$T_{\phi}z^{m}=\frac{d+|m|}{\lambda+|m|}z^{m}.$

Since $\lambda<d,$ this shows that $\left\|T_{\phi}\right\|>1,$ even though
$\left|\phi(z)\right|\,<1$ for all $z\in\mathbb{B}^{d}.$ Thus, Property 1
fails for $\lambda<d.$ (From this calculation it easily follows that if
$\phi(z)=(1-\left|z\right|^{2})/(\lambda-d),$ then $T_{\phi}$ is the bounded
operator $(\lambda I+N)^{-1},$ for all $\lambda\neq d.$)

For the second property, we let $\psi(z)=1-\phi(z)=1-|z|^{2}$ which is
positive. From the above calculation we obtain

$\langle
T_{\psi}z^{m},z^{m}\rangle_{H_{\lambda}}=\|z^{m}\|_{H_{\lambda}}^{2}-\left(\frac{d+|m|}{\lambda+|m|}\right)\|z^{m}\|_{H(\mathbb{B}^{d},\lambda)}^{2},$

which is negative if $0<\lambda<d$.

We now show that there is no constant $C$ such that $\|T_{\phi}\|\leq
C\sup_{z}|\phi(z)|$. Consider

$\displaystyle\phi_{k}(z)$
$\displaystyle:=(|z|^{2})^{k}=\left(\sum_{i=1}^{d}|z_{i}|^{2}\right)^{k}$
$\displaystyle=\sum_{|i|=k}\frac{k!}{i!}(|z_{1}|^{2})^{i_{1}}(|z_{2}|^{2})^{i_{2}}\cdots(|z_{d}|^{2})^{i_{d}}=\sum_{|i|=k}\frac{k!}{i!}\overline{z}^{i}z^{i}.$

Computing on the orthogonal basis in Theorem 4 we obtain

$T_{\phi_{k}}\mathbf{1}=\sum_{|i|=k}\frac{k!}{i!}(T_{\overline{z}^{i}z^{i}}\mathbf{1})=\sum_{|i|=k}\frac{k!}{i!}\frac{i!\Gamma(\lambda)}{\Gamma(\lambda+k)}\mathbf{1}=\mathcal{I}\frac{k!\Gamma(\lambda)}{\Gamma(\lambda+k)}\mathbf{1,}$

where $\mathbf{1}$ is the constant function. Here, $\mathcal{I}$ is the number
of multi-indices $i$ of length $d$ such that $|i|=k,$ which is equal to
${\binom{k+d-1}{d-1}}$. Thus

$T_{\phi_{k}}\mathbf{1}=\frac{(k+d-1)!}{(d-1)!}\frac{\Gamma(\lambda)}{\Gamma(\lambda+k)}\mathbf{1}=\frac{(d+k-1)\cdots(d)}{(\lambda+k-1)\cdots(\lambda)}\mathbf{1}=\prod_{j=0}^{k-1}\frac{d+j}{\lambda+j}\mathbf{1}.$

Consider
$\prod_{j=0}^{k-1}\frac{d+j}{\lambda+j}=\prod_{j=0}^{k-1}\left(1+\frac{d-\lambda}{\lambda+j}\right)$.
Since $d>\lambda,$ the terms $\frac{d-\lambda}{\lambda+j}$ are positive and
$\sum_{j=0}^{\infty}\frac{d-\lambda}{\lambda+j}$ diverges. This implies
$\prod_{j=0}^{\infty}\frac{d+j}{\lambda+j}=\infty$. Since
$\sup_{z}|\phi_{k}(z)|=1$ for all $k$, there is no a constant $C$ such that
$\|T_{\phi}\|\leq C\sup_{z}|\phi(z)|$. ∎

######  Remark 8.

For $\lambda<d,$ there does not exist any positive measure $\nu$ on
$\mathbb{B}^{d}$ such that
$\left\|f\right\|_{\lambda}=\left\|f\right\|_{L^{2}(\mathbb{B}^{d},\nu)}$ for
all $f$ in $H(\mathbb{B}^{d},\lambda).$ If such a $\nu$ did exist, then the
argument in the first part of the proof of Proposition 7 would show that
Properties 1 and 2 in the proposition hold.

##  4\. Bounded Toeplitz operators

In this section, we will consider a class of symbols $\phi$ for which we will
be able to define a Toeplitz operator $T_{\phi}$ as a bounded operator on
$H(\mathbb{B}^{d},\lambda)$ for all $\lambda>0.$ Our definition of $T_{\phi}$
will agree (for the relevant class of symbols) with the usual “multiply and
project” definition for $\lambda>d.$ In light of the examples in the previous
section, we cannot expect boundedness of $\phi$ to be sufficient to define
$T_{\phi}$ as a bounded operator. Instead, we will consider functions $\phi$
for which $\phi$ and a certain number of derivatives of $\phi$ are bounded.

Our strategy is to use integration by parts to give an alternative expression
for the matrix entries of a Toeplitz operator with sufficiently regular
symbol, in the case $\lambda>d.$ We then take this expression as our
definition of Toeplitz operator in the case $0<\lambda\leq d.$

######  Theorem 9.

Assume $\lambda>d$ and fix a positive integer $n.$ Let $\phi$ be a function
that is $2n$ times continuously differentiable and for which
$\bar{N}^{k}N^{l}\phi$ is bounded for all $0\leq k,l\leq n.$ Then

$\left\langle
f,T_{\phi}g\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=c_{\lambda+2n}\int_{\mathbb{B}^{d}}C\left[\left(\overline{f(z)}\phi(z)g(z)\right)\right]\left(1-|z|^{2}\right)^{\lambda+2n}d\tau(z)$

for all $f,g\in\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ where $C$ is
the operator given by

$C=\left(I+\frac{\bar{N}}{\lambda+2n-1}\right)\cdots\left(I+\frac{\bar{N}}{\lambda+n}\right)\left(I+\frac{N}{\lambda+n-1}\right)\cdots\left(I+\frac{N}{\lambda}\right).$
(4.1)

Thus, there exist constants $A_{jklm}$ (depending on $n$ and $\lambda$) such
that

$\left\langle
f,T_{\phi}g\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=\sum_{j,k,l,m=1}^{n}A_{jklm}\left\langle
N^{j}f,\left(\bar{N}^{k}N^{l}\phi\right)N^{m}g\right\rangle_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}.$
(4.2)

###### Proof.

Assume at first that $f$ and $g$ are polynomials, so that $f$ and $g$ and all
of their derivatives are bounded. We use (3.1) and apply the first equality in
Lemma 1 with $\psi=\bar{f}\phi g.$ We then apply the first equality in the
lemma again with $\psi=(I+N/\lambda)[\bar{f}\phi g].$ We continue on in this
fashion until we have applied the first equality in Lemma 1 $n$ times and the
second equality $n$ times. This establishes the desired equality in the case
that $f$ and $g$ are polynomials. For general $f$ and $g$ in
$\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}),$ we approximate by sequences
$f_{a}$ and $g_{a}$ of polynomials. From Theorem 3 we can see that convergence
of $f_{a}$ and $g_{a}$ in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$
implies convergence of $N^{j}f_{a}$ and $N^{k}g_{a}$ to $N^{j}f$ and $N^{k}g,$
so that applying (4.2) to $f_{a}$ and $g_{a}$ and taking a limit establishes
the desired result for $f$ and $g.$ ∎

######  Definition 10.

Assume $0<\lambda\leq d$ and fix a positive integer $n$ such that
$\lambda+2n>d.$ Let $\phi$ be a function that is $2n$ times continuously
differentiable and for which $\bar{N}^{k}N^{l}\phi$ is bounded for all $0\leq
k,l\leq n.$ Then we define the Toeplitz operator $T_{\phi}$ to be the unique
bounded operator on $H(\mathbb{B}^{d},\lambda)$ whose matrix entries are given
by

$\left\langle
f,T_{\phi}g\right\rangle_{H(\mathbb{B}^{d},\lambda)}=c_{\lambda+2n}\int_{\mathbb{B}^{d}}C\left[\left(\overline{f(z)}\phi(z)g(z)\right)\right]\left(1-|z|^{2}\right)^{\lambda+2n}dz,$
(4.3)

where $C$ is given by (4.1).

Note that from Theorem 4, $N^{j}f$ and $N^{m}g$ belong to
$L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})$ for all $0\leq j,m\leq n,$ for all
$f$ and $g$ in $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$ Furthermore,
$\left\|N^{j}f\right\|_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}$ and
$\left\|N^{m}g\right\|_{L^{2}(\mathbb{B}^{d},\mu_{\lambda+2n})}$ are bounded
by constants times $\left\|f\right\|_{\lambda}$ and
$\left\|g\right\|_{\lambda}$, respectively. Thus, the right-hand side of (4.3)
is a continuous sesquilinear form on $H(\mathbb{B}^{d},\lambda),$ which means
that there is a unique bounded operator $T_{\phi}$ whose matrix entries are
given by (4.3).

If $\lambda=d,$ then (as discussed in the introduction) the Hilbert space
$H(\mathbb{B}^{d},\lambda)$ is the Hardy space of holomorphic functions that
are square-integrable over the boundary. In that case, the Toeplitz operator
$T_{\phi}$ will be the zero operator whenever $\phi$ is identically zero on
the boundary of $\mathbb{B}^{d}.$ If $\lambda=d-1,$ $d-2,$ $\ldots,$ then the
inner product on $H(\mathbb{B}^{d},\lambda)$ can be related to the inner
product on the Hardy space. It is not hard to see that in these cases,
$T_{\phi}$ will be the zero operator if $\phi$ and enough of its derivatives
vanish on the boundary of $\mathbb{B}^{d}.$

Let us consider the case in which $\phi(z)=\overline{\psi_{1}(z)}\psi_{2}(z),$
where $\psi_{1}$ and $\psi_{2}$ are holomorphic functions such that the
function and the first $n$ derivatives are bounded. Then when applying $C$ to
$\overline{f(z)}\phi(z)g(z),$ all the $N$-factors go onto the expression
$\psi_{2}(z)g(z)$ and all the $\bar{N}$-factors go onto
$\overline{f(z)}\overline{\psi_{1}(z)}.$ Recalling from Theorem 4 the formula
for the inner product on $H(\mathbb{B}^{d},\lambda)$, we see that

$\left\langle
f,T_{\phi}g\right\rangle_{\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})}=\left\langle\psi_{1}f,\psi_{2}g\right\rangle_{H(\mathbb{B}^{d},\lambda)},$

as expected. This means that in this case,
$T_{\bar{\psi}_{1}\psi_{2}}=(M_{\psi_{1}})^{\ast}(M_{\psi_{2}}),$ as in the
case $\lambda>d.$ In particular, Definition 10 agrees with the definition we
used in Section 3 in the case that $\phi$ is a polynomial in $z$ and
$\bar{z}.$

##  5\. Hilbert–Schmidt Toeplitz operators

###  5.1. Statement of results

In this section, we will give sufficient conditions under which a Toeplitz
operator $T_{\phi}$ can be defined as a Hilbert–Schmidt operator on
$H(\mathbb{B}^{d},\lambda).$ Specifically, if $\phi$ belongs to
$L^{2}(\mathbb{B}^{d},\tau)$ then $T_{\phi}$ can be defined as a
Hilbert–Schmidt operator, provided that $\lambda>d/2.$ Meanwhile, if $\phi$
belongs to $L^{1}(\mathbb{B}^{d},\tau),$ then $T_{\phi}$ can be defined as a
Hilbert–Schmidt operator for all $\lambda>0.$ In both cases, we define
$T_{\phi}$ in such a way that for all bounded functions $f$ and $g$ in
$H(\mathbb{B}^{d},\lambda),$ we have

$\left\langle
f,T_{\phi}g\right\rangle_{\lambda}=c_{\lambda}\int_{\mathbb{B}^{d}}\overline{f(z)}\phi(z)g(z)(1-|z|^{2})^{\lambda}~{}d\tau(z),$
(5.1)

where $c_{\lambda}$ is defined by
$c_{\lambda}=\Gamma(\lambda)/(\pi^{d}\Gamma(\lambda-d)).$ This expression is
identical to (3.1) in the case $\lambda>d.$ The value of $c_{\lambda}$ should
be interpreted as 0 when $\lambda-d=0,-1,-2,\ldots$. This means that for
$\phi$ in $L^{2}(\mathbb{B}^{d},\tau)$ or $L^{1}(\mathbb{B}^{d},\tau)$ (but
not for other classes of symbols!), $T_{\phi}$ is the zero operator when
$\lambda=d,d-1,\ldots.$ This strange phenomenon is discussed in the next
subsection. Note that we are not claiming $T_{\phi}=0$ for arbitrary symbols
when $\lambda=d,d-1,\ldots,$ but only for symbols that are integrable or
square-integrable with respect to the hyperbolic volume measure $\tau.$ Such
functions must have reasonable rapid decay (in an average sense) near the
boundary of $\mathbb{B}^{d}.$

In the case $\phi\in L^{2}(\mathbb{B}^{d},\tau),$ the restriction
$\lambda>d/2$ is easy to explain: the function $(1-|z|^{2})^{\lambda}$ belongs
to $L^{2}(\mathbb{B}^{d},\tau)$ if and only if $\lambda>d/2.$ Thus, if $f$ and
$g$ are bounded and $\phi$ is in $L^{2}(\mathbb{B}^{d},\tau),$ then (5.1) is
absolutely convergent for $\lambda>d/2.$

In this subsection, we state our results; in the next subsection, we discuss
some unusual properties of $T_{\phi}$ for $\lambda<d$; and in the last
subsection of this section we give the proofs.

We begin by considering symbols $\phi$ in $L^{2}(\mathbb{B}^{d},\tau).$

######  Theorem 11.

Fix $\lambda>d/2$ and let
$c_{\lambda}=\Gamma(\lambda)/(\pi^{d}\Gamma(\lambda-d)).$ (We interpret
$c_{\lambda}$ to be zero if $\lambda$ is an integer and $\lambda\leq d$.) Then
the operator $A_{\lambda}$ given by

$A_{\lambda}\phi(z)=c_{\lambda}^{2}\int_{\mathbb{B}^{d}}\left[\frac{(1-|z|^{2})(1-|w|^{2})}{(1-w\cdot\bar{z})(1-\bar{w}\cdot
z)}\right]^{\lambda}\phi(w)\,d\tau(w)$

is a bounded operator from $L^{2}(\mathbb{B}^{d},\tau)$ to itself.

######  Theorem 12.

Fix $\lambda>d/2.$ Then for each $\phi\in L^{2}(\mathbb{B}^{d},\tau)$, there
is a unique Hilbert–Schmidt operator on $H(\mathbb{B}^{d},\lambda),$ denoted
$T_{\phi},$ with the property that

$\left\langle
f,T_{\phi}g\right\rangle_{\lambda}=c_{\lambda}\int_{\mathbb{B}^{d}}\overline{f(z)}\phi(z)g(z)(1-|z|^{2})^{\lambda}~{}d\tau(z)$
(5.2)

for all bounded holomorphic functions $f$ and $g$ in
$H(\mathbb{B}^{d},\lambda).$ The Hilbert–Schmidt norm of $T_{\phi}$ is given
by

$\left\|T_{\phi}\right\|_{HS}^{2}=\left\langle\phi,A_{\lambda}\phi\right\rangle_{L^{2}(\mathbb{B}^{d},\tau)}.$

If $\lambda>d$ and $\phi\in L^{2}(\mathbb{B}^{d},\tau)\cap
L^{\infty}(\mathbb{B}^{d},\tau)$, then the definition of $T_{\phi}$ in Theorem
12 agrees with the “multiply and project” definition; compare (3.1).

Applying Lemma 5 with $\lambda_{1}=\lambda_{2}=\lambda$ and $\lambda>d/2,$ we
see that for all $f$ and $g$ in $H(\mathbb{B}^{d},\lambda),$ the function
$z\rightarrow\overline{f(z)}g(z)(1-\left|z\right|^{2})^{\lambda}$ is in
$L^{2}(\mathbb{B}^{d},\tau).$ This means that the integral on the right-hand
side of (5.2) is absolutely convergent for all $f,g\in
H(\mathbb{B}^{d},\lambda).$ It is then not hard to show that (5.2) holds for
all $f,g\in H(\mathbb{B}^{d},\lambda).$

The operator $A_{\lambda}$ coincides, up to a constant, with the Berezin
transform. Let $\chi_{z}^{\lambda}(w):=K_{\lambda}(z,w)$ be the coherent state
at the point $z,$ which satisfies
$f(z)=\left\langle\chi_{z}^{\lambda},f\right\rangle_{\lambda}$ for all $f\in
H(\mathbb{B}^{d},\lambda).$ Then one standard definition of the Berezin
transform $B_{\lambda}$ is

$B_{\lambda}\phi=\frac{\left\langle\chi_{z}^{\lambda},T_{\phi}\chi_{z}^{\lambda}\right\rangle_{\lambda}}{\left\langle\chi_{z}^{\lambda},\chi_{z}^{\lambda}\right\rangle_{\lambda}}.$

The function $B_{\lambda}\phi$ may be thought of as the Wick-ordered symbol of
$T_{\phi},$ where $T_{\phi}$ is thought of as the anti-Wick-ordered
quantization of $\phi.$ Using the formula (Theorem 4) for the reproducing
kernel along with (5.2), we see that $A_{\lambda}=c_{\lambda}B_{\lambda}.$
(Note that $\chi_{z}^{\lambda}(w)$ is a bounded function of $w$ for each fixed
$z\in\mathbb{B}^{d}$ and that
$\left\langle\chi_{z}^{\lambda},\chi_{z}^{\lambda}\right\rangle_{\lambda}=K_{\lambda}(z,z).$)

Note that $\tau$ is an infinite measure, which means that if $\phi$ is in
$L^{2}(\mathbb{B}^{d},\tau)$ or $L^{1}(\mathbb{B}^{d},\tau),$ then $\phi$ must
tend to zero at the boundary of $\mathbb{B}^{d},$ at least in an average
sense. This decay of $\phi$ is what allows (5.2) to be a convergent integral.
If, for example, we want to take $\phi(z)\equiv 1,$ then we cannot use (5.2)
to define $T_{\phi},$ but must instead use the definition from Section 3 or
Section 4.

Note also that the space of Hilbert–Schmidt operators on
$H(\mathbb{B}^{d},\lambda)$ may be viewed as the quantum counterpart of
$L^{2}(\mathbb{B}^{d},\tau).$ It is thus natural to investigate the question
of when the Berezin–Toeplitz quantization maps $L^{2}(\mathbb{B}^{d},\tau)$
into the Hilbert–Schmidt operators.

We now show that if one considers a symbol $\phi$ in
$L^{1}(\mathbb{B}^{d},\tau),$ then one obtains a Hilbert–Schmidt Toeplitz
operator $T_{\phi}$ for all $\lambda>0.$

######  Theorem 13.

Fix $\lambda>0$ and let $c_{\lambda}$ be as in Theorem 12. Then for each
$\phi\in L^{1}(\mathbb{B}^{d},\tau),$ there exists a unique Hilbert–Schmidt
operator on $H(\mathbb{B}^{d},\lambda),$ denoted $T_{\phi},$ with the property
that

$\left\langle
f,T_{\phi}g\right\rangle_{\lambda}=c_{\lambda}\int_{\mathbb{B}^{d}}\overline{f(z)}\phi(z)g(z)(1-|z|^{2})^{\lambda}~{}d\tau(z)$
(5.3)

for all bounded holomorphic functions $f$ and $g$ in
$H(\mathbb{B}^{d},\lambda).$ The Hilbert–Schmidt norm of $T_{\phi}$ satisfies

$\left\|T_{\phi}\right\|_{HS}\leq
c_{\lambda}\left\|\phi\right\|_{L^{1}(\mathbb{B}^{d},\tau)}.$

Using the pointwise bounds on elements of $H(\mathbb{B}^{d},\lambda)$ coming
from the reproducing kernel, we see immediately that for all $f,g\in
H(\mathbb{B}^{d},\lambda),$ the function
$z\rightarrow\overline{f(z)}g(z)(1-|z|^{2})^{\lambda}$ is bounded. It is then
not hard to show that (5.3) holds for all $f,g\in H(\mathbb{B}^{d},\lambda).$

We have already remarked that the definition of $T_{\phi}$ given in this
section agrees with the “multiply and project” definition when $\lambda>d$
(and $\phi$ is bounded). It is also easy to see that the definition of
$T_{\phi}$ given in this section agrees with the one in Section 4, when $\phi$
falls under the hypotheses of both Definition 10 and either Theorem 12 or
Theorem 13. For some positive integer $n,$ consider the set of $\lambda$’s for
which $\lambda+2n>d$ and $\lambda>d/2,$ i.e., $\lambda>\max(d-2n,d/2).$ Now
suppose that $\phi$ belongs to $L^{2}(\mathbb{B}^{d},\tau)$ and that
$N^{k}\bar{N}^{l}\phi$ is bounded for all $0\leq k,l\leq n.$ It is easy to see
that the matrix entries $\left\langle f,T_{\phi}g\right\rangle_{\lambda}$
depend real-analytically on $\lambda$ for fixed polynomials $f$ and $g,$
whether $T_{\phi}$ is defined by Definition 10 or by Theorem 12. For
$\lambda>d,$ the two matrix entries agree because both definitions of
$T_{\phi}$ agree with the “multiply and project” definition. The matrix
entries therefore must agree for all $\lambda>\max(d-2n,d/2).$ Since
polynomials are dense in $H(\mathbb{B}^{d},\lambda)$ and both definitions of
$T_{\phi}$ give bounded operators, the two definitions of $T_{\phi}$ agree.
The same reasoning shows agreement of Definition 10 and Theorem 13.

###  5.2. Discussion

Before proceeding on with the proof, let us make a few remarks about the way
we are defining Toeplitz operators in this section. For $\lambda>d,$
$c_{\lambda}$ is the normalization constant that makes the measure
$\mu_{\lambda}$ a probability measure, which can be computed to have the value
$\Gamma(\lambda)/(\pi^{d}\Gamma(\lambda-d)).$ For $\lambda\leq d,$ although
the measure $(1-|z|^{2})^{\lambda}~{}d\tau(z)$ is an infinite measure, we
simply use the same formula for $c_{\lambda}$ in terms of the gamma function.
We understand this to mean that $c_{\lambda}=0$ whenever $\lambda$ is an
integer in the range $(0,d].$ It also means that $c_{\lambda}$ is negative
when $d-1<\lambda<d$ and when $d-3<\lambda<d-2,$ etc.

In the cases where $c_{\lambda}=0$, we have that $T_{\phi}=0$ for all $\phi$
in $L^{1}(\mathbb{B}^{d},\tau)$ or $L^{2}(\mathbb{B}^{d},\tau).$ This first
occurs when $\lambda=d.$ Recall that for $\lambda=d,$ the space
$H(\mathbb{B}^{d},\lambda)$ can be identified with the Hardy space of
holomorphic functions square-integrable over the boundary. Meanwhile, having
$\phi$ being integrable or square-integrable with respect to $\tau$ means that
$\phi$ tends to zero (in an average sense) at the boundary, in which case it
is reasonable that $T_{\phi}$ should be zero as an operator on the Hardy
space. For other integer values of $\lambda\leq d,$ the inner product on
$H(\mathbb{B}^{d},\lambda)$ can be expressed using the methods of Section 2 in
terms of integration over the boundary, but involving the functions and their
derivatives. In that case, we expect $T_{\phi}$ to be zero if $\phi$ has
sufficiently rapid decay at the boundary, and it is reasonable to think that
having $\phi$ in $L^{1}$ or $L^{2}$ with respect to $\tau$ constitutes
sufficiently rapid decay. Note, however, that the conclusion that $T_{\phi}=0$
when $c_{\lambda}=0$ applies only when $\phi$ is in $L^{1}$ or $L^{2}$; for
other classes of symbols, such as polynomials, $T_{\phi}$ is not necessarily
zero. For example, $T_{z^{m}}$ is equal to $M_{z^{m}},$ which is certainly a
nonzero operator on $H(\mathbb{B}^{d},\lambda),$ for all $\lambda>0.$

Meanwhile, if $c_{\lambda}<0,$ then we have the curious situation that if
$\phi$ is positive and in $L^{1}$ or $L^{2}$ with respect to $\tau,$ then the
operator $T_{\phi}$ is actually a negative operator. This is merely a dramatic
example of a phenomenon we have already noted: for $\lambda<d,$ non-negative
symbols do not necessarily give rise to non-negative Toeplitz operators.
Again, though, the conclusion that $T_{\phi}$ is negative for $\phi$ positive
applies only when $\phi$ belongs to $L^{1}$ or $L^{2}.$ For example, the
constant function $\mathbf{1}$ always maps to the (positive!) identity
operator, regardless of the value of $\lambda.$

###  5.3. Proofs

As motivation, we begin by computing the Hilbert–Schmidt norm of Toeplitz
operators in the case $\lambda>d.$ For any bounded measurable $\phi,$ we
extend the Toeplitz operator $T_{\phi}$ to all of
$L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ by making it zero on the orthogonal
complement of the holomorphic subspace. This extension is given by the formula
$P_{\lambda}M_{\phi}P_{\lambda}.$ Then the Hilbert–Schmidt norm of the
operator $T_{\phi}$ on $\mathcal{H}L^{2}(\mathbb{B}^{d},\mu_{\lambda})$ is the
same as the Hilbert–Schmidt norm of the operator
$P_{\lambda}M_{\phi}P_{\lambda}$ on $L^{2}(\mathbb{B}^{d},\mu_{\lambda}).$
Since $P_{\lambda}$ is computed as integration against the reproducing kernel,
we may compute that

$P_{\lambda}M_{\phi}P_{\lambda}f(z)=\int_{\mathbb{B}^{d}}\mathcal{K}_{\phi}(z,w)f(w)\,d\mu_{\lambda}(w),$

where

$\mathcal{K}_{\phi}(z,w)=\int_{\mathbb{B}^{d}}K(z,u)\phi(u)K(u,w)\,d\mu_{\lambda}(u).$

If we can show that $\mathcal{K}_{\phi}$ is in
$L^{2}(\mathbb{B}^{d}\times\mathbb{B}^{d},\mu_{\lambda}\times\mu_{\lambda})$,
then it will follow by a standard result that $T_{\phi}$ is Hilbert–Schmidt,
with Hilbert–Schmidt norm equal to the $L^{2}$ norm of $\mathcal{K}_{\phi}.$
For sufficiently nice $\phi,$ we can compute the $L^{2}$ norm of
$\mathcal{K}_{\phi}$ by rearranging the order of integration and using twice
the reproducing identity $\int K(z,w)K(w,u)~{}d\mu_{\lambda}(w)=K(z,u).$ (This
identity reflects that $P_{\lambda}^{2}=P_{\lambda}.$) This yields

$\int_{\mathbb{B}^{d}\times\mathbb{B}^{d}}\left|\mathcal{K}_{\phi}(z,w)\right|^{2}~{}d\mu_{\lambda}(z)~{}d\mu_{\lambda}=\left\langle\phi,A\phi\right\rangle_{L^{2}(\mathbb{B}^{d},\tau)},$

where $A_{\lambda}$ is the integral operator given by

$\displaystyle A_{\lambda}\phi(z)$
$\displaystyle=c_{\lambda}^{2}\int_{\mathbb{B}^{d}}\left|K(z,w)\right|^{2}(1-|z|^{2})^{\lambda}(1-|w|^{2})^{\lambda}\phi(w)~{}d\tau(w)$
$\displaystyle=c_{\lambda}^{2}\int_{\mathbb{B}^{d}}\left[\frac{(1-|z|^{2})(1-|w|^{2})}{(1-\bar{w}\cdot
z)(1-\bar{z}\cdot w)}\right]^{\lambda}\phi(w)\,d\tau(w).$ (5.4)

In the case $d/2<\lambda\leq d,$ it no longer makes sense to express
$T_{\phi}$ as $P_{\lambda}M_{\phi}P_{\lambda}.$ Nevertheless, we can consider
an operator $A_{\lambda}$ defined by (5.4). Our goal is to show that for all
$\lambda>d/2,$ (1) $A_{\lambda}$ is a bounded operator on
$L^{2}(\mathbb{B}^{d},\tau)$ and (2) if we define $T_{\phi}$ by (5.1), then
the Hilbert–Schmidt norm of $T_{\phi}$ is given by
$\left\langle\phi,A_{\lambda}\phi\right\rangle_{L^{2}(\mathbb{B}^{d},\tau)}.$
We will obtain similar results for all $\lambda>0$ if $\phi\in
L^{1}(\mathbb{B}^{d},\tau).$

###### Proof of Theorem 11.

We give two proofs of this result; the first generalizes more easily to other
bounded symmetric domains, whereas the second relates $A_{\lambda}$ to the
Laplacian for $\mathbb{B}^{d}$ (compare [13]).

First Proof. We let

$F_{\lambda}(z,w)=c_{\lambda}^{2}\left[\frac{(1-|z|^{2})(1-|w|^{2})}{(1-\bar{w}\cdot
z)(1-\bar{z}\cdot w)}\right]^{\lambda}\text{;}$

i.e., $F_{\lambda}$ is the integral kernel of the operator $A_{\lambda}.$ A
key property of $F_{\lambda}$ is its invariance under automorphisms:
$F_{\lambda}(\psi(z),\psi(w))=F_{\lambda}(z,w)$ for each automorphism
(biholomorphism) $\psi$ of $\mathbb{B}^{d}$ and all $z,w\in\mathbb{B}^{d}.$ To
establish the invariance of $F_{\lambda},$ let

$f_{\lambda}(z)=c_{\lambda}^{2}(1-|z|^{2})^{\lambda}.$ (5.5)

According to Lemma 1.2 of [Z2], $F_{\lambda}(z,w)=f_{\lambda}(\phi_{w}(z)),$
where $\phi_{w}$ is an automorphism of $\mathbb{B}^{d}$ taking $0$ to $w$ and
satisfying $\phi_{w}^{2}=I.$ Now, if $\psi$ is any automorphism, the
classification of automorphisms (Theorem 1.4 of [Z2]) implies that
$\psi\circ\phi_{w}=\phi_{\psi(w)}\circ U$ for some unitary matrix $U.$ From
this we can obtain $\phi_{\psi(w)}=U\circ\phi_{w}\circ\psi^{-1},$ and so

$f_{\lambda}(\phi_{\psi(w)}(\psi(z)))=f_{\lambda}(U(\phi_{w}(\psi^{-1}(\psi(z))))=f_{\lambda}(\phi_{w}(z)),$

i.e., $F_{\lambda}(\psi(z),\psi(w))=F_{\lambda}(z,w).$

The invariance of $F_{\lambda}$ under automorphisms means that
$A_{\lambda}\phi$ can be thought of as a convolution (over the automorphism
group $PSU(d,1)$) of $\phi$ with the function $f_{\lambda}.$ What this means
is that

$A_{\lambda}\phi(z)=\int_{G}f_{\lambda}(gh^{-1}\cdot 0)\phi(h\cdot 0)~{}dh,$

where $g\in G$ is chosen so that $g\cdot 0=z.$ Here $G=PSU(d,1)$ is the group
of automorphisms of $\mathbb{B}^{d}$ (given by fractional linear
transformations) and $dh$ is an appropriately normalized Haar measure on $G.$
Furthermore, $L^{2}(\mathbb{B}^{d},\tau)$ can be identified with the
right-$K$-invariant subspace of $L^{2}(G,dg)$, where $K:=U(d)$ is the
stabilizer of $0.$

If $\lambda>d,$ then $f_{\lambda}$ is in $L^{1}(\mathbb{B}^{d},\tau),$ in
which case it is easy to prove that $A_{\lambda}$ is bounded; see, for
example, Theorem 2.4 in [5]. This argument does not work if $\lambda\leq d.$
Nevertheless, if $\lambda>d/2,$ an easy computation shows that $f_{\lambda}$
belongs to $L^{2}(\mathbb{B}^{d},\tau)$ and also to
$L^{p}(\mathbb{B}^{d},\tau)$ for some $p<2.$ We could at this point appeal to
a general result known as the Kunze–Stein phenomenon [24]. The result states
that on connected semisimple Lie groups $G$ with finite center (including
$PSU(d,1)$), convolution with a function in $L^{p}(G,dg),$ $p<2,$ is a bounded
operator from $L^{2}(G,dg)$ to itself. (See [11] for a proof in this
generality.) However, the proof of this result is simpler in the case we are
considering, where the function in $L^{p}(G,dg)$ is bi-$K$-invariant and the
other function is right-$K$-invariant. (In our case, the function in
$L^{p}(G,dg)$ is the function $g\rightarrow f_{\lambda}(g\cdot 0)$ and the
function in $L^{2}(G,dg)$ is $g\rightarrow\phi(g\cdot 0).$) Using the Helgason
Fourier transform along with its behavior under convolution with a
bi-$K$-invariant function ([19, Lemma III.1.4]), we need only show that the
spherical Fourier transform of $f_{\lambda}$ is bounded. (Helgason proves
Lemma III.1.4 under the assumption that the functions are continuous and of
compact support, but the proof also applies more generally.) Meanwhile,
standard estimates show that for every $\varepsilon>0,$ the spherical
functions are in $L^{2+\varepsilon}(G/K),$ with $L^{2+\varepsilon}(G/K)$ norm
bounded independent of the spherical function. (Specifically, in the notation
of [18, Sect. IV.4], for all $\lambda\in\mathfrak{a}^{\ast},$ we have
$\left|\phi_{\lambda}(g)\right|\leq\phi_{0}(g),$ and estimates on $\phi_{0}$
(e.g., [1, Prop. 2.2.12]) show that $\phi_{0}$ is in $L^{2+\varepsilon}$ for
all $\varepsilon>0.$)

Choosing $\varepsilon$ so that $1/p+1/(2+\varepsilon)=1$ establishes the
desired boundedness.

Second proof. If $c_{\lambda}=0$ (i.e., if $\lambda\in\mathbb{Z}$ and
$\lambda\leq d$), then there is nothing to prove. Thus we assume $c_{\lambda}$
is nonzero, in which case $c_{\lambda+1}$ is also nonzero. The invariance of
$F_{\lambda}$ under automorphisms together with the square-integrability of
the function $(1-|z|^{2})^{\lambda}$ for $\lambda>d/2$ show that the integral
defining $A_{\lambda}f(z)$ is absolutely convergent for all $z.$

We introduce the (hyperbolic) Laplacian $\Delta$ for $\mathbb{B}^{d},$ given
by

$\Delta=(1-|z|^{2})\sum_{j,k=1}^{d}(\delta_{jk}-\bar{z}_{j}z_{k})\frac{\partial^{2}}{\partial\bar{z}_{j}\partial
z_{k}}.$ (5.6)

(This is a negative operator.) This operator commutes with the automorphisms
of $\mathbb{B}^{d}.$ It is known (e.g., [28]) that $\Delta$ is an unbounded
self-adjoint operator on $L^{2}(\mathbb{B}^{d},\tau),$ on the domain
consisting of those $f$’s in $L^{2}(\mathbb{B}^{d},\tau)$ for which $\Delta f$
in the distribution sense belongs to $L^{2}(\mathbb{B}^{d},\tau).$ In
particular, if $f\in L^{2}(\mathbb{B}^{d},\tau)$ is $C^{2}$ and $\Delta f$ in
the ordinary sense belongs to $L^{2}(\mathbb{B}^{d},\tau),$ then $f\in
Dom(\Delta).$

We now claim that

$\Delta_{z}F_{\lambda}(z,w)=\lambda(\lambda-d)(F_{\lambda}(z,w)-F_{\lambda+1}(z,w)),$
(5.7)

where $\Delta_{z}$ indicates that $\Delta$ is acting on the variable $z$ with
$w$ fixed. Since $\Delta$ commutes with automorphisms, it again suffices to
check this when $w=0,$ in which case it is a straightforward algebraic
calculation. Suppose, then, that $\phi$ is a $C^{\infty}$ function of compact
support. In that case, we are free to differentiate under the integral to
obtain

$\Delta
A_{\lambda}\phi=\lambda(\lambda-d)A_{\lambda}\phi-\lambda(\lambda-d)A_{\lambda+1}\phi.$
(5.8)

Now, the invariance of $F_{\lambda}$ tells us that
$L^{2}(\mathbb{B}^{d},\tau)$ norm of $F_{\lambda}(z,w)$ as a function of $z$
is finite for all $w$ and independent of $w.$ Putting the $L^{2}$ norm inside
the integral then shows that $A_{\lambda}\phi$ and $A_{\lambda+1}\phi$ are in
$L^{2}(\mathbb{B}^{d},\tau).$ This shows that $A_{\lambda}\phi$ is in
$Dom(\Delta).$ Furthermore, the condition $\lambda>d/2$ implies that
$\lambda(\lambda-d/2)>-d^{2}/4.$ It is known that the $L^{2}$ spectrum of
$\Delta$ is $(-\infty,-d^{2}/4].$ For general symmetric space of the
noncompact type, the $L^{2}$ spectrum of the Laplacian is
$(-\infty,-\left\|\rho\right\|^{2}],$ where $\rho$ is half the sum of the
positive (restricted) roots for $G/K,$ counted with their multiplicity. In our
case, there is one positive root $\alpha$ with multiplicity $(2d-2)$ and
another positive root $2\alpha$ with multiplicity 1. (See the entry for “A IV”
in Table VI of Chapter X of [17].) Thus, $\rho=d\alpha.$ It remains only to
check that if the metric is normalized so that the Laplacian comes out as in
(5.6), then $\left\|\alpha\right\|^{2}=1/4.$ This is a straightforward but
unilluminating computation, which we omit.

Since $\lambda(\lambda-d)$ is in the resolvent set of $\Delta,$ we may rewrite
(5.8) as

$A_{\lambda}\phi=-\lambda(\lambda-d)[\Delta-\lambda(\lambda-d)I]^{-1}A_{\lambda+1}\phi.$

Suppose now that $\lambda+1>d,$ so that (as remarked above) $A_{\lambda+1}$ is
bounded. Since $[\Delta-cI]^{-1}$ is a bounded operator for all $c$ in the
resolvent of $\Delta,$ we see that $A_{\lambda}$ has a bounded extension from
$C_{c}^{\infty}(\mathbb{B}^{d})$ to $L^{2}(\mathbb{B}^{d},\tau).$ Since the
integral computing $A_{\lambda}\phi(z)$ is a continuous linear functional on
$L^{2}(\mathbb{B}^{d},\tau)$ (integration against an element of
$L^{2}(\mathbb{B}^{d},\tau)$), it is easily seen that this bounded extension
coincides with the original definition of $A_{\lambda}.$

The above argument shows that $A_{\lambda}$ is bounded if $\lambda>d/2$ and
$\lambda+1>d.$ Iteration of the argument then shows boundedness for all
$\lambda>d/2.$ ∎

###### Proof of Theorem 12.

We wish to show that for all $\lambda>d/2,$ if $\phi$ is in
$L^{2}(\mathbb{B}^{d},\tau),$ then there is a unique Hilbert–Schmidt operator
$T_{\phi}$ with matrix entries given in (5.1) for all polynomials, and
furthermore,
$\left\|T_{\phi}\right\|_{HS}^{2}=\left\langle\phi,A_{\lambda}\phi\right\rangle_{\lambda}.$
At the beginning of this section, we had an calculation of
$\left\|T_{\phi}\right\|$ in terms of $A_{\lambda},$ but this argument relied
on writing $T_{\phi}$ as $P_{\lambda}M_{\phi}P_{\lambda},$ which does not make
sense for $\lambda\leq d.$

We work with an orthonormal basis for $H(\mathbb{B}^{d},\lambda)$ consisting
of normalized monomials, namely,

$e_{m}(z)=z^{m}\sqrt{\frac{\Gamma(\lambda+|m|)}{m!\Gamma(\lambda)}},$

for each multi-index $m.$ Then we want to establish the existence of a
Hilbert–Schmidt operator whose matrix entries in this basis are given by

$a_{lm}:=c_{\lambda}\int_{\mathbb{B}^{d}}\overline{e_{l}(z)}\phi(z)e_{m}(z)(1-|z|^{2})^{\lambda}~{}d\tau(z).$
(5.9)

There will exist a unique such operator provided that
$\sum_{l,m}|a_{lm}|^{2}<\infty.$

If we assume, for the moment, that Fubini’s Theorem applies, we obtain

$\displaystyle\sum_{l,m}\left|a_{lm}\right|^{2}$
$\displaystyle=c_{\lambda}^{2}\int_{\mathbb{B}^{d}}\int_{\mathbb{B}^{d}}\sum_{l,m}\frac{\Gamma(\lambda+|l|)}{l!\Gamma(\lambda)}\frac{\Gamma(\lambda+|m|)}{m!\Gamma(\lambda)}\bar{z}^{l}w^{l}z^{m}\bar{w}^{m}$
$\displaystyle\times\phi(z)\overline{\phi(w)}(1-|z|^{2})^{\lambda}(1-|w|^{2})^{\lambda}~{}d\tau(z)~{}d\tau(w),$
(5.10)

where $l$ and $m$ range over all multi-indices of length $d.$

We now apply the binomial series

$\frac{1}{(1-r)^{\lambda}}=\sum_{k=0}^{\infty}\binom{\lambda+k-1}{k}r^{k}$

for $r\in\mathbb{C}$ with $\left|r\right|<1,$ where

$\binom{\lambda+k-1}{k}=\frac{\Gamma(\lambda+k)}{k!\Gamma(\lambda)}.$

(This is the so-called negative binomial series.) We apply this with
$r=\sum_{j}\bar{z}_{j}w_{j},$ and we then apply the (finite) multinomial
series to the computation of $(\bar{z}\cdot w)^{k}.$ The result is that

$\sum_{l}\frac{\Gamma(\lambda+|l|)}{l!\Gamma(\lambda)}\bar{z}^{l}w^{l}=\frac{1}{(1-\bar{z}\cdot
w)^{\lambda}},$ (5.11)

where the sum is over all multi-indices $l.$ Applying this result, (5.10)
becomes

$\sum_{l,m}\left|a_{lm}\right|^{2}=\left\langle\phi,A_{\lambda}\phi\right\rangle_{\lambda},$
(5.12)

which is what we want to show.

Assume at first that $\phi$ is “nice,” say, continuous and supported in a ball
of radius $r<1.$ This ball has finite measure and $\phi$ is bounded on it.
Thus, if we put absolute values inside the sum and integral on the right-hand
side of (5.10), finiteness of the result follows from the absolute convergence
of the series (5.11). Thus, Fubini’s Theorem applies in this case.

Now for a general $\phi\in L^{2}(\mathbb{B}^{d},\tau),$ choose $\phi_{j}$
converging to $\phi$ with $\phi_{j}$ “nice.” Then (5.12) tells us that
$T_{\phi_{j}}$ is a Cauchy sequence in the space of Hilbert–Schmidt operators,
which therefore converges in the Hilbert–Schmidt norm to some operator $T.$
The matrix entries of $T_{\phi_{j}}$ in the basis $\\{e_{m}\\}$ are by
construction given by the integral in (5.9). The matrix entries of $T$ are the
limit of the matrix entries of $T_{\phi_{j}},$ hence also given by (5.9),
because $e_{l}$ and $e_{m}$ are bounded and $(1-|z|^{2})^{\lambda}$ belongs to
$L^{2}(\mathbb{B}^{d},\tau)$ for $\lambda>d/2.$

We can now establish that (5.2) in Theorem 12 holds for all bounded
holomorphic functions $f$ and $g$ in $H(\mathbb{B}^{d},\lambda)$ by
approximating these functions by polynomials. ∎

###### Proof of Theorem 13.

In the proof of Theorem 12, we did not use the assumption $\lambda>d/2$ until
the step in which we approximated arbitrary functions in
$L^{2}(\mathbb{B}^{d},\tau)$ by “nice” functions. In particular, if $\phi$ is
nice, then (5.9) makes sense for all $\lambda>0,$ and (5.12) still holds. Now,
since $F_{\lambda}(z,w)=f_{\lambda}(\phi_{w}(z)),$ where $f_{\lambda}$ is
given by (5.5), we see that $\left|F_{\lambda}(z,w)\right|\leq
c_{\lambda}^{2}$ for all $z,w\in\mathbb{B}^{d}.$ Thus,

$\left\langle\phi,A_{\lambda}\phi\right\rangle_{\lambda}\leq
c_{\lambda}^{2}\left\|\phi\right\|_{L^{1}(\mathbb{B}^{d},\tau)}^{2}$

for all nice $\phi.$ An easy approximation argument then establishes the
existence of a Hilbert–Schmidt operator with the desired matrix entries for
all $\phi\in L^{1}(\mathbb{B}^{d},\tau),$ with the desired estimate on the
Hilbert–Schmidt norm. ∎

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