# Helton-Howe Trace, the Connes-Chern character and Quantization

Xiang Tang, Yi Wang , and Dechao Zheng Department of Mathematics and
Statistics, Washington University, St. Louis, MO, U.S.A., 63130,
<EMAIL_ADDRESS>of Mathematics, Vanderbilt University,
Nashville, TN, U.S.A., 37240<EMAIL_ADDRESS>of
Mathematics, Vanderbilt University, Nashville, TN, U.S.A., 37240,
<EMAIL_ADDRESS>
###### Abstract

We study the Helton-Howe trace and the Connes-Chern character for Toeplitz
operators on weighted Bergman spaces via the idea of quantization. We prove a
local formula for the large $t$-limit of the Connes-Chern character as the
weight goes to infinity. And we show that the Helton-Howe trace of Toeplitz
operators is independent of the weight $t$ and obtain a local formula for the
Helton-Howe trace for all weighted Bergman spaces using harmonic analysis and
quantization.

Keywords: Connes-Chern character, Helton-Howe trace, quantization, Toeplitz
operator, weighted Bergman space

##  1 Introduction

Toeplitz extensions are fundamental objects in noncommutative geometry. They
are natural examples of finite summable Fredholm modules and define elements
in the corresponding $K$-homology group. Trace on Toeplitz operators has been
well studied with many interesting results, c.f. [56]. Since the 70s, trace
has been employed to extract geometric information of Toeplitz extension. In
particular, Connes [16, Sec. 2., Theorem 5] used trace on products of semi-
commutators to define the Connes-Chern character of finite summable
extensions.

For the unit disk $\mathbb{D}$ in $\mathbb{C}$, let $L^{2}_{a}(\mathbb{D})$ be
the Bergman space of $L^{2}$ analytic functions on $\mathbb{D}$. Given $f\in
C^{\infty}(\overline{\mathbb{D}})$, let $T^{(0)}_{f}$ be the Toeplitz operator
on $L^{2}_{a}(\mathbb{D})$ associated to the symbol $f$. The commutator
$[T^{(0)}_{f},T^{(0)}_{g}]:=T^{(0)}_{f}T^{(0)}_{g}-T^{(0)}_{g}T^{(0)}_{f}$ for
$f,g\in\mathscr{C}^{\infty}(\overline{\mathbb{D}})$ is a trace class operator.
Helton and Howe [32] discovered an interesting formula for the commutator

$\mathrm{Tr}\big{(}[T^{(0)}_{f},T^{(0)}_{g}]\big{)}=\frac{1}{2\pi\sqrt{-1}}\int_{\mathbb{D}}\mathrm{d}f\wedge\mathrm{d}g.$

The above result is deeply connected to the Pincus function for a pair of
noncommuting selfadjoint operators, c.f. [10, 11, 45].

For $\mathbb{B}_{n}$, the commutator $[T^{(0)}_{f},T^{(0)}_{g}]$ for two
Toeplitz operators with smooth symbols $f,g$ on $L^{2}_{a}(\mathbb{B}_{n})$ is
a Schatten-$p$ class operator for $p>n$. Suppose
$f_{1},...,f_{2n}\in\mathscr{C}^{\infty}(\overline{\mathbb{B}_{n}})$. Then the
product of the commutators

$[T^{(0)}_{f_{1}},T^{(0)}_{f_{2}}]\cdots[T^{(0)}_{f_{2n-1}},T^{(0)}_{2n}]$

is not a trace class operator. Helton and Howe [33, 35] made a breakthrough by
considering the antisymmetric sum of $T_{f_{1}},...,T_{f_{2n}}$ defined by

$[T^{(0)}_{f_{1}},...,T^{(0)}_{f_{2n}}]:=\sum_{\tau\in
S_{2n}}\operatorname{sgn}(\tau)T^{(0)}_{f_{\tau(1)}}T^{(0)}_{f_{\tau(2)}}...T^{(0)}_{f_{\tau(2n)}},$

where $S_{2n}$ is the permutations group of $2n$ elements and
$\operatorname{sgn}$ is the sign of the permutation $\tau$. The following is a
remarkable generalization of the Helton-Howe trace formula for the commutator
of two Toeplitz operators on $L^{2}_{a}(\mathbb{D})$.

######  Theorem 1.1 (Helton-Howe).

On the Bergman space $L^{2}_{a}(\mathbb{B}_{n})$ (and the Hardy space
$H^{2}(\mathbb{S}_{n})$), the antisymmetric sum
$[T^{(0)}_{f_{1}},...,T^{(0)}_{f_{2n}}]$ is a trace class operator, and

$\operatorname{Tr}\big{(}[T^{(0)}_{f_{1}},...,T^{(0)}_{f_{2n}}]\big{)}=\frac{n!}{(2\pi\sqrt{-1})^{n}}\int_{\mathbb{B}_{n}}\mathrm{d}f_{1}\wedge\mathrm{d}f_{2}\wedge\cdots\wedge\mathrm{d}f_{2n}.$
(1.1)

We observe that by Stoke’s theorem, the above integral only depends on the
value of $f_{1},...,f_{2n}$ on the unit sphere
$\mathbb{S}^{2n-1}=\partial\overline{\mathbb{B}_{n}}$, i.e.

$\frac{n!}{(2\pi\sqrt{-1})^{n}}\int_{\mathbb{B}_{n}}\mathrm{d}f_{1}\wedge\mathrm{d}f_{2}\wedge\cdots\wedge\mathrm{d}f_{2n}=\frac{n!}{(2\pi\sqrt{-1})^{n}}\int_{\mathbb{S}^{2n-1}}f_{1}\mathrm{d}f_{2}\wedge\cdots\wedge\mathrm{d}f_{2n}.$

The above idea of Schatten-$p$ commutators was revolutionized by Connes [16]
into a fundamental concept in noncommutative geometry as $p$-summable Fredholm
modules. And the Helton-Howe trace formula in Theorem 1.1 inspired Connes to
his ingenious discovery of cyclic cohomology and the Chern character for
$p$-summable Fredholm modules. The building block of the Connes-Chern
character is the semi-commutator

$\sigma_{t}(f,g)=T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}.$

Modulo constant, the Connes-Chern character for the Toeplitz extension is
defined to be

$\tau_{t}(f_{0},\cdots,f_{2p-1}):=\operatorname{Tr}\big{(}\sigma_{t}(f_{0},f_{1})...\sigma_{t}(f_{2p-2},f_{2p-1})\big{)}-\operatorname{Tr}\big{(}\sigma_{t}(f_{1},f_{2})...\sigma_{t}(f_{2p-1},f_{0})\big{)},$
(1.2)

for $p>n$. And the Helton-Howe trace, Equation (1.1), is the top degree
component of the above Connes-Chern character.

Connes observed that the above cocycle in general is not local, i.e. the value
of $\tau_{t}(f_{0},\cdots,f_{2p-1})$ can not be expressed by the germ of
$f_{0}\otimes\cdots\otimes f_{2p-1}$ on the diagonal in

$\underbrace{\mathbb{B}_{n}\times\cdots\times\mathbb{B}_{n}}_{2p}.$

Connes [17, 15] improved the Chern character, Equation (1.2), by employing the
Dixmier trace on the operator ideal $L^{1,\infty}$. In a series of works,
Engliš and his coauthors, e.g. [24, 25, 27, 28], studied a generalization of
the Helton-Howe trace formula by considering the Dixmier trace on the product

$[T^{(0)}_{f_{1}},T^{(0)}_{f_{2}}]\cdots[T^{(0)}_{f_{2n-1}},T^{(0)}_{f_{2n}}].$

They expressed the Dixmier trace of the above product as an integral of the
product of Poisson brackets between $f_{2k-1}$ and $f_{2k}$, $k=1,\cdots,n$.

In this article, we take a different approach to study the Connes-Chern
character (1.2) and the Helton-Howe trace, Theorem 1.1. Our main idea is to
put the Bergman space and Hardy space into the family of weighted Bergman
spaces $L^{2}_{a}(\mathbb{B}_{n},\lambda_{t}):=L_{a,t}^{2}(\mathbb{B}_{n})$
for the measure

$\mathrm{d}\lambda_{t}(z)=\frac{(n-1)!}{\pi^{n}B(n,t+1)}(1-|z|^{2})^{t}\mathrm{d}m(z),$

where $B(n,t+1)$ is the Beta function. Let $T^{(t)}_{f}$ be the associated
Toeplitz operator on $L_{a,t}^{2}(\mathbb{B}_{n})$ with symbol $f$. We study
the large $t$ behavior of the Connes-Chern character and the Helton-Howe
trace.

Our first result is about the Helton-Howe trace for
$T^{(t)}_{f_{1}},\ldots,T^{(t)}_{f_{2n}}$ for functions
$f_{1},\ldots,f_{2n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$, which
generalizes Theorem 1.1 to all weighted Bergman space.

######  Theorem 1.2.

(Theorem 11.1) Suppose
$f_{1},f_{2},\ldots,f_{2n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$ and
$t\geq-1$.

  1. 1.

$[T^{(t)}_{f_{1}},T^{(t)}_{f_{2}},\ldots,T^{(t)}_{f_{2n}}]$ is in the trace
class $\mathcal{S}^{1}$.

  2. 2.

$\mathrm{Tr}[T^{(t)}_{f_{1}},T^{(t)}_{f_{2}},\ldots,T^{(t)}_{f_{2n}}]=\frac{n!}{(2\pi
i)^{n}}\int_{\mathbb{B}_{n}}\mathrm{d}f_{1}\wedge\mathrm{d}f_{2}\wedge\ldots\wedge\mathrm{d}f_{2n},$
(1.3)

which is independent of $t$.

Thanks to the use of pseudodifferential calculus and its generalization in the
proof, Helton-Howe’s original theorem needs to assume that that the Toeplitz
operators to have smooth symbols. In this paper, we develop a new approach to
study the trace formula using harmonic analysis. As a result, we obtain an
improvement of the Helton-Howe trace formula for Toeplitz operators with
$\mathscr{C}^{2}$ symbols.

Our second result is about the Connes-Chern character
$\tau_{t}(f_{0},\ldots,f_{2p-1})$ for
$f_{0},\ldots,f_{2p-1}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$.
Different from the Helton-Howe trace, $\tau_{t}$ vanishes as $t$ goes to
$\infty$. In the following theorem, we identify the leading term of $\tau_{t}$
as $t\to\infty$.

######  Theorem 1.3.

(Theorem 11.6) Suppose $p\geq n+1$ is an integer and
$f_{0},f_{1},\ldots,f_{2p-1}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$.
Then

$\begin{split}&\lim_{t\to\infty}t^{p-n}\tau_{t}(f_{0},f_{1},\ldots,f_{2p-1})\\\
=&\frac{n^{p}}{\pi^{n}}\int_{\mathbb{B}_{n}}\bigg{(}\prod_{j=0}^{p-1}C_{1}(f_{2j},f_{2j+1})(z)-\prod_{j=0}^{p-1}C_{1}(f_{2j+1},f_{2j+2})(z)\bigg{)}\frac{\mathrm{d}m(z)}{(1-|z|^{2})^{n+1}},\end{split}$

where $C_{1}(f,g)$ is defined as follows,

$C_{1}(f,g)(z)=-\frac{1}{n}(1-|z|^{2})\bigg{[}\sum_{i=1}^{n}\partial_{i}f(z)\bar{\partial}_{i}g(z)-Rf(z)\bar{R}g(z)\bigg{]},\
R=\sum_{i=1}^{n}z_{i}\partial_{z_{i}},\
\overline{R}=\sum_{i=1}^{n}\bar{z}_{i}\overline{\partial}_{z_{i}}.$ (1.4)

Our approach to the above two main theorems is heavily influenced by the idea
of quantization, [6, 8, 9, 13, 19, 20, 21, 22]. Geometrically the defining
function $\psi=1-|z|^{2}$ on $\mathbb{B}_{n}$ defines the Bergman metric in
the following way.

$\begin{split}\omega:=&i\frac{-\psi\partial\bar{\partial}\psi+\partial\psi\wedge\bar{\partial}\psi}{\psi^{2}}\\\
=&i\frac{(1-|z|^{2})\sum_{j=1}^{n}\partial
z_{j}\wedge\bar{\partial}\bar{z}_{j}+(\sum_{j}\bar{z}_{j}\partial
z_{j})\wedge(\sum_{j^{\prime}}z_{j^{\prime}}\bar{\partial}\bar{z}_{j^{\prime}})}{(1-|z|^{2})^{2}}\end{split}$

defines a symplectic form on $\mathbb{B}_{n}$, c.f. [39, Prop. 2.6]. The
Toeplitz operator $T^{(t)}_{f}$ gives a quantization of the symplectic form
$i\omega$ , e.g. [22, Theorem 3], satisfying

$||T^{(t)}_{f}T^{(t)}_{g}-\sum_{j=0}^{k}t^{-j}T^{(t)}_{C_{j}(f,g)}||=O(t^{-k-1}),\
t\to\infty,$ (1.5)

where $C_{j}$ is a bilinear operator on
$\mathscr{C}^{\infty}(\overline{\mathbb{B}_{n}})$ and $C_{1}$ is defined in
Equation (1.4). The asymptotic expansion formula (1.5) provides the key tool
to study the semi-commutator,

$\sigma_{t}(f,g)=T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}.$

The asymptotic expansion formula (1.5) in the literature, e.g. [22], was well
studied for estimates on the operator norm. Estimates about the Schatten-$p$
norm in the expansion (1.5) are needed in our applications to the tracial
property in Theorem 11.1 and 11.6. We prove these estimates in Theorem 9.1. As
we need to study Toeplitz operators with $\mathscr{C}^{2}$ symbols in Theorem
11.1 and 11.6 and an estimate on Schatten-$p$ norm in Theorem 9.1, we need a
new method to develop the asymptotic estimate in Theorem 9.1 different from
the classical method via pseudodifferential/Toeplitz operator calculus [7, 8,
20, 22, 24, 25, 27, 28, 37], which requires to work with smooth symbols. Our
main tool comes from integration formulas developed in Section 7. Theorem 11.6
follows from the Schatten-$p$ estimate of the semi-commutator
$\sigma_{t}(f,g)$. As a byproduct, our method also provides an explicit
algorithm to compute the bilinear differential operator $C_{j}$ in the
asymptotic expansion (1.5), which is in general hard to compute.

A crucial fact used in our estimate is the different behavior of the
quantization in complex normal and complex tangential directions (see Remark
9.4 and Corollaries 9.7, 9.8). Roughly speaking, the expansion of the integral
kernel in the complex tangential direction creates improvement of
$\frac{1}{2}$ in the Schatten-$p$ membership, whereas the expansion of the
integral kernel in the complex normal direction creates improvement of $1$ in
the Schatten-$p$ membership. Essentially, this allows us to reduce our
estimates to the complex tangential direction. In contrast, in
pseudodifferential calculous, symbol functions for order 0 operators are
constant along the radial direction far away from the zero section of the
cotangent bundle, which simplifies the corresponding estimates. And the
difference between tangential and radial estimates suggests a deep link of our
study with the Heisenberg calculus for contact manifolds, e.g. [5, 34, 46,
50].

Instead of a direct computation as in [33], we prove Theorem 11.1 in two
steps. Suppose
$f_{1},f_{2},\ldots,f_{2n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$ and
$t\geq-1$.

  1. 1.

$\mathrm{Tr}[T^{(t)}_{f_{1}},T^{(t)}_{f_{2}},\ldots,T^{(t)}_{f_{2n}}]=\mathrm{Tr}[T^{(t+1)}_{f_{1}},T^{(t+1)}_{f_{2}},\ldots,T^{(t+1)}_{f_{2n}}].$
(1.6)

  2. 2.

$\lim_{s\to\infty}\mathrm{Tr}[T^{(s)}_{f_{1}},\ldots,T^{(s)}_{f_{2n}}]=\frac{n!}{(2\pi
i)^{n}}\int\mathrm{d}f_{1}\wedge\ldots\wedge\mathrm{d}f_{2n}.$ (1.7)

Equation (1.6) is proved using the decomposition formula
$T^{(t)}_{f_{i}}=A_{i}+B_{i}$ in Section 8, Lemmas 8.4 and 8.6. Here the
operator $A_{i}=T^{(t+1,t)}_{f_{i}}$ is the restriction of $T^{(t+1)}_{f_{i}}$
on $L_{a,t}^{2}(\mathbb{B}_{n})$. We observe that $A_{i}$ is the “principal
part” of $T^{(t)}_{f_{i}}$ for
$f_{i}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$. In Section 8 we show
that

$\mathrm{Tr}\big{\\{}[T^{(t)}_{f_{1}},T^{(t)}_{f_{2}},\ldots,T^{(t)}_{f_{2n}}]-[A_{1},A_{2},\ldots,A_{2n}]\big{\\}}=0.$
(1.8)

We point out that the “minor part” $B_{i}$ does not live in Schatten class
$\mathcal{S}^{p}$ for $p$ small enough. This fact prevents us from proving
(1.8) only using operator-theoretic tools. In Section 8, we develop Hypotheses
A which handles the operator-theoretic part of the proof of (1.8), and
Hypotheses B, where the rest is handled. As
$[T^{(t+1,t)}_{f_{1}},\ldots,T^{(t+1,t)}_{f_{2n}}]$ is the restriction of
$[T^{(t+1)}_{f_{1}},T^{(t+1)}_{f_{2}},\ldots,T^{(t+1)}_{f_{2n}}]$ on
$L_{a,t}^{2}(\mathbb{B}_{n})$, Lemma 2.6 gives

$\mathrm{Tr}[T^{(t+1,t)}_{f_{1}},\ldots,T^{(t+1,t)}_{f_{2n}}]=\mathrm{Tr}[T^{(t+1)}_{f_{1}},\ldots,T^{(t+1)}_{f_{2n}}].$
(1.9)

We obtain Equation (1.6) by combining Equations (1.8) and (1.9).

Equation (1.7) is proved by improving Theorem 11.6. In Theorem 11.6, $p$ is
assumed to be greater than or equal to $n+1$, while there are only $2n$
functions in Equation (1.7). We observe that in general the product
$\sigma_{t}(f_{1},f_{2})\ldots\sigma_{t}(f_{2n-1},f_{2n})$ is not a trace
class operator. This leads us to introduce even and odd partial
antisymmetrizations in Section 10. For $f_{1},...,f_{n},g_{1},...,g_{n}\in
L^{\infty}(\mathbb{B}_{n})$, define

$[f_{1},g_{1},\ldots,f_{n},g_{n}]^{\mathrm{odd}}_{t}=\sum_{\tau\in
S_{n}}\mathrm{sgn}(\tau)\sigma_{t}(f_{\tau(1)},g_{1})\ldots\sigma_{t}(f_{\tau(n)},g_{n}),$

and

$[f_{1},g_{1},\ldots,f_{n},g_{n}]^{\mathrm{even}}_{t}=\sum_{\tau\in
S_{n}}\mathrm{sgn}(\tau)\sigma_{t}(f_{1},g_{\tau(1)})\ldots\sigma_{t}(f_{n},g_{\tau(n)}).$

Using the asymptotic expansion of semi-commutator in Section 9, we improve
Theorem 11.6 to the even and odd partial antisymmetrizations.

######  Theorem 1.4.

(Theorem 10.3) Suppose $t\geq-1$ and
$f_{1},g_{1},\ldots,f_{n},g_{n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$.
Then the partial antisymmetrizations
$[f_{1},g_{1},\ldots,f_{n},g_{n}]_{t}^{\mathrm{odd}}$ and
$[f_{1},g_{1},\ldots,f_{n},g_{n}]_{t}^{\mathrm{even}}$ are in the trace class.
Moreover,

$\begin{split}\lim_{t\to\infty}\mathrm{Tr}[f_{1},g_{1},\ldots,f_{n},g_{n}]_{t}^{\mathrm{odd}}=&\lim_{t\to\infty}\mathrm{Tr}[f_{1},g_{1},\ldots,f_{n},g_{n}]_{t}^{\mathrm{even}}=\frac{1}{(2\pi
i)^{n}}\int_{\mathbb{B}_{n}}\partial
f_{1}\wedge\bar{\partial}g_{1}\wedge\ldots\wedge\partial
f_{n}\wedge\bar{\partial}g_{n}.\end{split}$

As a corollary of Theorem 10.3, we obtain Equation (1.7) by further
antisymmetrization of the even (odd) antisymmetric sums.

Our solution to the generalized Helton-Howe trace formula through quantization
is closely related to the method developed in [9] to a solution to the Atiyah-
Weinstein conjecture for quantized contact transform. Such a similarity
suggests that our developments can be generalized to strongly pseudoconvex
domains, egg domains, Fock spaces, submodules and their quotient modules of
$L^{2}_{a,t}(\mathbb{B}_{n})$, e.g. [29], and the Dury-Arveson spaces. More
generally, we hope that our study will shed a light on constructing new cyclic
cocycles beyond the Helton-Howe traces, which could have applications in
noncommutative geometry.

The paper is organized as follows. In Section 2, we recall the definitions of
weighted Bergman spaces and Hardy spaces together with their basic properties
and properties of Schatten $p$-class operators. In Section 3, we explain the
main theorems in the special case of unit disk. In particular, we obtain the
two main theorems by proving an explicit formula for the semi-commutator
$\sigma_{t}(f,g)$ in Theorem 3.1. An interesting application to the estimate
of Hilbert-Schmidt norm of a Hankel operator is discussed. In Section 4, we
introduce two sets of functions which arise naturally in our study of
integration formulas in Section 7. In Section 5, we develop criteria for
integral operators to be bounded between different weighted spaces
$L^{2}_{a,t}(\mathbb{B}_{n})$. We develop some useful estimate for integral
operators to belong to Schatten-$p$ class in Section 6. Section 7, we
establish an integration by parts formula which is crucial in our development
of the asymptotic expansion of semi-commutator of Toeplitz operators. In
Section 8, we prove Equation (1.8). We develop the asymptotic expansion
formula, Theorem 9.1, in Section 9. We introduce the even and odd
antisymmetrization and prove Theorem 10.3 and Equation (1.7) in Section 10.
The two main theorems, Theorem 11.1 and 11.6, are proved in Section 11. And a
proof of the generalized Bochner-Martinelli formula is included in the
appendix.

Acknowledgment: We would like to thank Mohammad Jabbari, Richard Rochberg, and
Jingbo Xia for inspiring discussions. Tang is partially supported by NSF
Grants DMS 1800666, 1952551; Wang is partially supported by NSF Grant 2101370.

##  2 Preliminaries

In this section, we recall some basic definitions and properties about
weighted Bergman spaces and Schatten-$p$ class operators.

###  2.1 Spaces on $\mathbb{B}_{n}$

Let $\mathbb{B}_{n}$ be the open unit ball of $\mathbb{C}^{n}$ and
$\mathbb{S}_{n}=\partial\mathbb{B}_{n}$ the unit sphere. Let $m$ be the
Lebesgue measure on $\mathbb{B}_{n}$ and $\sigma$ be the surface measure on
$\mathbb{S}_{n}$. Denote
$\sigma_{2n-1}=\sigma(\mathbb{S}_{n})=\frac{2\pi^{n}}{(n-1)!}$.

Hardy Space: The Hardy space $H^{2}(\mathbb{S}_{n})$ is the Hilbert space of
holomorphic functions on $\mathbb{B}_{n}$ with the norm

$\|f\|_{H^{2}(\mathbb{S}_{n})}^{2}=\sup_{0<r<1}\int_{\mathbb{S}_{n}}|f(rz)|^{2}\frac{\mathrm{d}\sigma(z)}{\sigma_{2n-1}}.$

Equivalently, $H^{2}(\mathbb{S}_{n})$ is the closure of analytic polynomials
in
$L^{2}(\mathbb{S}_{n}):=L^{2}(\mathbb{S}_{n},\frac{\mathrm{d}\sigma}{\sigma_{2n-1}})$.
The Hardy space is a reproducing kernel Hilbert space on $\mathbb{B}_{n}$ and
the reproducing kernel is

$K_{w}(z)=\frac{1}{(1-\langle z,w\rangle)^{n}},\quad\forall
w\in\mathbb{B}_{n}.$

For any $f\in L^{\infty}(\mathbb{S}_{n})$, the Toeplitz operator on
$H^{2}(\mathbb{S}_{n})$ with symbol $f$ is defined to be the compression

$T_{f}=PM_{f}|_{H^{2}(\mathbb{S}_{n})},$

where $M_{f}$ is the pointwise multiplication on $L^{2}(\mathbb{S}_{n})$, and
$P$ is the orthogonal projection from $L^{2}(\mathbb{S}_{n})$ onto
$H^{2}(\mathbb{S}_{n})$. Using the reproducing kernel, we can write $T_{f}$ as
an integral operator. For $h\in H^{2}(\mathbb{S}_{n})$,

$T_{f}h(z)=\int_{\mathbb{S}_{n}}f(w)h(w)K_{w}(z)\frac{\mathrm{d}\sigma(z)}{\sigma_{2n-1}},\quad\forall
z\in\mathbb{B}_{n}.$

Our discussion will also involve Hankel operators. The Hankel operator with
symbol $f$ is $H_{f}=(I-P)M_{f}P$ from $H^{2}(\mathbb{S}_{n})$ to
$L^{2}(\mathbb{S}_{n})$.

Weighted Bergman Spaces: For $t>-1$, define the probability measure on
$\mathbb{B}_{n}$:

$\mathrm{d}\lambda_{t}(z)=\frac{(n-1)!}{\pi^{n}B(n,t+1)}(1-|z|^{2})^{t}\mathrm{d}m(z).$

Here $B(n,t+1)$ is the Beta function. The weighted Bergman space
$L_{a,t}^{2}(\mathbb{B}_{n})$ is the subspace of
$L^{2}(\mathbb{B}_{n},\lambda_{t})$ consisting of holomorphic functions on
$\mathbb{B}_{n}$. The reproducing kernel of $L_{a,t}^{2}(\mathbb{B}_{n})$ is

$K^{(t)}_{w}(z)=\frac{1}{(1-\langle z,w\rangle)^{n+1+t}},\quad\forall
w\in\mathbb{B}_{n}.$

For any $f\in L^{\infty}(\mathbb{B}_{n})$, the Toeplitz operator $T^{(t)}_{f}$
is the compression

$T^{(t)}_{f}=P^{(t)}M^{(t)}_{f}|_{L_{a,t}^{2}(\mathbb{B}_{n})},$

where $P^{(t)}$ is the orthogonal projection from
$L^{2}(\mathbb{B}_{n},\lambda_{t})$ onto $L_{a,t}^{2}(\mathbb{B}_{n})$, and
$M^{(t)}_{f}$ is the multiplication operator on
$L^{2}(\mathbb{B}_{n},\lambda_{t})$. The Hankel operator with symbol $f$ is

$H^{(t)}_{f}=(I-P^{(t)})M_{f}^{(t)}P^{(t)}.$

Using the reproducing kernels, we can write $T^{(t)}_{f},H^{(t)}_{f}$ as
integral operators. For $h\in L_{a,t}^{2}(\mathbb{B}_{n})$, we have the
following expressions,

$T^{(t)}_{f}h(z)=\int_{\mathbb{B}_{n}}f(w)h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),\quad\forall
z\in\mathbb{B}_{n}.$
$H^{(t)}_{f}h(z)=\int_{\mathbb{B}_{n}}\big{(}f(z)-f(w)\big{)}h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),\quad\forall
z\in\mathbb{B}_{n}.$

It is sometimes understood that the Hardy space is the weighted Bergman space
at $t=-1$. This is partially justified by the form of its reproducing kernel.

An important tool on $\mathbb{B}_{n}$ is the Möbius transform.

######  Definition 2.1.

For $z\in\mathbb{B}_{n}$, $z\neq 0$, the Möbius transform $\varphi_{z}$ is the
biholomorphic mapping on $\mathbb{B}_{n}$ defined as follows.

$\varphi_{z}(w)=\frac{z-P_{z}(w)-(1-|z|^{2})^{1/2}Q_{z}(w)}{1-\langle
w,z\rangle},\quad\forall w\in\overline{\mathbb{B}_{n}}.$

Here $P_{z}$ and $Q_{z}$ denote the orthogonal projection from
$\mathbb{C}^{n}$ onto $\mathbb{C}z$ and $z^{\perp}$, respectively. Define

$\varphi_{0}(w)=-w,\quad\forall w\in\overline{\mathbb{B}_{n}}.$

It is well-known that $\varphi_{z}$ is an automorphism of $\mathbb{B}_{n}$
satisfying $\varphi_{z}\circ\varphi_{z}=Id$. Also, the two variable function
$\rho(z,w):=|\varphi_{z}(w)|=|\varphi_{w}(z)|$ defines a metric on
$\mathbb{B}_{n}$. Moreover, $\beta(z,w):=\tanh^{-1}\rho(z,w)$ coincides with
the Bergman metric on $\mathbb{B}_{n}$.

We list some lemmas that serve as basic tools on $\mathbb{B}_{n}$. Most of the
following can be found in [47, 55]. A proof will be provided when necessary.

For non-negative values $A,B$, by $A\lesssim B$ we mean that there is a
constant $C$ such that $A\leq CB$. Sometimes, to emphasize that the constant
$C$ depends on some parameter $a$, we write $A\lesssim_{a}B$. The notations
$\gtrsim,\gtrsim_{a},\approx,\approx_{a}$ are defined similarly.

######  Lemma 2.2.

Suppose $z,w,\zeta\in\mathbb{B}_{n}$.

  * (1)

$\frac{1}{1-\langle\varphi_{\zeta}(z),\varphi_{\zeta}(w)\rangle}=\frac{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,w\rangle)}{(1-|\zeta|^{2})(1-\langle
z,w\rangle)}$.

  * (2)

$1-|\varphi_{z}(w)|^{2}=\frac{(1-|z|^{2})(1-|w|^{2})}{|1-\langle
z,w\rangle|^{2}}$.

  * (3)

For any $R>0$ there exists $C>1$ such that whenever $\beta(z,w)<R$,

$\frac{1}{C}\leq\frac{1-|z|^{2}}{1-|w|^{2}}\leq
C,\quad\frac{1}{C}\leq\frac{|1-\langle z,\zeta\rangle|}{|1-\langle
w,\zeta\rangle|}\leq C.$

  * (4)

The real Jacobian of $\varphi_{z}$ is $\frac{(1-|z|^{2})^{n+1}}{|1-\langle
z,\cdot\rangle|^{2n+2}}$ on $\mathbb{B}_{n}$ and
$\frac{(1-|z|^{2})^{n}}{|1-\langle z,\cdot\rangle|^{2n}}$ on $\mathbb{S}_{n}$.

  * (5)

For $z\in\mathbb{B}_{n}$,

$z-\varphi_{z}(w)=\frac{(1-|z|^{2})P_{z}(w)+(1-|z|^{2})^{1/2}Q_{z}(w)}{1-\langle
w,z\rangle}:=\frac{A_{z}w}{1-\langle w,z\rangle},$

where $A_{z}=[a_{z}^{ij}]$ is an $n\times n$ matrix depending on $z$, and $w$
is viewed as a column vector.

  * (6)

There exists $C>0$ such that for any $z\in\mathbb{B}_{n}$, $z\neq 0$,

$|z-P_{z}(w)|\leq|\varphi_{z}(w)||1-\langle z,w\rangle|,\quad|Q_{z}(w)|\leq
C|\varphi_{z}(w)||1-\langle z,w\rangle|^{1/2},$ (2.1)

and

$|z-w|\leq C|\varphi_{z}(w)||1-\langle z,w\rangle|^{1/2}.$ (2.2)

In contrast, if $n=1$, then $|z-w|=|\varphi_{z}(w)||1-z\bar{w}|$.

  * (7)

$1-|z|^{2}\leq 2|1-\langle z,w\rangle|$.

###### Proof.

Most of the above are either well-known (cf. [47, 55]) or straightforward to
verify. The only part that requires some clarification is the second estimate
in (6), i.e.,

$|Q_{z}(w)|\lesssim|\varphi_{z}(w)||1-\langle z,w\rangle|^{1/2}.$ (2.3)

On the one hand, from the definition of $\varphi_{z}(w)$, we easily get the
follow inequality,

$|Q_{z}(w)|\leq|\varphi_{z}(w)|\frac{|1-\langle
z,w\rangle|}{(1-|z|^{2})^{1/2}}.$

If $|\varphi_{z}(w)|\leq\frac{1}{2}$, then by (3), $|1-\langle
z,w\rangle|\approx 1-|z|^{2}$. From this (2.3) follows.

On the other hand, since

$2|1-\langle z,w\rangle|\geq 2-2\mathrm{Re}\langle
z,w\rangle\geq|z|^{2}+|w|^{2}-2\mathrm{Re}\langle z,w\rangle=|z-w|^{2},$

for $|\varphi_{z}(w)|>\frac{1}{2}$, we obtain the following estimates,

$|Q_{z}(w)|=|Q_{z}(w-z)|\leq|z-w|\lesssim|1-\langle
z,w\rangle|^{\frac{1}{2}}<2|\varphi_{z}(w)||1-\langle
z,w\rangle|^{\frac{1}{2}}.$

Thus we get (2.3) in both cases. This completes the proof of Lemma 2.2. ∎

######  Lemma 2.3.

([47, Proposition 5.1.2]) The two variable function $d(z,w)=|1-\langle
z,w\rangle|^{\frac{1}{2}}$ on $\overline{\mathbb{B}_{n}}$ satisfies the
triangle inequality, i.e.,

$d(z,w)\leq d(z,\xi)+d(\xi,w),\quad\forall
z,w,\xi\in\overline{\mathbb{B}_{n}}.$

######  Lemma 2.4 (Rudin-Forelli type estimates).

  * (1)

Suppose $t>-1$, $c\in\mathbb{R}$. Then there exists $C>0$ such that for any
$z\in\mathbb{B}_{n}$,

$\int_{\mathbb{B}_{n}}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{n+1+t+c}}\mathrm{d}m(w)\leq\begin{cases}C(1-|z|^{2})^{-c},&c>0,\\\
C\ln\frac{1}{1-|z|^{2}},&c=0,\\\ C,&c<0,\end{cases}$ (2.4)
$\int_{\mathbb{S}_{n}}\frac{1}{|1-\langle
z,w\rangle|^{n+c}}\mathrm{d}\sigma(w)\leq\begin{cases}C(1-|z|^{2})^{-c},&c>0,\\\
C\ln\frac{1}{1-|z|^{2}},&c=0,\\\ C,&c<0.\end{cases}$ (2.5)

  * (2)

Suppose $t>-1$, $a,b,c>0$, $a\geq c,b\geq c$, and $a+b<n+1+t+c$. Then there
exists $C>0$ such that for any $z,\xi\in\mathbb{B}_{n}$,

$\int_{\mathbb{B}_{n}}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{a}|1-\langle w,\xi\rangle|^{b}}\mathrm{d}m(w)\leq
C\frac{1}{|1-\langle z,\xi\rangle|^{c}}.$ (2.6)

  * (3)

Suppose $\phi:(0,1)\to[0,\infty)$ is measurable. Suppose $a>-n$,
$b\in\mathbb{R}$, and

$\phi(s)\lesssim s^{a}(1-s)^{b},\quad s\in(0,1).$

Then for any $t>-1-b$, $c>-b$ there exists $C>0$ such that for any
$z\in\mathbb{B}_{n}$,

$\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{n+1+t+c}}\mathrm{d}m(w)\leq C(1-|z|^{2})^{-c}.$ (2.7)

###### Proof.

The estimates in (1) are standard Rudin-Forelli estimates. See [47,
Proposition 1.4.10] for a proof. Let

$A=\\{w\in\mathbb{B}_{n}:|1-\langle z,w\rangle|\leq|1-\langle w,\xi|\\};\quad
B=\\{w\in\mathbb{B}_{n}:|1-\langle z,w\rangle|>|1-\langle w,\xi\rangle|\\}.$

By Lemma 2.3, we have the following equality,

$|1-\langle z,\xi\rangle|^{1/2}\leq|1-\langle z,w\rangle|^{1/2}+|1-\langle
w,\xi\rangle|^{1/2}.$

Then we obtain the following bounds,

$|1-\langle z,w\rangle|\geq\frac{1}{4}|1-\langle z,\xi\rangle|,\forall w\in
B;\quad|1-\langle w,\xi\rangle|\geq\frac{1}{4}|1-\langle z,\xi\rangle|,\forall
w\in A.$

By assumption, $a\geq c$, $b\geq c$, $a+b-c<n+1+t$, therefore by the standard
Rudin-Forelli estimate, we compute the integral as follows,

$\displaystyle\int_{\mathbb{B}_{n}}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{a}|1-\langle w,\xi\rangle|^{b}}\mathrm{d}m(w)$ $\displaystyle=$
$\displaystyle\int_{A}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{a}|1-\langle
w,\xi\rangle|^{b}}\mathrm{d}m(w)+\int_{B}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{a}|1-\langle w,\xi\rangle|^{b}}\mathrm{d}m(w)$
$\displaystyle\lesssim$ $\displaystyle\frac{1}{|1-\langle
z,\xi\rangle|^{c}}\int_{A}\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{a+b-c}}\mathrm{d}m(w)+\frac{1}{|1-\langle
z,\xi|^{c}}\int_{B}\frac{(1-|w|^{2})^{t}}{|1-\langle
w,\xi\rangle|^{b+a-c}}\mathrm{d}m(w)$ $\displaystyle\lesssim$
$\displaystyle\frac{1}{|1-\langle z,\xi\rangle|^{c}},$

This proves (2).

To prove (3), make the change of variable $w=\varphi_{z}(\xi)$ in the left
hand side of (2.7). We compute the integral as follows,

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{n+1+t+c}}\mathrm{d}m(w)$ $\displaystyle=$
$\displaystyle\frac{1}{(1-|z|^{2})^{c}}\int_{\mathbb{B}_{n}}\phi(|\xi|^{2})\frac{(1-|\xi|^{2})^{t}}{|1-\langle
z,\xi\rangle|^{n+1+t-c}}\mathrm{d}m(\xi)$ $\displaystyle=$
$\displaystyle\frac{1}{(1-|z|^{2})^{c}}\int_{0}^{1}\phi(r^{2})r^{2n-1}(1-r^{2})^{t}\int_{\mathbb{S}_{n}}\frac{1}{|1-\langle
z,r\eta\rangle|^{n+1+t-c}}\mathrm{d}\sigma(\eta)\mathrm{d}r$
$\displaystyle\lesssim$
$\displaystyle\frac{1}{(1-|z|^{2})^{c}}\int_{0}^{1}\phi(r^{2})r^{2n-1}(1-r^{2})^{m}\mathrm{d}r,$

where $m=c-1$ when $1+t-c>0$, $m=t$ when $1+t-c<0$, and when $1+t-c=0$, we
take $m=t-\epsilon$ for a sufficiently small $\epsilon>0$. With our assumption
it is easy to see that $m>-b-1$ and therefore the integral above is finite.
This proves (3) and completes the proof of Lemma 2.3. ∎

The following lemma is well-known. See [56, Theorem 6.4] for a proof at $n=1$.
The same proof works for general $n$.

######  Lemma 2.5.

Suppose $t>-1$ and $T$ is a trace class operator on
$L_{a,t}^{2}(\mathbb{B}_{n})$. Then

$\mathrm{Tr}T=\int_{\mathbb{B}_{n}}\langle
TK^{(t)}_{z},K^{(t)}_{z}\rangle\mathrm{d}\lambda_{t}(z).$

######  Lemma 2.6.

Suppose $s>t\geq-1$ and $T$ is a bounded operator on
$L_{a,s}^{2}(\mathbb{B}_{n})$. Suppose $L_{a,t}^{2}(\mathbb{B}_{n})$ is
invariant under $T$. Denote $\hat{T}$ its restriction to
$L_{a,t}^{2}(\mathbb{B}_{n})$. By the closed graph theorem, $\hat{T}$ is
bounded on $L_{a,t}^{2}(\mathbb{B}_{n})$. Assume that
$T\in\mathcal{S}^{1}\big{(}L_{a,s}^{2}(\mathbb{B}_{n})\big{)}$ and
$\hat{T}\in\mathcal{S}^{1}\big{(}L_{a,t}^{2}(\mathbb{B}_{n})\big{)}$. Then

$\mathrm{Tr}T=\mathrm{Tr}\hat{T}.$

###### Proof.

The set $\\{z^{\alpha}\\}_{\alpha\in\mathbb{N}_{0}^{n}}$ forms an orthogonal
basis of both $L_{a,s}^{2}(\mathbb{B}_{n})$ and $L_{a,t}^{2}(\mathbb{B}_{n})$.
For $\alpha\in\mathbb{N}_{0}^{n}$, we write

$Tz^{\alpha}=\hat{T}z^{\alpha}=\sum_{\beta\in\mathbb{N}_{0}^{n}}a_{\alpha,\beta}z^{\beta}.$

Compute the traces of $T$ and $\hat{T}$ as follows.

$\displaystyle\mathrm{Tr}T=$
$\displaystyle\sum_{\alpha\in\mathbb{N}_{0}^{n}}\frac{\langle
Tz^{\alpha},z^{\alpha}\rangle_{L_{a,s}^{2}(\mathbb{B}_{n})}}{\|z^{\alpha}\|_{L_{a,s}^{2}(\mathbb{B}_{n})}^{2}}$
$\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}_{0}^{n}}\frac{a_{\alpha,\alpha}\langle
z^{\alpha},z^{\alpha}\rangle_{L_{a,s}^{2}(\mathbb{B}_{n})}}{\|z^{\alpha}\|^{2}_{L_{a,s}^{2}(\mathbb{B}_{n})}}$
$\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}_{0}^{n}}a_{\alpha,\alpha}$
$\displaystyle=$
$\displaystyle\sum_{\alpha\in\mathbb{N}_{0}^{n}}\frac{\langle\hat{T}z^{\alpha},z^{\alpha}\rangle_{L_{a,t}^{2}(\mathbb{B}_{n})}}{\|z^{\alpha}\|_{L_{a,t}^{2}(\mathbb{B}_{n})}^{2}}=\mathrm{Tr}\hat{T}.$

This completes the proof of Lemma 2.6. ∎

For $p>0$, a bounded operator $T$ on a Hilbert space $\mathcal{H}$ is said to
be in the Schatten-$p$ class $\mathcal{S}^{p}$ if $|T|^{p}$ belongs to the
trace class. The Schatten-$p$ class operators $\mathcal{S}^{p}$ are analogues
of $l^{p}$ spaces in the operator-theoretic setting. Conventionally,
$\mathcal{S}^{\infty}$ denotes the space of compact operators. The following
two lemmas will be used constantly.

######  Lemma 2.7.

([49, Theorem 2.8]) Suppose $A$, $B$ are bounded operators on a Hilbert space,
and $1\leq p,q,r\leq\infty$, $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$. If
$A\in\mathcal{S}^{p}$ and $B\in\mathcal{S}^{q}$, then

$AB\in\mathcal{S}^{r},\quad\text{ and
}\quad\|AB\|_{\mathcal{S}^{r}}\leq\|A\|_{\mathcal{S}^{p}}\|B\|_{\mathcal{S}^{q}}.$

######  Lemma 2.8.

([49, Corollary 3.8]) Suppose $A,B$ are bounded operators on a Hilbert space.
If both $AB$ and $BA$ are in the trace class then

$\mathrm{Tr}[A,B]=0.$

######  Remark 2.9.

Suppose $n>1$ and $X_{1},X_{2},\ldots,X_{n}$ are bounded operators on a
Hilbert space such that

$X_{i}\in\mathcal{S}^{p},\quad\forall p>n,$

and there exists $j\in\\{1,\ldots,n\\}$ such that $X_{j}$ is in
$\mathcal{S}^{p}$ for some $p<n$. Then we can choose
$p_{1},\ldots,p_{j-1},p_{j+1},\ldots,p_{n}>n$ and $1\leq p_{j}<n$ such that

$X_{i}\in\mathcal{S}^{p_{i}},i=1,\ldots,n,\quad\text{ and
}\quad\frac{1}{p_{1}}+\frac{1}{p_{2}}+\ldots+\frac{1}{p_{n}}=1.$

By an inductive application of Lemma 2.7 we have

$X_{1}X_{2}\ldots X_{n}\in\mathcal{S}^{1}.$

Moreover, by Lemma 2.8, for any $k=1,\ldots,n$,

$\mathrm{Tr}X_{1}X_{2}\ldots X_{n}=\mathrm{Tr}X_{k}X_{k+1}\ldots
X_{n}X_{1}\ldots X_{k-1}.$

This will be used repeatedly in this paper.

##  3 An Example: Trace Formulas on the Disk

As a demonstration of some of the key techniques used in this paper, in this
section, we study the example of unit disc.

Suppose $f$ and $g$ are two Lipschitz functions on $\mathbb{D}$. It is well-
known that for any $t>-1$, the semi-commutator
$T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}$ is in the trace class (cf. [55]).

A natural question is how to compute the trace of these semi-commutators. For
this paper we restrict our discussion to symbols with some smoothness
condition. It turns out that the following trace formula holds.

######  Theorem 3.1.

Suppose $t>-1$ and $f,g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$. Then

$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}=\frac{1}{2\pi
i}\int_{\mathbb{D}}\partial
f\wedge\bar{\partial}g+\int_{\mathbb{D}^{2}}\rho_{t}(|\varphi_{z}(w)|^{2})\Delta
f(z)\Delta g(w)\mathrm{d}m(z,w).$ (3.1)

Here $\rho_{t}$ is defined as below and is strictly positive on $(0,1)$.

$\rho_{t}(s)=\frac{t+1}{16\pi^{2}}\int_{s}^{1}(1-x)^{t}x^{-1}F(s,x)\mathrm{d}x,$

where

$F(s,x)=-\bigg{[}x\ln\frac{s}{x}+(1-x)\ln\frac{1-s}{1-x}\bigg{]}.$

Moreover,

$\lim_{t\to\infty}\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}=\frac{1}{2\pi
i}\int_{\mathbb{D}}\partial f\wedge\bar{\partial}g.$ (3.2)

######  Remark 3.2.

We can show that in the case of Hardy space, the following analogous result
holds. Suppose $f,g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$. Then

$\mathrm{Tr}\bigg{(}T_{f}T_{g}-T_{fg}\bigg{)}=\frac{1}{2\pi
i}\int_{\mathbb{D}}\partial
f\wedge\bar{\partial}g+\int_{\mathbb{D}^{2}}\rho_{-1}(|\varphi_{z}(w)|^{2})\Delta
f(z)\Delta g(w)\mathrm{d}m(z,w),$

where

$\rho_{-1}(s)=-\frac{1}{16\pi^{2}}\ln s.$

It follows that for fixed $f,g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$, the
traces $\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$ are
continuous in $t$.

######  Remark 3.3.

From the identity

$T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}=-H^{(t)*}_{\bar{f}}H^{(t)}_{g},$

we have the following equation of traces,

$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}=-\mathrm{Tr}\bigg{(}H^{(t)*}_{\bar{f}}H^{(t)}_{g}\bigg{)}=-\langle
H^{(t)}_{g},H^{(t)}_{\bar{f}}\rangle_{\mathcal{S}^{2}}.$

Thus the problem of finding a trace formula for semi-commutators is equivalent
to the problem of finding a formula for the Hilbert-Schmidt norm for Hankel
operators. For $h\in L_{a,t}^{2}(\mathbb{B}_{n})$,

$\displaystyle H^{(t)}_{g}h(z)=$
$\displaystyle(I-P^{(t)})(gh)(z)=\int_{\mathbb{D}}\big{(}g(z)-g(w)\big{)}h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$ $\displaystyle g(z)\langle
h,K^{(t)}_{z}\rangle_{L^{2}(\lambda_{t})}-\langle
gh,K^{(t)}_{z}\rangle_{L^{2}(\lambda_{t})}$ $\displaystyle=$ $\displaystyle
g(z)\langle h,K^{(t)}_{z}\rangle_{L^{2}(\lambda_{t})}-\langle
h,\bar{g}K^{(t)}_{z}\rangle_{L^{2}(\lambda_{t})}.$

In the case when $\bar{g}$ is analytic we have $\bar{g}K^{(t)}_{z}\in
L_{a,t}^{2}(\mathbb{B}_{n})$ and therefore the above integral vanishes for all
$h\in L^{2}(\lambda_{t})\ominus L_{a,t}^{2}(\mathbb{B}_{n})$. Thus in the case
when $\bar{g}$ is analytic, $H^{(t)}_{g}$ is exactly the integral operator on
$L^{2}(\lambda_{t})$ defined by the formula above. It is well-known that the
Hilbert-Schmidt norm of an integral operator is given by the $L^{2}$ norm of
its integral kernel. In other words, for conjugate analytic symbol $g$, we
already have

$\|H^{(t)}_{g}\|_{\mathcal{S}^{2}}^{2}=\int_{\mathbb{D}^{2}}|g(z)-g(w)|^{2}|K^{(t)}_{w}(z)|^{2}\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z).$

A similar argument works for the Hardy space.

Notice also that $H^{(t)}_{g}=0$ whenever $g$ is analytic. Thus the problem of
finding $\|H^{(t)}_{g}\|_{\mathcal{S}^{2}}$ is somewhat trivial for
$g=g_{1}+g_{2}$, with $g_{1}$ analytic and $g_{2}$ conjugate analytic. In the
case of the Hardy space $H^{2}(\mathbb{T})$, all symbols $g\in
L^{\infty}(\mathbb{T})$ are of this form. Thus we already have a formula for
$\|H_{g}\|_{\mathcal{S}^{2}}$ on the Hardy space. A refined formula was given
by Janson, Upmeier and Wallstén in [36], which holds also for some other
Schatten-classes (see Remark 11.5). We can also see this from a different
point of view. Notice that $L^{2}(\mathbb{T})$ has a natural orthonormal basis
$\\{e^{ik\theta}\\}_{k\in\mathbb{Z}}$. Using the matrix expression of the
Hankel operator $H_{g}$ under this basis gives

$\mathrm{Tr}H_{g}^{*}H_{g}=\|H_{g}\|_{\mathcal{S}^{2}}^{2}=\sum_{k=1}^{\infty}k|\widehat{g}(-k)|^{2}.$

Here $\widehat{g}$ denotes the Fourier transform of $g\in
L^{\infty}(\mathbb{T})$. In the case of the weighted Bergman spaces, the
Hankel operator $H^{(t)}_{g}$ depends on the value of $g$ inside $\mathbb{D}$,
and in general $g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$ does not divide
into analytic and conjugate analytic parts. Also, the ambient space
$L^{2}(\lambda_{t})$ does not have a standard orthonormal basis. Therefore our
formula in Theorem 3.1 is nontrivial. In this sense, the formula (3.1) can be
viewed as a generalization of the various formulas on $H^{2}(\mathbb{D})$.

Since $|\varphi_{z}(w)|=|\varphi_{w}(z)|$, it follows immediately that the
second term in (3.1) is symmetric in the symbols $f$ and $g$. As a
consequence, the following trace formula for commutators of Toeplitz operators
holds. For the case when $t=0$, this result is well-known(cf. [32, 54]).

######  Corollary 3.4.

Suppose $t\geq-1$ and $f,g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$. Then

$\mathrm{Tr}[T^{(t)}_{f},T^{(t)}_{g}]=\frac{1}{2\pi
i}\int_{\mathbb{D}}\mathrm{d}f\wedge\mathrm{d}g.$ (3.3)

As in Remark 3.3, by the identity
$T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}=-H^{(t)*}_{\bar{f}}H^{(t)}_{g}$, we have
the following equation of traces,

$\mathrm{Tr}(T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg})=-\mathrm{Tr}(H^{(t)*}_{\bar{f}}H^{(t)}_{g})=-\langle
H^{(t)}_{g},H^{(t)}_{\bar{f}}\rangle_{\mathcal{S}^{2}}.$

Thus (3.1) leads to a formula for the inner product of Hankel operators in the
Hilbert-Schmidt class. In particular, it leads to a formula for the Hilbert-
Schmidt norm of Hankel operators.

######  Corollary 3.5.

Suppose $t>-1$ and $g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$. Then

$\|H^{(t)}_{g}\|_{\mathcal{S}^{2}}^{2}=\frac{1}{\pi}\int_{\mathbb{D}}|\bar{\partial}g|^{2}\mathrm{d}m-\int_{\mathbb{D}^{2}}\rho_{t}(|\varphi_{z}(w)|^{2})\Delta\bar{g}(z)\Delta
g(w)\mathrm{d}m(z,w).$

where $\rho_{t}$ is defined as in Theorem 3.1. In particular,

$\lim_{t\to\infty}\|H^{(t)}_{g}\|_{\mathcal{S}^{2}}^{2}=\frac{1}{\pi}\int_{\mathbb{D}}|\bar{\partial}g|^{2}\mathrm{d}m.$

For Hankel operators with real subharmonic symbols, the second term is non-
negative. So the following holds.

######  Corollary 3.6.

Suppose $t\geq-1$ and $g\in\mathscr{C}^{2}(\overline{\mathbb{D}})$ is real-
valued and subharmonic in $\mathbb{D}$. Then

$\|H^{(t)}_{g}\|_{\mathcal{S}^{2}}^{2}\leq\frac{1}{\pi}\int_{\mathbb{D}}|\bar{\partial}g|^{2}\mathrm{d}m,$
(3.4)

with equality holds if and only if $g$ is harmonic in $\mathbb{D}$.

For the rest of this section, we give the proof of Theorem 3.1. The main
ingredient is an integral formula coming from the Cauchy formula.

######  Definition 3.7.

Suppose $t>-1$ and $\phi:(0,1)\to[0,\infty)$ is measurable. Define the
operations on $\phi$:

$\mathcal{F}^{(t)}\phi(s)=\int_{s}^{1}\phi(r)(1-r)^{t}\mathrm{d}r,\quad\mathcal{G}^{(t)}\phi(s)=s^{-1}(1-s)^{-t-1}\mathcal{F}^{(t)}\phi(s).$

######  Lemma 3.8.

Suppose that $t>-1$, and $\phi:(0,1)\mapsto[0,\infty)$ is a measurable
function, and $v\in\mathscr{C}^{1}(\overline{\mathbb{D}})$. Assume that the
two integrals

$\int_{\mathbb{D}}\phi(|z|^{2})v(z)\mathrm{d}\lambda_{t}(z),\quad\int_{\mathbb{D}}(1-|z|^{2})\mathcal{G}^{(t)}\phi(|z|^{2})\bar{z}\bar{\partial}v(z)\mathrm{d}\lambda_{t}(z)$

both converge absolutely. Then

$\displaystyle\int_{\mathbb{D}}\phi(|z|^{2})v(z)\mathrm{d}\lambda_{t}(z)$
(3.5) $\displaystyle=$
$\displaystyle\begin{cases}(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(0)+\int_{\mathbb{D}}(1-|z|^{2})\mathcal{G}^{(t)}\phi(|z|^{2})\bar{z}\bar{\partial}v(z)\mathrm{d}\lambda_{t}(z),&v(0)\neq
0,\mathcal{F}^{(t)}\phi(0)<\infty,\\\
\int_{\mathbb{D}}(1-|z|^{2})\mathcal{G}^{(t)}\phi(|z|^{2})\bar{z}\bar{\partial}v(z)\mathrm{d}\lambda_{t}(z),&v(0)=0,\mathcal{F}^{(t)}\phi(0)\leq\infty.\end{cases}$

###### Proof.

Assuming first that $v(0)=0$. For any $0<r<1$, denote $\sigma_{r}$ the
Euclidean surface measure on $r\mathbb{T}$. If we apply the Cauchy Formula to
$\Omega=r\mathbb{D}\subset\mathbb{C}$ and
$v\in\mathscr{C}^{1}(\overline{\mathbb{D}})$, then we have the following
equation of integrals,

$\int_{r\mathbb{T}}v(z)\mathrm{d}\sigma_{r}(z)=2r\int_{r\mathbb{D}}\frac{\bar{\partial}v(z)}{z}\mathrm{d}m(z).$

By assumption the left hand side of (3.5) is absolutely integrable, so we
compute the integral as follows,

$\displaystyle\int_{\mathbb{D}}\phi(|z|^{2})v(z)\mathrm{d}\lambda_{t}(z)=$
$\displaystyle\frac{(t+1)}{\pi}\int_{0}^{1}\phi(r^{2})(1-r^{2})^{t}\bigg{\\{}\int_{r\mathbb{T}}v(z)\mathrm{d}\sigma_{r}(z)\bigg{\\}}\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{(t+1)}{\pi}\int_{0}^{1}\phi(r^{2})(1-r^{2})^{t}\bigg{\\{}2r\int_{r\mathbb{D}}\frac{\bar{\partial}v(z)}{z}\mathrm{d}m(z)\bigg{\\}}\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{(t+1)}{\pi}\int_{\mathbb{D}}\int_{|z|}^{1}2r\phi(r^{2})(1-r^{2})^{t}\mathrm{d}r\cdot\frac{\bar{\partial}v(z)}{z}\mathrm{d}m(z)$
$\displaystyle=$
$\displaystyle\frac{(t+1)}{\pi}\int_{\mathbb{D}}\bigg{[}\int_{|z|^{2}}^{1}\phi(s)(1-s)^{t}\mathrm{d}s\bigg{]}\frac{\bar{\partial}v(z)}{z}\mathrm{d}m(z)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{D}}(1-|z|^{2})\mathcal{G}^{(t)}\phi(|z|^{2})\bar{z}\bar{\partial}v(z)\mathrm{d}\lambda_{t}(z).$

The absolute convergence of the integral in the right hand side of (3.5)
ensures the third equality above. This proves the second case.

Now assume that $v(0)\neq 0$, and $\mathcal{F}^{(t)}\phi(0)<\infty$. We notice
that

$\int_{\mathbb{D}}\phi(|z|^{2})\mathrm{d}\lambda_{t}(z)=2\pi\cdot\frac{t+1}{\pi}\int_{0}^{1}r\phi(r^{2})(1-r^{2})^{t}\mathrm{d}r=(t+1)\mathcal{F}^{(t)}\phi(0).$

Then we get the following estimate

$\int_{\mathbb{D}}\bigg{|}\phi(|z|^{2})\big{(}v(z)-v(0)\big{)}\bigg{|}\mathrm{d}\lambda_{t}(z)\leq\int_{\mathbb{D}}\bigg{|}\phi(|z|^{2})v(z)\bigg{|}\mathrm{d}\lambda_{t}(z)+(t+1)\mathcal{F}^{(t)}\phi(0)|v(0)|<\infty.$

Applying the second case to $v(z)-v(0)$ and reorganizing the terms gives the
first case. This completes the proof of Lemma 3.8. ∎

######  Lemma 3.9.

Suppose that $t>-1$, and $\phi:(0,1)\to[0,\infty)$ is a measurable function,
and $v\in\mathscr{C}^{1}(\overline{\mathbb{D}})$.

  * (1)

Assuming that $z\in\mathbb{D}$, and the integrals

$\displaystyle\int_{\mathbb{D}}\phi(|\varphi_{z}(w)|^{2})v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),$
$\displaystyle\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})\overline{(z-w)}}{1-w\bar{z}}\bar{\partial}v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$

converge absolutely, then

$\displaystyle\int_{\mathbb{D}}\phi(|\varphi_{z}(w)|^{2})v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
(3.6) $\displaystyle=$
$\displaystyle\begin{cases}(t+1)\mathcal{F}^{(t)}\phi(0)\cdot v(z)&v(z)\neq
0,\mathcal{F}^{(t)}\phi(0)<\infty,\\\
~{}~{}-\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})\overline{(z-w)}}{1-w\bar{z}}\bar{\partial}v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),\\\
\\\
-\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})\overline{(z-w)}}{1-w\bar{z}}\bar{\partial}v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),&v(z)=0,\mathcal{F}^{(t)}\phi(0)\leq\infty.\end{cases}$

  * (2)

Assuming that $w\in\mathbb{D}$, and the integrals

$\displaystyle\int_{\mathbb{D}}\phi(|\varphi_{z}(w)|^{2})v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z),\qquad\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\partial
v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z)$

converge absolutely. Then

$\displaystyle\int_{\mathbb{D}}\phi(|\varphi_{z}(w)|^{2})v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z)$
(3.7) $\displaystyle=$
$\displaystyle\begin{cases}(t+1)\mathcal{F}^{(t)}\phi(0)\cdot v(w)&v(w)\neq
0,\mathcal{F}^{(t)}\phi(0)<\infty,\\\
~{}~{}+\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\partial
v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z),\\\ \\\
\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\partial
v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z),&v(w)=0,\mathcal{F}^{(t)}\phi(0)\leq\infty.\end{cases}$

###### Proof.

First, we prove case (1). The formula is obtained from (3.5) by taking Möbius
transforms. By Lemma 2.2 (4) it is easy to verify the following equation,

$K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\xlongequal[\xi=\varphi_{z}(w)]{w=\varphi_{z}(\xi)}K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(\xi).$
(3.8)

By Lemma 2.2 (1)(2), Lemma 3.8 and the above equality, if
$\mathcal{F}^{(t)}\phi(0)<\infty$, then we compute the integral as follows.

$\displaystyle\int_{\mathbb{D}}\phi(|\varphi_{z}(w)|^{2})v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle\xlongequal{w=\varphi_{z}(\xi)}$
$\displaystyle\int_{\mathbb{D}}\phi(|\xi|^{2})v\circ\varphi_{z}(\xi)K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle\xlongequal{\eqref{eqn: formula D 0}}$
$\displaystyle(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(z)+\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\xi|^{2})(1-|\xi|^{2})\bar{\xi}\bar{\partial}_{\xi}\bigg{(}v\circ\varphi_{z}(\xi)K^{(t)}_{z}(\xi)\bigg{)}\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle=$ $\displaystyle(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(z)+\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\xi|^{2})(1-|\xi|^{2})\bar{\xi}\bar{\partial}v(\varphi_{z}(\xi))\overline{\varphi_{z}^{\prime}(\xi)}K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle=$ $\displaystyle(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(z)-\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\xi|^{2})(1-|\xi|^{2})\bar{\xi}\bar{\partial}v(\varphi_{z}(\xi))\frac{1-|z|^{2}}{(1-z\bar{\xi})^{2}}K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle\xlongequal{\xi=\varphi_{z}(w)}$
$\displaystyle(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(z)-\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})\overline{(z-w)}}{1-w\bar{z}}\bar{\partial}v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w).$

The case when $\mathcal{F}^{(t)}\phi(0)=\infty,v(z)=0$ is proved by the same
equations as above, but with the term “$(t+1)\mathcal{F}^{(t)}\phi(0)\cdot
v(z)$” removed. This proves (1). To prove (2), apply (3.6) to
$\overline{v(z)}$, then swap $z$ and $w$, then take conjugate over the
equation. Note that the equations $|\varphi_{z}(w)|=|\varphi_{w}(z)|$ and
$\overline{K^{(t)}_{z}(w)}=K^{(t)}_{w}(z)$ is used here. This completes the
proof of Lemma 3.9. ∎

###### Proof of Theorem 3.1.

By Lemma 2.5, we have the following expression of
$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$,

$\displaystyle\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}=\int_{\mathbb{D}}\langle\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}K^{(t)}_{\xi},K^{(t)}_{\xi}\rangle\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}^{2}}(f(z)-f(w))g(w)K^{(t)}_{\xi}(w)K^{(t)}_{w}(z)K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(\xi).$

The rest of the proof is simply iterating (3.6) and (3.7) on the integral
above. For each fixed $\xi\in\mathbb{D}$, we calculate the inner product
$\langle\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}K^{(t)}_{\xi},K^{(t)}_{\xi}\rangle$
as follows,

$\displaystyle\langle\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}K^{(t)}_{\xi},K^{(t)}_{\xi}\rangle$
$\displaystyle=$
$\displaystyle\int_{\mathbb{D}^{2}}(f(z)-f(w))g(w)K^{(t)}_{\xi}(w)K^{(t)}_{w}(z)K^{(t)}_{z}(\xi)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\big{[}(f(z)-f(w))g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(w)$
$\displaystyle\xlongequal{\eqref{eqn: formula D z d'}}$
$\displaystyle\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\mathcal{G}^{(t)}1(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\big{[}\partial
f(z)g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(w).$

In the above, we apply (3.7) with $\phi=1$ and
$v(z)=\big{[}(f(z)-f(w))g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}$. Here, by
direct computation, we obtain the following equation

$\big{(}\mathcal{G}^{(t)}1\big{)}(s)=\frac{1}{(t+1)s}.$

Since $\xi\in\mathbb{D}$ is fixed, and $f,g$ are $\mathscr{C}^{1}$ to the
boundary, the term $\big{[}\partial
f(z)g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}$ is bounded. By (2.7), the
two-fold integral in the above converge absolutely. Applying Fubini’s Theorem,
and then (3.6) with

$\phi=\mathcal{G}^{(t)}1,\quad
v(w)=\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\big{[}\partial
f(z)g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]},$

the above integral is computed as follows,

$\displaystyle\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\mathcal{G}^{(t)}1(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}\big{[}\partial
f(z)g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\bigg{\\}}\mathrm{d}\lambda_{t}(z)$
$\displaystyle\xlongequal{\eqref{eqn: formula D z d''}}$
$\displaystyle-\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\big{(}\mathcal{G}^{(t)}\big{)}^{2}1(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})\overline{(z-w)}}{1-w\bar{z}}\cdot\frac{(1-|z|^{2})(z-w)}{1-w\bar{z}}$
$\displaystyle\qquad\qquad\cdot\big{[}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\bigg{\\}}\mathrm{d}\lambda_{t}(z)$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\big{(}\mathcal{G}^{(t)}\big{)}^{2}1(|\varphi_{z}(w)|^{2})(1-|\varphi_{z}(w)|^{2})|\varphi_{z}(w)|^{2}(1-z\bar{w})^{2}$
$\displaystyle\qquad\qquad\cdot\big{[}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\bigg{\\}}\mathrm{d}\lambda_{t}(z)$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{D}^{2}}\phi_{1}(|\varphi_{z}(w)|^{2})(1-z\bar{w})^{2}\big{[}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z).$

Here we have the following bound of $\phi_{1}$,

$\phi_{1}(s)=\big{(}\mathcal{G}^{(t)}\big{)}^{2}1(s)(1-s)s=\frac{\int_{s}^{1}r^{-1}(1-r)^{t}\mathrm{d}r}{(t+1)(1-s)^{t}}\lesssim_{t}s^{-1/2}(1-s).$
(3.9)

Using (2.6) and (2.7), we can show that the integrand above is absolutely
integrable with the measure
$\mathrm{d}\lambda_{t}(\xi)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)$.
Therefore by Lemma 2.5 and Fubini’s Theorem, we compute
$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$ as follows,

$\displaystyle\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$
$\displaystyle=$
$\displaystyle\int_{\mathbb{D}}\bigg{\\{}-\int_{\mathbb{D}^{2}}\phi_{1}(|\varphi_{z}(w)|^{2})(1-z\bar{w})^{2}\big{[}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{\xi}(w)K^{(t)}_{z}(\xi)\big{]}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(\xi)$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{D}^{2}}\phi_{1}(|\varphi_{z}(w)|^{2})(1-z\bar{w})^{2}\partial
f(z)\bar{\partial}g(w)|K^{(t)}_{w}(z)|^{2}\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z).$

To obtain (3.1), we apply (3.7) again, with $z,w$ reversed,

$\phi=\phi_{1},\qquad v(w)=\big{[}(1-z\bar{w})^{2}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}.$

We compute the above integral as follows.

$\displaystyle\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\phi_{1}(|\varphi_{z}(w)|^{2})\big{[}(1-z\bar{w})^{2}\partial
f(z)\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}K^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(w)\bigg{\\}}\mathrm{d}\lambda_{t}(z)$
$\displaystyle\xlongequal{\eqref{eqn: formula D z d'}}$
$\displaystyle-\int_{\mathbb{D}}\bigg{\\{}(t+1)\mathcal{F}^{(t)}\phi_{1}(0)\cdot(1-|z|^{2})^{2}\partial
f(z)\bar{\partial}g(z)K^{(t)}_{z}(z)+\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi_{1}(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})(w-z)}{1-z\bar{w}}$
$\displaystyle\hskip 199.16928pt\big{[}(1-z\bar{w})^{2}\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}K^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(w)\bigg{\\}}\mathrm{d}\lambda_{t}(z)$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\int_{\mathbb{D}}\partial
f\wedge\bar{\partial}g-\int_{\mathbb{D}^{2}}\phi_{2}(|\varphi_{z}(w)|^{2})\big{[}(1-|w|^{2})(w-z)(1-z\bar{w})\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}K^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z).$

Here we have the following expression

$\mathcal{F}^{(t)}\phi_{1}(0)=\frac{\int_{0}^{1}\int_{s}^{1}r^{-1}(1-r)^{t}\mathrm{d}r\mathrm{d}s}{t+1}=\frac{\int_{0}^{1}\int_{0}^{r}r^{-1}(1-r)^{t}\mathrm{d}s\mathrm{d}r}{t+1}=\frac{\int_{0}^{1}(1-r)^{t}\mathrm{d}r}{t+1}=(t+1)^{-2},$
(3.10)

and

$\phi_{2}(s)=\mathcal{G}^{(t)}\phi_{1}(s)=\frac{\int_{s}^{1}\phi_{1}(r)(1-r)^{t}\mathrm{d}r}{s(1-s)^{t+1}}=\frac{\int_{s}^{1}\int_{r}^{1}x^{-1}(1-x)^{t}\mathrm{d}x\mathrm{d}r}{(t+1)s(1-s)^{t+1}}\lesssim_{t}s^{-1/2}(1-s).$
(3.11)

Again, applying Fubini’s Theorem and (3.6), with $z,w$ reversed,
$\phi=\phi_{2}$, $v(z)=\big{[}(1-|w|^{2})(w-z)(1-z\bar{w})\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}$, we compute the second
integral in the above computation of
$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}$ as follows.

$\displaystyle-\int_{\mathbb{D}^{2}}\phi_{2}(|\varphi_{z}(w)|^{2})\big{[}(1-|w|^{2})(w-z)(1-z\bar{w})\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}$ $\displaystyle\hskip
284.52756ptK^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)$
$\displaystyle=$
$\displaystyle-\int_{\mathbb{D}}\qquad\bigg{\\{}\int_{\mathbb{D}}\phi_{2}(|\varphi_{z}(w)|^{2})\big{[}(1-|w|^{2})(w-z)(1-z\bar{w})\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}K^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(w)$
$\displaystyle\xlongequal{\eqref{eqn: formula D z d''}}$
$\displaystyle\int_{\mathbb{D}}\bigg{\\{}\int_{\mathbb{D}}\mathcal{G}^{(t)}\phi_{2}(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})\overline{(w-z)}}{1-z\bar{w}}$
$\displaystyle~{}~{}\cdot\big{[}(1-|w|^{2})(w-z)(1-z\bar{w})\bar{\partial}\partial
f(z)\partial\bar{\partial}g(w)K^{(t)}_{w}(z)\big{]}K^{(t)}_{z}(w)\mathrm{d}\lambda_{t}(z)\bigg{\\}}\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$
$\displaystyle\frac{(t+1)^{2}}{16\pi^{2}}\int_{\mathbb{D}^{2}}\mathcal{G}^{(t)}\phi_{2}(|\varphi_{z}(w)|^{2})\frac{(1-|z|^{2})^{t+1}(1-|w|^{2})^{t+1}|w-z|^{2}}{|1-z\bar{w}|^{4+2t}}\Delta
f(z)\Delta g(w)|\mathrm{d}m(z,w)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{D}^{2}}\rho_{t}(|\varphi_{z}(w)|^{2})\Delta
f(z)\Delta g(w)\mathrm{d}m(z,w).$

Here, by Lemma 2.2, we have

$\displaystyle\rho_{t}(s)=$
$\displaystyle\frac{(t+1)^{2}}{16\pi^{2}}\mathcal{G}^{(t)}\phi_{2}(s)(1-s)^{t+1}s=\frac{(t+1)^{2}}{16\pi^{2}}\int_{s}^{1}\phi_{2}(s_{1})(1-s_{1})^{t}\mathrm{d}s_{1}$
$\displaystyle=$
$\displaystyle\frac{(t+1)}{16\pi^{2}}\int_{s}^{1}\int_{s_{1}}^{1}\int_{s_{2}}^{1}s_{1}^{-1}(1-s_{1})^{-1}s_{3}^{-1}(1-s_{3})^{t}\mathrm{d}s_{3}\mathrm{d}s_{2}\mathrm{d}s_{1}.$
(3.12)

Therefore we have reached the following identity

$\mathrm{Tr}\bigg{(}T^{(t)}_{f}T^{(t)}_{g}-T^{(t)}_{fg}\bigg{)}=\frac{1}{2\pi
i}\int_{\mathbb{D}}\partial
f\wedge\bar{\partial}g+\int_{\mathbb{D}^{2}}\rho_{t}(|\varphi_{z}(w)|^{2})\Delta
f(z)\Delta g(w)\mathrm{d}m(z,w).$ (3.13)

Let us simplify the expression of $\rho_{t}$.

$\displaystyle\int_{s}^{1}\int_{s_{1}}^{1}\int_{s_{2}}^{1}s_{1}^{-1}(1-s_{1})^{-1}s_{3}^{-1}(1-s_{3})^{t}\mathrm{d}s_{3}\mathrm{d}s_{2}\mathrm{d}s_{1}$
$\displaystyle=$
$\displaystyle\iiint_{\\{(s_{1},s_{2},s_{3}):s<s_{1}<s_{2}<s_{3}<1\\}}s_{1}^{-1}(1-s_{1})^{-1}s_{3}^{-1}(1-s_{3})^{t}\mathrm{d}s_{1}\mathrm{d}s_{2}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle\iint_{\\{(s_{1},s_{3}):s<s_{1}<s_{3}<1\\}}(s_{3}-s_{1})s_{1}^{-1}(1-s_{1})^{-1}s_{3}^{-1}(1-s_{3})^{t}\mathrm{d}s_{1}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle\iint_{\\{(s_{1},s_{3}):s<s_{1}<s_{3}<1\\}}\frac{(1-s_{3})^{t}}{s_{1}(1-s_{1})}-\frac{s_{3}^{-1}(1-s_{3})^{t}}{1-s_{1}}\mathrm{d}s_{1}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle\iint_{\\{(s_{1},s_{3}):s<s_{1}<s_{3}<1\\}}\frac{(1-s_{3})^{t}}{s_{1}}+\frac{(1-s_{3})^{t}}{1-s_{1}}-\frac{s_{3}^{-1}(1-s_{3})^{t}}{1-s_{1}}\mathrm{d}s_{1}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle\iint_{\\{(s_{1},s_{3}):s<s_{1}<s_{3}<1\\}}\frac{(1-s_{3})^{t}}{s_{1}}+\frac{(1-s_{3}^{-1})(1-s_{3})^{t}}{1-s_{1}}\mathrm{d}s_{1}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle\int_{s}^{1}(1-s_{3})^{t}\ln\frac{s_{3}}{s}+(1-s_{3}^{-1})(1-s_{3})^{t}\ln\frac{1-s}{1-s_{3}}\mathrm{d}s_{3}$
$\displaystyle=$
$\displaystyle-\int_{s}^{1}(1-s_{3})^{t}s_{3}^{-1}\bigg{[}s_{3}\ln\frac{s}{s_{3}}+(1-s_{3})\ln\frac{1-s}{1-s_{3}}\bigg{]}\mathrm{d}s_{3}.$

Therefore we obtain the following expression of $\rho_{t}$ in Equation (3.13),

$\rho_{t}(s)=\frac{t+1}{16\pi^{2}}\int_{s}^{1}(1-x)^{t}x^{-1}F(s,x)\mathrm{d}x,$

where

$F(s,x)=-\bigg{[}x\ln\frac{s}{x}+(1-x)\ln\frac{1-s}{1-x}\bigg{]}.$

Since the function $-\ln$ is strictly concave, we have the following property
for $F$,

$F(s,x)>-\ln\bigg{[}x\cdot\frac{s}{x}+(1-x)\cdot\frac{1-s}{1-x}\bigg{]}=0,\quad\forall
0<x<1.$

Thus $\rho_{t}$ is strictly positive on $(0,1)$. Now we have completed the
proof of Equation (3.1). It remains to prove (3.2). In other words, the second
term of (3.13) tends to zero as $t$ tends to infinity. Clearly the absolute
value of the second term has the following bound,

$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{D}}\int_{\mathbb{D}}\rho_{t}(|\varphi_{z}(w)|^{2})\mathrm{d}m(w)\mathrm{d}m(z)=\int_{\mathbb{D}}\int_{\mathbb{D}}\rho_{t}(|\zeta|^{2})\frac{(1-|z|^{2})^{2}}{|1-\zeta\bar{z}|^{4}}\mathrm{d}m(\zeta)\mathrm{d}m(z)$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{D}}\rho_{t}(|\zeta|^{2})\ln\frac{1}{1-|\zeta|^{2}}\mathrm{d}m(\zeta)\lesssim\int_{\mathbb{D}}\rho_{t}(|\zeta|)(1-|\zeta|^{2})^{-1/2}\mathrm{d}m(\zeta)\approx\int_{0}^{1}\rho_{t}(s)(1-s)^{-1/2}\mathrm{d}s$
$\displaystyle\approx$
$\displaystyle(t+1)\int_{0}^{1}\int_{s}^{1}(1-s)^{-1/2}F(s,x)x^{-1}(1-x)^{t}\mathrm{d}x\mathrm{d}s$
$\displaystyle=$
$\displaystyle(t+1)\int_{0}^{1}\bigg{[}\int_{0}^{x}(1-s)^{-1/2}F(s,x)\mathrm{d}s\bigg{]}x^{-1}(1-x)^{t}\mathrm{d}x.$

We claim that $\int_{0}^{x}(1-s)^{-1/2}F(s,x)\mathrm{d}s\lesssim x^{2}$ which
implies that the above is controlled by $(t+1)B(2,t+1)\approx(t+1)^{-1}\to 0$
as $t\to\infty$. Because $\ln(1-x)<0$, we have the following bound.

$\displaystyle\int_{0}^{x}(1-s)^{-1/2}F(s,x)\mathrm{d}s=-\int_{0}^{x}(1-s)^{-1/2}\bigg{[}x\ln\frac{s}{x}+(1-x)\ln\frac{1-s}{1-x}\bigg{]}\mathrm{d}s$
$\displaystyle=$
$\displaystyle-x^{2}\int_{0}^{x}(1-s)^{-1/2}\ln\frac{s}{x}\frac{\mathrm{d}s}{x}-(1-x)\int_{0}^{x}(1-s)^{-1/2}\ln(1-s)\mathrm{d}s$
$\displaystyle\hskip
170.71652pt+(1-x)\ln(1-x)\int_{0}^{x}(1-s)^{-1/2}\mathrm{d}s$ $\displaystyle<$
$\displaystyle-x^{2}\int_{0}^{x}(1-s)^{-1/2}\ln\frac{s}{x}\frac{\mathrm{d}s}{x}-(1-x)\int_{0}^{x}(1-s)^{-1/2}\ln(1-s)\mathrm{d}s.$

For the first term, take the change of variable $r=\frac{s}{x}$, then compute

$-x^{2}\int_{0}^{x}(1-s)^{-1/2}\ln\frac{s}{x}\frac{\mathrm{d}s}{x}=-x^{2}\int_{0}^{1}(1-xr)^{-1/2}\ln
r\mathrm{d}r\leq-x^{2}\int_{0}^{1}(1-r)^{-1/2}\ln r\mathrm{d}r\lesssim x^{2}.$

Since $-\ln(1-s)\lesssim s$, we obtain the following bound,

$-(1-x)\int_{0}^{x}(1-s)^{-1/2}\ln(1-s)\mathrm{d}s\lesssim(1-x)\int_{0}^{x}(1-s)^{-1/2}s\mathrm{d}s\leq(1-x)^{1/2}\int_{0}^{x}s\mathrm{d}s\lesssim
x^{2}.$

Altogether, this proves our claim, which in turn shows that the second term in
(3.13) vanishes as $t\to\infty$. This completes the proof of Theorem 3.1. ∎

##  4 Auxiliary Functions and Operations

In Section 3, the integral operations $\mathcal{F}^{(t)},\mathcal{G}^{(t)}$
simplify our computation. To work in higher dimensions, it is necessary to
extend those integral operations and establish some basic properties. This is
the goal of the current section.

######  Definition 4.1.

For $t\in\mathbb{R}$, denote

$\phi_{t}(s)=(1-s)^{t}.$

Suppose $\phi:(0,1)\to[0,\infty)$ is a measurable function. For a positive
integer $m$ and any $t>-1$, define the operations on $\phi$

$\mathcal{F}^{(t)}_{m}\phi(s)=\int_{s}^{1}r^{m-1}\phi(r)(1-r)^{t}\mathrm{d}r\in[0,\infty],$
(4.1)

and

$\mathcal{G}^{(t)}_{m}\phi(s)=\frac{1}{s^{m}\phi_{t+1}(s)}\mathcal{F}^{(t)}_{m}\phi(s)=\frac{\int_{s}^{1}r^{m-1}\phi(r)(1-r)^{t}\mathrm{d}r}{s^{m}(1-s)^{t+1}}\in[0,\infty].$
(4.2)

For any $t>-1$, inductively define the functions

$\Phi_{n,0}^{(t)}\equiv
1,\quad\Phi_{n,k+1}^{(t)}=M_{\phi_{1}}\big{(}\mathcal{G}^{(t)}_{n+k}\big{)}^{2}\Phi_{n,k}^{(t)}.$

Equivalently,

$\Phi_{n,k}^{(t)}=M_{\phi_{1}}\big{(}\mathcal{G}^{(t)}_{n+k-1}\big{)}^{2}\ldots
M_{\phi_{1}}\big{(}\mathcal{G}^{(t)}_{n}\big{)}^{2}1.$ (4.3)

It is straightforward to verify that the following estimates hold.

######  Lemma 4.2.

Suppose $a>m$ is not an integer and $b\geq 0$. Suppose
$\phi:(0,1)\to[0,\infty)$ is measurable and

$\phi(s)\lesssim s^{-a}(1-s)^{b}.$

Then

$\mathcal{G}^{(t)}_{m}\phi(s)\lesssim s^{-a}(1-s)^{b}.$

As a consequence, the following hold.

######  Lemma 4.3.

For any $t>-1$ and integers $n>0,k\geq 0$,

$\Phi_{n,k}^{(t)}(s)\lesssim s^{-n-k+\frac{1}{2}}(1-s)^{k},$ (4.4)

and

$\mathcal{G}^{(t)}_{n+k}\Phi^{(t)}_{n,k}(s)\lesssim
s^{-n-k-\frac{1}{2}}(1-s)^{k}.$ (4.5)

######  Lemma 4.4.

For any $t>-1$ and positive integers $m,k$ and $\phi:(0,1)\to[0,\infty)$ we
have

$\mathcal{F}^{(t)}_{m+k}M_{\phi_{1}}\mathcal{G}^{(t)}_{m}\phi(0)=\frac{1}{k}\mathcal{F}^{(t)}_{m+k}\phi(0),$
(4.6)
$\mathcal{F}^{(t)}_{m+k}\mathcal{G}^{(t)}_{m}\phi(0)=\sum_{j=0}^{\infty}\frac{1}{k+j}\mathcal{F}^{(t)}_{m+k+j}\phi(0),$
(4.7)
$\mathcal{F}^{(t)}_{m}\phi=\mathcal{F}^{(t)}_{m}M_{\phi_{1}}\phi(0)+\mathcal{F}^{(t)}_{m+1}\phi,$
(4.8)

and

$\mathcal{F}^{(t)}_{m}1(0)=B(m,t+1).$ (4.9)

###### Proof.

The proof is a simple application of Fubini’s Theorem. By definition, we have
the following computation for
$\mathcal{F}^{(t)}_{m+k}M_{\phi_{1}}\mathcal{G}^{(t)}_{m}\phi(0)$,

$\displaystyle\mathcal{F}^{(t)}_{m+k}M_{\phi_{1}}\mathcal{G}^{(t)}_{m}\phi(0)$
$\displaystyle=$
$\displaystyle\int_{0}^{1}r^{m+k-1}(1-r)^{t+1}\mathcal{G}^{(t)}_{m}\phi(r)\mathrm{d}r$
$\displaystyle=$
$\displaystyle\int_{0}^{1}r^{k-1}\int_{r}^{1}s^{m-1}(1-s)^{t}\phi(s)\mathrm{d}s\mathrm{d}r$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\int_{0}^{s}r^{k-1}\mathrm{d}rs^{m-1}(1-s)^{t}\phi(s)\mathrm{d}s$
$\displaystyle=$
$\displaystyle\frac{1}{k}\int_{0}^{1}s^{m+k-1}(1-s)^{t}\phi(s)\mathrm{d}s$
$\displaystyle=$ $\displaystyle\frac{1}{k}\mathcal{F}^{(t)}_{m+k}\phi(0).$

This proves (4.6). We prove (4.7) by simply using the expansion

$\frac{1}{1-s}=\sum_{j=0}^{\infty}s^{j}.$

Finally, we arrive at the following equation,

$\mathcal{F}^{(t)}_{m}1(0)=\int_{0}^{1}r^{m-1}(1-r)^{t}\mathrm{d}r=B(m,t+1).$

This completes the proof of Lemma 4.4. ∎

######  Lemma 4.5.

Suppose $k$ is a nonnegative integer. Then there exist $C>c>0$ such that for
$t$ large enough,

$ct^{-n-k}\leq\mathcal{F}^{(t)}_{n+k}\Phi_{n,k}^{(t)}(0)\leq Ct^{-n-k},$
(4.10)

and

$\int_{0}^{1}\Phi_{n,k}^{(t)}(s)s^{n+k-1}(1-s)^{\frac{t}{4}}\mathrm{d}s\leq
Ct^{-n-k}.$ (4.11)

###### Proof.

The case when $k=0$ is simply a consequence of (4.9). We assume $k>0$.

From the Sterling’s asymptotic formula (cf. [2, Theorem 1.4.1]) and the well-
known identity $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ it follows that
for fixed $x>-1$,

$B(x,y)\approx_{x}y^{-x}$

for large $y$. By (4.6), (4.7) and (4.9), we compute
$\mathcal{F}^{(t)}_{n+k}\Phi_{n,k}^{(t)}(0)$ as follows,

$\displaystyle\mathcal{F}^{(t)}_{n+k}\Phi_{n,k}^{(t)}(0)$ $\displaystyle=$
$\displaystyle\mathcal{F}^{(t)}_{n+k}M_{\phi_{1}}\big{(}\mathcal{G}^{(t)}_{n+k-1}\big{)}^{2}\ldots
M_{\phi_{1}}\big{(}\mathcal{G}^{(t)}_{n}\big{)}^{2}1(0)$ $\displaystyle=$
$\displaystyle\sum_{j_{1},\ldots,j_{k}=0}^{\infty}\frac{B(n+k+j_{1}+\ldots+j_{k},t+1)}{1\cdot(1+j_{1})(2+j_{1})(2+j_{1}+j_{2})\ldots(k+j_{1}+\ldots+j_{k-1})(k+j_{1}+\ldots+j_{k})}$
$\displaystyle=$ $\displaystyle\sum_{0\leq s_{1}\leq\ldots\leq
s_{k}<\infty}\frac{B(n+k+s_{k},t+1)}{(1+s_{1})(2+s_{1})\ldots(k-1+s_{k-1})(k+s_{k-1})(k+s_{k})}$
$\displaystyle\leq$
$\displaystyle\sum_{s_{k}=0}^{\infty}\sum_{s_{1},\ldots,s_{k-1}=0}^{\infty}\frac{B(n+k+s_{k},t+1)}{(1+s_{1})(2+s_{1})\ldots(k-1+s_{k-1})(k+s_{k-1})(k+s_{k})}$
$\displaystyle=$
$\displaystyle\sum_{s_{k}=0}^{\infty}\frac{B(n+k+s_{k},t+1)}{(k-1)!(k+s_{k})}$
$\displaystyle=$
$\displaystyle\frac{1}{(k-1)!}\int_{0}^{1}\big{(}\sum_{s_{k}=0}^{\infty}\frac{x^{n-1+(k+s_{k})}}{k+s_{k}}\big{)}(1-x)^{t}\mathrm{d}x$
$\displaystyle\leq$
$\displaystyle\frac{1}{(k-1)!}\int_{0}^{1}\big{(}\sum_{s_{k}=0}^{\infty}x^{n-1}\bigg{(}\ln\frac{1}{1-x}\bigg{)}(1-x)^{t}\mathrm{d}x$
$\displaystyle\lesssim$
$\displaystyle\int_{0}^{1}x^{n-1}(1-x)^{t-1}\mathrm{d}x$ $\displaystyle=$
$\displaystyle B(n,t)$ $\displaystyle\lesssim$ $\displaystyle t^{-n-k},$

when $t$ is large. The other inequality
$\mathcal{F}^{(t)}_{n+k}\Phi_{n,k}^{(t)}(0)\gtrsim t^{-n-k}$ is also obvious
from the equation above. This proves (4.10).

By the expansion

$\frac{1}{(1-s)^{\frac{3t}{4}}}=\sum_{j=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+j)}{j!\Gamma(\frac{3t}{4})}s^{j},$

we directly compute the following integral,

$\displaystyle\int_{0}^{1}\Phi_{n,k}^{(t)}(s)s^{n+k-1}(1-s)^{\frac{t}{4}}\mathrm{d}s$
$\displaystyle=$
$\displaystyle\sum_{j=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+j)}{j!\Gamma(\frac{3t}{4})}\int_{0}^{1}\Phi_{n,k}^{(t)}(s)s^{n+k+j-1}(1-s)^{t}\mathrm{d}s$
$\displaystyle=$
$\displaystyle\sum_{j_{0}=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+j_{0})}{j_{0}!\Gamma(\frac{3t}{4})}\mathcal{F}^{(t)}_{n+k+j_{0}}\Phi_{n,k}^{(t)}(0).$

Similarly as the proof of (4.10), assuming $t$ is large enough, we estimate
the above term as follows,

$\displaystyle\leq$ $\displaystyle\sum_{0\leq s_{0}\leq
s_{k}<\infty}\frac{\Gamma(\frac{3t}{4}+s_{0})B(n+k+s_{k},t+1)}{(s_{0}+1)!\Gamma(\frac{3t}{4})(k-1)!(k+s_{k})}$
$\displaystyle=$
$\displaystyle\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+a)B(n+k+a+b,t+1)}{(a+1)!\Gamma(\frac{3t}{4})(k-1)!(k+a+b)}$
$\displaystyle\leq$
$\displaystyle\sum_{a=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+a)}{(a+1)!\Gamma(\frac{3t}{4})(k-1)!}\int_{0}^{1}\sum_{b=0}^{\infty}\frac{s^{b+1}}{k+a+b}s^{n+k+a-2}(1-s)^{t}\mathrm{d}s$
$\displaystyle\leq$
$\displaystyle\sum_{a=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+a)}{(a+1)!\Gamma(\frac{3t}{4})(k-1)!}\int_{0}^{1}\sum_{b=0}^{\infty}\frac{s^{b+1}}{b+1}s^{n+k+a-2}(1-s)^{t}\mathrm{d}s$
$\displaystyle=$
$\displaystyle\sum_{a=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+a)}{(a+1)!\Gamma(\frac{3t}{4})(k-1)!}\int_{0}^{1}s^{n+k+a-2}(1-s)^{t}\ln\frac{1}{1-s}\mathrm{d}s$
$\displaystyle\lesssim$
$\displaystyle\sum_{a=0}^{\infty}\frac{\Gamma(\frac{3t}{4}+a)}{(a+1)!\Gamma(\frac{3t}{4})(k-1)!}B(n+k+a,t)$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}s^{n+k-1}(1-s)^{t-1-\frac{3t}{4}}\mathrm{d}s$
$\displaystyle=$ $\displaystyle B(n+k,\frac{t}{4})$ $\displaystyle\lesssim$
$\displaystyle t^{-n-k}.$

This completes the proof of Lemma 4.5. ∎

##  5 Operators Between Weighted Spaces

For our proofs in this paper it is important to get criteria for integral
operators to be bounded between different weighted spaces. In this section, we
obtain such criteria. Most of the lemmas established in this section follow
from standard techniques from, for example, [47, 55]. One thing that is less
standard is that our integral kernels may involve functions of
$|\varphi_{z}(w)|^{2}$. Similar as in Section 3, these functions come from
applying integration by parts formulas that are analogous to Lemma 3.9.

######  Lemma 5.1.

Suppose $t>-1,s>-1$, $a>-n$, $b\geq 0$, and $c<t+1+b-\frac{s+1}{2}$. Suppose
$F(z,w)$ is measurable on $\mathbb{B}_{n}\times\mathbb{B}_{n}$, and

$|F(z,w)|\leq\frac{|\varphi_{z}(w)|^{2a}(1-|\varphi_{z}(w)|^{2})^{b}}{|1-\langle
z,w\rangle|^{n+1+t-c}},\quad\forall z,w\in\mathbb{B}_{n}.$

Then the integral operator

$Th(z)=\int_{\mathbb{B}_{n}}h(w)F(z,w)\mathrm{d}\lambda_{t}(w)$

is bounded from $L^{2}(\lambda_{s+2c})$ to $L^{2}(\lambda_{s})$.

###### Proof.

By assumption, we can take $x\in\mathbb{R}$ so that

$\max\\{-1-t-b,-c-s-1-b\\}<x<\min\\{b-c,t+b-2c-s\\}.$

Then we have the following inequalities,

$t+x>-1-b,\quad-x-c>-b,\quad x+c+s>-1-b,\quad t-x-2c-s>-b.$

Take $p(w)=(1-|w|^{2})^{x}$ and $q(z)=(1-|z|^{2})^{x+c}$. The integral kernel
of $T$ as an operator from $L^{2}(\lambda_{s+2c})$ to $L^{2}(\lambda_{s})$ is

$\frac{B(n,s+2c+1)}{B(n,t+1)}F(z,w)(1-|w|^{2})^{t-s-2c}.$

Then by the inequalities above and Lemma 2.4 (3), we have the following
estimates of integrals,

$\displaystyle\int_{\mathbb{B}_{n}}|F(z,w)|(1-|w|^{2})^{t-s-2c}p(w)\mathrm{d}\lambda_{s+2c}(w)$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{B}_{n}}|\varphi_{z}(w)|^{2a}(1-|\varphi_{z}(w)|^{2})^{b}\frac{(1-|w|^{2})^{t+x}}{|1-\langle
z,w\rangle|^{n+1+t-c}}\mathrm{d}m(w)$ $\displaystyle\lesssim$
$\displaystyle(1-|z|^{2})^{x+c}=q(z),$

and

$\displaystyle\int_{\mathbb{B}_{n}}|F(z,w)|(1-|w|^{2})^{t-s-2c}q(z)\mathrm{d}\lambda_{s}(z)$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{B}_{n}}|\varphi_{z}(w)|^{2a}(1-|\varphi_{z}(w)|^{2})^{b}\frac{(1-|w|^{2})^{t-s-2c}(1-|z|^{2})^{x+c+s}}{|1-\langle
z,w\rangle|^{n+1+t-c}}\mathrm{d}m(z)$ $\displaystyle\lesssim$
$\displaystyle(1-|w|^{2})^{x}=p(w).$

The conclusion follows from the Schur’s test (cf. [56, Theorem 3.6]). This
completes the proof of Lemma 5.1. ∎

######  Lemma 5.2.

Suppose $t>-1$ and $0<d<c<t+1$. Suppose $F(z,w)$ is piecewise continuous on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$, and

$|F(z,w)|\leq\frac{1}{|1-\langle z,w\rangle|^{n+1+t-c}},\quad\forall
z,w\in\mathbb{B}_{n}.$

For any $0<r<1$, define the integral operator

$T_{r}h(z)=\int_{\mathbb{B}_{n}}h(w)F(rz,w)\mathrm{d}\lambda_{t}(w),\quad
z\in\mathbb{S}_{n}.$

Then each $T_{r}$ defines a bounded operator from $L^{2}(\lambda_{-1+2d})$ to
$L^{2}(\mathbb{S}_{n})$. Moreover,

$\sup_{0<r<1}\|T_{r}\|_{L^{2}(\lambda_{-1+2d})\to
L^{2}(\mathbb{S}_{n})}<\infty.$

###### Proof.

Take $p(w)=(1-|w|^{2})^{-d}$, $q(z)=1$. The integral kernel of $T_{r}$ is

$T_{r}(z,w):=\frac{B(n,2d)}{B(n,t+1)}F(rz,w)(1-|w|^{2})^{t+1-2d}.$

As $t-d>-1$ and $d<c$, by Lemma 2.4, we have the following estimates of
integrals,

$\int_{\mathbb{B}_{n}}|T_{r}(z,w)|p(w)\mathrm{d}\lambda_{-1+2d}(w)\lesssim\int_{\mathbb{B}_{n}}\frac{(1-|w|^{2})^{t-d}}{|1-\langle
rz,w\rangle|^{n+1+t-c}}\mathrm{d}m(w)\lesssim 1=q(z),$

and

$\displaystyle\int_{\mathbb{S}_{n}}|T_{r}(z,w)|q(z)\frac{\mathrm{d}\sigma(z)}{\sigma_{2n-1}}$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{S}_{n}}\frac{(1-|w|^{2})^{t+1-2d}}{|1-\langle
z,rw\rangle|^{n+1+t-c}}\mathrm{d}\sigma(z)$ $\displaystyle\lesssim$
$\displaystyle(1-|w|^{2})^{c-2d}\lesssim(1-|w|^{2})^{-d}=p(w).$

The estimates are independent of $r$. Thus the conclusion follows from the
Schur’s test. This completes the proof of Lemma 5.2. ∎

######  Lemma 5.3.

Suppose $t>-1$ and $0<d<c<t+1$. Suppose $F(z,w)$ is piecewise continuous on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$, and

$|F(z,w)|\leq\frac{1}{|1-\langle z,w\rangle|^{n+1+t-c}},\quad\forall
z,w\in\mathbb{B}_{n}.$

Assume further that $F(z,w)$ is holomorphic in $z$ for each fixed $w$. Define
the operator

$Th(z)=\int_{\mathbb{B}_{n}}h(w)F(z,w)\mathrm{d}\lambda_{t}(w),\quad
z\in\mathbb{B}_{n}.$

Then $T$ defines a bounded operator from $L^{2}(\lambda_{-1+2d})$ to
$H^{2}(\mathbb{S}_{n})$.

###### Proof.

Since $F(z,w)$ is holomorphic in $z$, for every $h$, the function $Th$ is
holomorphic in $\mathbb{B}_{n}$. Denote

$M=\sup_{0<r<1}\|T_{r}\|_{L^{2}(\lambda_{-1+2d})\to
L^{2}(\mathbb{S}_{n})}<\infty.$

Then for any $h\in L^{2}(\lambda_{-1+2d})$, we verify the following identity

$\sup_{0<r<1}\int_{\mathbb{S}_{n}}|Th(rz)|^{2}\mathrm{d}\sigma(z)=\sup_{0<r<1}\|T_{r}h\|_{L^{2}(\mathbb{S}_{n})}^{2}\leq
M^{2}\|h\|_{L^{2}(\lambda_{-1+2d})}^{2}.$

Therefore $T$ is bounded. This completes the proof of Lemma 5.3. ∎

The following fact about embedding operators between different weighted spaces
is well-known to experts. In our proof, estimates of the operator norms and
Schatten norms of these embeddings are needed. We give the calculation for
completeness.

######  Lemma 5.4.

Suppose $t\geq-1$ and $c>0$. Then for any $p>\frac{2n}{c}$ the following hold.

  * (1)

The embedding map $E_{t,c}$ from $L_{a,t}^{2}(\mathbb{B}_{n})$ to
$L_{a,t+c}^{2}(\mathbb{B}_{n})$ (identifying $L_{a,-1}(\mathbb{B}_{n})$ with
$H^{2}(\mathbb{S}_{n})$) is in the Schatten $p$ class $\mathcal{S}^{p}$.

  * (2)

There exists $C>0$, independent of $t$, such that

$\|E_{t,c}\|=1,\quad\|E_{t,c}\|_{p}\leq C(t+n+1)^{\frac{n}{p}}.$

###### Proof.

For any multi-index $\alpha\in\mathbb{N}_{0}^{n}$, we have the norm

$\|z^{\alpha}\|_{L_{a,t}^{2}(\mathbb{B}_{n})}^{2}=\frac{\Gamma(n+t+1)\alpha!}{\Gamma(n+|\alpha|+t+1)}.$

Thus $E_{t,c}$ is unitarily equivalent to a diagonal operator with entry

$\frac{\|z^{\alpha}\|_{L_{a,t+c}^{2}(\mathbb{B}_{n})}}{\|z^{\alpha}\|_{L_{a,t}^{2}(\mathbb{B}_{n})}}=\sqrt{\frac{\Gamma(n+t+c+1)\Gamma(n+|\alpha|+t+1)}{\Gamma(n+|\alpha|+t+c+1)\Gamma(n+t+1)}}=\sqrt{\frac{B(n+|\alpha|+t+1,c)}{B(n+t+1,c)}}$

at $\alpha\in\mathbb{N}_{0}^{n}$. It follows immediately that $\|E_{t,c}\|=1$.
Thus it suffices to show

$\displaystyle\sum_{\alpha\in\mathbb{N}_{0}^{n}}\bigg{(}\frac{B(n+|\alpha|+t+1,c)}{B(n+t+1,c)}\bigg{)}^{\frac{p}{2}}$
$\displaystyle=$
$\displaystyle\sum_{d=0}^{\infty}\frac{(d+n-1)!}{d!(n-1)!}\cdot\bigg{(}\frac{B(n+d+t+1,c)}{B(n+t+1,c)}\bigg{)}^{\frac{p}{2}}\lesssim(t+n+1)^{n}.$

Since $B(n+d+t+1,c)$ is decreasing in $d$, and $\frac{(d+n-1)!}{d!}\approx
d^{n-1}$ for $d\geq 1$, we have the following inequalities,

$\displaystyle\sum_{d=0}^{\infty}\frac{(d+n-1)!}{d!(n-1)!}\cdot\bigg{(}\frac{B(n+d+t+1,c)}{B(n+t+1,c)}\bigg{)}^{\frac{p}{2}}$
$\displaystyle\lesssim$ $\displaystyle
1+\int_{0}^{\infty}x^{n-1}\bigg{(}\frac{B(n+x+t+1,c)}{B(n+t+1,c)}\bigg{)}^{p/2}\mathrm{d}x$
$\displaystyle\lesssim$ $\displaystyle
1+\int_{0}^{\infty}x^{n-1}\bigg{(}\frac{n+t+1}{n+x+t+1}\bigg{)}^{\frac{pc}{2}}\mathrm{d}x.$

If we take the change of variable $y=\frac{x}{n+t+1}$ then the integral above
has the following bounds,

$\lesssim
1+(n+t+1)^{n}\int_{0}^{\infty}y^{n-1}\bigg{(}\frac{1}{1+y}\bigg{)}^{\frac{pc}{2}}\mathrm{d}y\lesssim(n+t+1)^{n}$

when $n<\frac{pc}{2}$, i.e., $p>\frac{2n}{c}$. This completes the proof of
Lemma 5.4. ∎

##  6 Schatten Class Criteria

In this section, we obtain criteria for integral operators to belong to
Schatten class. As explained in the beginning of Section 5, the integral
kernels may involve functions of $|\varphi_{z}(w)|^{2}$. Thus we need to
modify some of the well-known results to fit our case.

A standard way of proving Schatten class membership of integral operators is
to use the following lemma. See [56, Lemma 8.26] for a proof when $n=1$, using
interpolation. The exact same proof works for general $n$.

######  Lemma 6.1.

Suppose $t>-1$ and $G(z,w)$ is measurable on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$. Suppose $p\geq 2$ and

$\int_{\mathbb{B}_{n}^{2}}|G(z,w)|^{p}|K^{(t)}_{w}(z)|^{2}\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)<\infty.$

Define the operator

$Th(z)=\int_{\mathbb{B}_{n}}h(w)G(z,w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w).$

Then $T$ defines a bounded operator on $L^{2}(\lambda_{t})$ that belongs to
$\mathcal{S}^{p}$.

######  Corollary 6.2.

Suppose $t>-1$ and $F(z,w)$ is measurable on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$. Suppose $c>0$ and

$|F(z,w)|\leq\frac{1}{|1-\langle z,w\rangle|^{n+1+t-c}},\quad
z,w\in\mathbb{B}_{n}.$

Define the integral operator

$Th(z)=\int_{\mathbb{B}_{n}}h(w)F(z,w)\mathrm{d}\lambda_{t}(w),\quad h\in
L^{2}(\lambda_{t}).$

Then $T$ is bounded on $L^{2}(\lambda_{t})$, and for any $p>\frac{n}{c}$ and
$p\geq 2$, $T\in\mathcal{S}^{p}$.

###### Proof.

Let

$G(z,w)=\frac{F(z,w)}{K^{(t)}_{w}(z)},\quad z,w\in\mathbb{B}_{n}.$

Then by assumption, we have the following bound,

$|G(z,w)|=|F(z,w)|\cdot|1-\langle z,w\rangle|^{n+1+t}\leq|1-\langle
z,w\rangle|^{c}.$

Choose $p$ so that $\frac{n}{c}<p<\frac{n+1+t}{c}$. Then by Lemma 2.4, we
compute the following integral.

$\displaystyle\int_{\mathbb{B}_{n}}\int_{\mathbb{B}_{n}}|G(z,w)|^{p}|K^{(t)}_{w}(z)|^{2}\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{B}_{n}}\int_{\mathbb{B}_{n}}\frac{(1-|z|^{2})^{t}(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{2(n+1+t)-cp}}\mathrm{d}m(w)\mathrm{d}m(z)$
$\displaystyle\lesssim$
$\displaystyle\int_{\mathbb{B}_{n}}(1-|z|^{2})^{t-(n+1+t-cp)}\mathrm{d}m(z)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{B}_{n}}(1-|z|^{2})^{-1+(cp-n)}\mathrm{d}m(z)<\infty.$

Since $G(z,w)$ is bounded, for $p\geq\frac{n+1+t}{c}$ the integral is also
finite. Therefore we have the following bound,

$\int_{\mathbb{B}_{n}}\int_{\mathbb{B}_{n}}|G(z,w)|^{p}|K^{(t)}_{w}(z)|^{2}\mathrm{d}\lambda_{t}(w)\mathrm{d}\lambda_{t}(z)<\infty,\quad\forall
p>\frac{n}{c}.$

Finally, by Lemma 6.1, $T\in\mathcal{S}^{p}$ for any $p>\frac{n}{c}$ and
$p\geq 2$. This completes the proof of Corollary 6.2. ∎

######  Corollary 6.3.

Let $t,c,F,T$ be as in Corollary 6.2. Then for any Lipschitz function $u$ on
$\mathbb{B}_{n}$ and any $p>\frac{n}{c+\frac{1}{2}}$ and $p\geq 2$, the
commutator

$[T,M_{u}]\in\mathcal{S}^{p}.$

###### Proof.

By definition, for $h\in L^{2}(\lambda_{t})$, we have the following
expression,

$[T,M_{u}]h(z)=\int_{\mathbb{B}_{n}}\big{(}u(w)-u(z)\big{)}F(z,w)h(w)\mathrm{d}\lambda_{t}(w).$

Since $u$ is Lipschitz, by Lemma 2.2 (6), we get the following bounds,

$\big{|}\big{(}u(w)-u(z)\big{)}F(z,w)\big{|}\lesssim|z-w||F(z,w)|\lesssim\frac{1}{|1-\langle
z,w\rangle|^{n+1+t-c-1/2}}.$

The corollary then follows from Corollary 6.2. This completes the proof of
Corollary 6.3. ∎

It is well-known that Hankel operators of Lipschitz symbols belong to
$\mathcal{S}^{p}$ for any $p>2n$. In fact, the Schatten-class membership for
Hankel operators is completely characterized (see [40], [56, Theorem 8.36]).

######  Corollary 6.4.

Suppose $t>-1$ and $u$ is a Lipschitz function on $\mathbb{B}_{n}$. Then the
Hankel operator $H^{(t)}_{u}=(I-P^{(t)})M_{u}P^{(t)}$ belongs to
$\mathcal{S}^{p}$ for any $p>2n$.

###### Proof.

The corollary follows from Corollary 6.3 and the equation

$H^{(t)}_{u}=(I-P^{(t)})M_{u}P^{(t)}=[M_{u},P^{(t)}]P^{(t)}.$

This completes the proof of Corollary 6.4. ∎

Corollaries 6.2 and 6.3 provide convenient criteria for an integral operator
to be in $\mathcal{S}^{p}$, when $p\geq 2$. However, in this paper we will
also need to deal with the case when $1\leq p<2$. Moreover, we will need to
consider integral operators of the form

$Th(z)=\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})F(z,w)h(w)\mathrm{d}\lambda_{t}(w),$

where $\phi$ is an unbounded function. For such $T$, if we take the double
integral as in Lemma 6.1, its integral is very likely to be infinite. In
application, it is enough for us to obtain Schatten-class membership of the
operators $TP^{(t)}$ or $P^{(t)}T$. An alternative way is to take advantage of
Lemma 5.4. In particular, the following lemma holds.

######  Lemma 6.5.

Suppose $t>-1$, $a>-n$, $b\geq 0$ and $c>0$. Suppose $F(z,w)$ is measurable on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$, $\phi:(0,1)\to[0,\infty)$ is measurable,
and

$\phi(s)\leq s^{a}(1-s)^{b},\quad s\in(0,1),$ $|F(z,w)|\leq\frac{1}{|1-\langle
z,w\rangle|^{n+1+t-c}},\quad\forall z,w\in\mathbb{B}_{n}.$

Define the integral operator on $L^{2}(\lambda_{t})$.

$Th(z)=\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})F(z,w)h(w)\mathrm{d}\lambda_{t}(w).$

Then both $P^{(t)}T$ and $TP^{(t)}$ belong to $\mathcal{S}^{p}$ for any
$p>\max\\{\frac{n}{c},\frac{n}{b+\frac{1+t}{2}}\\}$.

###### Proof.

Notice that $P^{(t)}T\in\mathcal{S}^{p}$ if and only if
$T^{*}P^{(t)}\in\mathcal{S}^{p}$, and that $T^{*}$ is an integral operator
with integral kernel satisfying the same estimate as $T$. Thus it suffices to
prove the statement for $TP^{(t)}$. For any
$q>\max\\{\frac{n}{c},\frac{n}{b+\frac{1+t}{2}}\\}$, let
$c^{\prime}=\frac{n}{q}$. Then $c^{\prime}<b+\frac{1+t}{2}$. Split the map
$TP^{(t)}$ as follows.

$TP^{(t)}:L^{2}(\lambda_{t})\xrightarrow{P^{(t)}}L_{a,t}^{2}(\mathbb{B}_{n})\xrightarrow{E_{t,2c^{\prime}}}L_{a,t+2c^{\prime}}^{2}(\mathbb{B}_{n})\xrightarrow{\hat{T}}L^{2}(\lambda_{t}).$

Here $\hat{T}:L_{a,t+2c^{\prime}}^{2}(\mathbb{B}_{n})\to L^{2}(\lambda_{t})$
is defined by the same integral formula as $T$. By Lemma 5.4,
$E_{t,2c^{\prime}}\in\mathcal{S}^{p}$ for any $p>\frac{n}{c^{\prime}}=q$. Also
by Lemma 5.1, $\hat{T}$ is bounded. Since $q$ is any number with
$q>\max\\{\frac{n}{c},\frac{n}{b+\frac{1+t}{2}}\\}$, we have
$TP^{(t)}\in\mathcal{S}^{p},\forall
p>\max\\{\frac{n}{c},\frac{n}{b+\frac{1+t}{2}}\\}$. This completes the proof
of Lemma 6.5. ∎

In the case when $\phi=s^{k}\Phi^{(t)}_{n,k}$, the following Schatten-norm
estimate holds.

######  Theorem 6.6.

Suppose $t>-1$, $c>0$ and $k$ is a non-negative integer. Suppose $F(z,w)$ is
measurable on $\mathbb{B}_{n}\times\mathbb{B}_{n}$, and

$|F(z,w)|\leq\frac{|\varphi_{z}(w)|^{2k}}{|1-\langle
z,w\rangle|^{n+1+t-c}},\quad\forall z,w\in\mathbb{B}_{n}.$

Define the integral operator on $L^{2}(\mathbb{B}_{n},\lambda_{t})$.

$Th(z)=\int_{\mathbb{B}_{n}}\Phi_{n,k}^{(t)}(|\varphi_{z}(w)|^{2})F(z,w)h(w)\mathrm{d}\lambda_{t}(w).$

Then both $P^{(t)}T$ and $TP^{(t)}$ belong to $\mathcal{S}^{p}$ for any
$p>\max\\{\frac{n}{c},\frac{n}{k+\frac{1+t}{2}}\\}$, $p\geq 1$. Moreover, for
$p>\frac{n}{c}$, and $p\geq 1$, and $t$ large enough, we have

$\|P^{(t)}T\|_{p}\lesssim t^{-k+\frac{n}{p}},\quad\|TP^{(t)}\|_{p}\lesssim
t^{-k+\frac{n}{p}}.$

We postpone the proof of Theorem 6.6 to the end of this section. As in the
proof of Corollary 6.3, Lemma 6.5 and Theorem 6.6 imply the following.

######  Corollary 6.7.

Suppose $t>-1$, $c>0$, $F(z,w)$ is measurable on
$\mathbb{B}_{n}\times\mathbb{B}_{n}$, and

$|F(z,w)|\leq\frac{1}{|1-\langle z,w\rangle|^{n+1+t-c}},\quad
z,w\in\mathbb{B}_{n}.$

Suppose $\phi:(0,1)\to[0,\infty)$ is measurable. Define the integral operator
on $L^{2}(\lambda_{t})$,

$Th(z)=\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})F(z,w)h(w)\mathrm{d}\lambda_{t}(w).$

Assume that $u$ is a Lipschitz function on $\mathbb{B}_{n}$.

  1. 1.

Suppose $a>-n$, $b\geq 0$ and $\phi(s)\leq s^{a}(1-s)^{b}$. Then

$[T,M_{u}]P^{(t)},\quad P^{(t)}[T,M_{u}]\in\mathcal{S}^{p},\quad\forall
p>\max\\{\frac{n}{c+\frac{1}{2}},\frac{n}{b+\frac{1+t}{2}}\\}.$

  2. 2.

If $\phi(s)=s^{k}\Phi^{(t)}_{n,k}$, then for $t$ large enough and
$p>\frac{n}{c+\frac{1}{2}}$,

$\big{\|}[T,M_{u}]P^{(t)}\big{\|}_{\mathcal{S}^{p}}\lesssim
t^{-k+\frac{n}{p}},\quad\big{\|}P^{(t)}[T,M_{u}]\big{\|}_{\mathcal{S}^{p}}\lesssim
t^{-k+\frac{n}{p}}.$

A trivial application of Theorem 6.6 gives the following.

######  Lemma 6.8.

Suppose $c>0$, $t>-1$ and $f\in L^{\infty}(\mathbb{B}_{n})$ satisfies

$|f(z)|\leq(1-|z|^{2})^{c},\quad\forall z\in\mathbb{B}_{n}.$

Then for any $p>\max\\{\frac{n}{c},\frac{2n}{1+t}\\},p\geq 1$,

$T^{(t)}_{f}\in\mathcal{S}^{p}.$

For any $p>\frac{n}{c}$, $p\geq 1$ and $t$ large enough,

$\|T^{(t)}_{f}\|_{\mathcal{S}^{p}}\lesssim_{p}t^{\frac{n}{p}}.$

###### Proof.

By definition, we have the following expression of $T^{(t)}_{f}$,

$T^{(t)}_{f}h(z)=\int_{\mathbb{B}_{n}}f(w)h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w).$

By the assumption, we have the following inequalities

$|f(w)K^{(t)}_{w}(z)|\leq\frac{(1-|z|^{2})^{c}}{|1-\langle
z,w\rangle|^{n+1+t}}\lesssim\frac{1}{|1-\langle z,w\rangle|^{n+1+t-c}}.$

Since $\Phi^{(t)}_{n,0}=1$, the conclusion follows directly from Theorem 6.6.
This completes the proof of Lemma 6.8. ∎

######  Lemma 6.9.

Suppose $\phi:(0,1)\to[0,\infty)$ is measurable. Then for any
$c,d\in\mathbb{R}$ there exist $C>0$ and $t_{0}>0$ such that whenever
$t>t_{0}$,

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})^{\frac{t}{2}+c}}{|1-\langle
z,w\rangle|^{n+1+t+d}}\mathrm{d}m(w)\leq
C\int_{0}^{1}\phi(s)s^{n-1}(1-s)^{\frac{t}{4}}\mathrm{d}s\cdot(1-|z|^{2})^{-\frac{t}{2}-d+c},\quad\forall
z\in\mathbb{B}_{n}.$ (6.1)

###### Proof.

The proof is almost a repetition of that of Lemma 2.4 (3). Make the change of
variable $w=\varphi_{z}(\xi)$. Using Lemma 2.2 we have the following equation,

$\frac{(1-|w|^{2})^{t}}{|1-\langle
z,w\rangle|^{n+1+t}}\mathrm{d}m(w)\xlongequal[\xi=\varphi_{z}(w)]{w=\varphi_{z}(\xi)}\frac{(1-|\xi|^{2})^{t}}{|1-\langle
z,\xi\rangle|^{n+1+t}}\mathrm{d}m(\xi).$

Then we have the following estimate of the left side of Equation (6.1),

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})\frac{(1-|w|^{2})^{\frac{t}{2}+c}}{|1-\langle
z,w\rangle|^{n+1+t+d}}\mathrm{d}m(w)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\xi|^{2})\bigg{(}\frac{(1-|z|^{2})(1-|\xi|^{2})}{|1-\langle
z,\xi\rangle|^{2}}\bigg{)}^{c-\frac{t}{2}}\bigg{(}\frac{|1-\langle
z,\xi\rangle|}{1-|z|^{2}}\bigg{)}^{d}\frac{(1-|\xi|^{2})^{t}}{|1-\langle
z,\xi\rangle|^{n+1+t}}\mathrm{d}m(\xi)$ $\displaystyle=$
$\displaystyle(1-|z|^{2})^{c-\frac{t}{2}-d}\int_{\mathbb{B}_{n}}\phi(|\xi|^{2})\frac{(1-|\xi|^{2})^{c+\frac{t}{2}}}{|1-\langle
z,\xi\rangle|^{n+1+2c-d}}\mathrm{d}m(\xi)$ $\displaystyle=$
$\displaystyle(1-|z|^{2})^{c-\frac{t}{2}-d}\int_{0}^{1}\phi(r^{2})(1-r^{2})^{c+\frac{t}{2}}r^{2n-1}\bigg{[}\int_{\mathbb{S}_{n}}\frac{1}{|1-\langle
rz,\zeta\rangle|^{n+1+2c-d}}\mathrm{d}\sigma(\zeta)\bigg{]}\mathrm{d}r$
$\displaystyle\lesssim$
$\displaystyle(1-|z|^{2})^{c-\frac{t}{2}-d}\int_{0}^{1}\phi(r^{2})(1-r^{2})^{c+\frac{t}{2}}r^{2n-1}(1-|rz|^{2})^{-a}\mathrm{d}r$
$\displaystyle\leq$
$\displaystyle(1-|z|^{2})^{c-\frac{t}{2}-d}\int_{0}^{1}\phi(r^{2})(1-r^{2})^{c-a+\frac{t}{2}}r^{2n-1}\mathrm{d}r$
$\displaystyle\xlongequal{s=r^{2}}$
$\displaystyle\frac{1}{2}(1-|z|^{2})^{c-\frac{t}{2}-d}\int_{0}^{1}\phi(s)(1-s)^{c-a+\frac{t}{2}}s^{n-1}\mathrm{d}s.$

Here $a=2c-d$ if $2c-d>0$; $a=\frac{1}{2}$ if $2c-d=0$; $a=0$ if $2c-d<0$. For
$t$ large enough, $c-a+\frac{t}{2}>\frac{t}{4}$. So we have the following
inequality,

$\int_{0}^{1}\phi(s)(1-s)^{c-a+\frac{t}{2}}s^{n-1}\mathrm{d}s\leq\int_{0}^{1}\phi(s)(1-s)^{\frac{t}{4}}s^{n-1}\mathrm{d}s.$

This completes the proof of Lemma 6.9. ∎

###### Proof of Theorem 6.6.

The Schatten class memberships of $TP^{(t)}$ and $P^{(t)}T$ are implied by
Lemma 6.5 and 4.3. It remains to prove the Schatten norm estimates. We may
assume that $t$ is large enough so that $0<c<k+\frac{1+t}{2}$. Also, as in the
proof of Lemma 6.5, it suffices to prove the statement for $TP^{(t)}$.

Split the operator $TP^{(t)}$ as

$L^{2}(\mathbb{B}_{n},\lambda_{t})\xrightarrow{P^{(t)}}L_{a,t}^{2}(\mathbb{B}_{n})\xrightarrow{E_{t,2c}}L_{a,t+2c}^{2}(\mathbb{B}_{n})\xrightarrow{\hat{T}}L^{2}(\mathbb{B}_{n},\lambda_{t}),$

where $\hat{T}$ is the same integral operator as $T$. By Lemma 5.4, it
suffices to show that $\hat{T}$ defines a bounded operator from
$L_{a,t+2c}^{2}(\mathbb{B}_{n})$ to $L^{2}(\mathbb{B}_{n},\lambda_{t})$ with
$\|\hat{T}\|_{L_{a,t+2c}^{2}(\mathbb{B}_{n})\to
L_{a,t}^{2}(\mathbb{B}_{n})}\lesssim t^{-k}$ for large $t$.

This is done by the Schur’s test. Since $c<k+\frac{t+1}{2}$, if we take
$x=c+\frac{t+1}{2}$, then $t+1+k>x>2c-k$. Take $p(w)=(1-|w|^{2})^{-x}$ and
$q(z)=(1-|z|^{2})^{c-x}$. The integral kernel of
$\hat{T}:L^{2}(\mathbb{B}_{n},\lambda_{t+2c})\to
L^{2}(\mathbb{B}_{n},\lambda_{t})$ is

$T(z,w):=\frac{B(n,t+2c+1)}{B(n,t+1)}\Phi_{n,k}^{(t)}(|\varphi_{z}(w)|^{2})F(z,w)(1-|w|^{2})^{-2c}.$

Then by (2.7), (6.1), (4.10) and (4.11), we have the following estimates,

$\displaystyle\int_{\mathbb{B}_{n}}|T(z,w)|p(w)\mathrm{d}\lambda_{t+2c}(w)$
$\displaystyle\approx$
$\displaystyle\frac{1}{B(n,t+1)}\int_{\mathbb{B}_{n}}\Phi^{(t)}_{n,k}(|\varphi_{z}(w)|^{2})|\varphi_{z}(w)|^{2k}\frac{(1-|w|^{2})^{t-x}}{|1-\langle
z,w\rangle|^{n+1+t-c}}\mathrm{d}m(w)$ $\displaystyle\lesssim$
$\displaystyle\frac{\int_{0}^{1}\Phi_{n,k}^{(t)}(s)s^{n+k-1}(1-s)^{\frac{t}{4}}\mathrm{d}s}{B(n,t+1)}q(z)\lesssim
t^{-k}q(z).$

Similarly, we have the following inequality,

$\int_{\mathbb{B}_{n}}|T(z,w)|q(z)\mathrm{d}\lambda_{t}(z)\lesssim
t^{-k}p(w).$

From this we have the following bound for $p>n/c$ and large $t$,

$\|T\|_{\mathcal{S}^{p}}\leq\|E_{t,2c}\|_{\mathcal{S}^{p}}\|\hat{T}\|\lesssim
t^{-k+n/p}.$

This completes the proof of Theorem 6.6. ∎

##  7 Integration by Parts

In this section we establish analogues of Lemma 3.8 and Lemma 3.9 in higher
dimensions. Similar integral formulas were discovered by Charpentier in [12].
Such integral formulas were used in [43] to study Bergman-Besov spaces and in
[18] to study the corona problem on the multiplier algebra of the Drury-
Arveson space.

Recall that $R=\sum_{i=1}^{n}z_{i}\partial_{z_{i}}$ is the radial derivative
operator, and $\bar{R}=\sum_{i=1}^{n}\bar{z}_{i}\bar{\partial}_{z_{i}}$. The
following lemma can be verified directly on $v(z)=z^{\gamma}\bar{z}^{\iota}$,
and then using approximation. In the appendix, we show that it is a special
case of a Bochner-Martinelli type formula.

######  Lemma 7.1.

Suppose $r>0$, $\alpha,\beta\in\mathbb{N}_{0}^{n}$, $|\alpha|\geq|\beta|$ and
$v\in\mathscr{C}^{1}(\overline{\mathbb{B}_{n}})$. Then

$\displaystyle\int_{r\mathbb{S}_{n}}z^{\alpha}\bar{z}^{\beta}v(z)\mathrm{d}\sigma_{r}(z)$
(7.1) $\displaystyle=$ $\displaystyle
a_{\alpha,\beta}\sigma_{2n-1}r^{2|\beta|+2n-1}v(0)+2r^{2|\beta|+2n-1}\int_{r\mathbb{B}_{n}}\frac{z^{\alpha}\bar{z}^{\beta}}{|z|^{2|\beta|+2n}}\bar{R}v(z)\mathrm{d}m(z).$

Here

$a_{\alpha,\beta}=\begin{cases}\frac{(n-1)!\alpha!}{(n-1+|\beta|)!},&\text{if
}\alpha\leq\beta,\\\ 0,&\text{otherwise}.\end{cases}$

Integrating both sides of (7.1) along $r$ with some weight gives the
following.

######  Lemma 7.2.

For any $t>-1$ the following hold.

  * (1)

For any $v\in\mathscr{C}^{1}(\overline{\mathbb{B}_{n}})$,

$\int_{\mathbb{B}_{n}}v(z)\mathrm{d}\lambda_{t}(z)=v(0)+\frac{t+1}{n+t+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{n}1(|z|^{2})\bar{R}v(z)\mathrm{d}\lambda_{t+1}(z).$
(7.2)

  * (2)

$\mathcal{G}^{(t)}_{n}1(s)\approx s^{-n}$ in a neighborhood of $0$, and
$\lim_{s\to 1^{-}}\mathcal{G}^{(t)}_{n}1(s)=\frac{1}{t+1}$.

  * (3)

$|\mathcal{G}^{(t)}_{n}1(s)-\frac{1}{t+1}|\lesssim 1-s$ for $s$ in a
neighborhood of $1$.

###### Proof.

The proof is almost the same as that of Lemma 3.8. For any $0<r<1$, by Lemma
7.1, the following equation holds,

$\int_{r\mathbb{S}_{n}}v(z)\mathrm{d}\sigma_{r}(z)=\sigma_{2n-1}r^{2n-1}v(0)+2r^{2n-1}\int_{r\mathbb{B}_{n}}|z|^{-2n}\bar{R}v(z)\mathrm{d}m(z).$
(7.3)

Integrating along the radial direction, we get the following computation,

$\displaystyle\int_{\mathbb{B}_{n}}v(z)\mathrm{d}\lambda_{t}(z)=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}(1-r^{2})^{t}\bigg{[}\int_{r\mathbb{S}_{n}}v(z)\mathrm{d}\sigma_{r}(z)\bigg{]}\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}(1-r^{2})^{t}\bigg{[}\sigma_{2n-1}r^{2n-1}v(0)+2r^{2n-1}\int_{r\mathbb{B}_{n}}|z|^{-2n}\bar{R}v(z)\mathrm{d}m(z)\bigg{]}\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{(n-1)!\sigma_{2n-1}}{\pi^{n}B(n,t+1)}\int_{0}^{1}(1-r^{2})^{t}r^{2n-1}\mathrm{d}r\cdot
v(0)$
$\displaystyle+\frac{2(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\bigg{[}\int_{|z|}^{1}(1-r^{2})^{t}r^{2n-1}\mathrm{d}r\bigg{]}|z|^{-2n}\bar{R}v(z)\mathrm{d}m(z)$
$\displaystyle=$ $\displaystyle
v(0)+\frac{t+1}{n+t+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{n}1(|z|^{2})\bar{R}v(z)\mathrm{d}\lambda_{t+1}(z).$

This proves (7.2). Finally, (2) and (3) can be proved easily using
L’Hospitial’s Rule. We omit the details. This completes the proof of Lemma
7.2. ∎

Equation (7.3) can be viewed as an integration by parts formula: comparing the
integrals on both sides, the part $v(z)$ is differentiated, and the part
$\mathrm{d}\lambda_{t}$ is improved into $\mathrm{d}\lambda_{t+1}$, which can
be viewed as integration. Lemma 7.2 will play an important role in Section 8.
For a Toeplitz operator $T^{(t)}_{f}$ with
$f\in\mathscr{C}^{k}(\overline{\mathbb{B}_{n}})$ for some positive integer
$k$, Lemma 7.2 allows us to give an integral representation of $T^{(t)}_{f}$
with improved kernels. (See Lemma 8.4.)

Next, we generalize Lemma 7.1 in another direction: In (7.1), the first term
in the right hand side subtracts the value of $v$ at $0$. Combing with the
Möbius transform, we are allowed subtract information at any point in
$\mathbb{B}_{n}$. The resulting formulas will be useful in Section 9. We start
with a few definitions.

Recall that by Lemma 2.2 (5), for $z\in\mathbb{B}_{n}$,

$(1-|z|^{2})P_{z}(w)+(1-|z|^{2})^{1/2}Q_{z}(w)=(1-\langle
w,z\rangle)(z-\varphi_{z}(w))=A_{z}w,$

where $A_{z}$ is an $n\times n$ matrix depending on $z$, and $w$ is treated as
a column vector.

######  Definition 7.3.

For multi-indices $\alpha,\beta\in\mathbb{N}_{0}^{n}$ and
$z\in\mathbb{B}_{n}$, define

$d_{\alpha,\beta}(z)=\int_{\mathbb{S}_{n}}\big{(}A_{z}\zeta\big{)}^{\alpha}\overline{\big{(}A_{z}\zeta\big{)}}^{\beta}\frac{\mathrm{d}\sigma(\zeta)}{\sigma_{2n-1}}.$

In particular, $d_{0,0}=1$, and

$d_{\alpha,\beta}(z)=\delta_{\alpha,\beta}(1-|z|^{2})^{\alpha_{1}+|\alpha|}\frac{(n-1)!\alpha!}{(n-1+|\alpha|)!},\quad\text{if
}z=(z_{1},0,\ldots,0).$ (7.4)

For multi-indices $\alpha,\beta\in\mathbb{N}_{0}^{n}$ and
$\zeta\in\mathbb{C}^{n}$, denote

$I^{\alpha,\beta}(\zeta)=\zeta^{\alpha}\bar{\zeta}^{\beta}.$

######  Lemma 7.4.

Suppose $\alpha,\beta\in\mathbb{N}_{0}^{n}$, and
$v\in\mathscr{C}^{1}(\overline{\mathbb{B}_{n}})$. Then the following hold.

  1. 1.

If $|\alpha|\geq|\beta|$, then

$\displaystyle\int_{\mathbb{S}_{n}}I^{\alpha,\beta}(z-w)v(w)K_{w}(z)\frac{\mathrm{d}\sigma(w)}{\sigma_{2n-1}}$
(7.5) $\displaystyle=$ $\displaystyle
d_{\alpha,\beta}(z)v(z)-\frac{1}{n}\sum_{j=1}^{n}\int_{\mathbb{B}_{n}}|\varphi_{z}(w)|^{-2|\beta|-2n}I^{\alpha,\beta+e_{j}}(z-w)\frac{\bar{\partial}_{j}\big{[}(1-\langle
z,w\rangle)^{|\beta|}v(w)\big{]}}{(1-\langle z,w\rangle)^{|\beta|}(1-\langle
w,z\rangle)}K_{w}(z)\mathrm{d}\lambda_{0}(w),$

  2. 2.

If $|\alpha|\leq|\beta|$, then

$\displaystyle\int_{\mathbb{S}_{n}}I^{\alpha,\beta}(z-w)v(z)K_{w}(z)\frac{\mathrm{d}\sigma(z)}{\sigma_{2n-1}}$
(7.6) $\displaystyle=$ $\displaystyle
d_{\beta,\alpha}(w)v(w)+\frac{1}{n}\sum_{i=1}^{n}\int_{\mathbb{B}_{n}}|\varphi_{z}(w)|^{-2|\alpha|-2n}I^{\alpha+e_{i},\beta}(z-w)\frac{\partial_{i}\big{[}(1-\langle
z,w\rangle)^{|\alpha|}v(z)\big{]}}{(1-\langle z,w\rangle)^{|\alpha|}(1-\langle
w,z\rangle)}K_{w}(z)\mathrm{d}\lambda_{0}(z),$

######  Lemma 7.5.

Suppose $t>-1$, $\alpha,\beta\in\mathbb{N}_{0}^{n}$. Suppose
$\phi:(0,1)\to[0,\infty)$ is measurable and
$v\in\mathscr{C}^{1}(\mathbb{B}_{n})$. Then the following hold.

  1. 1.

If $|\alpha|\geq|\beta|$ and all integrals converge absolutely, then

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
(7.7) $\displaystyle=$
$\displaystyle\begin{cases}\frac{d_{\alpha,\beta}(z)}{B(n,t+1)}\cdot\mathcal{F}^{(t)}_{n+|\beta|}\phi(0)v(z)-\sum_{j=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{|\beta|+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta+e_{j}}(z-w)S_{j}(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),&\\\
\hskip 284.52756pt\text{ if }\mathcal{F}^{(t)}_{n+|\beta|}\phi(0)<\infty,&\\\
&\\\
-\sum_{j=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{|\beta|+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta+e_{j}}(z-w)S_{j}(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w),&\\\
\hskip 284.52756pt\text{ if
}\mathcal{F}^{(t)}_{n+|\beta|}\phi(0)=\infty,&\end{cases}$

where

$S_{j}(w)=\frac{(1-|w|^{2})\bar{\partial}_{w_{j}}\big{[}(1-\langle
z,w\rangle)^{|\beta|}v(w)\big{]}}{(1-\langle w,z\rangle)(1-\langle
z,w\rangle)^{|\beta|}}.$

  2. 2.

If $|\alpha|\leq|\beta|$ and all integrals converge absolutely, then

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)v(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z)$
(7.8) $\displaystyle=$
$\displaystyle\begin{cases}\frac{d_{\beta,\alpha}(w)}{B(n,t+1)}\cdot\mathcal{F}^{(t)}_{n+|\alpha|}\phi(0)v(w)+\sum_{i=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{|\alpha|+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha+e_{i},\beta}(z-w)S^{\prime}_{i}(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z),&\\\
\hskip 284.52756pt\text{ if }\mathcal{F}^{(t)}_{n+|\alpha|}\phi(0)<\infty,&\\\
&\\\
\sum_{i=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{|\alpha|+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha+e_{i},\beta}(z-w)S^{\prime}_{i}(z)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(z),&\\\
\hskip 284.52756pt\text{ if
}\mathcal{F}^{(t)}_{n+|\alpha|}\phi(0)=\infty,&\end{cases}$

where

$S^{\prime}_{i}(z)=\frac{(1-|z|^{2})\partial_{z_{i}}\big{[}(1-\langle
z,w\rangle)^{|\alpha|}v(z)\big{]}}{(1-\langle w,z\rangle)(1-\langle
z,w\rangle)^{|\alpha|}}.$

Suppose $\\{e_{1},\ldots,e_{n}\\}$ and $\\{f_{1},\ldots,f_{n}\\}$ are two
orthonormal basis of $\mathbb{C}^{n}$. Suppose

$\sum_{i=1}^{n}\zeta_{i}e_{i}=\sum_{i=1}^{n}\xi_{i}f_{i};\quad\sum_{i=1}^{n}\lambda_{i}e_{i}=\sum_{i=1}^{n}\omega_{i}f_{i}.$

Then there is a unitary matrix $U=[u_{ij}]$ such that

$\xi=U\zeta,\quad\omega=U\lambda.$

Denote $U^{*}=[u_{ij}^{*}]$. Therefore

$\sum_{i=1}^{n}\frac{\partial
v}{\partial\bar{\zeta_{i}}}\bar{\lambda}_{i}=\sum_{i,j,k=1}^{n}\bigg{(}\frac{\partial
v}{\partial\overline{\xi}_{j}}\bar{u}_{ji}\bar{\zeta}_{i}\bigg{)}\overline{\big{(}u^{*}_{ik}\omega_{k}\big{)}}=\sum_{j,k=1}^{n}\frac{\partial
v}{\partial\bar{\xi}_{j}}\delta_{jk}\bar{w}_{k}=\sum_{j=1}^{n}\frac{\partial
v}{\partial\bar{\xi}_{j}}\bar{w}_{j}.$ (7.9)

In other words, the function

$\sum_{i=1}^{n}\bar{\partial}_{i}v\bar{\lambda}_{i}=\langle\bar{\partial}v,\lambda\rangle$

does not depend on the choice of a basis.

######  Lemma 7.6.

Suppose $v\in\mathscr{C}^{1}(\mathbb{B}_{n})$ and $z\in\mathbb{B}_{n}$. Then

$\langle\bar{\partial}_{\zeta}v(\zeta),\zeta\rangle=-\frac{\langle\bar{\partial}\big{(}v\circ\varphi_{z}\big{)}(\varphi_{z}(\zeta)),z-\varphi_{z}(\zeta)\rangle}{1-\langle
z,\zeta\rangle}.$ (7.10)

###### Proof.

By (7.9), both sides of (7.10) do not depend on the choice of basis. Thus we
may assume $z=(r,0,\ldots,0)$. In this case, we have the following expression

$\varphi_{z}(w)=\frac{1}{1-w_{1}r}\big{(}r-w_{1},~{}-(1-r^{2})^{1/2}w_{2},~{}\ldots,~{}-(1-r^{2})^{1/2}w_{n}\big{)}.$

Consequently, we compute the Jacobian

$\bigg{[}\frac{\partial(\varphi_{z})_{i}}{\partial
w_{j}}\bigg{]}=\begin{bmatrix}-\frac{1-r^{2}}{(1-w_{1}r)^{2}}&0&0&0&\ldots&0\\\
-\frac{(1-r^{2})^{1/2}w_{2}r}{(1-w_{1}r)^{2}}&-\frac{(1-r^{2})^{1/2}}{1-w_{1}r}&0&0&\ldots&0\\\
-\frac{(1-r^{2})^{1/2}w_{3}r}{(1-w_{1}r)^{2}}&0&-\frac{(1-r^{2})^{1/2}}{1-w_{1}r}&0&\ldots&0\\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
-\frac{(1-r^{2})^{1/2}w_{n}r}{(1-w_{1}r)^{2}}&0&0&0&\ldots&-\frac{(1-r^{2})^{1/2}}{1-w_{1}r}\end{bmatrix}.$

Therefore we have the following expression,

$\langle\bar{\partial}\big{(}v\circ\varphi_{z}\big{)}(w),z-w\rangle=\sum_{i,j=1}^{n}\bar{\partial}_{i}v\big{(}\varphi_{z}(w)\big{)}\overline{\bigg{(}\frac{\partial\big{(}\varphi_{z}\big{)}_{i}}{\partial
w_{j}}\bigg{)}}\overline{(z_{j}-w_{j})}=\langle\bar{\partial}v\big{(}\varphi_{z}(w)\big{)},\xi\rangle,$
(7.11)

where

$\xi_{i}=\sum_{j=1}^{n}\frac{\partial\big{(}\varphi_{z}\big{)}_{i}}{\partial
w_{j}}(z_{j}-w_{j})=\frac{\partial\big{(}\varphi_{z}\big{)}_{i}}{\partial
w_{1}}(r-w_{1})-\sum_{j=2}^{n}\frac{\partial\big{(}\varphi_{z}\big{)}_{i}}{\partial
w_{j}}w_{j}.$

By the above, set

$\xi_{1}=-\frac{(1-r^{2})(r-w_{1})}{(1-w_{1}r)^{2}},$

and for $i=2,\ldots,n$,

$\xi_{i}=-\frac{(1-r^{2})^{1/2}w_{i}r(r-w_{1})}{(1-w_{1}r)^{2}}+\frac{(1-r^{2})^{1/2}w_{i}}{1-w_{1}r}=\frac{(1-r^{2})^{1/2}}{(1-w_{1}r)^{2}}\big{(}w_{i}-w_{i}r^{2}\big{)}=\frac{(1-r^{2})^{3/2}}{(1-w_{1}r)^{2}}w_{i}.$

Thus we have

$\xi=-\frac{1-r^{2}}{1-w_{1}r}\varphi_{z}(w).$ (7.12)

If we plug in $w=\varphi_{z}(\zeta)$ then by (7.11), (7.12) and Lemma 2.2, we
obtain the equalities

$\langle\bar{\partial}\big{(}v\circ\varphi_{z}\big{)}(\varphi_{z}(\zeta)),z-\varphi_{z}(\zeta)\rangle=\langle\bar{\partial}v(\zeta),-\frac{1-|z|^{2}}{1-\langle\varphi_{z}(\zeta),z\rangle}\zeta\rangle=-(1-\langle
z,\zeta\rangle)\langle\bar{\partial}v(\zeta),\zeta\rangle.$

Equivalently, (7.10) holds. This completes the proof of Lemma 7.6. ∎

###### Proof of Lemma 7.5.

Let $t,\alpha,\beta,\phi,v$ be as in Lemma 7.5. Denote $|\alpha|=k$,
$|\beta|=l$. Then by assumption $k\geq l$. By Lemma 2.2 (4) it is easy to
verify the following identity,

$K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)\xlongequal[\zeta=\varphi_{z}(w)]{w=\varphi_{z}(\zeta)}K^{(t)}_{z}(\zeta)\mathrm{d}\lambda_{t}(\zeta).$
(7.13)

Also by Lemma 2.2 (5), we obtain the following equation,

$z-\varphi_{z}(\zeta)=\frac{A_{z}\zeta}{1-\langle\zeta,z\rangle}=(1-|z|^{2})P_{z}(\zeta)+(1-|z|^{2})Q_{z}(\zeta),\quad
z,\zeta\in\mathbb{B}_{n}.$

Therefore we compute the integral as follows,

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle\xlongequal{w=\varphi_{z}(\zeta)}$
$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\zeta|^{2})I^{\alpha,\beta}(z-\varphi_{z}(\zeta))v\circ\varphi_{z}(\zeta)K^{(t)}_{z}(\zeta)\mathrm{d}\lambda_{t}(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\phi\phi_{t}(|\zeta|^{2})I^{\alpha,\beta}(z-\varphi_{z}(\zeta))\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle\zeta,z\rangle)^{n+1+t}}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\phi\phi_{t}(|\zeta|^{2})I^{\alpha,\beta}(A_{z}\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta\rangle)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}m(\zeta).$

Since $A_{z}\zeta$ is linear on $\zeta$, we have the following expression,

$I^{\alpha,\beta}(A_{z}\zeta)=\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}I^{\kappa,\gamma}(\zeta)=I^{\alpha,\beta}(z-\varphi_{z}(\zeta))(1-\langle\zeta,z\rangle)^{k}(1-\langle
z,\zeta\rangle)^{l}.$

Then we continue the computation of the above integral.

$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\phi\phi_{t}(|\zeta|^{2})I^{\alpha,\beta}(A_{z}\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta\rangle)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(|\zeta|^{2})I^{\kappa,\gamma}(\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}\bigg{\\{}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(r^{2})\int_{r\mathbb{S}_{n}}I^{\kappa,\gamma}(\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}\sigma_{r}(\zeta)\bigg{\\}}\mathrm{d}r.$

By (7.1), for each pair of indices $\kappa,\gamma$, we have the following
identity,

$\displaystyle\int_{r\mathbb{S}_{n}}I^{\kappa,\gamma}(\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}\sigma_{r}(\zeta)$
$\displaystyle=$ $\displaystyle a_{\kappa,\gamma}\sigma_{2n-1}r^{2l+2n-1}v(z)$
$\displaystyle~{}~{}+2r^{2l+2n-1}\int_{r\mathbb{B}_{n}}|\zeta|^{-2l-2n}I^{\kappa,\gamma}(\zeta)\langle\bar{\partial}_{\zeta}\bigg{[}\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\bigg{]},\zeta\rangle\mathrm{d}m(\zeta).$

By Lemma 7.6, we obtain the following equation,

$\displaystyle\langle\bar{\partial}_{\zeta}\bigg{[}\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\bigg{]},\zeta\rangle$
$\displaystyle=$ $\displaystyle-\frac{1}{1-\langle
z,\zeta\rangle}\big{\langle}\bar{\partial}_{w}\bigg{[}\frac{v(w)}{(1-\langle
z,\varphi_{z}(w)\rangle)^{l}(1-\langle\varphi_{z}(w),z\rangle)^{n+1+t+k}}\bigg{]}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}$
$\displaystyle=$ $\displaystyle-\frac{1}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{[}\frac{v(w)}{(1-\langle
z,\varphi_{z}(w)\rangle)^{l}}\bigg{]}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}$
$\displaystyle=$ $\displaystyle-\frac{(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{[}v(w)(1-\langle
z,w\rangle)^{l}\bigg{]}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}.$

So we obtain the following equation of integrals,

$\displaystyle\int_{r\mathbb{S}_{n}}I^{\kappa,\gamma}(\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}\sigma_{r}(\zeta)$
$\displaystyle=$ $\displaystyle a_{\kappa,\gamma}\sigma_{2n-1}r^{2l+2n-1}v(z)$
$\displaystyle~{}~{}-2r^{2l+2n-1}\int_{r\mathbb{B}_{n}}\mathrm{d}m(\zeta)\frac{|\zeta|^{-2l-2n}I^{\kappa,\gamma}(\zeta)(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{[}v(w)(1-\langle
z,w\rangle)^{l}\bigg{]}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}.$

Therefore we continue to compute the integral as follows,

$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}\bigg{\\{}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(r^{2})\int_{r\mathbb{S}_{n}}I^{\kappa,\gamma}(\zeta)\frac{v\circ\varphi_{z}(\zeta)}{(1-\langle
z,\zeta)^{l}(1-\langle\zeta,z\rangle)^{n+1+t+k}}\mathrm{d}\sigma_{r}(\zeta)\bigg{\\}}\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}r\bigg{\\{}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(r^{2})\bigg{[}a_{\kappa,\gamma}\sigma_{2n-1}r^{2l+2n-1}v(z)-2r^{2l+2n-1}$
$\displaystyle\cdot\int_{r\mathbb{B}_{n}}\frac{|\zeta|^{-2l-2n}I^{\kappa,\gamma}(\zeta)(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\bigg{]}\mathrm{d}m(\zeta)\bigg{\\}}\mathrm{d}$
$\displaystyle=$ $\displaystyle
Cv(z)-\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}\int_{r\mathbb{B}_{n}}\bigg{\\{}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(r^{2})2r^{2l+2n-1}$
$\displaystyle\cdot\frac{|\zeta|^{-2l-2n}I^{\kappa,\gamma}(\zeta)(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\bigg{\\}}\mathrm{d}m(\zeta)\mathrm{d}r.$

Here $C$ is a number depending on $z$. By Fubini’s Theorem, the second part in
the above equals

$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\int_{|\zeta|}^{1}\bigg{\\{}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}\phi\phi_{t}(r^{2})2r^{2l+2n-1}$
$\displaystyle\cdot\frac{|\zeta|^{-2l-2n}I^{\kappa,\gamma}(\zeta)(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\bigg{\\}}\mathrm{d}r\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\bigg{(}\int_{|\zeta|}^{1}\phi\phi_{t}(r^{2})2r^{2l+2n-1}\mathrm{d}r\bigg{)}\bigg{(}\sum_{|\kappa|=k,|\gamma|=l}c_{\alpha,\beta,\kappa,\gamma,z}I^{\kappa,\gamma}(\zeta)\bigg{)}$
$\displaystyle\cdot\frac{|\zeta|^{-2l-2n}(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}\bigg{(}\int_{|\zeta|^{2}}^{1}\phi\phi_{t}(s)s^{l+n-1}\mathrm{d}s\bigg{)}I^{\alpha,\beta}(A_{z}\zeta)$
$\displaystyle\cdot\frac{|\zeta|^{-2l-2n}(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{\mathbb{B}_{n}}(1-|\zeta|^{2})^{t+1}\mathcal{G}^{(t)}_{l+n}\phi(|\zeta|^{2})I^{\alpha,\beta}(z-\varphi_{z}(\zeta))(1-\langle\zeta,z\rangle)^{k}(1-\langle
z,\zeta\rangle)^{l}$ $\displaystyle\cdot\frac{(1-|z|^{2})^{-l}}{(1-\langle
z,\zeta\rangle)(1-\langle\zeta,z\rangle)^{n+1+t+k}}\big{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\big{\rangle}\mathrm{d}m(\zeta)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{l+n}\phi(|\zeta|^{2})I^{\alpha,\beta}(z-\varphi_{z}(\zeta))(1-|\zeta|^{2})(1-|z|^{2})^{-l}(1-\langle
z,\zeta\rangle)^{l-1}$
$\displaystyle\cdot\bigg{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{|}_{w=\varphi_{z}(\zeta)},z-\varphi_{z}(\zeta)\bigg{\rangle}K^{(t)}_{z}(\zeta)\mathrm{d}\lambda_{t}(\zeta).$

Applying the change of variable $\zeta=\varphi_{z}(w)$, by (7.13), the above
equals

$\displaystyle\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{l+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)(1-|\varphi_{z}(w)|^{2})(1-|z|^{2})^{-l}(1-\langle
z,\varphi_{z}(w)\rangle)^{l-1}$
$\displaystyle\qquad\cdot\bigg{\langle}\bar{\partial}_{w}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)},z-w\bigg{\rangle}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{l+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)\frac{(1-|w|^{2})}{(1-\langle
w,z\rangle)(1-\langle z,w\rangle)^{l}}$
$\displaystyle\qquad\cdot\bigg{(}\sum_{j=1}^{n}\overline{(z_{j}-w_{j})}\bar{\partial}_{w_{j}}\bigg{(}v(w)(1-\langle
z,w\rangle)^{l}\bigg{)}\bigg{)}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{l+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta+e_{j}}(z-w)S_{j}(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w).$

Combining all the above, we have obtained the following equation,

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)v(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$ $\displaystyle
Cv(z)-\sum_{j=1}^{n}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{l+n}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta+e_{j}}(z-w)S_{j}(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w).$

It remains to compute the constant $C$. First we notice that by the proof,
$C=0$ if either $v(z)=0$ or $k\neq l$. Assuming $k=l$ and $v(z)\neq 0$. Take
$v(w)=\frac{1}{(1-\langle z,w\rangle)^{l}(1-\langle w,z\rangle)^{n+1+t+l}}$.
Then the second term equals zero. So we have the following expression,

$\displaystyle C(1-|z|^{2})^{-n-1-t-2l}$ $\displaystyle=$
$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\varphi_{z}(w)|^{2})I^{\alpha,\beta}(z-w)\frac{1}{|1-\langle
z,w\rangle|^{2(n+1+t+l)}}\mathrm{d}\lambda_{t}(w)$
$\displaystyle\xlongequal{w=\varphi_{z}(\zeta)}$
$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\zeta|^{2})I^{\alpha,\beta}(A_{z}\zeta)\frac{1}{(1-|z|^{2})^{n+1+2l+t}}\mathrm{d}\lambda_{t}(\zeta).$

Therefore we have the following expression for $C$,

$\displaystyle C=$
$\displaystyle\int_{\mathbb{B}_{n}}\phi(|\zeta|^{2})I^{\alpha,\beta}(A_{z}\zeta)\mathrm{d}\lambda_{t}(\zeta)$
$\displaystyle=$
$\displaystyle\frac{(n-1)!}{\pi^{n}B(n,t+1)}\int_{0}^{1}r^{2n+2l-1}\phi(r^{2})(1-r^{2})^{t}\int_{\mathbb{S}_{n}}I^{\alpha,\beta}(A_{z}\zeta)\mathrm{d}\sigma(\zeta)\mathrm{d}r$
$\displaystyle=$
$\displaystyle\frac{\mathcal{F}^{(t)}_{n+l}\phi(0)}{B(n,t+1)}\int_{\mathbb{S}_{n}}I^{\alpha,\beta}(A_{z}\zeta)\frac{\mathrm{d}\sigma(\zeta)}{\sigma_{2n-1}}.$

This completes the proof of Lemma 7.5. ∎

The formulas in Lemma 7.4 is obtained similarly by applying the Möbius
transform to (7.1) at $r=1$. We omit the details.

##  8 Traces on Different Weighted Bergman Spaces

As explained in the introduction, the goal of this section is to prove
Equation (1.8). More precisely, we will prove Lemma 8.1 and Lemma 8.2 stated
below.

Suppose $t>s\geq-1$. It is well-known that $L_{a,s}^{2}(\mathbb{B}_{n})\subset
L_{a,t}^{2}(\mathbb{B}_{n})$, and $L_{a,s}^{2}(\mathbb{B}_{n})$ is dense in
$L_{a,t}^{2}(\mathbb{B}_{n})$. For a Toeplitz operator $T^{(t)}_{f}$ on
$L_{a,t}^{2}(\mathbb{B}_{n})$, if the restriction of $T^{(t)}_{f}$ on
$L_{a,s}^{2}(\mathbb{B}_{n})$ defines a bounded operator on
$L_{a,s}^{2}(\mathbb{B}_{n})$, then we denote this restriction to be
$T^{(t,s)}_{f}$. On the other hand, if a Toeplitz operator $T^{(s)}_{f}$ on
$L_{a,s}^{2}(\mathbb{B}_{n})$ extends (uniquely) into a bounded operator on
$L_{a,t}^{2}(\mathbb{B}_{n})$, then we denote this operator to be
$T^{(s,t)}_{f}$. It follows from Lemma 5.1 that if $f\in
L^{\infty}(\mathbb{B}_{n})$ and $t>s>-1$, then the restricted operator
$T^{(t,s)}_{f}$ is well-defined.

######  Lemma 8.1.

Suppose
$f_{1},f_{2},\ldots,f_{2n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$ and
$t>-1$. Then

$[T^{(t)}_{f_{1}},T^{(t)}_{f_{2}},\ldots,T^{(t)}_{f_{2n}}]-[T^{(t+1,t)}_{f_{1}},T^{(t+1,t)}_{f_{2}},\ldots,T^{(t+1,t)}_{f_{2n}}]$

is a trace class operator on $L_{a,t}^{2}(\mathbb{B}_{n})$ with zero trace.

In the case of the Hardy space we show the following.

######  Lemma 8.2.

Suppose
$f_{1},f_{2},\ldots,f_{2n}\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$. Then

$[T_{f_{1}},T_{f_{2}},\ldots,T_{f_{2n}}]-[T^{(1,-1)}_{f_{1}},T^{(1,-1)}_{f_{2}},\ldots,T^{(1,-1)}_{f_{2n}}]$

is a trace class operator on $H^{2}(\mathbb{S}_{n})$ with zero trace.

The fact that each $T^{(1,-1)}_{f_{i}}$ is well-defined will be explained in
Remark 8.21. The proof of Lemma 8.1 (and Lemma 8.2) involves writing
$T^{(t)}_{f}$ (resp. $T_{f}$) as the sum of $T^{(t+1,t)}_{f}$ (resp.
$T^{(1,-1)}_{f}$) and some perturbation operators.

######  Definition 8.3.

For $t>-1$, define

$X^{(t)}_{w}(z)=\frac{t+1}{n+t+1}\mathcal{G}^{(t)}_{n}1(|w|^{2})\bar{R}_{w}K^{(t)}_{w}(z)-K^{(t+1)}_{w}(z),$
(8.1)

and

$Y^{(t)}_{w}(z)=\frac{t+1}{n+t+1}\mathcal{G}^{(t)}_{n}1(|w|^{2})K^{(t)}_{w}(z).$
(8.2)

For a symbol function $g$, formally define the integral operators

$X^{(t)}_{g}h(z)=\int_{\mathbb{B}_{n}}g(w)h(w)X^{(t)}_{w}(z)\mathrm{d}\lambda_{t+1}(w);\quad
Y^{(t)}_{g}h(z)=\int_{\mathbb{B}_{n}}g(w)h(w)Y^{(t)}_{w}(z)\mathrm{d}\lambda_{t+1}(w).$

It will be clear in subsequent proof that for
$f\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$, $X^{(t)}_{f}$ and
$Y^{(t)}_{\bar{R}f}$ define bounded operators on $L_{a,t}^{2}(\mathbb{B}_{n})$
(see Lemma 8.16 and Lemma 8.14).

######  Lemma 8.4.

Suppose $f\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$. Then for any $t>-1$,

$T^{(t)}_{f}=T^{(t+1,t)}_{f}+X^{(t)}_{f}+Y^{(t)}_{\bar{R}f}+f(0)E_{0},$ (8.3)

where $E_{0}h=h(0)$ is a rank $1$ operator.

###### Proof.

For any $h\in\mathrm{Hol}(\overline{\mathbb{B}_{n}})$, by Lemma 7.2, we
compute $T^{(t)}_{f}$ as follows,

$\displaystyle T^{(t)}_{f}h(z)=$
$\displaystyle\int_{\mathbb{B}_{n}}f(w)h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t}(w)$
$\displaystyle=$ $\displaystyle
f(0)h(0)+\frac{t+1}{n+t+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{n}1(|w|^{2})\bar{R}f(w)h(w)K^{(t)}_{w}(z)\mathrm{d}\lambda_{t+1}(w)$
$\displaystyle+\frac{t+1}{n+t+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(t)}_{n}1(|w|^{2})f(w)h(w)\bar{R}K^{(t)}_{w}(z)\mathrm{d}\lambda_{t+1}(w)$
$\displaystyle=$ $\displaystyle
f(0)\big{(}E_{0}f\big{)}(z)+Y^{(t)}_{\bar{R}f}h(z)+X^{(t)}_{f}h(z)+T^{(t+1,t)}_{f}h(z).$

This completes the proof of Lemma 8.4. ∎

In the case of the Hardy space, we need to apply integration by parts twice to
make our arguments work.

######  Definition 8.5.

Let $\phi(s)=s^{-n}$. Define

$X^{(-1)}_{w}(z)=\frac{1}{n(n+1)}\mathcal{G}_{n}^{(0)}\phi(|w|^{2})\bar{R}^{2}_{w}K_{w}(z)-K_{w}^{(1)}(z),$
$Y^{(-1)}_{w}(z)=\frac{2}{n(n+1)}\mathcal{G}_{n}^{(0)}\phi(|w|^{2})\bar{R}_{w}K_{w}(z),$

and

$Z^{(-1)}_{w}(z)=\frac{1}{n(n+1)}\mathcal{G}_{n}^{(0)}\phi(|w|^{2})K_{w}(z).$

Formally define the symboled integral operators

$X^{(-1)}_{f}h(z)=\int_{\mathbb{B}_{n}}h(w)f(w)X^{(-1)}_{w}(z)\mathrm{d}\lambda_{1}(w),$
$Y^{(-1)}_{f}h(z)=\int_{\mathbb{B}_{n}}h(w)f(w)Y^{(-1)}_{w}(z)\mathrm{d}\lambda_{1}(w),$

and

$Z^{(-1)}_{f}h(z)=\int_{\mathbb{B}_{n}}h(w)f(w)Z^{(-1)}_{w}(z)\mathrm{d}\lambda_{1}(w).$

Again, it will be clear later that for
$f\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$, $X^{(-1)}_{f}$,
$Y^{(-1)}_{\bar{R}f}$ and $Z^{(-1)}_{\bar{R}^{2}f}$ define bounded operators
on $H^{2}(\mathbb{S}_{n})$ (see Lemma 8.20).

######  Lemma 8.6.

Suppose $f\in\mathscr{C}^{2}(\overline{\mathbb{B}_{n}})$. Then

$T_{f}=T^{(1,-1)}_{f}+X^{(-1)}_{f}+Y^{(-1)}_{\bar{R}f}+Z^{(-1)}_{\bar{R}^{2}f}+f(0)E_{0}.$

Taking $\alpha=\beta=0$ and $z=0$ in Lemma 7.4, we get the following.

######  Lemma 8.7.

Suppose $v\in\mathscr{C}^{1}(\overline{\mathbb{B}_{n}})$. Then

$\int_{\mathbb{S}_{n}}v(w)\frac{\mathrm{d}\sigma(w)}{\sigma_{2n-1}}=v(0)+\frac{1}{n}\int_{\mathbb{B}_{n}}|w|^{-2n}\bar{R}v(w)\mathrm{d}\lambda_{0}(w).$
(8.4)

Also, taking $\alpha=\beta=0,t=0,z=0$ in Lemma 7.5 gives the following.

######  Lemma 8.8.

Suppose $\phi:(0,1)\to[0,\infty)$ is measurable and
$v\in\mathscr{C}^{1}(\mathbb{B}_{n})$. Then whenever all integrals converge
absolutely, we have

$\displaystyle\int_{\mathbb{B}_{n}}\phi(|w|^{2})v(w)\mathrm{d}\lambda_{0}(w)$
(8.5) $\displaystyle=$
$\displaystyle\begin{cases}\frac{\mathcal{F}^{(0)}_{n}\phi(0)}{B(n,t+1)}v(0)+\frac{1}{n+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(0)}_{n}\phi(|w|^{2})\bar{R}v(w)\mathrm{d}\lambda_{1}(w),&\mathcal{F}^{(0)}_{n}\phi(0)<\infty\\\
\frac{1}{n+1}\int_{\mathbb{B}_{n}}\mathcal{G}^{(0)}_{n}\phi(|w|^{2})\bar{R}v(w)\mathrm{d}\lambda_{1}(w),&\mathcal{F}^{(0)}_{n}\phi(0)\leq\infty,~{}v(0)=0\end{cases}.$

###### Proof of Lemma 8.6.

By Lemma 8.7 and Lemma 8.8, for $h\in\mathrm{Hol}(\overline{\mathbb{B}_{n}})$,
we compute $T_{f}$ as follows,

$\displaystyle T_{f}h(z)=$
$\displaystyle\int_{\mathbb{S}_{n}}h(w)f(w)K_{w}(z)\frac{\mathrm{d}\sigma(w)}{\sigma_{2n-1}}$
$\displaystyle\xlongequal{\eqref{eqn: formula Sn R 0}}$ $\displaystyle
f(0)h(0)+\frac{1}{n}\int_{\mathbb{B}_{n}}\phi(|w|^{2})\bar{R}\bigg{[}h(w)f(w)K_{w}(z)\bigg{]}\mathrm{d}\lambda_{0}(w)$
$\displaystyle\xlongequal{\eqref{eqn: formula Bn R 0}}$ $\displaystyle
f(0)h(0)+\frac{1}{n(n+1)}\int_{\mathbb{B}_{n}}\mathcal{G}_{n}^{(0)}\phi(|w|^{2})\bar{R}^{2}\bigg{[}h(w)f(w)K_{w}(z)\bigg{]}\mathrm{d}\lambda_{1}(w)$
$\displaystyle=$ $\displaystyle
f(0)h(0)+\frac{1}{n(n+1)}\int_{\mathbb{B}_{n}}\mathcal{G}_{n}^{(0)}\phi(|w|^{2})h(w)\mathrm{d}\lambda_{1}(w)$
$\displaystyle\qquad\qquad\qquad\bigg{[}\bar{R}^{2}f(w)K_{w}(z)+2\bar{R}f(w)\bar{R}_{w}K_{w}(z)+f(w)\bar{R}_{w}^{2}K_{w}(z)\bigg{]}$
$\displaystyle=$ $\displaystyle
f(0)\big{(}E_{0}f\big{)}(z)+T^{(1,-1)}_{f}h(z)+X^{(-1)}_{f}h(z)+Y^{(-1)}_{\bar{R}f}h(z)+Z^{(-1)}_{\bar{R}^{2}f}h(z).$

This completes the proof of Lemma 8.6. ∎

In view of (8.3), Lemma 8.1 essentially says that the trace

$\mathrm{Tr}[T^{(t+1,t)}_{f_{1}},T^{(t+1,t)}_{f_{2}},\ldots,T^{(t+1,t)}_{f_{2n}}]$

if exist, is invariant under the perturbations of $X^{(t)}_{f_{i}}$,
$Y^{(t)}_{\bar{R}f_{i}}$ and $f_{i}(0)E_{0}$. Similarly, Lemma 8.2 can also be
interpreted as the stability of trace under certain perturbations.

###  8.1 Hypotheses A

In this subsection, let us assume the following.

Hypotheses A: Suppose $A_{1},A_{2},\ldots,A_{2n},B_{1},B_{2},\ldots,B_{2n}$
are bounded linear operators on a Hilbert space $\mathcal{H}$. Denote
$C_{i}=A_{i}+B_{i},i=1,\ldots,2n$. The operators satisfy the following
properties.

  1. (1)

For any $i=1,\ldots,2n$,

$B_{i}\in\mathcal{S}^{p},\quad\forall p>n.$

  2. (2)

For any $i,j=1,\ldots,2n$,

$[A_{i},A_{j}]\in\mathcal{S}^{p},\quad\forall p>n.$

  3. (3)

For any $i,j=1,\ldots,2n,$

$[A_{i},B_{j}]\in\mathcal{S}^{p},\quad\text{ for some }p<n.$

The goal of this subsection is to prove the following.

######  Proposition 8.9.

Assume Hypotheses A. Then the operator

$[C_{1},C_{2},\ldots,C_{2n}]-[A_{1},A_{2},\ldots,A_{2n}]$

is in the trace class. Moreover, the operator

$[C_{1},C_{2},\ldots,C_{2n}]-[A_{1},A_{2},\ldots,A_{2n}]-\sum_{k=1}^{2n}[A_{1},\ldots,A_{k-1},B_{k},A_{k+1},\ldots,A_{2n}]$

is a trace class operator with zero trace.

######  Lemma 8.10.

Suppose $\\{X_{1},X_{2},\ldots,X_{2n}\\}$ is a subset of
$\\{A_{1},\ldots,A_{2n},B_{1},\ldots,B_{2n}\\}$, and at least two of
$X_{1},X_{2},\ldots,X_{2n}$ are in $\\{B_{1},B_{2},\ldots,B_{2n}\\}$. Then the
operator

$[X_{1},X_{2},\ldots,X_{2n}]$

is in the trace class of zero trace.

###### Proof.

The lemma can be restated as follows:

“if $B\in\\{B_{1},B_{2},\ldots,B_{2n}\\}$,
$\\{X_{1},\ldots,X_{2n-1}\\}\subset\\{A_{1},\ldots,A_{2n},B_{1},\ldots,B_{2n}\\}$,
and at least one of $X_{1},\ldots,X_{2n-1}$ is in $\\{B_{1},\ldots,B_{2n}\\}$,
then the operator

$[B,X_{1},\ldots,X_{2n-1}]$

is in the trace class with zero trace.”

Under the above assumption on $B$ and $X_{1},\cdots,X_{2n-1}$, we compute the
antisymmetrization $[B,X_{1},\ldots,X_{2n-1}]$ as follows.

$\displaystyle[B,X_{1},\ldots,X_{2n-1}]$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\bigg{(}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)X_{\tau_{1}}\ldots
X_{\tau_{2k-2}}BX_{\tau_{2k-1}}\ldots X_{\tau_{2n-1}}-\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)X_{\tau_{1}}\ldots
X_{\tau_{2k-1}}BX_{\tau_{2k}}\ldots X_{\tau_{2n-1}}\bigg{)}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{k=1}^{n}\bigg{(}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)[X_{\tau_{1}},X_{\tau_{2}}]\ldots
X_{\tau_{2k-2}}BX_{\tau_{2k-1}}\ldots X_{\tau_{2n-1}}$
$\displaystyle\qquad\qquad\qquad-\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)[X_{\tau_{1}},X_{\tau_{2}}]\ldots
X_{\tau_{2k-1}}BX_{\tau_{2k}}\ldots X_{\tau_{2n-1}}\bigg{)}$
$\displaystyle\ldots$ $\displaystyle=$ $\displaystyle
2^{-n+1}\sum_{k=1}^{n}\bigg{(}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle~{}~{}~{}~{}-\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]\bigg{)}$
$\displaystyle=$ $\displaystyle 2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)\bigg{(}[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle~{}~{}~{}~{}-[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]\bigg{)}.$

In this proof, we use the following notation.

######  Notation 8.11.

For operators $S,T$, write $S\sim T$ when $S-T$ is a trace class operator with
zero trace.

For each $k=1,\ldots,n$ and $\tau\in S_{2n-1}$, we claim that

$\displaystyle[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle\sim$ $\displaystyle
BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}][X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}],$

and

$\displaystyle[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle\sim$ $\displaystyle
B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}][X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}.$

If $X_{\tau_{2k-1}}\in\\{B_{1},B_{2},\ldots,B_{2n}\\}$, then we have

$B,X_{\tau_{2k-1}}\in\mathcal{S}^{p},\forall
p>n,\quad[X_{\tau_{i}},X_{\tau_{j}}]\in\mathcal{S}^{p},\forall p>n,\forall
i,j=1,\ldots,2n-1.$

By Remark 2.9 it is easy to see that the claim holds. If
$X_{\tau_{2k-1}}\in\\{A_{1},\ldots,A_{2n}\\}$, then at least one of
$X_{\tau_{1}},\ldots,X_{\tau_{2k-2}},X_{\tau_{2k}},\ldots,X_{\tau_{2n-1}}$ is
in $\\{B_{1},B_{2},\ldots,B_{2n}\\}$. Thus at least one of the commutators

$[X_{\tau_{1}},X_{\tau_{2}}],\ldots,[X_{\tau_{2k-3}},X_{\tau_{2k-2}}],[X_{\tau_{2k}},X_{\tau_{2k+1}}],\ldots,[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$

is in $\mathcal{S}^{p}$ for some $p<n$. Again, by Remark 2.9, the claim also
follows. Thus in both cases the claim holds. By the claim, we compute
$[B,X_{1},\ldots,X_{2n-1}]$.

$\displaystyle[B,X_{1},\ldots,X_{2n-1}]$ $\displaystyle=$ $\displaystyle
2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)\bigg{(}[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle~{}~{}~{}~{}-[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]\bigg{)}$
$\displaystyle\sim$ $\displaystyle 2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)BX_{\tau_{2k-1}}[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}][X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]$
$\displaystyle-2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)B[X_{\tau_{2k}},X_{\tau_{2k+1}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}][X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2k-3}},X_{\tau_{2k-2}}]X_{\tau_{2k-1}}$
$\displaystyle=$ $\displaystyle 2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)BX_{\tau_{1}}[X_{\tau_{2}},X_{\tau_{3}}]\ldots[X_{\tau_{2n-2}},X_{\tau_{2n-1}}]$
$\displaystyle-2^{-n+1}\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)B[X_{\tau_{1}},X_{\tau_{2}}]\ldots[X_{\tau_{2n-3}},X_{\tau_{2n-2}}]X_{\tau_{2n-1}}$
$\displaystyle=$ $\displaystyle\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)BX_{\tau_{1}}\ldots
X_{\tau_{2n-1}}-\sum_{k=1}^{n}\sum_{\tau\in
S_{2n-1}}\mathrm{sgn}(\tau)BX_{\tau_{1}}\ldots X_{\tau_{2n-1}}$
$\displaystyle=$ $\displaystyle 0.$

Here the third-to-last equality is because the antisymmetric sums
$\sum_{\tau\in S_{2n-1}}\mathrm{sgn}(\tau)\cdots$ are invariant under any even
permutation. This completes the proof of Lemma 8.10. ∎

A verbatim repetition of the proof of Lemma 8.10 also proves the following.

######  Corollary 8.12.

Assume in addition to Hypotheses A that

$B_{i}\in\mathcal{S}^{p},\quad\text{ for some}~{}p<n.$

Suppose
$\\{X_{1},X_{2},\ldots,X_{2n}\\}\subset\\{A_{1},\ldots,A_{2n},B_{1},\ldots,B_{2n}\\}$,
and at least one of $X_{1},X_{2},\ldots,X_{2n}\in\\{B_{1},\ldots,B_{2n}\\}$.
Then the operator

$[X_{1},X_{2},\ldots,X_{2n}]$

is in the trace class with zero trace. Consequently, the operator

$[C_{1},C_{2},\ldots,C_{2n}]-[A_{1},A_{2},\ldots,A_{2n}]$

is in the trace class with zero trace.

######  Remark 8.13.

Notice that the assumption of Corollary 8.12 is equivalent to the following.

  1. 1.

For $i,j=1,\ldots,2n$,

$[A_{i},A_{j}]\in\mathcal{S}^{p},\quad\forall p>n.$

  2. 2.

For $i=1,\ldots,2n$,

$B_{i}\in\mathcal{S}^{p},\quad\text{ for some }p<n.$

Thus Corollary 8.12 essentially says that the trace