# Toeplitz algebra and Symbol map via Berezin transform on
$H^{2}(\mathbb{D}^{n})$

Mo Javed  Indian Institute of Technology Roorkee, Department of Mathematics,
Roorkee-247 667, Uttarakhand, India<EMAIL_ADDRESS>and  Amit Maji
Indian Institute of Technology Roorkee, Department of Mathematics, Roorkee-247
667, Uttarakhand, India<EMAIL_ADDRESS><EMAIL_ADDRESS>(†Corresponding author)

###### Abstract.

Let $\mathscr{T}(L^{\infty}(\mathbb{T}))$ be the Toeplitz algebra, that is,
the $C^{*}$-algebra generated by the set $\\{T_{\varphi}:\varphi\in
L^{\infty}(\mathbb{T})\\}$. Douglas’s theorem on _symbol map_ states that
there exists a $C^{*}$-algebra homomorphism from
$\mathscr{T}(L^{\infty}(\mathbb{T}))$ onto $L^{\infty}(\mathbb{T})$ such that
$T_{\varphi}\mapsto\varphi$ and the kernel of the homomorphism coincides with
commutator ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}))$. In this paper, we
use the Berezin transform to study results akin to Douglas’s theorem for
operators on the Hardy space $H^{2}(\mathbb{D}^{n})$ over the open unit
polydisc $\mathbb{D}^{n}$ for $n\geq 1$. We further obtain a class of bigger
$C^{*}$-algebras than the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ for which the analog of _symbol map_
still holds true.

###### Key words and phrases:

Hardy space, Toeplitz operator, Hankel operator, Toeplitz algebra, Symbol map,
Berezin transform

###### 2010 Mathematics Subject Classification:

47B35, 47A13, 32A65

##  1\. Introduction

One of the major concerns of operator theory and function theory has generally
been the study of operators associated with the spaces of holomorphic
functions. We often come across linear operators induced by functions (usually
called the symbol of the operator), such as Toeplitz operators, Hankel
operators, and composition operators; on the other hand, we can investigate
functions that are induced by an operator. A great illustration of the
relationship between operators and functions is the Berezin transform.

The Berezin transform, which was first introduced in [3] as a tool in
quantization, is named after F. A. Berezin. Usually, it involves an open
subset $\mathcal{G}$ of $\mathbb{C}^{n}$ ($n\geq 1$) and a reproducing kernel
Hilbert space ${H}(\mathcal{G})$ with a reproducing kernel, say, $K_{\bm{z}}$
for $\bm{z}\in\mathcal{G}$. Then, for a bounded operator $A$ on
${H}(\mathcal{G})$, its Berezin transform, denoted as $\widetilde{A}$, is a
complex-valued function on $\mathcal{G}$ defined by

$\widetilde{A}(\bm{z})=\langle
Ak_{\bm{z}},k_{\bm{z}}\rangle_{{H}(\mathcal{G})}\quad(\bm{z}\in\mathcal{G}),$

where $k_{\bm{z}}={K_{\bm{z}}}/{\|K_{\bm{z}}\|_{{H}(\mathcal{G})}}$ is the
normalized reproducing kernel of ${H}(\mathcal{G})$. Thus every bounded linear
operator $A$ on ${H}(\mathcal{G})$ induces a bounded function $\widetilde{A}$
on $\mathcal{G}$. Indeed, $\|\widetilde{A}\|_{\infty}\leq\|A\|$. Therefore,
the Berezin transform of an operator yields a crucial information about the
operator. Beyond its original purpose, the Berezin transform has shown
usefulness in a number of contexts, including the Hardy space, the Bergman
space ([17], [1]), the Bargmann-Segal space [5]. It has also established
important relationships with the Bloch space and functions of bounded mean
oscillation, and the detailed investigation of these connections can be found
in [18]. It has also found many applications in many areas of mathematical
analysis and mathematical physics (see [4], [13], [19]).

Toeplitz operators emerge to be one of the most distinguishable classes of
concrete operators because of their immensely useful and widespread
applications including operator theory, function theory, operator algebras,
and associated disciplines. Furthermore, it is preferable to think of these
operators as elements within operator algebras, like the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}))$), that is, the $C^{*}$-algebra generated
by the set $\\{T_{\varphi}:\varphi\in L^{\infty}(\mathbb{T})\\}$ of Toeplitz
operators on the Hardy space $H^{2}(\mathbb{D})$ (see [9], [11]). The _symbol
map_ , which was initially discovered and utilized by Douglas, proves to be an
important tool in the study of the Toeplitz algebra. The classical result of
Douglas’s _symbol map_ [11, chapter 7] is as follows:

######  Theorem 1.1.

Let $\mathscr{T}(L^{\infty}(\mathbb{T}))$ denote the $C^{*}$-algebra generated
by the set $\\{T_{\varphi}:\varphi\in L^{\infty}(\mathbb{T})\\}$. Then there
exists a $C^{*}$-homomorphism

$\sigma:\mathscr{T}(L^{\infty}(\mathbb{T}))\to L^{\infty}(\mathbb{T})$

such that kernel of $\sigma$ is the commutator ideal, that is, the ideal in
$\mathscr{T}(L^{\infty}(\mathbb{T}))$ generated by the commutators $[A,B]=AB-
BA$; $A,B\in\mathscr{T}(L^{\infty}(\mathbb{T}))$, and
$\sigma(T_{\varphi})=\varphi$ for all $\varphi\in L^{\infty}(\mathbb{T})$.

The $C^{*}$-homomorphism $\sigma$ is called the _symbol map_. However,
Douglas’s proof of Theorem 1.1 was mere existential and much abstract in
nature until 1982, when Barría and Halmos in [2] presented a constructive
proof of Theorem 1.1. More specifically, Barría and Halmos obtained the symbol
of an operator $A\in\mathscr{T}(L^{\infty}(\mathbb{T}))$ as the symbol of the
Toeplitz operator in the SOT-limit of the sequence
$\\{T_{z}^{*m}AT_{z}^{m}\\}_{m=1}^{\infty}$, where $T_{z}$ denotes the
unilateral shift on the Hardy space $H^{2}(\mathbb{D})$. However, due to the
nuances of limits in the strong operator topology, it is still very difficult
to find the symbol of any operator $A$ in
$\mathscr{T}(L^{\infty}(\mathbb{T}))$. The Berezin transform is a useful tool
in this way to obtain the symbol of an operator
$A\in\mathscr{T}(L^{\infty}(\mathbb{T}))$ more concretely. Indeed, the Berezin
transform of a Toeplitz operator on Hardy space reduces to the Poisson
integral of the symbol of that Toeplitz operator (see Theorem 3.2 below). In
particular, Engliš [12] developed a technique using the Berezin transform in
which he obtained the symbol of an operator
$A\in\mathscr{T}(L^{\infty}(\mathbb{T}))$ as the radial boundary function of
the Berezin transform of $A$. The key observation in his proof is that

$T_{\varphi}K_{\xi}=\varphi(\xi)K_{\xi},$

for the densely defined Toeplitz operator $T_{\varphi}$ with symbol
$\varphi\in H^{2}(\mathbb{D})^{\perp}$ and $K_{\xi}$ is the reproducing kernel
of $H^{2}(\mathbb{D})$. It uses the fact that $\overline{\varphi}\in
H^{2}(\mathbb{D})$ whenever $\varphi\in H^{2}(\mathbb{D})^{\perp}$. However,
this technique fails in the case $\mathbb{D}^{n}$ for $n>1$, for an example:
if we take $\varphi(z_{1},z_{2})=\overline{z_{1}}z_{2}$ then both $\varphi$
and $\overline{\varphi}$ are in $H^{2}(\mathbb{D}^{2})^{\perp}$, and
$T_{\varphi}K_{\bm{\xi}}\neq\varphi(\bm{\xi})K_{\bm{\xi}}$ for
$\bm{\xi}\in\mathbb{D}^{2}$, where
$K_{\bm{\xi}}(\bm{z})=\dfrac{1}{(1-\overline{\xi_{1}}z_{1})(1-\overline{\xi_{2}}z_{2})}$
is a reproducing kernel of $H^{2}(\mathbb{D}^{2})$. Thus the natural question
arises:

Question: Is the Douglas’s symbol map valid in the setting of the open unit
polydisc $\mathbb{D}^{n}$ for $n>1$?

In this paper we present a comprehensive study of symbols of operators on
$H^{2}(\mathbb{D}^{n})$. The main contribution of this article is to address
the above question with the help of the Berezin transform. More precisely, we
obtain the following result (see Theorem 4.3 below):

######  Theorem.

Let $T\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, the Toeplitz algebra of
operators on $H^{2}(\mathbb{D}^{n})$ for $n\geq 1$. Then
$\widetilde{T}\to\varphi$ radially for some $\varphi\in
L^{\infty}(\mathbb{T}^{n})$. Moreover, the map

$\Sigma:\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))\to
L^{\infty}(\mathbb{T}^{n}),\qquad T\mapsto\varphi$

is a $C^{*}$-algebra homomorphism with ker$(\Sigma)=\mathscr{S}$, the _semi-
commutator ideal_ in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ and
$\Sigma(T_{\varphi})=\varphi$ for any Toeplitz operator $T_{\varphi}$.

The rest of the paper is outlined as follows: Section 2 is devoted to
establishing the necessary background, and along the way, we fix some
notations and state some properties of Poison integral and Berezin transform.
In Section 3, we prove an important result concerning the radial limits, which
serves as a key ingredient in the proof of the _symbol map_. Section 4
concerns with the main result, that is, the _symbol map_ (see Theorem 4.3),
and we depict through an example that Theorem 4.3 can not be extended to the
whole of $\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$, the $C^{*}$-algebra of all
bounded operators on $H^{2}(\mathbb{D}^{n})$. In section 5, we present a
family of larger $C^{*}$-algebras than the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ for which the analog of symbol map
still holds true. In section 6, we conclude with some open questions.

##  2\. Preliminaries

Let $\mathbb{Z}_{+}$ denote the set of all non-negative integers,
$\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$ be the open unit disc in the complex
plane $\mathbb{C}$, $\mathbb{T}=\\{z\in\mathbb{C}:|z|=1\\}$ the unit circle in
$\mathbb{C}$, and $\mathfrak{B}(\mathcal{H})$ denote the set of all bounded
linear operators on a Hilbert space $\mathcal{H}$. For $n\geq 1$, the
Cartesian product of $n$ copies of $\mathbb{D}$, denoted by $\mathbb{D}^{n}$,
is the unit polydisc in $\mathbb{C}^{n}$ and the $n$-torus $\mathbb{T}^{n}$
denotes the distinguished boundary of $\mathbb{D}^{n}$. Then
$H^{2}(\mathbb{D}^{n})$ denotes as the Hilbert space of all analytic functions
$f$ on $\mathbb{D}^{n}$ for which

$\|f\|_{H^{2}(\mathbb{D}^{n})}:=\underset{0\leq
r<1}{\text{sup}}\left(\int_{\mathbb{T}^{n}}\left|f(r\bm{\zeta})\right|^{2}\,d\mathfrak{m}_{n}(\bm{\zeta})\right)^{1/2}<\infty,$

where $\mathfrak{m}_{n}$ stands for the normalised Lebesgue measure on
$\mathbb{T}^{n}$. We shall write bold faced $\bm{\xi}$ and $\bm{\zeta}$,
respectively as vectors $\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$
and $\bm{\zeta}=(\zeta_{1},\dots,\zeta_{n})\in\mathbb{T}^{n}$ for $n>1$. For
$1\leq p<\infty$,
$L^{p}(\mathbb{T}^{n}):=L^{p}(\mathbb{T}^{n},d\mathfrak{m}_{n})$ is the Banach
space of equivalence classes of Lebesgue complex measurable functions $f$ such
that $|f|^{p}$ is integrable on $\mathbb{T}^{n}$ with respect to
$\mathfrak{m}_{n}$ and $p=2$ gives the Hilbert space $L^{2}(\mathbb{T}^{n})$
of square-integrable equivalence classes of functions with respect to
$\mathfrak{m}_{n}$. Also $L^{\infty}(\mathbb{T}^{n})$ denotes the commutative
$C^{*}$-algebra of essentially bounded complex measurable functions on
$\mathbb{T}^{n}$ with respect to $\mathfrak{m}_{n}$, and
$H^{\infty}(\mathbb{D}^{n})$ stands for the space of bounded holomorphic
functions on $\mathbb{D}^{n}$. Let
$\\{e_{\bm{k}}\\}_{\bm{k}\in\mathbb{Z}^{n}}$ denote the standard orthonormal
basis for $L^{2}(\mathbb{T}^{n})$, where
$e_{\bm{k}}(\bm{\zeta})=\zeta_{1}^{k_{1}}\dots\zeta_{n}^{k_{n}}$ for
$\bm{k}=(k_{1},\dots,k_{n})\in\mathbb{Z}^{n}$. Consider the closed subspace
$H^{2}(\mathbb{T}^{n})$ of $L^{2}(\mathbb{T}^{n})$ defined by

$H^{2}(\mathbb{T}^{n}):=\\{f\in
L^{2}(\mathbb{T}^{n}):{\langle}f,e_{\bm{k}}{\rangle}_{L^{2}(\mathbb{T}^{n})}=0\;\mbox{for
all~{}}\bm{k}\in\mathbb{Z}^{n}\setminus\mathbb{Z}_{+}^{n}\\}.$

We then identify the Hardy space $H^{2}(\mathbb{D}^{n})$ (via radial limits)
with $H^{2}(\mathbb{T}^{n})$ by the unitary map

$H^{2}(\mathbb{D}^{n})\ni\widetilde{f}=\sum_{\bm{k}\in\mathbb{Z}_{+}^{n}}a_{\bm{k}}\bm{z}^{\bm{k}}\mapsto\sum_{\bm{k}\in\mathbb{Z}_{+}^{n}}a_{\bm{k}}e_{\bm{k}}=f\in
H^{2}(\mathbb{T}^{n}).$

Now onwards, without any ambiguity, we shall interchangeably write
$H^{2}(\mathbb{D}^{n})$ (respectively, its elements $\widetilde{f}$) in place
of $H^{2}(\mathbb{T}^{n})$ (respectively, its elements $f$) and vice-versa.
Furthermore, for $\widetilde{f},\widetilde{g}\in H^{2}(\mathbb{D}^{n})$ and
$f,g\in H^{2}(\mathbb{T}^{n})$, we have

$\langle\widetilde{f},\widetilde{g}\rangle_{H^{2}(\mathbb{D}^{n})}=\langle
f,g\rangle_{L^{2}(\mathbb{T}^{n})}.$

For $\varphi\in L^{\infty}(\mathbb{T}^{n})$, the Toeplitz operator
$T_{\varphi}$ is a bounded operator on $H^{2}(\mathbb{D}^{n})$ given by

$T_{\varphi}f=P_{H^{2}(\mathbb{D}^{n})}(\varphi f)\qquad(f\in
H^{2}(\mathbb{D}^{n})),$

and the Hankel operator $H_{\varphi}$ with symbol $\varphi\in
L^{\infty}(\mathbb{T}^{n})$ is a bounded operator from $H^{2}(\mathbb{D}^{n})$
to ${H^{2}(\mathbb{D}^{n})^{\perp}}$ defined by

$H_{\varphi}f=P_{H^{2}(\mathbb{D}^{n})^{\perp}}(\varphi f)\qquad(f\in
H^{2}(\mathbb{D}^{n})),$

where $P_{H^{2}(\mathbb{D}^{n})}$ is the orthogonal projection of
$L^{2}(\mathbb{T}^{n})$ onto $H^{2}(\mathbb{D}^{n})$ and
$P_{H^{2}(\mathbb{D}^{n})^{\perp}}=I_{L^{2}(\mathbb{T}^{n})}-P_{H^{2}(\mathbb{D}^{n})}$.
Clearly, the adjoint operator $T_{\varphi}^{*}=T_{\bar{\varphi}}$, where
${\bar{\varphi}}$ is the complex conjugate of $\varphi$. It is easy to check
that $[T_{\varphi},T_{\psi})=H_{\overline{\varphi}}^{*}H_{\psi}$ for
$\varphi,\psi\in L^{\infty}(\mathbb{T}^{n})$.

We frequently use the identification of the Hardy space over the polydisc
$H^{2}(\mathbb{D}^{n})$ with $H^{2}(\mathbb{D})\otimes\cdots\otimes
H^{2}(\mathbb{D})$, the $n$-fold ($n>1$) tensor product of Hilbert space
$H^{2}(\mathbb{D})$ via the canonical unitary map $z_{1}^{m_{1}}\dots
z_{n}^{m_{n}}\mapsto z^{m_{1}}\otimes\cdots\otimes z^{m_{n}}$, where
$m_{i}\in\mathbb{Z}_{+}$. Thus, we shall often write
$(f_{1}\otimes\cdots\otimes f_{n})=f_{1}\cdots f_{n}$, where $f_{j}\in
H^{2}(\mathbb{D})$. The Hardy space over the polydisc $H^{2}(\mathbb{D}^{n})$
is a reproducing kernel Hilbert space on $\mathbb{D}^{n}$ for $n\geq 1$. For
$\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$, the (Szegö) kernel
$K_{\bm{\xi}}$ of $H^{2}(\mathbb{D}^{n})$ is defined as

$K_{\bm{\xi}}(\bm{z})=\prod_{j=1}^{n}\frac{1}{(1-\overline{\xi_{j}}z_{j})},\qquad(\bm{z}\in\mathbb{D}^{n})$

and it reproduces the value of each $f\in H^{2}(\mathbb{D}^{n})$, that means
$f(\bm{\xi})=\langle f,K_{\bm{\xi}}\rangle$. The normalised reproducing kernel
is denoted by

$k_{\bm{\xi}}(\bm{z})=\frac{K_{\bm{\xi}}(\bm{z})}{\|K_{\bm{\xi}}\|}=\frac{\sqrt{(1-|\xi_{1}|^{2})\cdots(1-|\xi_{n}|^{2})}}{(1-\overline{\xi_{1}}z_{1})\cdots(1-\overline{\xi_{n}}z_{n})},\qquad(\bm{z}\in\mathbb{D}^{n}).$

In particular, the reproducing kernel $K_{\xi}$ ($\xi\in\mathbb{D}$) of the
Hardy space over the disc $H^{2}(\mathbb{D})$ is given by

$K_{\xi}(z)=\frac{1}{1-\overline{\xi}z}\quad(z\in\mathbb{D}),$

and $k_{\xi}(z)=\frac{\sqrt{1-|\xi|^{2}}}{1-\overline{\xi}z}$ is the
normalised reproducing kernel of $H^{2}(\mathbb{D})$.

Let
$\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n},\bm{\zeta}=(\zeta_{1},\dots,\zeta_{n})\in\mathbb{T}^{n}$,
$\xi_{j}=r_{j}e^{i\theta_{j}}$ and $\zeta_{j}=e^{it_{j}}$ where $0\leq
r_{j}<1$ and $\theta_{j},t_{j}\in[0,2\pi)$ for all $1\leq j\leq n$. The
Poisson kernel $P(\bm{\xi},\bm{\zeta})$ for the unit polydisc $\mathbb{D}^{n}$
is defined as

$\displaystyle P(\bm{\xi},\bm{\zeta})$
$\displaystyle=\prod_{j=1}^{n}\frac{1-|\xi_{j}|^{2}}{|1-\xi_{j}\overline{\zeta_{j}}|^{2}}$
$\displaystyle=P_{r_{1}}(\theta_{1}-t_{1})\cdots P_{r_{n}}(\theta_{n}-t_{n}),$

where $P_{r}(\theta)=\frac{1-r^{2}}{1-2r\cos\theta+r^{2}}$ is the Poisson
kernel for the unit disc $\mathbb{D}$. For a complex Borel measure $\mu$ on
$\mathbb{T}^{n}$, its Poisson integral, denoted by $P[d\mu]$, is the integral
given by

$P[d\mu](\bm{\xi})=\int_{\mathbb{T}^{n}}P(\bm{\xi},\bm{\zeta})\,d\mu(\bm{\zeta})\qquad(\bm{\xi}\in\mathbb{D}^{n}).$

For $\varphi\in L^{1}(\mathbb{T}^{n})$, its harmonic extension
$\widetilde{\varphi}$ into $\mathbb{D}^{n}$ is defined by the Poisson integral
formula

$\widetilde{\varphi}(\bm{\xi})=P[\varphi
d\mathfrak{m}_{n}](\bm{\xi})=\int_{\mathbb{T}^{n}}P(\bm{\xi},\bm{\zeta})\varphi(\bm{\zeta})\,d\mathfrak{m}_{n}(\bm{\zeta})\qquad(\bm{\xi}\in\mathbb{D}^{n}).$

Let $\Phi:\mathbb{D}^{n}\to\mathbb{C}$ and
$\varphi:\mathbb{T}^{n}\to\mathbb{C}$ be any two complex functions. Define
$\hat{\Phi}$ as

$\hat{\Phi}(\bm{\zeta})=\lim_{r\to 1^{-}}\Phi(r\bm{\zeta})$

for all $\bm{\zeta}\in\mathbb{T}^{n}$ at which the (radial) limit on the right
exists. We say that ‘$\Phi$ approaches $\varphi$ radially’, denoted by
‘$\Phi\to\varphi$ _radially_ ’ or ‘$\Phi(\bm{\xi})\to\varphi$ _radially_ ’ if

$\hat{\Phi}(\bm{\zeta})=\varphi(\bm{\zeta})\quad\text{for a.e.~{}
}\bm{\zeta}\in\mathbb{T}^{n}.$

######  Theorem 2.1.

(See Theorem 2.3.1 in [16]) Let $\varphi\in L^{1}(\mathbb{T}^{n})$, and $\nu$
be a measure on $\mathbb{T}^{n}$ such that $\nu$ is singular with respect to
$\mathfrak{m}_{n}$. If $\Phi=P[\varphi d\mathfrak{m}_{n}+d\nu]$, then

$\Phi\to\varphi\;\text{ radially }.$

In particular, if $\nu=0$, then $\widetilde{\varphi}\to\varphi$ radially.

The Berezin transform of $T\in\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$ is a
complex function on $\mathbb{D}^{n}$ defined by

$\widetilde{T}(\bm{\xi})=\langle Tk_{\bm{\xi}},k_{\bm{\xi}}\rangle\quad\text{
for all }\bm{\xi}\in\mathbb{D}^{n}.$

Here $k_{\bm{\xi}}$ is the normalised reproducing kernel of
$H^{2}(\mathbb{D}^{n})$.

The following notions are important and used throughout the paper:

  1. (1)

For $\varphi$, $\psi\in L^{\infty}(\mathbb{T}^{n})$, the _semicommutator_ is
defined as

$[T_{\varphi},T_{\psi})=T_{\varphi\psi}-T_{\varphi}T_{\psi}.$

  2. (2)

$\mathscr{T}(L^{\infty}(\mathbb{T}^{n})):=C^{*}$-algebra generated by the set
$\\{T_{\varphi}:\varphi\in L^{\infty}(\mathbb{T}^{n})\\}$, called the Toeplitz
algebra of operators on $H^{2}(\mathbb{D}^{n})$.

  3. (3)

$\mathscr{C}:=$ closed ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$
generated by the set $\\{[A,B]=AB-
BA:A,B\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))\\},$ called the _commutator
ideal_.

  4. (4)

$\mathscr{S}:=$ closed ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$
generated by the set $\\{[T_{\varphi},T_{\psi}):\varphi,\psi\in
L^{\infty}(\mathbb{T}^{n})\\},$ called the _semicommutator ideal_.

##  3\. On Radial Limits

In this section we shall discuss about some key results on radial limits which
will be used in the next section. Before moving into our main result, we will
fix some notations first.

Suppose $H^{2}(\mathbb{D})^{1}:=H^{2}(\mathbb{D})$ and
$H^{2}(\mathbb{D})^{-1}:=H^{2}(\mathbb{D})^{\perp}$. For
$\alpha=(\alpha_{1},\dots,\alpha_{n})\in\\{-1,1\\}^{n}$ and
$\bm{k}=(k_{1},\dots,k_{n})\in\mathbb{Z}^{n}$, we write

$H^{2}(\mathbb{D}^{n})^{\alpha}:=\bigvee\\{e_{\bm{k}}:k_{j}<0\text{ if
}\alpha_{j}=-1\text{ and }k_{j}\geq 0\text{ if }\alpha_{j}=1\text{ for }1\leq
j\leq n\\}.$

It is easy to check that $H^{2}(\mathbb{D}^{n})^{\alpha}$ is a closed subspace
of $L^{2}(\mathbb{T}^{n})$ for all $\alpha\in\\{-1,1\\}^{n}$. We identify
$H^{2}(\mathbb{D}^{n})^{\alpha}$ with the $n$-fold tensor product of Hibert
spaces $H^{2}(\mathbb{D})^{\alpha_{j}}$ for $1\leq j\leq n$ by the unitary map
$e_{\bm{k}}\mapsto e_{k_{1}}\otimes\cdots\otimes e_{k_{n}}$ for all $k_{j}<0$
(provided $\alpha_{j}=-1$) and $k_{j}\geq 0$ (provided $\alpha_{j}=1$), where

$(e_{k_{1}}\otimes\cdots\otimes
e_{k_{n}})(\bm{\zeta})=\zeta_{1}^{k_{1}}\dots\zeta_{n}^{k_{n}},$

for all $\bm{\zeta}\in\mathbb{T}^{n}$, $k_{j}<0$ (provided $\alpha_{j}=-1$)
and $k_{j}\geq 0$ (provided $\alpha_{j}=1$). We shall simply write
$\otimes_{j=1}^{n}f_{j}=\prod_{j=1}^{n}f_{j}$ for $f_{j}\in
H^{2}(\mathbb{D})^{\alpha_{j}}$, and also write bold faced
$\bm{1}=(1,\dots,1)\in\\{-1,1\\}^{n}$.

The following lemma is an easy consequence of the Poisson integral:

######  Lemma 3.1.

Let $f,g\in L^{2}(\mathbb{T}^{n})$. Then

${\langle}fk_{\bm{\xi}},gk_{\bm{\xi}}{\rangle}_{L^{2}(\mathbb{T}^{n})}=(\widetilde{f\overline{g}})(\bm{\xi}),\qquad(\bm{\xi}\in\mathbb{D}^{n}).$

Moreover,
${\langle}fk_{\bm{\xi}},gk_{\bm{\xi}}{\rangle}_{L^{2}(\mathbb{T}^{n})}\rightarrow
f\overline{g}$ radially.

###### Proof.

Suppose $f,g\in L^{2}(\mathbb{T}^{n})$. Now

$\displaystyle{\langle}fk_{\bm{\xi}},gk_{\bm{\xi}}{\rangle}_{L^{2}(\mathbb{T}^{n})}$
$\displaystyle=\int_{\mathbb{T}^{n}}(fk_{\bm{\xi}})(\bm{\zeta})(\overline{gk_{\bm{\xi}}})(\bm{\zeta})\,d\mathfrak{m}_{n}(\bm{\zeta})$
(3.1)
$\displaystyle=\int_{\mathbb{T}^{n}}(f\overline{g})(\bm{\zeta})|k_{\bm{\xi}}(\bm{\zeta})|^{2}\,d\mathfrak{m}_{n}(\bm{\zeta}).$

Note that

$|k_{\bm{\xi}}(\bm{\zeta})|^{2}=\frac{(1-|\xi_{1}|^{2})\cdots(1-|\xi_{n}|^{2})}{\left|1-\overline{\xi_{1}}\zeta_{1}\right|^{2}\cdots\left|1-\overline{\xi_{n}}\zeta_{n}\right|^{2}}=P(\bm{\xi},\bm{\zeta}),$

where $\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$ and
$\bm{\zeta}=(\zeta_{1},\dots,\zeta_{n})\in\mathbb{T}^{n}$. Therefore, the
equation (3.1) yields

${\langle}fk_{\bm{\xi}},gk_{\bm{\xi}}{\rangle}_{L^{2}(\mathbb{T}^{n})}=\widetilde{(f\overline{g})}(\bm{\xi}).$

Since $f,g\in L^{2}(\mathbb{T}^{n})$, $f\overline{g}\in
L^{1}(\mathbb{T}^{n})$. Hence, Theorem 2.1 infers that $\langle
fk_{\bm{\xi}},gk_{\bm{\xi}}\rangle_{L^{2}(\mathbb{T}^{n})}\to f\overline{g}$
radially.

The following theorem establishes the fact that the Berezin transform of
$T_{\varphi}$ is the harmonic extension of $\varphi$:

######  Theorem 3.2.

For $\varphi\in L^{\infty}(\mathbb{T}^{n})$,
$\widetilde{T_{\varphi}}(\bm{\xi})=\widetilde{\varphi}(\bm{\xi})$ for all
$\bm{\xi}\in\mathbb{D}^{n}$. Furthermore, $\widetilde{T_{\varphi}}\to\varphi$
radially.

###### Proof.

Let $\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$. Then

$\displaystyle\widetilde{T_{\varphi}}(\bm{\xi})=\langle
T_{\varphi}k_{\bm{\xi}},k_{\bm{\xi}}\rangle=\langle
P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}}),k_{\bm{\xi}}\rangle=\langle\varphi
k_{\bm{\xi}},k_{\bm{\xi}}\rangle_{L^{2}(\mathbb{T}^{n})}=\widetilde{\varphi}(\bm{\xi}),$

where the last equality follows from Lemma 3.1.

Now Theorem 2.1 gives $\widetilde{T_{\varphi}}\to\varphi$ radially.

######  Lemma 3.3.

If $x\in H^{2}(\mathbb{D})^{\perp}$, then
$P_{H^{2}(\mathbb{D})}(xk_{\xi})=\widetilde{x}(\xi)k_{\xi}$ for all
$\xi\in\mathbb{D}$.

###### Proof.

Let $z\in\mathbb{D}$. Then

$\displaystyle\langle
P_{H^{2}(\mathbb{D})}(xK_{\xi}),K_{z}\rangle_{H^{2}(\mathbb{D})}$
$\displaystyle=\langle xK_{\xi},K_{z}\rangle_{L^{2}(\mathbb{T})}$
$\displaystyle=\langle K_{\xi},\overline{x}K_{z}\rangle_{H^{2}(\mathbb{D})}.$

Since $x\in H^{2}(\mathbb{D})^{\perp}$, $\overline{x}\in H^{2}(\mathbb{D})$,
and hence $\overline{x}K_{z}\in H^{2}(\mathbb{D})$. Thus, by reproducing
property

$\displaystyle\langle
P_{H^{2}(\mathbb{D})}(xK_{\xi}),K_{z}\rangle_{H^{2}(\mathbb{D})}$
$\displaystyle=\overline{\widetilde{\overline{x}}(\xi)K_{z}(\xi)}$
$\displaystyle=\widetilde{x}(\xi)\langle
K_{\xi},K_{z}\rangle_{H^{2}(\mathbb{D})}$
$\displaystyle=\langle\widetilde{x}(\xi)K_{\xi},K_{z}\rangle_{H^{2}(\mathbb{D})}\qquad(\text{for
all }z\in\mathbb{D}).$

Therefore, $P_{H^{2}(\mathbb{D})}(xk_{\xi})=\widetilde{x}(\xi)k_{\xi}$.

We now state the main result of this section.

######  Theorem 3.4.

Let $f\in H^{2}(\mathbb{D}^{n})^{\alpha}$ and $g\in
H^{2}(\mathbb{D}^{n})^{\beta}$ for
$\alpha=(\alpha_{1},\dots,\alpha_{n}),\beta=(\beta_{1},\dots,\beta_{n})\in\\{-1,1\\}^{n}$.
Then

$\left\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\right\rangle\to
f\overline{g}\;\text{ radially},$

that is,

$\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(gk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle\to
f\left(e^{i\theta_{1}},\dots,e^{i\theta_{n}}\right)\overline{g}\left(e^{i\theta_{1}},\dots,e^{i\theta_{n}}\right)$

as $r\to 1^{-}$ a.e.
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}$.

###### Proof.

Suppose that $f\in H^{2}(\mathbb{D}^{n})^{\alpha}$ and $g\in
H^{2}(\mathbb{D}^{n})^{\beta}$ for
$\alpha=(\alpha_{1},\dots,\alpha_{n}),\beta=(\beta_{1},\dots,\beta_{n})\in\\{-1,1\\}^{n}$.
If either $\alpha=\bm{1}$ or $\beta=\bm{1}$, then

$\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle=\langle
fk_{\bm{\xi}},gk_{\bm{\xi}}\rangle,$

and hence $\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle\to
f\overline{g}$ radially by using Lemma 3.1.

We now assume that $\alpha\neq\bm{1}$ and $\beta\neq\bm{1}$. Firstly, consider

$I=\\{j:\alpha_{j}=-1\\}\quad\mbox{and}\quad J=\\{j:\beta_{j}=-1\\}.$

Let $f=f_{1}\otimes\cdots\otimes f_{n}$ and $g=g_{1}\otimes\cdots\otimes
g_{n}$, where $f_{j}\in H^{2}(\mathbb{D})^{\alpha_{j}}$ and $g_{j}\in
H^{2}(\mathbb{D})^{\beta_{j}}$ for all $1\leq j\leq n$. Then

$\displaystyle\langle
P_{H^{2}(\mathbb{D}^{n})}(fK_{\bm{\xi}}),K_{\bm{z}}\rangle$
$\displaystyle=\langle
fK_{\bm{\xi}},K_{\bm{z}}\rangle_{L^{2}(\mathbb{T}^{n})}$
$\displaystyle=\langle(f_{1}\otimes\cdots\otimes
f_{n})(K_{\xi_{1}}\otimes\dots\otimes
K_{\xi_{n}}),K_{z_{1}}\otimes\cdots\otimes
K_{z_{n}}\rangle_{L^{2}(\mathbb{T}^{n})}$ $\displaystyle=\langle
f_{1}K_{\xi_{1}}\otimes\cdots\otimes
f_{n}K_{\xi_{n}},K_{z_{1}}\otimes\cdots\otimes
K_{z_{n}}\rangle_{L^{2}(\mathbb{T}^{n})}$
$\displaystyle=\prod_{j=1}^{n}\langle
f_{j}K_{\xi_{j}},K_{z_{j}}\rangle_{L^{2}(\mathbb{T})}$
$\displaystyle=\prod_{j=1}^{n}\langle
P_{H^{2}(\mathbb{D})}(f_{j}K_{\xi_{j}}),K_{z_{j}}\rangle_{H^{2}(\mathbb{D})}$
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left(\prod_{j=1}^{n}\langle
P_{H^{2}(\mathbb{D})}(f_{j}^{\prime}K_{\xi_{j}}),K_{z_{j}}\rangle_{H^{2}(\mathbb{D})}\right),$

where $f_{j}^{\prime}=\begin{cases}f_{j}&\text{ if }j\notin I;\\\ 1&\text{ if
}j\in I.\end{cases}$, and the last equality follows from Lemma 3.3. Therefore,

$\displaystyle\langle
P_{H^{2}(\mathbb{D}^{n})}(fK_{\bm{\xi}}),K_{\bm{z}}\rangle$
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left(\prod_{j=1}^{n}\langle
f_{j}^{\prime}K_{\xi_{j}},K_{z_{j}}\rangle_{H^{2}(\mathbb{D})}\right)$
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left\langle\bigotimes_{j=1}^{n}f_{j}^{\prime}K_{\xi_{j}},\bigotimes_{j=1}^{n}K_{z_{j}}\right\rangle$
$\displaystyle=\left\langle\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left(\bigotimes_{j=1}^{n}f_{j}^{\prime}K_{\xi_{j}}\right),K_{\bm{z}}\right\rangle,$

for all $\bm{z}\in\mathbb{D}^{n}$. Hence

$P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}})=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left(\bigotimes_{j=1}^{n}f_{j}^{\prime}k_{\xi_{j}}\right),~{}\text{where
}f_{j}^{\prime}=\begin{cases}f_{j}&\text{ if }j\notin I;\\\ 1&\text{ if }j\in
I.\end{cases}$

Similarly, we have

$P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})=\left(\prod_{j\in
J}\widetilde{g_{j}}(\xi_{j})\right)\left(\bigotimes_{j=1}^{n}g_{j}^{\prime}k_{\xi_{j}}\right),~{}\text{where
}g_{j}^{\prime}=\begin{cases}g_{j}&\text{ if }j\notin J;\\\ 1&\text{ if }j\in
J.\end{cases}$

Now

$\displaystyle\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle$
$\displaystyle=\left\langle\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\left(\bigotimes_{j=1}^{n}f_{j}^{\prime}k_{\xi_{j}}\right),\left(\prod_{j\in
J}\widetilde{g_{j}}(\xi_{j})\right)\left(\bigotimes_{j=1}^{n}g_{j}^{\prime}k_{\xi_{j}}\right)\right\rangle$
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\overline{\left(\prod_{j\in
J}\widetilde{g_{j}}(\xi_{j})\right)}\left\langle\left(\bigotimes_{j=1}^{n}f_{j}^{\prime}k_{\xi_{j}}\right),\left(\bigotimes_{j=1}^{n}g_{j}^{\prime}k_{\xi_{j}}\right)\right\rangle$
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\overline{\left(\prod_{j\in
J}\widetilde{g_{j}}(\xi_{j})\right)}\prod_{j=1}^{n}\langle
f_{j}^{\prime}k_{\xi_{j}},g_{j}^{\prime}k_{\xi_{j}}\rangle$ (By Lemma 3.1)
$\displaystyle=\left(\prod_{j\in
I}\widetilde{f_{j}}(\xi_{j})\right)\overline{\left(\prod_{j\in
J}\widetilde{g_{j}}(\xi_{j})\right)}\prod_{j=1}^{n}\widetilde{(f_{j}^{\prime}\overline{g_{j}^{\prime}})}(\xi_{j})$
(By Theorem 2.1) $\displaystyle\rightarrow\left(\prod_{j\in
I}f_{j}\right)\overline{\left(\prod_{j\in
J}g_{j}\right)}\left(\prod_{j=1}^{n}f_{j}^{\prime}\overline{g_{j}^{\prime}}\right)$
$\displaystyle=\left(\prod_{j\in I}f_{j}\right)\overline{\left(\prod_{j\in
J}g_{j}\right)}\left(\prod_{j=1}^{n}f_{j}^{\prime}\right)\left(\prod_{j=1}^{n}\overline{g_{j}^{\prime}}\right)$
$\displaystyle=\left(\prod_{j\in I}f_{j}\right)\left(\prod_{j\in
J}\overline{g_{j}}\right)\left(\prod_{j\in I^{c}}f_{j}\right)\left(\prod_{j\in
J^{c}}\overline{g_{j}}\right)$
$\displaystyle=\left(\prod_{j=1}^{n}f_{j}\right)\left(\prod_{j=1}^{n}\overline{g_{j}}\right)$
$\displaystyle=f\overline{g}.$

This proves that

$\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle\rightarrow
f\overline{g}~{}\text{ radially}.$

Suppose that $f$ is a finite sum of elements of the form
$f_{1}\otimes\cdots\otimes f_{n}$, where $f_{j}\in
H^{2}(\mathbb{D})^{\alpha_{j}}$, and $g$ is a finite sum of elements of the
form $g_{1}\otimes\cdots\otimes g_{n}$, where $g_{j}\in
H^{2}(\mathbb{D})^{\beta_{j}}$ for all $1\leq j\leq n$. That is,

$f=\sum_{p=1}^{l}(f_{1p}\otimes\cdots\otimes f_{np})\;\text{ and
}\;g=\sum_{q=1}^{l^{\prime}}(g_{1q}\otimes\cdots\otimes g_{nq}),$

where $f_{jp}\in H^{2}(\mathbb{D})^{\alpha_{j}}$ and $g_{jq}\in
H^{2}(\mathbb{D})^{\beta_{j}}$ for all $1\leq j\leq n$. Therefore,

$\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle$

$\displaystyle=\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(\left(\sum_{p=1}^{l}(f_{1p}\otimes\cdots\otimes
f_{np})\right)k_{\bm{\xi}}\right),P_{H^{2}(\mathbb{D}^{n})}\left(\left(\sum_{q=1}^{l^{\prime}}(g_{1q}\otimes\cdots\otimes
g_{nq})\right)k_{\bm{\xi}}\right)\right\rangle$
$\displaystyle=\left\langle\sum_{p=1}^{l}P_{H^{2}(\mathbb{D}^{n})}((f_{1p}\otimes\cdots\otimes
f_{np})k_{\bm{\xi}}),\sum_{q=1}^{l^{\prime}}P_{H^{2}(\mathbb{D}^{n})}((g_{1q}\otimes\cdots\otimes
g_{nq})k_{\bm{\xi}})\right\rangle$
$\displaystyle=\sum_{p=1}^{l}\sum_{q=1}^{l^{\prime}}\langle
P_{H^{2}(\mathbb{D}^{n})}((f_{1p}\otimes\cdots\otimes
f_{np})k_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}((g_{1q}\otimes\cdots\otimes
g_{nq})k_{\bm{\xi}})\rangle.$

From the above, we obtain that

$\displaystyle\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle$
$\displaystyle\to\sum_{p=1}^{l}\sum_{q=1}^{l^{\prime}}(f_{1p}\otimes\cdots\otimes
f_{np})\overline{(g_{1q}\otimes\cdots\otimes g_{nq})}=f\overline{g}~{}\text{
radially. }$

Finally, assume that $f\in H^{2}(\mathbb{D})^{\alpha_{1}}\otimes\cdots\otimes
H^{2}(\mathbb{D})^{\alpha_{n}}$ and $g\in
H^{2}(\mathbb{D})^{\beta_{1}}\otimes\cdots\otimes
H^{2}(\mathbb{D})^{\beta_{n}}$. Then we shall show that

$\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\rangle\to
f\overline{g}\;\text{ radially. }$

Now there exist sequences $(f^{(m)})_{m=1}^{\infty}$ and
$(g^{(m)})_{m=1}^{\infty}$ such that $\|f^{(m)}-f\|_{L^{2}(\mathbb{T}^{n})}\to
0$ and $\|g^{(m)}-g\|_{L^{2}(\mathbb{T}^{n})}\to 0$ as $m\to\infty$, where
each $f^{(m)}$ and $g^{(m)}$ are the finite sums of elements of the form
$f_{1}\otimes\cdots\otimes f_{n}$ and $g_{1}\otimes\cdots\otimes g_{n}$,
respectively, and $f_{j}\in H^{2}(\mathbb{D})^{\alpha_{j}}$ and $g_{j}\in
H^{2}(\mathbb{D})^{\beta_{j}}$ for all $1\leq j\leq n$. Since $f^{(m)}\to f$
in $L^{2}(\mathbb{T}^{n})$, there exists a subsequence
$\left(f^{(m_{t})}\right)_{t=1}^{\infty}$ of
$\left(f^{(m)}\right)_{m=1}^{\infty}$ which converges to $f$ pointwise a.e. on
$\mathbb{T}^{n}$. Similarly, there exists a subsequence
$\left(g^{(m_{t})}\right)_{t=1}^{\infty}$ of
$\left(g^{(m)}\right)_{m=1}^{\infty}$ such that
$\left(g^{(m_{t})}\right)_{t=1}^{\infty}$ converges to $g$ pointwise a.e. on
$\mathbb{T}^{n}$. Thus

$f^{(m_{t})}\overline{g^{(m_{t})}}\to f\overline{g}\text{ pointwise a.e.\
}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}.$

Suppose $E_{0}$ is a measurable null subset of $\mathbb{T}^{n}$ such that

$f^{(m_{t})}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m_{t})}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\to
f(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})$

as $t\to\infty$ for all
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}\setminus E_{0}$. We
have also shown that for each $m\geq 1$,

$\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m)}k_{\bm{\xi}}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m)}k_{\bm{\xi}}\right)\right\rangle\to
f^{(m)}\overline{g^{(m)}}\;\text{ radially. }$

That means there exists a measurable null subset $E_{m}$ of $\mathbb{T}^{n}$
such that

$\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m)}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m)}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle\to
f^{(m)}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m)}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})$

as $r\to 1^{-}$, and
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}\setminus E_{m}$. Let
$E=\bigcup_{m=0}^{\infty}E_{m}$. Then $E$ is clearly a measurable null subset
of $\mathbb{T}^{n}$. Also $f^{(m)}k_{\bm{\xi}}\to fk_{\bm{\xi}}$ and
$g^{(m)}k_{\bm{\xi}}\to gk_{\bm{\xi}}$ in $L^{2}(\mathbb{T}^{n})$ because
$k_{\bm{\xi}}\in L^{\infty}(\mathbb{T}^{n})$. Hence

$P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m)}k_{\bm{\xi}}\right)\to
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{\bm{\xi}}\right)\;\text{ and
}\;P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m)}k_{\bm{\xi}}\right)\to
P_{H^{2}(\mathbb{D}^{n})}\left(gk_{\bm{\xi}}\right)\text{ in
}H^{2}(\mathbb{D}^{n}).$

as $m\to\infty$. Let $\varepsilon>0$ be any number and fix
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}\setminus E$. Then
there exists $N_{1}\in\mathbb{N}$ such that for $m\geq N_{1}$

$|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(gk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle-$

$\hskip 142.26378pt\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m)}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m)}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle|<\frac{\varepsilon}{3}.$

Now choose $M>N_{1}$ such that

$\left|f^{(m_{M})}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m_{M})}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})-f(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\right|<\frac{\varepsilon}{3}.$

Also there exists $s\in(0,1)$ such that

$|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle-$

$\hskip
142.26378ptf^{(m_{M})}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m_{M})}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})|<\frac{\varepsilon}{3}$

for all $r\in(s,1)$. Note that

$\left|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(gk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle-f(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\right|$

$\displaystyle\leq|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(gk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle$
$\displaystyle\hskip 28.45274pt-\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle|$
$\displaystyle+|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(f^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(g^{(m_{M})}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle$
$\displaystyle\hskip
28.45274pt-f^{(m_{M})}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m_{M})}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})|$
$\displaystyle+\left|f^{(m_{M})}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g^{(m_{M})}}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})-f(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\right|.$

Hence

$\displaystyle|\left\langle
P_{H^{2}(\mathbb{D}^{n})}\left(fk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right),P_{H^{2}(\mathbb{D}^{n})}\left(gk_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\right)\right\rangle$
$\displaystyle-f(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\overline{g}(e^{i\theta_{1}},\dots,e^{i\theta_{n}})|<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon$

whenever $r\in(s,1)$.

This completes the proof.

We have an immediate consequence of the above result as follows.

######  Corollary 3.5.

If $f,g\in L^{2}(\mathbb{T}^{n})$, then

$\left\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\right\rangle\to
f\overline{g}\;\text{ radially.}$

###### Proof.

Note that

$L^{2}(\mathbb{T}^{n})=\bigoplus_{\alpha\in\\{-1,1\\}^{n}}\negthickspace\negthickspace
H^{2}(\mathbb{D}^{n})^{\alpha}.$

Let $f,g\in L^{2}(\mathbb{T}^{n})$ such that

$f=\bigoplus_{\alpha\in\\{-1,1\\}^{n}}\negthickspace\negthickspace
f_{\alpha}\quad\text{ and }\quad
g=\bigoplus_{\beta\in\\{-1,1\\}^{n}}\negthickspace\negthickspace g_{\beta},$

where $f_{\alpha}\in H^{2}(\mathbb{D}^{n})^{\alpha}$ and $g_{\beta}\in
H^{2}(\mathbb{D}^{n})^{\beta}$ for all $\alpha,\beta\in\\{-1,1\\}^{n}$.
Therefore,

$\displaystyle\left\langle
P_{H^{2}(\mathbb{D}^{n})}(fk_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(gk_{\bm{\xi}})\right\rangle$
$\displaystyle=\sum_{\alpha\in\\{-1,1\\}^{n}}\sum_{\beta\in\\{-1,1\\}^{n}}\left\langle
P_{H^{2}(\mathbb{D}^{n})}(f_{\alpha}k_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(g_{\beta}k_{\bm{\xi}})\right\rangle$
$\displaystyle\to\sum_{\alpha\in\\{-1,1\\}^{n}}\sum_{\beta\in\\{-1,1\\}^{n}}f_{\alpha}\overline{g_{\beta}}=f\overline{g}\quad\text{
radially},$

which follows from Theorem 3.4.

##  4\. Symbol map for $H^{2}(\mathbb{D}^{n})$

In this section we shall discuss about the _symbol map_ on the Toeplitz
algebra on $H^{2}(\mathbb{D}^{n})$. Before that, we need some results in the
sequel.

######  Theorem 4.1.

For $\varphi\in L^{\infty}(\mathbb{T}^{n})$, $\|H_{\varphi}k_{\bm{\xi}}\|\to
0$ radially. Moreover, if $\varphi,\psi\in L^{\infty}(\mathbb{T}^{n})$, then
$\widetilde{[T_{\varphi},T_{\psi})}\to 0$ radially.

###### Proof.

Suppose $\varphi\in L^{\infty}(\mathbb{T}^{n})$. Then by definition

$\displaystyle H_{\varphi}k_{\bm{\xi}}$
$\displaystyle=P_{H^{2}(\mathbb{D}^{n})^{\perp}}(\varphi k_{\bm{\xi}})$
$\displaystyle=P_{H^{2}(\mathbb{D}^{n})^{\perp}}\left((P_{H^{2}(\mathbb{D}^{n})}\varphi+P_{H^{2}(\mathbb{D}^{n})^{\perp}}\varphi)k_{\bm{\xi}}\right)$
$\displaystyle=P_{H^{2}(\mathbb{D}^{n})^{\perp}}\left((P_{H^{2}(\mathbb{D}^{n})}\varphi)k_{\bm{\xi}}\right)+P_{H^{2}(\mathbb{D}^{n})^{\perp}}\left((P_{H^{2}(\mathbb{D}^{n})^{\perp}}\varphi)k_{\bm{\xi}}\right).$

Since $k_{\bm{\xi}}\in H^{\infty}(\mathbb{D}^{n})$, clearly
$(P_{H^{2}(\mathbb{D}^{n})}\varphi)k_{\bm{\xi}}\in H^{2}(\mathbb{D}^{n})$.
Therefore,

$H_{\varphi}k_{\bm{\xi}}=P_{H^{2}(\mathbb{D}^{n})^{\perp}}\left((P_{H^{2}(\mathbb{D}^{n})^{\perp}}\varphi)k_{\bm{\xi}}\right).$

Set $y=P_{H^{2}(\mathbb{D}^{n})^{\perp}}\varphi$. Then

$H_{\varphi}k_{\bm{\xi}}=P_{H^{2}(\mathbb{D}^{n})^{\perp}}(yk_{\bm{\xi}}).$

Again

$yk_{\bm{\xi}}=P_{H^{2}(\mathbb{D}^{n})}(yk_{\bm{\xi}})+P_{H^{2}(\mathbb{D}^{n})^{\perp}}(yk_{\bm{\xi}}).$

Thus
$\|yk_{\bm{\xi}}\|^{2}=\|P_{H^{2}(\mathbb{D}^{n})}(yk_{\bm{\xi}})\|^{2}+\|P_{H^{2}(\mathbb{D}^{n})^{\perp}}(yk_{\bm{\xi}})\|^{2}$,
and hence

$\|H_{\varphi}k_{\bm{\xi}}\|^{2}=\|yk_{\bm{\xi}}\|^{2}-\|P_{H^{2}(\mathbb{D}^{n})}(yk_{\bm{\xi}})\|^{2}.$

Now Lemma 3.1 yields

$\|yk_{\bm{\xi}}\|^{2}=\langle
yk_{\bm{\xi}},yk_{\bm{\xi}}\rangle=\widetilde{|y|^{2}}(\bm{\xi}).$

Therefore,

(4.1)
$\|H_{\varphi}k_{\bm{\xi}}\|^{2}=\widetilde{|y|^{2}}(\bm{\xi})-\|P_{H^{2}(\mathbb{D}^{n})}(yk_{\bm{\xi}})\|^{2}.$

Now Corollary 3.5 yields

$\|P_{H^{2}(\mathbb{D}^{n})}(yk_{\bm{\xi}})\|^{2}\to|y|^{2}\quad\text{radially}.$

Hence, from equation (4.1), we conclude that

$\|H_{\varphi}k_{\bm{\xi}}\|\to 0\quad\text{ radially}.$

Note that

$\displaystyle\left|\widetilde{[T_{\varphi},T_{\psi})}(\bm{\xi})\right|=\left|\langle
H_{\overline{\varphi}}^{*}H_{\psi}k_{\bm{\xi}},k_{\bm{\xi}}\rangle\right|=\left|\langle
H_{\psi}k_{\bm{\xi}},H_{\overline{\varphi}}k_{\bm{\xi}}\rangle\right|\leq\|H_{\psi}k_{\bm{\xi}}\|\,\|H_{\overline{\varphi}}k_{\bm{\xi}}\|\to
0\quad\text{ radially.}$

This finishes the proof of the second part.

######  Theorem 4.2.

Let $S\in\mathscr{S}$. Then $\widetilde{S}\to 0$ radially.

###### Proof.

Note that the set of all finite sums of operators of the form
$A[T_{\varphi},T_{\psi})B$, where
$A,B\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, forms a dense subset of
$\mathscr{S}$. Now the set of all finite sums of operators of the form
$T_{\varphi_{1}}T_{\varphi_{2}}\cdots T_{\varphi_{n}}$ is dense in
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ and hence the span of the operators
of the form

$T_{\varphi_{1}}T_{\varphi_{2}}\cdots
T_{\varphi_{n}}[T_{\varphi},T_{\psi})T_{\psi_{1}}T_{\psi_{2}}\cdots
T_{\psi_{m}}$

form a dense subset of $\mathscr{S}$. Therefore, it suffices to prove for
$S=T_{\varphi_{1}}T_{\varphi_{2}}\cdots
T_{\varphi_{n}}[T_{\varphi},T_{\psi})T_{\psi_{1}}T_{\psi_{2}}\cdots
T_{\psi_{m}}$, where $\varphi,\varphi_{i},\psi,\psi_{i}\in
L^{\infty}(\mathbb{T}^{n})$. Again for any symbol $u,v,w\in
L^{\infty}(\mathbb{T}^{n})$,

$\displaystyle T_{w}[T_{u},T_{v})$ $\displaystyle=T_{w}T_{uv}-T_{w}T_{u}T_{v}$
$\displaystyle=(T_{w}T_{uv}-T_{wuv})+(T_{wuv}-T_{wu}T_{v})+(T_{wu}T_{v}-T_{w}T_{u}T_{v})$
$\displaystyle=-[T_{w},T_{uv})+[T_{wu},T_{v})+[T_{w},T_{u})T_{v}.$

Thus, it is enough to assume that $S\in\mathscr{S}$ is of the form

$S=[T_{\varphi},T_{\psi})A=H_{\overline{\varphi}}^{*}H_{\psi}A\qquad(A\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n})),\varphi,\psi\in
L^{\infty}(\mathbb{T}^{n})).$

Therefore,

$\displaystyle\left|\widetilde{S}(\bm{\xi})\right|$
$\displaystyle=\left|\langle
H_{\overline{\varphi}}^{*}H_{\psi}Ak_{\bm{\xi}},k_{\bm{\xi}}\rangle\right|$
$\displaystyle=\left|\langle
H_{\psi}Ak_{\bm{\xi}},H_{\overline{\varphi}}k_{\bm{\xi}}\rangle\right|$
$\displaystyle=\|H_{\psi}Ak_{\bm{\xi}}\|\,\|H_{\overline{\varphi}}k_{\bm{\xi}}\|$
$\displaystyle\leq\|H_{\psi}A\|\,\|H_{\overline{\varphi}}k_{\bm{\xi}}\|\to
0\quad\text{ radially, }$

which follows from Theorem 4.1.

We are now in a position to state the main result of this section.

######  Theorem 4.3.

Let $T\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$. Then
$\widetilde{T}\to\varphi$ radially for some $\varphi\in
L^{\infty}(\mathbb{T}^{n})$. Moreover, the map

$\Sigma:\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))\to
L^{\infty}(\mathbb{T}^{n}),\qquad T\mapsto\varphi$

is a $C^{*}$-algebra homomorphism with ker$(\Sigma)=\mathscr{S}$ and
$\Sigma(T_{\varphi})=\varphi$ for any Toeplitz operator $T_{\varphi}$.

###### Proof.

We shall prove by induction that

$\mbox{for any $m\geq 2$ if~{}}\varphi_{1},\dots,\varphi_{m}\in
L^{\infty}(\mathbb{T}^{n}),~{}\mbox{then~{}}T_{\varphi_{1}}\cdots
T_{\varphi_{m}}-T_{\varphi_{1}\cdots\varphi_{m}}\in\mathscr{S}.$

Recall that for any two symbols $\varphi_{1},\varphi_{2}\in
L^{\infty}(\mathbb{T}^{n})$,
$T_{\varphi_{1}}T_{\varphi_{2}}-T_{\varphi_{1}\varphi_{2}}=-[T_{\varphi_{1}},T_{\varphi_{2}})\in\mathscr{S}$.

Suppose now $T_{\varphi_{1}}\cdots
T_{\varphi_{m}}-T_{\varphi_{1}\cdots\varphi_{m}}\in\mathscr{S}$ for some $m>2$
and $\varphi_{1},\dots,\varphi_{m}\in L^{\infty}(\mathbb{T}^{n}).$

Note that

$\displaystyle T_{\varphi_{1}}\cdots
T_{\varphi_{m}}T_{\varphi_{m+1}}-T_{\varphi_{1}\cdots\varphi_{m}\varphi_{m+1}}$
$\displaystyle=(T_{\varphi_{1}}\cdots
T_{\varphi_{m}}T_{\varphi_{m+1}}-T_{\varphi_{1}\cdots\varphi_{m}}T_{\varphi_{m+1}})$
$\displaystyle+(T_{\varphi_{1}\cdots\varphi_{m}}T_{\varphi_{m+1}}-T_{(\varphi_{1}\cdots\varphi_{m})\varphi_{m+1}})$
$\displaystyle=(T_{\varphi_{1}}\cdots
T_{\varphi_{m}}-T_{\varphi_{1}\cdots\varphi_{m}})T_{\varphi_{m+1}}$
$\displaystyle-[T_{\varphi_{1}\cdots\varphi_{m}},T_{\varphi_{m+1}}).$

Since $\mathscr{S}$ is a closed ideal in
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, we have

$T_{\varphi_{1}}\cdots
T_{\varphi_{m}}T_{\varphi_{m+1}}-T_{\varphi_{1}\cdots\varphi_{m}\varphi_{m+1}}\in\mathscr{S}.$

Now the collection of all finite sums of operators of the form
$T_{\varphi_{1}}\cdots T_{\varphi_{m}}$ is dense in
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$. Thus operators of the form

$T=T_{\varphi}+S\qquad(\varphi\in
L^{\infty}(\mathbb{T}^{n}),~{}S\in\mathscr{S})$

is a dense subset of $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$. Consider

$\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))=\\{\,T_{\varphi}+S:\varphi\in
L^{\infty}(\mathbb{T}^{n}),\,S\in\mathscr{S}\,\\}.$

Suppose $T=T_{\varphi}+S$ for some $\varphi\in L^{\infty}(\mathbb{T}^{n})$ and
$S\in\mathscr{S}$. Then, Theorem 3.2 and Theorem 4.2 yield
$\widetilde{T}\to\varphi$ radially. Define a map
$\Sigma_{0}:\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))\to
L^{\infty}(\mathbb{T}^{n})$ as

(4.2)
$\displaystyle\Sigma_{0}(T)=\varphi\quad(T=T_{\varphi}+S\in\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))).$

We shall show that ${\Sigma_{0}}$ is well-defined. Suppose
$T_{\varphi_{1}}+S_{1}=T_{\varphi_{2}}+S_{2}$. Then

$\displaystyle T_{\varphi_{1}}-T_{\varphi_{2}}=S_{2}-S_{1}\implies
T_{\varphi_{1}-\varphi_{2}}=S_{2}-S_{1}.$

Now Theorem 4.2 gives
$\widetilde{T}_{\varphi_{1}-\varphi_{2}}=\widetilde{S_{2}-S_{1}}\to 0\text{
radially}$. Again Theorem 3.2 infers $\varphi_{1}-\varphi_{2}=0\text{ a.e.}$
and hence $\varphi_{1}=\varphi_{2}\text{ a.e.}$.

Suppose
$T_{\varphi_{1}}+S_{1},T_{\varphi_{2}}+S_{2}\in\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))$
and $\alpha\in\mathbb{C}$. Then

$\displaystyle\Sigma_{0}\left(\alpha(T_{\varphi_{1}}+S_{1})+(T_{\varphi_{2}}+S_{2})\right)$
$\displaystyle=\Sigma_{0}(T_{\alpha\varphi_{1}+\varphi_{2}}+\alpha
S_{1}+S_{2})$ (By (4.2)) $\displaystyle=\alpha\varphi_{1}+\varphi_{2}$
$\displaystyle=\alpha\Sigma_{0}(T_{\varphi_{1}}+S_{1})+\Sigma_{0}(T_{\varphi_{2}}+S_{2}).$

Again if $T\in\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$, then

$\left|\widetilde{T}(\bm{\xi})\right|=\left|\langle
Tk_{\bm{\xi}},k_{\bm{\xi}}\rangle\right|\leq\|T\|\qquad(\bm{\xi}\in\mathbb{D}^{n}).$

Suppose $T=T_{\varphi}+S$ for $\varphi\in L^{\infty}(\mathbb{T}^{n})$ and
$S\in\mathscr{S}$. Then

$\left|\widetilde{T}(re^{i\theta_{1}},\dots,re^{i\theta_{n}})\right|\leq\|T\|\quad\mbox{for
all~{}}0\leq r<1,\,(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}.$

Letting $r\to 1^{-}$, we obtain
$\left|\varphi(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\right|\leq\|T\|$ a.e.
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}$. Hence

$\|\Sigma_{0}(T)\|_{L^{\infty}(\mathbb{T}^{n})}=\|\varphi\|_{L^{\infty}(\mathbb{T}^{n})}\leq\|T\|.$

Therefore, $\Sigma_{0}$ is a linear and bounded map from
$\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))$ to $L^{\infty}(\mathbb{T}^{n})$.

We now extend $\Sigma_{0}$ to
$\overline{\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))}=\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$,
denoting its extension by $\Sigma$. Thus
$\Sigma:\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))\to L^{\infty}(\mathbb{T}^{n})$
is a linear and bounded map. Suppose
$T\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ and set $\varphi=\Sigma(T)$.
Then there exists a sequence $\\{T_{m}\\}_{m=1}^{\infty}$, where each
$T_{m}=T_{\varphi_{m}}+S_{m}\in\mathscr{T}_{0}(L^{\infty}(\mathbb{T}^{n}))$,
such that $\|T_{m}-T\|\to 0$ as $m\to\infty$. Therefore,

$\|T_{\varphi_{m}}-T_{\varphi}\|=\|T_{\varphi_{m}-\varphi}\|=\|\varphi_{m}-\varphi\|_{L^{\infty}(\mathbb{T}^{n})}=\|\Sigma(T_{m})-\Sigma(T)\|_{L^{\infty}(\mathbb{T}^{n})}\to
0~{}~{}\mbox{~{}as~{}}m\to\infty.$

Thus

$S_{m}=T_{m}-T_{\varphi_{m}}\to T-T_{\varphi}\text{ as }m\to\infty.$

Let $S=T-T_{\varphi}$. Then $S\in\mathscr{S}$ as $\\{S_{m}\\}$ in
$\mathscr{S}$ and $\mathscr{S}$ is closed, and hence if
$T\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, then $T=T_{\varphi}+S$, where
$\varphi\in L^{\infty}(\mathbb{T}^{n})$ and $S\in\mathscr{S}$. Thus

$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))=\\{\,T_{\varphi}+S:\varphi\in
L^{\infty}(\mathbb{T}^{n}),\,S\in\mathscr{S}\,\\}.$

Therefore, Theorem 3.2 and Theorem 4.2 give $\widetilde{T}\to\varphi$
radially.

To show $\Sigma$ is a $C^{*}$-algebra homomorphism, let $T=T_{\varphi}+S$,
where $\varphi\in L^{\infty}(\mathbb{T}^{n})$ and $S\in\mathscr{S}$. Note that

$\displaystyle\widetilde{(T_{\varphi}+S)^{*}}$
$\displaystyle=\widetilde{T_{\overline{\varphi}}+S^{*}}=\widetilde{T_{\overline{\varphi}}}+\widetilde{S^{*}}=\widetilde{T_{\overline{\varphi}}}+\overline{\widetilde{S}}\to\overline{\varphi}\quad
radially.$

Therefore, $\Sigma(T^{*})=\overline{\Sigma(T)}$. Suppose
$T_{1}=T_{\varphi_{1}}+S_{1}$ and $T_{2}=T_{\varphi_{2}}+S_{2}$, where
$\varphi_{1},\varphi_{2}\in L^{\infty}(\mathbb{T}^{n})$ and
$S_{1},S_{2}\in\mathscr{S}$. Then

$\displaystyle T_{1}T_{2}$
$\displaystyle=T_{\varphi_{1}}T_{\varphi_{2}}+T_{\varphi_{1}}S_{2}+S_{1}T_{\varphi_{2}}+S_{1}S_{2}$
$\displaystyle=T_{\varphi_{1}\varphi_{2}}+(T_{\varphi_{1}}T_{\varphi_{2}}-T_{\varphi_{1}\varphi_{2}})+(T_{\varphi_{1}}S_{2}+S_{1}T_{\varphi_{2}}+S_{1}S_{2}).$

Since $\mathscr{S}$ is an ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$,
clearly
$(T_{\varphi_{1}}T_{\varphi_{2}}-T_{\varphi_{1}\varphi_{2}})+(T_{\varphi_{1}}S_{2}+S_{1}T_{\varphi_{2}}+S_{1}S_{2})\in\mathscr{S}$.
Hence

$\widetilde{T_{1}T_{2}}\to\varphi_{1}\varphi_{2}~{}~{}\mbox{radially}.$

It follows that $\Sigma(T_{1}T_{2})=\Sigma(T_{1})\Sigma(T_{2})$ for any
$T_{1},T_{2}\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$.

Finally, it is clear from the definition that ker($\Sigma$)$=\mathscr{S}$, the
semicommutator ideal $\mathscr{S}$ in
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, and $\Sigma(T_{\varphi})=\varphi$
for all $\varphi\in L^{\infty}(\mathbb{T}^{n})$. This finishes the proof.

The next result readily follows from the above theorem.

######  Corollary 4.4.

For $T\in\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$, $\widetilde{[T^{*},T]}\to
0$ radially.

We shall now show that the symbol map can not be extended to the whole of
$\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$ in the following example (see [12] for
$n=1$ case).

######  Example 4.5.

Consider the operator $T$ on $H^{2}(\mathbb{D})$, defined by

$Tz^{k}=z^{2k+1}\qquad(k\in\mathbb{Z}_{+}),$

where $\\{z^{k}\\}_{k\in\mathbb{Z}_{+}}$ denotes the standard orthonormal
basis for $H^{2}(\mathbb{D})$. Now

$\langle Tz^{k},Tz^{l}\rangle=\langle
z^{2k+1},z^{2l+1}\rangle=\delta_{k,l}=\langle
z^{k},z^{l}\rangle,\qquad(k,l\in\mathbb{Z}_{+}).$

Therefore, $T$ is an isometry on $H^{2}(\mathbb{D})$. Thus,
$T^{*}T=I_{H^{2}(\mathbb{D})}$ and $TT^{*}=P_{\mathcal{R}(T)}$, the orthogonal
projection onto the range
$\mathcal{R}(T)=\bigvee_{k=0}^{\infty}\\{z^{2k+1}\\}$. Hence

$(T^{*}T-TT^{*})z^{k}=(I-P_{\mathcal{R}(T)})z^{k}=\begin{cases}0\quad&\text{if
$k$ is odd;}\\\ z^{k}\quad&\text{if $k$ is even.}\end{cases}$

That means $[T^{*},T]$ is an orthogonal projection onto
$\mathcal{M}=\bigvee\\{z^{2k}:k\in\mathbb{Z}_{+}\\}$. Therefore, for
$\xi\in\mathbb{D}$

$\displaystyle\widetilde{[T^{*},T]}(\xi)=\|P_{\mathcal{M}}k_{\xi}\|^{2}$
$\displaystyle=(1-|\xi|^{2})\|P_{\mathcal{M}}K_{\xi}\|^{2}$
$\displaystyle=(1-|\xi|^{2})\left\|P_{\mathcal{M}}\left(\sum_{k=0}^{\infty}\overline{\xi}^{k}z^{k}\right)\right\|^{2}$
$\displaystyle=(1-|\xi|^{2})\left\|\sum_{k=0}^{\infty}\overline{\xi}^{2k}z^{2k}\right\|^{2}$
$\displaystyle=(1-|\xi|^{2})\sum_{k=0}^{\infty}|\xi|^{4k}$
$\displaystyle=\frac{(1-|\xi|^{2})}{(1-|\xi|^{4})}=\frac{1}{1+|\xi|^{2}}\to\frac{1}{2}\text{
radially.}$

Now we consider the operator $S$ on $H^{2}(\mathbb{D}^{n})$ as

$S=\overbrace{T\otimes\cdots\otimes T}^{n\text{-times}}.$

Then

$\displaystyle S^{*}S-SS^{*}$ $\displaystyle=(T\otimes\cdots\otimes
T)^{*}(T\otimes\cdots\otimes T)-(T\otimes\cdots\otimes
T)(T\otimes\cdots\otimes T)^{*}$ $\displaystyle=(T^{*}T\otimes
T^{*}T\otimes\cdots\otimes T^{*}T)-(TT^{*}\otimes TT^{*}\otimes\cdots\otimes
TT^{*})$ $\displaystyle=I_{H^{2}(\mathbb{D}^{n})}-(TT^{*}\otimes
TT^{*}\otimes\cdots\otimes TT^{*}).$

Thus for $\bm{k}=(k_{1},\dots,k_{n})\in\mathbb{Z}_{+}^{n}$,

$[S^{*},S]\bm{z}^{\bm{k}}=\begin{cases}\bm{z}^{\bm{k}}\quad&\text{if $k_{j}$
is even for some $j$;}\\\ 0\quad&\text{if $k_{j}$ is odd for all
$j$.}\end{cases}$

Hence $[S^{*},S]$ is an orthogonal projection onto

$\mathcal{N}=\bigvee\\{\bm{z}^{\bm{k}}:\bm{k}=(k_{1},\dots,k_{n})\in\mathbb{Z}_{+},k_{j}\text{
is even for some }1\leq j\leq n\\},$

that is, $[S^{*},S]=P_{\mathcal{N}}$.

Now for $\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$,

$\displaystyle\widetilde{[S^{*},S]}(\bm{\xi})=\|P_{\mathcal{N}}k_{\bm{\xi}}\|^{2}$
$\displaystyle=\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left\|P_{\mathcal{N}}\left(\sum_{\bm{k}\in\mathbb{Z}_{+}^{n}}\overline{\bm{\xi}}^{\bm{k}}\bm{z}^{\bm{k}}\right)\right\|^{2}$
$\displaystyle=\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left\|\sum_{\underset{\text{some
$k_{j}$
even}}{\bm{k}\in\mathbb{Z}_{+}^{n}}}\overline{\bm{\xi}}^{\bm{k}}\bm{z}^{\bm{k}}\right\|^{2}$
$\displaystyle=\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left(\sum_{\text{some
$k_{j}$ even}}|\xi_{1}|^{2k_{1}}\dots|\xi_{n}|^{2k_{n}}\right)$
$\displaystyle\geq\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left(\sum_{\text{all
$k_{j}$ even}}|\xi_{1}|^{2k_{1}}\dots|\xi_{n}|^{2k_{n}}\right)$
$\displaystyle=\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left(\sum_{k_{1}\text{
even}}|\xi_{1}|^{2k_{1}}\right)\dots\left(\sum_{k_{n}\text{
even}}|\xi_{n}|^{2k_{n}}\right)$
$\displaystyle=\left(\prod_{j=1}^{n}(1-|\xi_{j}|^{2})\right)\left(\sum_{m=0}^{\infty}|\xi_{1}|^{4m}\right)\dots\left(\sum_{m=0}^{\infty}|\xi_{n}|^{4m}\right)$
$\displaystyle=\frac{\prod_{j=1}^{n}(1-|\xi_{j}|^{2})}{\prod_{j=1}^{n}(1-|\xi_{j}|^{4})}={\prod_{j=1}^{n}(1+|\xi_{j}|^{2})^{-1}}\geq\frac{1}{2^{n}}>0.$

Hence, by corollary 4.4 we conclude that the symbol calculus can not be
extended to any $C^{*}$-algebra that contains $S$.

##  5\. Extension of Toeplitz Algebra

In this section, we present a family of $C^{*}$-algebras containing the
Toeplitz algebra $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ and the analogs of
Theorem 4.3 still hold.

Let $\psi\in L^{\infty}(\mathbb{T}^{n})$ and $S\in\mathscr{S}$. Consider

(5.1)
$\mathscr{B}(\psi,S)=\\{A\in\mathfrak{B}(H^{2}(\mathbb{D}^{n})):\|A(T_{\psi}+S)k_{\bm{\xi}}\|\text{
and }\|A^{*}(T_{\psi}+S)k_{\bm{\xi}}\|\to 0\text{ radially}\\}.$

In particular, if $\psi=1$ a.e. on $\mathbb{T}^{n}$ and $S=\bm{0}$, where
$\bm{0}$ denotes the zero operator on $H^{2}(\mathbb{D}^{n})$, then we simply
write $\mathscr{B}=\mathscr{B}(1,\bm{0})$.

######  Lemma 5.1.

$\mathscr{B}(\psi,\bm{0})$ is a $C^{*}$-algebra for all $\psi\in
L^{\infty}(\mathbb{T}^{n})$.

###### Proof.

Let $A_{1},A_{2}\in\mathscr{B}(\psi,\bm{0})$ and $c\in\mathbb{C}$. Then
$cA_{1}+A_{2},A_{1}A_{2}\in\mathscr{B}(\psi,\bm{0})$ as

$\displaystyle\|(cA_{1}+A_{2})T_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq|c|\|A_{1}T_{\psi}k_{\bm{\xi}}\|+\|A_{2}T_{\psi}k_{\bm{\xi}}\|,$
$\displaystyle\|(cA_{1}+A_{2})^{*}T_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq|c|\|A_{1}^{*}T_{\psi}k_{\bm{\xi}}\|+\|A_{2}^{*}T_{\psi}k_{\bm{\xi}}\|,$
$\|A_{1}A_{2}T_{\psi}k_{\bm{\xi}}\|\leq\|A_{1}\|\|A_{2}T_{\psi}k_{\bm{\xi}}\|\quad\text{and}\quad\|(A_{1}A_{2})^{*}T_{\psi}k_{\bm{\xi}}\|\leq\|A_{2}^{*}\|\|A_{1}^{*}T_{\psi}k_{\bm{\xi}}\|.$

From the definition, it follows that if $A\in\mathscr{B}(\psi,\bm{0})$, then
$A^{*}\in\mathscr{B}(\psi,\bm{0})$.

To see that $\mathscr{B}(\psi,\bm{0})$ is closed, let $(A_{m})$ be a sequence
in $\mathscr{B}(\psi,\bm{0})$ such that $\|A-A_{m}\|\to 0$ as $m\to\infty$ for
some $A\in\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$. Now for any $\varepsilon>0$
choose $N\in\mathbb{N}$ such that
$\|A-A_{N}\|<\frac{\varepsilon}{2(1+\|T_{\psi}\|)}$. Therefore,

$\displaystyle\|AT_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq\|(A-A_{N})T_{\psi}k_{\bm{\xi}}\|+\|A_{N}T_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq\|A-A_{N}\|\|T_{\psi}\|+\|A_{N}T_{\psi}k_{\bm{\xi}}\|$ (5.2)
$\displaystyle<\varepsilon/2+\|A_{N}T_{\psi}k_{\bm{\xi}}\|.$

Since $\|A_{m}T_{\psi}k_{\bm{\xi}}\|\to 0$ radially for each $m\in\mathbb{N}$,
there exists a measure zero subset $E_{m}$ of $\mathbb{T}^{n}$ such that

$\|A_{m}T_{\psi}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\|\to
0\quad\text{as $r\to 1^{-}$},$

for all $(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}\setminus
E_{m}$. Consider $E=\cup_{m=1}^{\infty}E_{m}$ of $\mathbb{T}^{n}$, and hence
$E$ is a measure zero subset. Thus for a fixed
$(e^{i\theta_{1}},\dots,e^{i\theta_{n}})\in\mathbb{T}^{n}\setminus E$, we
choose $0<s<1$ such that
$\|A_{N}T_{\psi}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\|<\varepsilon/2$
whenever $s<r<1$. Hence the equation (5.2) implies that

$s<r<1\implies\|AT_{\psi}k_{(re^{i\theta_{1}},\dots,re^{i\theta_{n}})}\|<\varepsilon.$

Therefore, $\|AT_{\psi}k_{\bm{\xi}}\|\to 0$ radially. Now
$\|A_{m}^{*}-A^{*}\|=\|A_{m}-A\|\to 0$ as $m\to\infty$ and
$\|A_{m}^{*}T_{\psi}k_{\bm{\xi}}\|\to 0$ radially for all $m\in\mathbb{N}$. It
follows that $\|A^{*}T_{\psi}k_{\bm{\xi}}\|\to 0$ radially. Thus
$A\in\mathscr{B}(\psi,\bm{0})$ and hence $\mathscr{B}(\psi,\bm{0})$ is closed.

######  Lemma 5.2.

$\mathscr{C}\subseteq\mathscr{S}\subseteq\mathscr{B}$.

###### Proof.

Firstly, it is easy to see that $\mathscr{C}\subseteq\mathscr{S}$. Let
$\varphi,\psi\in L^{\infty}(\mathbb{T}^{n})$. Then

$\displaystyle\|[T_{\varphi},T_{\psi})k_{\bm{\xi}}\|$
$\displaystyle=\|H_{\overline{\varphi}}^{*}H_{\psi}k_{\bm{\xi}}\|\leq\|H_{\overline{\varphi}}^{*}\|\|H_{\psi}k_{\bm{\xi}}\|,$
$\displaystyle\text{ and }\|[T_{\varphi},T_{\psi})^{*}k_{\bm{\xi}}\|$
$\displaystyle=\|(H_{\overline{\varphi}}^{*}H_{\psi})^{*}k_{\bm{\xi}}\|\leq\|H_{\psi}^{*}\|\|H_{\overline{\varphi}}k_{\bm{\xi}}\|.$

Now Theorem 4.1 infers that $[T_{\varphi},T_{\psi})\in\mathscr{B}$, and hence
$\mathscr{C}\subseteq\mathscr{S}\subseteq\mathscr{B}$.

######  Lemma 5.3.

$\mathscr{B}(\psi,S)=\mathscr{B}(\psi,\bm{0})$ for all $\psi\in
L^{\infty}(\mathbb{T}^{n})$ and $S\in\mathscr{S}$.

###### Proof.

It follows easily from Lemma 5.2 that
$\mathscr{B}(\psi,\bm{0})\subseteq\mathscr{B}(\psi,S)$.

Suppose $\psi\in L^{\infty}(\mathbb{T}^{n})$ and $S\in\mathscr{S}$, and let
$A\in\mathscr{B}(\psi,S)$. Then $\|Sk_{\bm{\xi}}\|\to 0$ radially, and we also
have

$\displaystyle\|AT_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq\|A(T_{\psi}+S)k_{\bm{\xi}}\|+\|A\|\|Sk_{\bm{\xi}}\|\to 0,$
$\displaystyle\|A^{*}T_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq\|A^{*}(T_{\psi}+S)k_{\bm{\xi}}\|+\|A\|\|Sk_{\bm{\xi}}\|\to
0.$

Therefore, $A\in\mathscr{B}(\psi,\bm{0})$.

######  Theorem 5.4.

Let $\varphi\in L^{2}(\mathbb{T}^{n})$. Then
$\|P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}})-\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|\to 0$ radially. In
particular,
$\|T_{\varphi}k_{\bm{\xi}}-\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|\to 0$
radially for $\varphi\in L^{\infty}(\mathbb{T}^{n})$.

###### Proof.

First, assume that $\varphi\in L^{2}(\mathbb{T}^{n})$. Then we have

$\displaystyle\|P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}})-\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|^{2}$
$\displaystyle=\|P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}})\|^{2}+\|\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|^{2}-{\langle}P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}}),\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}{\rangle}-{\langle}\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}},P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}}){\rangle}$ $\displaystyle=\|P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}})\|^{2}+\|\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|^{2}-\overline{\widetilde{\varphi}(\bm{\xi})}{\langle}P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}}),k_{\bm{\xi}}{\rangle}-\widetilde{\varphi}(\bm{\xi}){\langle}k_{\bm{\xi}},P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}}){\rangle}.$

Now Corollary 3.5 yields

$\|P_{H^{2}(\mathbb{D}^{n})}(\varphi
k_{\bm{\xi}})-\widetilde{\varphi}(\bm{\xi})k_{\bm{\xi}}\|\to 0\quad\text{
radially}.$

Finally, if $\varphi\in L^{\infty}(\mathbb{T}^{n})$, then
$P_{H^{2}(\mathbb{D}^{n})}(\varphi k_{\bm{\xi}})=T_{\varphi}k_{\bm{\xi}}$ and
hence the required result follows.

######  Proposition 5.5.

If $\varphi\in L^{\infty}(\mathbb{T}^{n})$ and $A\in\mathscr{B}(\psi,\bm{0})$,
then $T_{\varphi}A,AT_{\varphi}\in\mathscr{B}(\psi,\bm{0})$.

###### Proof.

To show $T_{\varphi}A$ and $AT_{\varphi}$ are in $\mathscr{B}(\psi,\bm{0})$
for $A\in\mathscr{B}(\psi,\bm{0})$ and $\varphi\in
L^{\infty}(\mathbb{T}^{n})$, it suffices to check that
$\|AT_{\varphi}T_{\psi}k_{\bm{\xi}}\|\to 0$ radially. Note that

$\displaystyle\|AT_{\varphi}T_{\psi}k_{\bm{\xi}}\|$
$\displaystyle\leq\|AT_{\varphi}T_{\psi}k_{\bm{\xi}}-\widetilde{\varphi}(\bm{\xi})AT_{\psi}k_{\bm{\xi}}\|+\|\widetilde{\varphi}(\bm{\xi})AT_{\psi}k_{\bm{\xi}}\|$
(5.3)
$\displaystyle\leq\|A\|\|T_{\varphi}T_{\psi}k_{\bm{\xi}}-\widetilde{\varphi}(\bm{\xi})T_{\psi}k_{\bm{\xi}}\|+\|\varphi\|_{\infty}\|AT_{\psi}k_{\bm{\xi}}\|.$

Also

$\|T_{\varphi}T_{\psi}k_{\bm{\xi}}-\widetilde{\varphi}(\bm{\xi})T_{\psi}k_{\bm{\xi}}\|^{2}=(T_{\psi}^{*}T_{\varphi}^{*}T_{\varphi}T_{\psi}\widetilde{)}(\bm{\xi})+|\widetilde{\varphi}(\bm{\xi})|^{2}\|T_{\psi}k_{\bm{\xi}}\|^{2}-2\text{Re}\left(\overline{\widetilde{\varphi}(\bm{\xi})}(T_{\psi}^{*}T_{\varphi}T_{\psi}\widetilde{)}(\bm{\xi})\right).$

Applying Theorems 2.1, 4.3, and 3.2, we obtain

$\|T_{\varphi}T_{\psi}k_{\bm{\xi}}-\widetilde{\varphi}(\bm{\xi})T_{\psi}k_{\bm{\xi}}\|^{2}\to|\varphi\psi|^{2}+|\varphi\psi|^{2}-2|\varphi\psi|^{2}=0\quad\text{radially.}$

Hence $\|AT_{\varphi}T_{\psi}k_{\bm{\xi}}\|\to 0$ radially follows from (5.3).

An immediate consequence of Proposition 5.5 is the following.

######  Lemma 5.6.

Let $\psi_{1}$ and $\psi_{2}$ be two inner functions on $\mathbb{D}^{n}$ and
let $\psi=\overline{\psi_{1}}\psi_{2}$. Then
$\mathscr{B}(\psi,\bm{0})=\mathscr{B}$.

###### Proof.

Follows from Proposition 5.5.

Now we record the following properties of the collection $\mathscr{B}$ defined
in (5.1):

######  Remark 5.7.

Suppose $\varphi\in L^{\infty}(T^{n})$ and
$T_{\varphi}\in\mathscr{B}(\psi,\bm{0})$, where $|\psi|>0$ a.e. on
$\mathbb{T}^{n}$. Then

$|\widetilde{T_{\varphi\psi}}(\bm{\xi})|\leq\|T_{\varphi\psi}k_{\bm{\xi}}\|\leq\|[T_{\varphi},T_{\psi})k_{\bm{\xi}}\|+\|T_{\varphi}T_{\psi}k_{\bm{\xi}}\|\to
0\quad\text{radially.}$

Now Theorem 3.2 yields that $\varphi\psi\equiv 0$ a.e. on $\mathbb{T}^{n}$.
But $|\psi|>0$ a.e. on $\mathbb{T}^{n}$, hence $\varphi\equiv 0$ a.e. on
$\mathbb{T}^{n}$. Thus, any Toeplitz operator in $\mathscr{B}(\psi,\bm{0})$
corresponds to the symbol $0$.

For $\psi\in L^{\infty}(\mathbb{T}^{n})$ with $|\psi|=1$ a.e. on
$\mathbb{T}^{n}$, consider

$\mathscr{A}_{\psi}:=\\{T_{\varphi}+A:\varphi\in
L^{\infty}(\mathbb{T}^{n}),A\in\mathscr{B}(\psi,\bm{0})\\}.$

######  Theorem 5.8.

The collection $\mathscr{A}_{\psi}$ satisfies the following:

  1. (1)

If $T\in\mathscr{A}_{\psi}$, then $\widetilde{T}\to\varphi$ radially for some
$\varphi\in L^{\infty}(\mathbb{T}^{n})$.

  2. (2)

$\mathscr{A}_{\psi}$ is a $C^{*}$-algebra.

  3. (3)

Define

$\Sigma_{\psi}:\mathscr{A}_{\psi}\to
L^{\infty}(\mathbb{T}^{n})\quad\text{by}\quad\Sigma_{\psi}(T)=\varphi,$

where $\varphi$ is as in (1). Then $\Sigma_{\psi}$ induces a $C^{*}$-algebra
homomorphism from $\mathscr{A}_{\psi}$ onto $L^{\infty}(\mathbb{T}^{n})$ such
that $\Sigma_{\psi}(T_{\varphi})=\varphi$ for all $\varphi\in
L^{\infty}(\mathbb{T}^{n})$.

###### Proof.

  1. (1)

Let $T\in\mathscr{A}_{\psi}$. Then $T=T_{\varphi}+A$ for some $\varphi\in
L^{\infty}(\mathbb{T}^{n}),A\in\mathscr{B}(\psi,\bm{0})$. Now Theorem 3.2
implies that

$\widetilde{T}=\widetilde{T_{\varphi}}+\widetilde{A}\to\varphi\quad\text{radially}.$

Also

$\displaystyle|\widetilde{A}(\bm{\xi})|$
$\displaystyle=|{\langle}Ak_{\bm{\xi}},k_{\bm{\xi}}{\rangle}|$
$\displaystyle=|{\langle}\psi Ak_{\bm{\xi}},\psi k_{\bm{\xi}}{\rangle}|$
$\displaystyle\leq|{\langle}P_{H^{2}(\mathbb{D}^{n})}(\psi
Ak_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})}(\psi
k_{\bm{\xi}}){\rangle}|+|{\langle}P_{H^{2}(\mathbb{D}^{n})^{\perp}}(\psi
Ak_{\bm{\xi}}),P_{H^{2}(\mathbb{D}^{n})^{\perp}}(\psi k_{\bm{\xi}}){\rangle}|$
$\displaystyle=|{\langle}T_{\psi}Ak_{\bm{\xi}},T_{\psi}k_{\bm{\xi}}{\rangle}|+|{\langle}H_{\psi}Ak_{\bm{\xi}},H_{\psi}k_{\bm{\xi}}{\rangle}|$
$\displaystyle\leq\|A^{*}T_{\psi}^{*}T_{\psi}k_{\bm{\xi}}\|+\|H_{\psi}A\|\|H_{\psi}k_{\bm{\xi}}\|.$

Thus, Proposition 5.5 and Theorem 4.1 infer that $\widetilde{A}(\bm{\xi})\to
0$ radially.

  2. (2)

It is immediate from Lemma 5.1 that $\mathscr{A}_{\psi}$ is linear space and
closed under adjoint. Let $\varphi_{1},\varphi_{2}\in
L^{\infty}(\mathbb{T}^{n})$, and $A_{1},A_{2}\in\mathscr{B}(\psi,\bm{0})$.
Then

$\displaystyle(T_{\varphi_{1}}+A_{1})(T_{\varphi_{2}}+A_{2})$
$\displaystyle=T_{\varphi_{1}}T_{\varphi_{2}}+T_{\varphi_{1}}A_{2}+A_{1}T_{\varphi_{2}}+A_{1}A_{2}$
(5.4)
$\displaystyle=T_{\varphi_{1}\varphi_{2}}-([T_{\varphi_{1}},T_{\varphi_{2}})-T_{\varphi_{1}}A_{2}-A_{1}T_{\varphi_{2}}-A_{1}A_{2}).$

Using Lemma 5.1, Lemma 5.2, and Proposition 5.5, we have $\mathscr{A}_{\psi}$
is closed under multiplication.

To show that $\mathscr{A}_{\psi}$ is closed, we show
$\mathscr{A}_{\psi}=\overline{\mathscr{A}_{\psi}}$. Consider

$\eta_{0}:\mathscr{A}_{\psi}\to L^{\infty}(\mathbb{T}^{n})\quad\text{defined
by}\quad\eta_{0}(T_{\varphi}+A)=\varphi.$

It is easily seen that $\eta_{0}$ is a well-defined bounded linear map from
$\mathscr{A}_{\psi}$ onto $L^{\infty}(\mathbb{T}^{n}).$ Suppose $\eta$ is the
bounded linear extension of $\eta_{0}$ from $\overline{\mathscr{A}_{\psi}}$
onto $L^{\infty}(\mathbb{T}^{n})$. Assume that
$T\in\overline{\mathscr{A}_{\psi}}$. Then there exists a sequence
$\\{T_{\varphi_{m}}+A_{m}\\}$ in $\mathscr{A}_{\psi}$ such that
$\|(T_{\varphi_{m}}+A_{m})-T\|\to 0$ as $m\to\infty$, and hence
$\varphi_{m}=\eta(T_{\varphi_{m}}+A_{m})\to\eta(T)$ in
$L^{\infty}(\mathbb{T}^{n}).$ Thus

$\|T_{\varphi_{m}}-T_{\eta(T)}\|=\|\varphi_{m}-\eta(T)\|_{\infty}\to
0\quad\mbox{as}~{}m\to\infty.$

Therefore, $A_{m}=(T_{\varphi_{m}}+A_{m})-T_{\varphi_{m}}\to T-T_{\eta(T)}$ as
$m\to\infty$. Since $\mathscr{B}(\psi,\bm{0})$ is closed,
$T-T_{\eta(T)}\in\mathscr{B}(\psi,\bm{0})$. Therefore,
$T=T_{\eta(T)}+(T-T_{\eta(T)})\in\mathscr{A}_{\psi}$.

  3. (3)

It is easy to check that the mapping $\Sigma_{\psi}$ is well-defined linear,
preserves conjugation. Also (5.4) yields that the map is multiplicative too.
Thus $\Sigma_{\psi}$ is a $C^{*}$-algebra homomorphism from
$\mathscr{A}_{\psi}$ onto $L^{\infty}(\mathbb{T}^{n})$, and also
$\Sigma_{\psi}(T_{\varphi})=\varphi$ with
ker$(\Sigma_{\psi})=\mathscr{B}(\psi,\bm{0})$.

It is obvious from Lemma 5.2 that
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))\subseteq\mathscr{A}_{\psi}$.
Following example shows that Theorem 5.8 is indeed an extension of Theorem
4.3.

######  Example 5.9.

Let $P=P_{\mathcal{M}}$ be the orthogonal projection of $H^{2}(\mathbb{D})$
onto the closed subspace $\mathcal{M}$, where
$\mathcal{M}:=\bigvee_{m=0}^{\infty}\\{z^{2^{m}}\\}$. Then for $m\geq 1$

$\|(T_{z}P-PT_{z})z^{2^{m}}\|=\|z^{2^{m}+1}-Pz^{2^{m}+1}\|=1.$

But $z^{2^{m}}\to 0$ weakly as $m\to\infty$, and hence $T_{z}P-PT_{Z}$ is not
a compact operator. Therefore $P\notin\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$
(see [2]). Also it is shown in [12, Example 8] that $\|Pk_{re^{i\theta}}\|\to
0$ as $r\to 1^{-}$ for a.e. $e^{i\theta}\in\mathbb{T}.$ Consider an operator
$A$ on $H^{2}(\mathbb{D}^{n})$ defined as

$A=\overbrace{P\otimes I\otimes\dots\otimes I}^{n\text{-times}},$

where $I$ denotes the identity operator on $H^{2}(\mathbb{D})$. Now for
$\bm{\xi}=(\xi_{1},\dots,\xi_{n})\in\mathbb{D}^{n}$,

$\displaystyle\|Ak_{\bm{\xi}}\|$ $\displaystyle=\|(P\otimes
I\otimes\cdots\otimes I)(k_{\xi_{1}}\otimes k_{\xi_{2}}\otimes\cdots\otimes
k_{\xi_{n}})\|$ $\displaystyle=\|Pk_{\xi_{1}}\otimes
k_{\xi_{2}}\otimes\cdots\otimes k_{\xi_{n}}\|$
$\displaystyle=\|Pk_{\xi_{1}}\|\|k_{\xi_{2}}\|\cdots\|k_{\xi_{n}}\|$
$\displaystyle=\|Pk_{\xi_{1}}\|.$

It follows that $\|Ak_{\bm{\xi}}\|\to 0$ radially, and
$A\in\mathscr{A}_{\psi}$, in fact $A\in\mathscr{B}$. Let $T_{z_{i}}$ denote
the multiplication operator on $H^{2}(\mathbb{D}^{n})$ by the $i$-th
coordinate function $z_{i}$ for $1\leq i\leq n$, that is,
$(T_{z_{i}}f)(\bm{\xi})=\xi_{i}f(\bm{\xi})$ for all
$\bm{\xi}\in\mathbb{D}^{n},f\in H^{2}(\mathbb{D}^{n})$. Consider

$\mathbb{E}:=\\{T\in\mathfrak{B}(H^{2}(\mathbb{D}^{n})):T_{z_{i}}^{*}TT_{z_{i}}-T\text{
is compact for all }1\leq i\leq n\\}.$

Then $\mathbb{E}$ is a norm-closed algebra. Now if $T_{\varphi}$ is Toeplitz
operator on $H^{2}(\mathbb{D}^{n})$, then
$T_{z_{i}}^{*}T_{\varphi}T_{z_{i}}=T_{\varphi}$ for all $1\leq i\leq n$ (see
[15]). Hence $\mathbb{E}$ contains all Toeplitz operators on
$H^{2}(\mathbb{D}^{n})$. We show that
$A\notin\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ by showing that
$A\notin\mathbb{E}$. Now

$T_{z_{1}}^{*}AT_{z_{1}}-A=(T_{z}^{*}PT_{z}-P)\otimes I\otimes\cdots\otimes
I,$

and $T_{z}^{*}PT_{z}-P$ is not compact. Therefore, $T_{z_{1}}^{*}AT_{z_{1}}-A$
is not compact (see [14]) and hence $A\notin\mathbb{E}$.

Denote

$\mathscr{A}=\\{T_{\varphi}+A:\varphi\in
L^{\infty}(\mathbb{T}^{n}),A\in\mathscr{B}\\}.$

Note that if $A\in\mathscr{B}$ then Proposition 5.5 gives that
$A\in\mathscr{B}(\psi,\bm{0})$ for each $\psi\in L^{\infty}(\mathbb{T}^{n})$,
and hence $\mathscr{A}\subseteq\mathscr{A}_{\psi}$. In the next result we
prove that the class $\mathscr{A}$ is maximal onto which the _symbol map_
exists and $T_{\varphi}\to\varphi$, provided that the symbol of an operator is
obtained as the radial limit of its Berezin transform.

######  Proposition 5.10.

Let $\mathcal{A}$ be any closed $C^{*}$-subalgebra of
$\mathfrak{B}(H^{2}(\mathbb{D}^{n}))$ containing Toeplitz operators such that
there exists a $C^{*}$-homomorphism $\mathfrak{S}$ from $\mathcal{A}$ onto
$L^{\infty}(\mathbb{T}^{n})$ with $\mathfrak{S}(T_{\varphi})=\varphi$ for all
$\varphi\in L^{\infty}(\mathbb{T}^{n})$, and for each $T\in\mathcal{A}$,
$\mathfrak{S}(T)$ is the radial limit of $\widetilde{T}$. Then
$\mathcal{A}\subseteq\mathscr{A}.$ In particular,
$\mathscr{A}_{\psi}=\mathscr{A}$, where $\mathscr{A}_{\psi}$ is as in Theorem
5.8.

###### Proof.

Let $T\in\mathcal{A}$. Suppose $\mathfrak{S}(T)=\varphi$ for some $\varphi$ in
$L^{\infty}(\mathbb{T}^{n})$. Then $T-T_{\varphi}\in ker(\mathfrak{S})$ and
hence

$\mathcal{A}=\\{T_{\varphi}+A:\varphi\in L^{\infty}(\mathbb{T}^{n}),A\in
ker(\mathfrak{S})\\}.$

Now if $A\in ker(\mathfrak{S})$, then $A^{*}A$ is so. Therefore,

$\|Ak_{\bm{\xi}}\|^{2}=\langle Ak_{\bm{\xi}},Ak_{\bm{\xi}}\rangle=\langle
A^{*}Ak_{\bm{\xi}},k_{\bm{\xi}}\rangle=\widetilde{(A^{*}A)}(\bm{\xi})\to
0\text{ radially.}$

In the similar fashion, it can be shown that $\|A^{*}k_{\bm{\xi}}\|\to 0$
radially. Thus $ker(\mathfrak{S})\subseteq\mathscr{B}$, and hence
$\mathcal{A}\subseteq\mathscr{A}$.

Following Theorem 5.8, in particular, we have $\mathscr{A}_{\psi}=\mathscr{A}$
for all $\psi\in L^{\infty}(\mathbb{T}^{n})$ with $|\psi|=1$ a.e. on
$\mathbb{T}^{n}$.

We shall conclude this section with the following application of our _symbol
map_ result, which gives a partial answer in some sense to the question raised
by Barría and Halmos in [2]:

Which projections belong to the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}))$?

Using the _symbol map_ , that is, Theorem 4.3, we shall show that if $P$ is a
projection in $\mathscr{T}(L^{\infty}(\mathbb{T}))$ which is a finite rank
perturbation of a Toeplitz operator, then the rank of $P$ is finite or
$I_{H^{2}(\mathbb{D})}-P$ is finite rank, that is, co-finite. However, for
$n>1$, if a projection in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ is compact
perturbation of a Toeplitz operator, then its rank is either finite or co-
finite. More precisely, we have the following result:

######  Theorem 5.11.

Let $P$ be an orthogonal projection in the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$. For $n=1$, if $P=T_{\varphi}+F$,
where $\varphi\in L^{\infty}(\mathbb{T})$ and $F$ is of finite rank, then
either $P$ or $I_{H^{2}(\mathbb{D})}-P$ is of finite rank. For $n>1$, if
$P=T_{\varphi}+K$, where $\varphi\in L^{\infty}(\mathbb{T}^{n})$ and $K$ is
compact, then either $P$ or $I_{H^{2}(\mathbb{D}^{n})}-P$ is of finite rank.

###### Proof.

Suppose $P$ is an orthogonal projection in the Toeplitz algebra
$\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ such that $P$ is a either finite or
compact perturbation of a Toeplitz operator $T_{\varphi}$ for $n\geq 1$. Then
Theorem 4.3 yields $\overline{\varphi}=\varphi$ and $\varphi^{2}=\varphi$ a.e.
on $\mathbb{T}^{n}$. Hence $\varphi=\chi_{E}$, where
$E=\\{\bm{\zeta}\in\mathbb{T}^{n}:\varphi(\bm{\zeta})=1\\}$, a measurable
subset of $\mathbb{T}^{n}$.

For $n=1$: If $P=T_{\chi_{E}}+F$, where $\varphi\in L^{\infty}(\mathbb{T})$
and $F$ is of finite rank, then $P^{2}=P$ gives

$T_{\chi_{E}}-T^{2}_{\chi_{E}}=F^{2}-F+T_{\chi_{E}}F+FT_{\chi_{E}},$

and hence $T_{\chi_{E}}T_{\chi_{E^{c}}}=T_{\chi_{E}}-T^{2}_{\chi_{E}}$ is of
finite rank. Therefore, either $T_{\chi_{E}}=0$ or $T_{\chi_{E^{c}}}=0$ by [7,
Corollary 2.2]. Thus, either $P$ or $I_{H^{2}(\mathbb{D})}-P$ is of finite
rank.

For $n>1$: we shall get

$T_{\chi_{E}}-T^{2}_{\chi_{E}}=K^{2}-K+T_{\chi_{E}}K+KT_{\chi_{E}},$

where $K$ is a compact operator. But
$T_{\chi_{E}}T_{\chi_{E^{c}}}=T_{\chi_{E}}-T^{2}_{\chi_{E}}=H_{\chi_{E}}^{*}H_{\chi_{E}}$
which is compact, and hence $H_{\chi_{E}}$ is compact. Now [6, Corollary 5],
gives all compact Hankel operators are zero whenever $n>1$. Thus
$T_{\chi_{E}}T_{\chi_{E^{c}}}=0$ implies either $T_{\chi_{E}}=0$ or
$T_{\chi_{E^{c}}}=0$ by Theorem 2.1 in [7]. Therefore, $P=T_{\chi_{E}}+K$
gives either $P$ or $I_{H^{2}(\mathbb{D}^{n})}-P$ is compact. Since both $P$
and $I_{H^{2}(\mathbb{D}^{n})}-P$ are orthogonal projections and all compact
orthogonal projections are of finite rank; therefore either $P$ or
$I_{H^{2}(\mathbb{D}^{n})}-P$ is of finite rank.

This completes the proof.

##  6\. Concluding Remarks

The Toeplitz algebra $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ provides an
effective way to study the Toeplitz operators as elements of a
$C^{*}$-algebra. More specifically, Theorem 4.3 shows that the study of the
Toeplitz algebra $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ is largely reduces
to that of the semi-commutator ideal $\mathscr{S}$. Let $\theta$ be an inner
function on $\mathbb{D}^{n}$, and
$\mathcal{K}_{\theta}=H^{2}(\mathbb{D}^{n})\ominus\theta
H^{2}(\mathbb{D}^{n})$. Suppose $P_{\theta}$ denotes the orthogonal projection
of $H^{2}(\mathbb{D}^{n})$ onto $\mathcal{K}_{\theta}$. Then

$P_{\theta}=I_{H^{2}(\mathbb{D}^{n})}-T_{\theta}T_{\overline{\theta}}=[T_{\theta},T_{\overline{\theta}})\in\mathscr{S}.$

In fact, if $\varphi=\theta_{1}\overline{\eta_{1}}$ and
$\psi=\theta_{2}\overline{\eta_{2}}$, where $\theta_{i},\eta_{i}$ are inner
functions, then

$\displaystyle[T_{\varphi},T_{\psi})$
$\displaystyle=T_{\varphi\psi}-T_{\varphi}T_{\psi}$
$\displaystyle=T_{\theta_{1}\overline{\eta_{1}}\theta_{2}\overline{\eta_{2}}}-T_{\theta_{1}\overline{\eta_{1}}}T_{\theta_{2}\overline{\eta_{2}}}$
$\displaystyle=T_{\overline{\eta_{1}}}T_{\overline{\eta_{2}}}T_{\theta_{1}}T_{\theta_{2}}-T_{\overline{\eta_{1}}}T_{\theta_{1}}T_{\overline{\eta_{2}}}T_{\theta_{2}}$
$\displaystyle=T_{\overline{\eta_{1}}}T_{\overline{\eta_{2}}}T_{\theta_{1}}(I-T_{\eta_{2}}T_{\overline{\eta_{2}}})T_{\theta_{2}}$
$\displaystyle=T_{\overline{\eta_{1}}}T_{\overline{\eta_{2}}}T_{\theta_{1}}P_{\eta_{2}}T_{\theta_{2}}.$

In the case for $n=1$, the semi-commutator ideal coincides with the commutator
ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}))$ due to the availability of
Douglas-Rudin theorm; again from the above calculation the semi-commutator
ideal coincides with the closed ideal in $\mathscr{T}(L^{\infty}(\mathbb{T}))$
generated by all orthogonal projections on the model space. Thus, Douglas-
Rudin theorem (see [10]) plays an important role in this study. We now
conclude this paper by raising some questions which naturally arise in our
paper:

Question 1. Is it true that the class
$\\{\overline{\varphi}\psi:\varphi,\psi\in H^{\infty}(\mathbb{T}^{n})\\}$ is
norm-dense in $L^{\infty}(\mathbb{T}^{n})$ for $n>1$?

Question 2. Does the semi-commutator ideal coincide with the commutator ideal
in $\mathscr{T}(L^{\infty}(\mathbb{T}^{n}))$ for $n>1$?

It is noteworthy to point out that the affirmative answer to Question 1 gives
an affirmative answer to Question 2, which we shall try to carry out in the
future paper.

Data availability: Data sharing is not applicable to this article as no data
sets were generated or analysed during the current study.

Declarations

Conflict of interest: The authors have no competing interests to declare.

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