# Dirichlet type spaces in the unit bidisc

Monojit Bhattacharjee  ,  Rajeev Gupta  and  Vidhya Venugopal  Department of
Mathematics
Birla Institute of Technology and Science K K Birla Goa Campus, India  School
of Mathematics and Computer Science
Indian Institute of Technology Goa, India  Department of Mathematics
Birla Institute of Technology and Science K K Birla Goa Campus, India

###### Abstract.

In this article, we prove that an analytic $2$-isometric left-inverse
commuting pair on a Hilbert space always possess wandering subspace property.
Along with this, we define Dirichlet-type space
$\mathcal{D}^{2}(\boldsymbol{\mu})$ over the bidisc $\mathbb{D}^{2}$ for any
measure $\boldsymbol{\mu}\in\mathcal{P}\mathcal{M}_{+}(\mathbb{T}^{2}).$ We
prove that the pair $(M_{z_{1}},M_{z_{2}})$ of multiplication by the
coordinate functions on $\mathcal{D}^{2}(\boldsymbol{\mu})$ is an analytic
$2$-isometric left-inverse commuting pair and satisfies $\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\boldsymbol{\mu}}=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}(\mu_{1})}\langle
z_{2}^{n},z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})},$ for any
$m,n,p,q\in\mathbb{Z}_{\geqslant 0}.$ Furthermore, the unitary equivalent
class of the commuting analytic $2$-isometric pair $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\mu})$ is identified using left-inverse
commutativity and the above property of inner product.

###### Key words and phrases:

$m$-isometry, $m$-concave operators, wandering subspace property, Wold-type
decomposition, Dirichlet-type spaces

###### 2010 Mathematics Subject Classification:

Primary 46E20, 47B32, 47B38, Secondary 47A50, 31C25

##  1\. Introduction

Let $\mathbb{N}$ and $\mathbb{Z}$ denote the set of all positive integers and
the set of all integers respectively. For each $n\in\mathbb{N},$ let
$\mathbb{Z}^{n}_{\geqslant 0}$ be the set of $n$-tuples of non-negative
integers. We shall denote the unit circle and the open unit disc in the
complex plane $\mathbb{C}$ by $\mathbb{T}$ and $\mathbb{D}$ respectively. For
any set $X,$ the notation $X^{n}$ shall be used to denote the $n$-times
cartesian product of $X$ with itself.

Let $\boldsymbol{T}=(T_{1},\ldots,T_{n})$ be an $n$-tuple of commuting bounded
linear operators on a Hilbert space $\mathcal{H}$. A closed subspace
$\mathcal{W}\subseteq\mathcal{H}$ is said to be a wandering subspace of
$\boldsymbol{T}$ if

$\mathcal{W}\perp T_{1}^{k_{1}}\cdots
T_{n}^{k_{n}}\mathcal{W}\quad\quad\text{for
all}~{}(k_{1},\ldots,k_{n})\in\mathbb{Z}_{\geqslant 0}^{n}$

and, following Halmos [7], is said to be a generating wandering subspace for
$\boldsymbol{T}$ if in addition

$\mathcal{H}=\overline{\text{span}}\\{T_{1}^{k_{1}}\cdots
T_{n}^{k_{n}}\mathcal{W}~{}:~{}(k_{1},\ldots,k_{n})\in\mathbb{Z}_{\geqslant
0}^{n}\\}.$

We say an $n$-tuple of operators $\boldsymbol{T}=(T_{1},\ldots,T_{n})$ acting
on a Hilbert space $\mathcal{H}$ has wandering subspace property provided
$\boldsymbol{T}$ has a generating wandering subspace in $\mathcal{H}$. We say
that an $n$-tuple of operators $\boldsymbol{T}=(T_{1},\ldots,T_{n})$ acting on
a Hilbert space $\mathcal{H}$ is cyclic if there exists a vector
$x_{0}\in\mathcal{H}$ such that
$\mathcal{H}=\overline{\text{span}}\\{T_{1}^{k_{1}}\cdots
T_{n}^{k_{n}}x_{0}~{}:~{}(k_{1},\ldots,k_{n})\in\mathbb{Z}_{\geqslant
0}^{n}\\}.$

One of the major motivation behind studying wandering subspaces is their
natural connection with invariant subspaces. The existence of wandering
subspace property for the restriction of multiplication by the coordinate
function $M_{z}$ to any invariant subspace of the Hardy space
$H^{2}(\mathbb{D})$, the Bergman space $A^{2}(\mathbb{D})$ and the Dirichlet-
type space $\mathcal{D}_{\mu}(\mathbb{D})$ for any finite positive Borel
measure $\mu$ on the unit circle $\mathbb{T}$ are shown by Beurling in [2], by
Alemann, Richter and Sundberg in [1], and by Richter in [10], respectively.
Later on, Shimorin proved the existence of this property for an abstract class
of left-invertible operators (see [19]).

In multi-variable operator theory, characterization of invariant subspaces for
$n$-tuples of operators is one of the interesting and demanding topic from the
last few decades. In continuation with that, characterization of $n$-tuples of
commuting bounded linear operators on a Hilbert space having wandering
subspace property turns out to be one of the most important and interesting
questions in this regard. In [17], Rudin shows that the existence of the
wandering subspace property of general invariant subspace fails even for a
pair of commuting isometries. In particular, he constructs an invariant
subspace of Hardy space over bidisc $\mathbb{D}^{2}$ not having wandering
subspace property. For analogous kind of examples on analytic functions over
Euclidean unit ball, we refer the reader to [4]. Therefore, an immediate
generalization of the single variable result is not possible. Meanwhile, the
existence of wandering subspace property for several class of $n$-tuple of
commuting bounded linear operators $(T_{1},\ldots,T_{n})$ acting on a Hilbert
space $\mathcal{H}$ has been proven under the assumption of doubly
commutativity, that is, $T_{i}^{*}T_{j}=T_{j}T_{i}^{*}$ for all $1\leqslant
i<j\leqslant n$. For more details, we refer the reader to [8, 9, 13, 5] and
references therein.

In this article, we have significantly extended the class of commuting pairs
of bounded linear operators on a Hilbert space that possess the wandering
subspace property. In particular, we prove a multi-variable generalization of
Shimorin’s result [19], which turns out to be a generalization of all
aforesaid results in this regard. Along with this, we develop a natural bidisc
$\mathbb{D}^{2}$ counter part of the existing theory of the Dirichlet space
over $\mathbb{D}$ introduced by Richter [10, 12] with an explicit description
of the norm. We also show that it becomes a model space for a class of pair of
cyclic analytic $2$-isometries. With the help of the model theorem we
establish a connection between the doubly commuting pairs of analytic
$2$-isometries and the class of analytic $2$-isometries introduced in this
article.

We now describe our main results more precisely. For a pair of commuting
bounded linear operators $\boldsymbol{T}=(T_{1},T_{2})$ on a Hilbert space
$\mathcal{H}$, we say $\boldsymbol{T}$ is analytic $2$-isometry if for each
$i=1,2$, $T_{i}$ is analytic $2$-isometry. For the definition and basic
properties of analyticity and $2$-isometry, we refer the reader to [12].

Let $\mathcal{M}_{+}(\mathbb{T})$ denote the set of all positive finite Borel
measures on the unit circle $\mathbb{T}.$ Let
$\mathcal{PM}_{+}(\mathbb{T}^{2})$ denote the set of all measures
$\boldsymbol{\mu}$ on $\mathbb{T}^{2}$ such that
$\boldsymbol{\mu}=\mu_{1}\times\mu_{2},$ where $\mu_{1}$ and $\mu_{2}$ are
positive finite Borel measures on $\mathbb{T},$ i.e.

$\mathcal{PM}_{+}(\mathbb{T}^{2}):=\left\\{\boldsymbol{\mu}=\mu_{1}\times\mu_{2}:\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T})\right\\}.$

We define the Poisson kernel over $\mathbb{D}^{2}$ as a product of Poisson
kernels over $\mathbb{D}$ see also [17, Chapter 2]. More precisely, for any
$z=(z_{1},z_{2})\in\mathbb{D}^{2}$ and
$\xi=(\xi_{1},\xi_{2})\in\mathbb{T}^{2}$,
$\boldsymbol{P}(z,\xi):=P(z_{1},\xi_{1})~{}P(z_{2},\xi_{2}),$ where

$P(z_{i},\xi_{i})=\frac{(1-|z_{i}|^{2})}{|\xi_{i}-z_{i}|^{2}},\quad i=1,2.$

For any $\boldsymbol{\mu}\in\mathcal{PM}_{+}(\mathbb{T}^{2})$ and
$(z_{1},z_{2})\in\mathbb{D}^{2}$, we define the Poisson integral
$\boldsymbol{P}_{\boldsymbol{\mu}}(z_{1},z_{2}):=P_{\mu_{1}}(z_{1})$
$P_{\mu_{2}}(z_{2}),$ where

$P_{\mu_{i}}(z_{i})=\frac{1}{2\pi}\int_{0}^{2\pi}P(z_{i},e^{it})d\mu_{i}(t),\quad
i=1,2.$

Let $\mathcal{O}(\mathbb{D}^{2})$ denote the set of all complex valued
holomorphic functions on the unit bidisc $\mathbb{D}^{2}.$ For any
$f\in\mathcal{O}(\mathbb{D}^{2})$ and for
$\boldsymbol{\mu}\in\mathcal{PM}_{+}(\mathbb{T}^{2}),$ define

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f)$
$\displaystyle:=\lim_{r\rightarrow{1}^{-}}\frac{1}{2\pi}\iint_{\mathbb{D}\times\mathbb{T}}\left|\frac{\partial{f}}{\partial{z_{1}}}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dA(z_{1})~{}dt,$
(1) $\displaystyle D_{\boldsymbol{\mu},2}^{(2)}(f)$
$\displaystyle:=\lim_{r\rightarrow{1}^{-}}\frac{1}{2\pi}\iint_{\mathbb{T}\times\mathbb{D}}\left|\frac{\partial{f}}{\partial{z_{2}}}(re^{it},z_{2})\right|^{2}P_{\mu_{2}}(z_{2})~{}dA(z_{2})~{}dt,$
(2)

and

$\displaystyle D_{\boldsymbol{\mu},3}^{(2)}(f)$
$\displaystyle:=\iint_{\mathbb{D}^{2}}\left|\frac{\partial^{2}{f}}{\partial{z_{1}}\partial{z_{2}}}(z_{1},z_{2})\right|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(z_{1},z_{2})~{}dA(z_{1})dA(z_{2}),$
(3)

where $dA$ denotes the normalized Lebesgue area measure on the unit disc
$\mathbb{D}$. Wherever needed, we shall use the short hand notations
$\partial_{1}f,\partial_{2}f$ and $\partial_{1}\partial_{2}f$ in place of
$\frac{\partial{f}}{\partial{z_{1}}},\frac{\partial{f}}{\partial{z_{2}}}$ and
$\frac{\partial^{2}{f}}{\partial{z_{1}}\partial{z_{2}}}$ respectively.
Whenever it is convenient, we shall denote $\boldsymbol{z}=(z_{1},z_{2})$ and
$d\boldsymbol{A}(\boldsymbol{z})=dA(z_{1})dA(z_{2}).$ We define the Dirichlet-
type integral over $\mathbb{D}^{2}$ as

$\displaystyle
D^{(2)}_{\boldsymbol{\mu}}(f):=D_{\boldsymbol{\mu},1}^{(2)}(f)+D_{\boldsymbol{\mu},2}^{(2)}(f)+D_{\boldsymbol{\mu},3}^{(2)}(f).$

We define the Dirichlet-type space on the unit bidisc with respect to the non-
zero measure $\boldsymbol{\mu}$ to be the following space of analytic
functions

$\displaystyle\mathcal{D}^{2}(\boldsymbol{\mu}):=\\{f\in\mathcal{O}(\mathbb{D}^{2}):D^{(2)}_{\boldsymbol{\mu}}(f)<\infty\\}.$

For $\boldsymbol{\mu}=0,$ we set
$\mathcal{D}^{2}(\boldsymbol{\mu)}=H^{2}(\mathbb{D}^{2}),$ the Hardy space
over the unit bidisc. The function $D^{(2)}_{\boldsymbol{\mu}}(\cdot)$ defines
a semi-norm on $\mathcal{D}^{2}(\boldsymbol{\mu}).$ In Lemma 2.1, we prove
that Dirichlet-type space over unit bidisc with respect to any product
positive measure $\boldsymbol{\mu}$ on $\mathbb{T}^{2}$ is contained in the
Hardy space over unit bidisc as a subset. With this information, we can
convert the semi-norm $D^{(2)}_{\boldsymbol{\mu}}(\cdot)$ into a norm in a
traditional way by adding the norm borrowed from $H^{2}(\mathbb{D}^{2})$ to
it. More precisely, for any $f\in\mathcal{D}^{2}(\boldsymbol{\mu}),$ we define
the norm $\|f\|_{\boldsymbol{\mu}}$ by setting

$\displaystyle\|f\|_{\boldsymbol{\mu}}^{2}=\|f\|_{{H}^{2}(\mathbb{D}^{2})}^{2}+D^{(2)}_{\boldsymbol{\mu}}(f).$

It is observed in Theorem 2.2 that for any
$\boldsymbol{\mu}\in\mathcal{PM}_{+}(\mathbb{T}^{2})$, the space
$(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$ is a
reproducing kernel Hilbert space. When $\boldsymbol{\mu}$ is the Lebesgue
measure on $\mathbb{T}^{2}$, then we denote
$\mathcal{D}^{2}(\boldsymbol{\mu})$ by $\mathcal{D}^{2}$. The associated
kernel function $K_{\mathcal{D}^{2}}$ turns out to be

$K_{\mathcal{D}^{2}}\big{(}(z_{1},z_{2}),(w_{1},w_{2})\big{)}=K_{\mathcal{D}}(z_{1},w_{1})K_{\mathcal{D}}(z_{2},w_{2}),\quad(z_{1},z_{2}),(w_{1},w_{2})\in\mathbb{D}^{2},$
(4)

where $K_{\mathcal{D}}$ is the reproducing kernel for the classical Dirichlet
space $\mathcal{D}$.

In a recent article [3], for any
$\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T})$ the Dirichlet-type space
$\mathcal{D}(\mu_{1},\mu_{2})$ over the bidisc is defined. In their
formulation of $\mathcal{D}(\mu_{1},\mu_{2})$, they consider the semi-norm
$\mathcal{D}_{\mu_{1},\mu_{2}}(\cdot)$, which is determined solely by the
terms specified in equations (1) and (2). The primary objective of their
article was to analyze and propose a suitable model for toral $2$-isometries.
By adopting their chosen norm, the authors in [3] achieve a squared norm for a
function $f$ in their version of the classical Dirichlet space over the bidisc
given by $\sum_{m,n}|\hat{f}(m,n)|^{2}(n+m+1)$. This differs from that of
$\mathcal{D}\otimes\mathcal{D}$, which might be a very natural choice for the
classical Dirichlet space over the bidisc. Our motivation to introduce the
quantity in equation (3) is to restore this property, as indicated in equation
(4). Consequently, the class of commuting tuples $(T_{1},T_{2})$ that we
investigate in this article differs significantly from those studied in [3].
The specific class of commuting tuples we examine, which we term ‘left-inverse
commuting pairs’, is defined as follows. Throughout this article, we
occasionally refer to results obtained in [3] or incorporate some of their
arguments. In such instances, we explicitly acknowledge the source.

######  Definition 1.1 (left-inverse commuting tuple).

A tuple $(T_{1},\ldots,T_{n})$ of left invertible commuting operators is said
to be a left-inverse commuting tuple if

$\displaystyle L_{i}T_{j}=T_{j}L_{i}\quad\mbox{for }i\neq j,$ (5)

where $L_{j}$ denotes the left inverse of $T_{j}$ with the property
$\ker(L_{j})=\ker T_{j}^{*}.$

The multiplication by coordinate functions $z_{1}$ and $z_{2}$ are bounded
analytic $2$-isometries on $\mathcal{D}^{2}(\boldsymbol{\mu}).$ These
operators further satisfy (5). It turns out that the set of polynomials
densely sits in the Hilbert space
$(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$. Moreover,
we observe in Lemma 2.18 that

$\displaystyle\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\boldsymbol{\mu}}=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}(\mu_{1})}\langle
z_{2}^{n},z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})},$

for any $m,n,p,q\in\mathbb{Z}_{\geqslant 0}.$ Here $\mathcal{D}(\mu_{j})$
denotes the Dirichlet-type space over the unit disc as defined by Richter in
[12]. In Theorem 3.2, we prove that if $(T_{1},T_{2})$ is a left-inverse
commuting pair of analytic $2$-isometries on a Hilbert space $\mathcal{H}$
then $\ker T_{1}^{*}\cap\ker T_{2}^{*}$ serves as a wandering subspace for the
pair $(T_{1},T_{2}).$ In [5, Corollary 2.2], authors have proved that if
$(T_{1},T_{2})$ is a doubly commuting pair of bounded analytic $2$-isometries
then $\ker T_{1}^{*}\cap\ker T_{2}^{*}$ serves as a wandering subspace for
$(T_{1},T_{2}).$ Note that if a pair $(T_{1},T_{2})$ is doubly commuting then
it necessarily satisfies (5). In this sense, Theorem 3.2 generalizes [5,
Corollary 2.2]. In Example 4.3, we present an example of analytic
$2$-isometric left-inverse commuting pair $(T_{1},T_{2})$ which is not doubly
commuting. Furthermore, in Theorem 4.1, we prove that if $(T_{1},T_{2})$ is a
left-inverse commuting pair of analytic $2$-isometries with

$\displaystyle\langle
T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=\langle
T_{1}^{m}x_{0},T_{1}^{p}x_{0}\rangle\langle
T_{2}^{n}x_{0},T_{2}^{q}x_{0}\rangle,\quad m,n,p,q\in\mathbb{Z}_{\geqslant
0},$

for some unit vector $x_{0}\in\ker T_{1}^{*}\cap\ker T_{2}^{*},$ then the pair
$(T_{1},T_{2})$ is unitarily equivalent to $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\mu)}$ for some
$\boldsymbol{\mu}\in\mathcal{P}\mathcal{M}_{+}(\mathbb{T}^{2}).$

##  2\. A few properties of $\mathcal{D}^{2}(\boldsymbol{\mu})$

In what follows, it will be useful to set the following notations: for any
$R,R_{1},R_{2}\in(0,1),$ set

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f,R)$
$\displaystyle:=\lim_{r\rightarrow{1}^{-}}\frac{1}{2\pi}\iint_{R\mathbb{D}\times\mathbb{T}}\left|\partial_{1}{f}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dA(z_{1})~{}dt$
$\displaystyle D_{\boldsymbol{\mu},2}^{(2)}(f,R)$
$\displaystyle:=\lim_{r\rightarrow{1}^{-}}\frac{1}{2\pi}\iint_{\mathbb{T}\times
R\mathbb{D}}\left|\partial_{2}{f}(re^{it},z_{2})\right|^{2}P_{\mu_{2}}(z_{2})~{}dA(z_{2})~{}dt$
$\displaystyle D_{\boldsymbol{\mu},3}^{(2)}(f,R_{1},R_{2})$
$\displaystyle:=\iint_{R_{1}\mathbb{D}\times
R_{2}\mathbb{D}}\left|\partial_{1}\partial_{2}{f}(z_{1},z_{2})\right|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(z_{1},z_{2})~{}dA(z_{1})dA(z_{2}).$

If $R_{1}=R_{2}=R,$ then $D_{\boldsymbol{\mu},3}^{(2)}(f,R_{1},R_{2})$ will
simply be denoted by $D_{\boldsymbol{\mu},3}^{(2)}(f,R).$

The norm on $\mathcal{D}^{2}(\boldsymbol{\mu})$ satisfies the parallelogram
identity, i.e.,
$\|f+g\|^{2}_{\boldsymbol{\mu}}+\|f-g\|^{2}_{\boldsymbol{\mu}}=2(\|f\|^{2}_{\boldsymbol{\mu}}+\|g\|^{2}_{\boldsymbol{\mu}}),$
and therefore
$(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$ is an inner
product space. We shall denote the inner product of
$\mathcal{D}^{2}(\boldsymbol{\mu})$ by
$\langle\cdot,\cdot\rangle_{\boldsymbol{\mu}}$. Thus for
$f,g\in\mathcal{D}^{2}(\boldsymbol{\mu})$, we set

$\langle f,g\rangle_{\boldsymbol{\mu}}=\langle
f,g\rangle_{H^{2}(\mathbb{D}^{2})}+\langle
f,g\rangle_{\boldsymbol{\mu},1}+\langle
f,g\rangle_{\boldsymbol{\mu},2}+\langle f,g\rangle_{\boldsymbol{\mu},3},$

where we choose to denote

$\displaystyle\langle f,g\rangle_{{\boldsymbol{\mu},1}}:$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\iint_{\mathbb{D}\times\mathbb{T}}\partial_{1}f(z_{1},re^{it})~{}\overline{\partial_{1}g(z_{1},re^{it})}~{}P_{\mu_{1}}(z_{1})dA(z_{1})dt$
$\displaystyle\langle f,g\rangle_{\boldsymbol{\mu},2}:$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\iint_{\mathbb{T}\times\mathbb{D}}\partial_{2}f(re^{it},z_{2})~{}\overline{\partial_{2}g(re^{it},z_{2})}~{}P_{\mu_{2}}(z_{2})dA(z_{2})dt$
$\displaystyle\langle f,g\rangle_{\boldsymbol{\mu},3}:$
$\displaystyle=\iint_{\mathbb{D}^{2}}\partial_{1}\partial_{2}f(\boldsymbol{z})~{}\overline{\partial_{1}\partial_{2}g(\boldsymbol{z})}~{}\boldsymbol{P}_{\boldsymbol{\mu}}(\boldsymbol{z})d\boldsymbol{A}(\boldsymbol{z}).$

We start with the following important lemma.

######  Lemma 2.1.

For any $\boldsymbol{\mu}\in
PM_{+}(\mathbb{T}^{2}),~{}~{}\mathcal{D}^{2}(\boldsymbol{\mu})\subseteq
H^{2}(\mathbb{D}^{2}).$

###### Proof.

Let
$f(z_{1},z_{2})=\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}\in\mathcal{D}^{2}(\boldsymbol{\mu}).$
Using monotone convergence theorem, we get

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f)\geqslant\lim_{R\to
1^{-}}\lim_{r\to
1^{-}}\frac{1}{2\pi}\displaystyle\iint_{R\mathbb{D}\times\mathbb{T}}\left|\partial_{1}f(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})dtdA(z_{1}).$

Since
$P_{\mu_{1}}(z_{1})\geqslant\frac{\mu_{1}(\mathbb{T})}{8\pi}(1-|z_{1}|^{2})$
(see [18, pg 236]), we get that,

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f)$
$\displaystyle\geqslant\lim_{r\to 1^{-}}\lim_{R\to
1^{-}}\frac{1}{2\pi}\iint_{R\mathbb{D}\times\mathbb{T}}\bigg{|}\sum_{m=1,n=0}^{\infty}ma_{m,n}z_{1}^{m-1}(re^{it})^{n}\bigg{|}^{2}~{}\frac{\mu_{1}(\mathbb{T})}{8\pi}(1-|z_{1}|^{2})dt~{}dA(z_{1})$
$\displaystyle=\lim_{r\to 1^{-}}\lim_{R\to
1^{-}}\frac{\mu_{1}(\mathbb{T})}{4\pi}\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}m^{2}|a_{m,n}|^{2}r^{2n}R^{2m}\left\\{\frac{1}{2m}-\frac{R^{2}}{2m+2}\right\\}$
$\displaystyle=\lim_{r\to
1^{-}}\frac{\mu_{1}(\mathbb{T})}{8\pi}\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}|a_{m,n}|^{2}\frac{m}{m+1}r^{2n}.$

Since $\frac{m}{m+1}\geqslant\frac{1}{2},$ it follows that

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f)\geqslant\lim_{r\to
1^{-}}\frac{\mu_{1}(\mathbb{T})}{16\pi}\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}|a_{m,n}|^{2}r^{2n}.$
(6)

Similarly we also get that,

$D_{\boldsymbol{\mu},2}^{(2)}(f)\geqslant\lim_{r\to
1^{-}}\frac{\mu_{2}(\mathbb{T})}{16\pi}\sum_{m=0}^{\infty}\sum_{n=1}^{\infty}|a_{m,n}|^{2}r^{2m}.$
(7)

Since $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$, the left hand sides and
consequently the right hand sides of (6) and (7) are finite. Now, it follows
from the Dominated Convergence Theorem that $f$ belongs to
$H^{2}(\mathbb{D}^{2}).$ ∎

In the following theorem, we note that it is, in fact, a reproducing kernel
Hilbert space.

######  Theorem 2.2.

For any $\boldsymbol{\mu}\in\mathcal{PM}_{+}(\mathbb{T}^{2}),$ the Dirichlet-
type space $(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$
is a reproducing kernel Hilbert space.

###### Proof.

From Lemma 2.1, we have that $\mathcal{D}^{2}(\boldsymbol{\mu})\subseteq
H^{2}(\mathbb{D}^{2}).$ This implies that
$(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$ is
contained contractively in $H^{2}(\mathbb{D}^{2}).$ Hence
$\mathcal{D}^{2}(\boldsymbol{\mu})$ is a reproducing kernel Hilbert space
provided $\mathcal{D}^{2}(\boldsymbol{\mu})$ is a Hilbert space with respect
to $\|\cdot\|_{\boldsymbol{\mu}}.$ To prove the completeness of the space
$(\mathcal{D}^{2}(\boldsymbol{\mu}),\|\cdot\|_{\boldsymbol{\mu}})$, let
$(f_{n})$ be a Cauchy sequence in $\mathcal{D}^{2}(\boldsymbol{\mu}).$ This in
turn implies that $(f_{n})$ is a Cauchy sequence in $H^{2}(\mathbb{D}^{2}).$
Since $H^{2}(\mathbb{D}^{2})$ is a Hilbert space, there exists an $f\in
H^{2}(\mathbb{D}^{2})$ such that $f_{n}\rightarrow f$ as $n\rightarrow\infty$
in $H^{2}(\mathbb{D}^{2})$. Fix an $\epsilon>0.$ There exists an
$N_{1}\in\mathbb{N}$ such that
$\|f_{n}-f\|_{H^{2}(\mathbb{D}^{2})}<\frac{\epsilon}{2}$ for all $n\geqslant
N_{1}.$ Note that

$\displaystyle
D^{(2)}_{\boldsymbol{\mu}}(f_{n}-f)=D_{\boldsymbol{\mu},1}^{(2)}(f_{n}-f)+D_{\boldsymbol{\mu},2}^{(2)}(f_{n}-f)+D_{\boldsymbol{\mu},3}^{(2)}(f_{n}-f)$
$\displaystyle=\lim_{r\to 1^{-}}\lim_{r_{1}\to
1^{-}}\iint_{r_{1}\mathbb{D}\times\mathbb{T}}\left|{\partial_{1}{(f_{n}-f)}}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})+$
$\displaystyle\lim_{r\to 1^{-}}\lim_{r_{1}\to 1^{-}}\iint_{\mathbb{T}\times
r_{1}\mathbb{D}}\left|{\partial_{2}{(f_{n}-f)}}(re^{it},z_{2})\right|^{2}P_{\mu_{2}}(z_{2})~{}dA(z_{1})~{}dt+$
$\displaystyle\lim_{r_{1}\to
1^{-}}\iint_{(r_{1}\mathbb{D})^{2}}\left|{\partial_{1}\partial_{2}(f_{n}-f)}(\boldsymbol{z})\right|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(\boldsymbol{z})d\boldsymbol{A}(\boldsymbol{z}).$

Since, for each $n\in\mathbb{N},$ $f_{n}$ is an analytic map, therefore
$f_{n},{\partial_{1}f_{n}},$ ${\partial_{2}f_{n}},$ and
${\partial_{1}\partial_{2}f_{n}}$ will converge uniformly on any compact
subset of bidisc $\mathbb{D}^{2}$ respectively to $f,{\partial_{1}f},$
${\partial_{2}f},$ and ${\partial_{1}\partial_{2}f}$. For each
$n\in\mathbb{N},$ let
$g_{n}(z_{1},re^{it}):=\partial_{1}(f_{n}-f)(z_{1},re^{it}).$ Then we have,

$\displaystyle\iint_{r_{1}\mathbb{D}\times\mathbb{T}}\displaystyle\lim_{n\to\infty}\left|g_{n}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})=0.$

Therefore, by the Monotone Convergence theorem, there exists
$N_{2}\in\mathbb{N}$ such that

$\displaystyle\lim_{r\to 1^{-}}\lim_{r_{1}\to
1^{-}}\iint_{r_{1}\mathbb{D}\times\mathbb{T}}\left|g_{n}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})<\frac{\epsilon}{6},\quad
n\geqslant N_{2}.$

Therefore, we get

$\lim_{r\to
1^{-}}\iint_{\mathbb{D}\times\mathbb{T}}\left|{\partial_{1}{(f_{n}-f)}}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})<\frac{\epsilon}{6},\quad
n\geqslant N_{2}.$

Similarly, we get that for some $N_{3},N_{4}\in\mathbb{N},$

$\displaystyle\lim_{r\to
1^{-}}\iint_{\mathbb{T}\times\mathbb{D}}\left|{\partial_{2}{(f_{n}-f)}}(re^{it},z_{2})\right|^{2}P_{\mu_{2}}(z_{2})~{}dA(z_{1})dt~{}<\frac{\epsilon}{6},\quad
n\geqslant N_{3},$

and

$\displaystyle\lim_{r\rightarrow
1^{-}}\iint_{(r_{1}\mathbb{D})^{2}}\left|{\partial_{1}\partial_{2}(f_{n}-f)}(z_{1},z_{2})\right|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(\boldsymbol{z})d\boldsymbol{A}(\boldsymbol{z})<\frac{\epsilon}{6},\quad
n\geqslant N_{4}.$

By choosing $N=\max\\{N_{1},N_{2},N_{3},N_{4}\\},$ we observe that
$D^{(2)}_{\boldsymbol{\mu}}(f_{n}-f)<\frac{\epsilon}{2},$ for all $n\geqslant
N.$ Since $(f_{n})$ is Cauchy in $\mathcal{D}^{2}(\boldsymbol{\mu}),$ it is a
bounded sequence. That is, there exists a $K>0$ such that,
$D^{(2)}_{\boldsymbol{\mu}}(f{{}_{n}})<K$ for all $n\in\mathbb{N}.$ Using the
inequality
$\|f\|_{\boldsymbol{\mu}}\leqslant\|(f-f_{N})\|_{\boldsymbol{\mu}}+\|f_{N}\|_{\boldsymbol{\mu}},$
we conclude that $f\in\mathcal{D}^{2}(\boldsymbol{\mu}).$ Since, for all
$n\geqslant N,$ we have
$D^{(2)}_{\boldsymbol{\mu}}(f_{n}-f)<\frac{\epsilon}{2}$ and
$\|f_{n}-f\|_{H^{2}(\mathbb{D}^{2})}<\frac{\epsilon}{2},$ it follows that
Using this and the fact that $f_{n}\to f$ in
$\mathcal{D}^{2}(\boldsymbol{\mu})$ it follows that, $(f_{n})$ converges to
$f$ in $\mathcal{D}^{2}(\boldsymbol{\mu}).$ This completes the proof of the
theorem. ∎

When $\boldsymbol{\mu}$ is the Lebesgue measure on $\mathbb{T}^{2}$, then we
denote $\mathcal{D}^{2}(\boldsymbol{\mu})$ by $\mathcal{D}^{2}$ and call it
classical Dirichlet space over bidisc. We note that

$\mathcal{D}^{2}=\bigg{\\{}f(z_{1},z_{2})=\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}:\sum_{m,n=0}^{\infty}(m+1)(n+1)|a_{m,n}|^{2}<\infty\bigg{\\}},$

and consequently norm of any element $f\in\mathcal{D}^{(2)}$ with power series
representation $\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}$, is given by

$\|f\|_{\mathcal{D}^{2}}^{2}=\sum_{m,n=0}^{\infty}(m+1)(n+1)|a_{m,n}|^{2}.$

The associated kernel function $K_{\mathcal{D}^{2}}$ at
$(z_{1},z_{2}),(w_{1},w_{2})\in\mathbb{D}^{2}$ therefore turns out to be

$K_{\mathcal{D}^{2}}\big{(}(z_{1},z_{2}),(w_{1},w_{2})\big{)}=\sum_{m,n=0}^{\infty}\frac{z_{1}^{m}\overline{w}_{1}^{m}~{}~{}z_{2}^{n}\overline{w}_{2}^{n}}{(m+1)(n+1)}=K_{\mathcal{D}}(z_{1},w_{1})K_{\mathcal{D}}(z_{2},w_{2}),$

where $K_{\mathcal{D}}$ is the reproducing kernel for the classical Dirichlet
space $\mathcal{D}$. For any $f\in\mathcal{O}(\mathbb{D}^{2})$ and fixed
$(w_{1},w_{2})\in\mathbb{D}^{2},$ we define $f_{w_{1}}(z):=f(w_{1},z)$ and
$f_{w_{2}}(z):=f(z,w_{2})$ for any $z\in\mathbb{D}.$ We may refer $f_{w_{1}}$
and $f_{w_{2}}$ as $w_{1}$-slice and $w_{2}$-slice of $f$ respectively. The
next proposition tells us that for any $f\in\mathcal{D}^{2},$ its
$w_{1}$-slice as well as $w_{2}$-slice belong to Dirichlet space
$\mathcal{D}.$

######  Proposition 2.3.

Suppose $(w_{1},w_{2})\in\mathbb{D}^{2}$ and $f$ is a function in
$\mathcal{D}^{2}.$ Then $f_{w_{1}}$ and $f_{w_{2}}$ are in the classical
Dirichlet space $\mathcal{D}.$

###### Proof.

Let $f(z_{1},z_{2})$ be represented by the power series
$\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}$. Fix
$(w_{1},w_{2})\in\mathbb{D}^{2}.$ For each $m\in\mathbb{Z}_{+}$, define
$b_{m}=\sum_{n=0}^{\infty}a_{m,n}w_{2}^{n}.$ In these notations, we have

$\displaystyle f_{w_{2}}(z)=\sum_{m=0}^{\infty}b_{m}z^{m}.$

Note that

$\displaystyle\sum_{m=0}^{\infty}(m+1)|b_{m}|^{2}\leqslant\frac{1}{1-|w_{2}|^{2}}\sum_{m,n=0}^{\infty}(m+1)|a_{m,n}|^{2}\leqslant\frac{1}{1-|w_{2}|^{2}}\sum_{m,n=0}^{\infty}(m+1)(n+1)|a_{m,n}|^{2}.$

This shows that $f_{w_{2}}\in\mathcal{D}.$ Similarly, one can show that
$f_{w_{1}}\in\mathcal{D}$. ∎

Proposition 2.3 is true for general
$\boldsymbol{\mu}\in\mathcal{P}\mathcal{M}_{+}(\mathbb{T}^{2})$ when
$(w_{1},w_{2})=0.$ We shall need this in what follows, e.g. in Lemma 2.11.
Moreover, it says that $f(0,z_{2})$ and $f(z_{1},0),$ when treated as
functions on bidisc, belong to Dirichlet-type space
$\mathcal{D}^{2}(\boldsymbol{\mu}).$

Throughout the paper, we will consistently adhere to the following notations:
for $m,p\in\mathbb{Z}_{\geqslant 0}$, define $m\wedge p:=\min\\{m,p\\}$ and
$m\vee p:=\max\\{m,p\\}$. Also, we choose $S$ to denote the open square
$(0,2\pi)\times(0,2\pi)$.

######  Lemma 2.4.

For $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$, the functions $f(0,z_{2})$ and
$f(z_{1},0)\in\mathcal{D}^{2}(\boldsymbol{\mu})$. Moreover,
$D_{\boldsymbol{\mu},j}^{(2)}(f)\geqslant
D_{\boldsymbol{\mu},j}^{(2)}(f(0,z_{2}))$ and
$D_{\boldsymbol{\mu},j}^{(2)}(f)\geqslant
D_{\boldsymbol{\mu},j}^{(2)}(f(z_{1},0)),$ for $j=1,2,3.$

###### Proof.

Consider $f(z_{1},z_{2})=\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}$. If
we let $z_{1}=0$, then the resulting function $f_{2}=f(0,z_{2})$ will be
$f(0,z_{2})=\sum_{n=0}^{\infty}a_{0,n}z_{2}^{n}$. So it is enough to check
$D^{(2)}_{\boldsymbol{\mu},2}(f)<\infty$. The same computations as above show
that

$\displaystyle
D_{\boldsymbol{\mu},2}^{(2)}(f_{2},R)=\frac{1}{2\pi}\sum_{n,q=1}^{\infty}(n\wedge
q)a_{0,n}\overline{a_{0,q}}~{}\widehat{\mu}_{2}(q-n)~{}R^{2(n\vee q)}.$

Thus from (12), it can be seen that

$\displaystyle D_{\boldsymbol{\mu},2}^{(2)}(f,R)$ $\displaystyle=$
$\displaystyle\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\sum_{m=0}^{\infty}\sum_{n,q=1}^{\infty}(n\wedge
q)a_{m,n}\overline{a_{m,q}}~{}\widehat{\mu}_{2}(q-n)~{}R^{2(n\vee q)}r^{2m}$
$\displaystyle=$ $\displaystyle
D_{\boldsymbol{\mu},2}^{(2)}(f_{2},R)+D_{\boldsymbol{\mu},2}^{(2)}\Big{(}\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n},R\Big{)}.$

Taking limit $R$ tending to $1,$ this implies that
$D_{\boldsymbol{\mu},2}^{(2)}(f)\geqslant
D_{\boldsymbol{\mu},2}^{(2)}(f_{2})$. As
$D_{\boldsymbol{\mu},1}^{(2)}(f_{2})=0=D_{\boldsymbol{\mu},3}^{(2)}(f_{2}),$
the proof of the lemma is completed. ∎

Now, we proceed towards proving that the functions $z_{1}$ and $z_{2}$ are
multipliers of $\mathcal{D}^{2}(\boldsymbol{\mu}).$ To this end, few notations
and a number of lemmas are in order.

For any $f\in\mathcal{O}(\mathbb{D}^{2})$ with power series representation
$\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n},$ define

$(T_{k_{1},k_{2}}f)(z_{1},z_{2}):=\sum_{m=k_{1}}^{\infty}\sum_{n=k_{2}}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n},\quad
k_{1},k_{2}\in\mathbb{Z}_{\geqslant 0}.$

Following Richter’s paper [12, page 330], note that for any $R\in(0,1),$ and
$t\in[0,2\pi],$

$\frac{1}{2\pi}\int_{0}^{2\pi}e^{i(n-m)s}P(Re^{it},e^{is})ds=R^{|n-m|}e^{i(n-m)t},\quad
m,n\in\mathbb{Z}_{\geqslant 0}.$ (8)

In what follows, we shall be using the fact that for any $R\in[0,1],$
$t,s\in[0,2\pi],$ one gets $P(Re^{it},e^{is})=P(Re^{is},e^{it})$.

Let $\nu$ be a finite positive Borel measure on the unit circle $\mathbb{T}$
and $n\in\mathbb{Z},$ the $n^{\it{th}}$-Fourier coefficient of $\nu$ is
defined by,

$\displaystyle\widehat{\nu}(n):=\int_{0}^{2\pi}e^{-int}d\nu(t).$

At many occasions we shall be dealing with quantities of the form

$\displaystyle\sum_{m,p=1}^{\infty}\sum_{n=0}^{\infty}(m\wedge
p)a_{m,n}\overline{a}_{p,n}\widehat{\nu}(p-m)\mbox{ and
}\sum_{n,q=1}^{\infty}\sum_{m=0}^{\infty}(n\wedge
q)a_{m,n}\overline{a}_{m,q}\widehat{\nu}(q-n).$ (9)

We shall find it convenient to adopt the following notation to denote the
quantities in (9): For any measure $\nu\in\mathcal{M}_{+}(\mathbb{T}),$ define
the matrix $M_{\nu}:=\big{(}\\!\big{(}(j\wedge
k)\widehat{\nu}(k-j)\big{)}\\!\big{)}.$ This is a matrix of size infinite and
we shall denote it’s $(j,k)$-th entry by $M_{\nu}(j,k).$ The quantity in (9)
can be denoted as

$\displaystyle\sum_{n=0}^{\infty}\langle M_{\nu}\boldsymbol{a}_{\cdot
n},\boldsymbol{a}_{\cdot n}\rangle\mbox{ and }\sum_{m=0}^{\infty}\langle
M_{\nu}\boldsymbol{a}_{m\cdot},\boldsymbol{a}_{m\cdot}\rangle$

respectively, where $\boldsymbol{a}_{\cdot n}$ denotes the column
$(a_{1n},a_{2n},\ldots)^{\tt T}$ and $\boldsymbol{a}_{m\cdot}$ denotes the
column $(a_{m1},a_{m2},\ldots)^{\tt T}$. Note that for any
$\nu\in\mathcal{M}_{+}(\mathbb{T}),$ the matrix $M_{\nu}$ is formally positive
definite. We shall use this fact in what follows, e.g. in Lemma 2.14.

We have the following lemmas which are analogous to the results of Richter in
[12].

######  Lemma 2.5.

Let $0<R<1.$ If $f$ is analytic on $\mathbb{D}^{2}$, then

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f,R)=\lim_{r\to
1^{-}}\sum_{k_{1}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\left|(T_{k_{1},0}f)(Re^{is},re^{it})\right|^{2}P_{\mu_{1}}(Re^{is})dsdt.$

###### Proof.

For any $f(z_{1},z_{2})\in\mathcal{O}(\mathbb{D}^{2})$ with power series
representation $\sum_{m,n=0}^{\infty}a_{m,n}z_{1}^{m}z_{2}^{n}$ and
$k_{1}\in\mathbb{Z}_{\geqslant 0}.$ Since
$\int_{0}^{2\pi}e^{i(m-n)t}dt=\delta_{m,n},$ it follows that

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f,R)$ $\displaystyle=\lim_{r\to
1^{-}}\frac{1}{2\pi}\iint_{R\mathbb{D}\times\mathbb{T}}\left|\partial_{1}{f}(z_{1},re^{it})\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})$
$\displaystyle=\lim_{r\to
1^{-}}\frac{1}{2\pi}\iint_{R\mathbb{D}\times\mathbb{T}}\left|\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}ma_{m,n}z_{1}^{m-1}(re^{it})^{n}\right|^{2}P_{\mu_{1}}(z_{1})~{}dt~{}dA(z_{1})$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\frac{1}{\pi}\sum_{m,n,p=0}^{\infty}mpa_{m,n}\overline{a_{p,n}}\frac{R^{2(m\vee
p)}}{2(m\vee p)}\widehat{\mu}_{1}(p-m)~{}2\pi~{}r^{2n}.$

Therefore we get that

$D_{\boldsymbol{\mu},1}^{(2)}(f,R)=\lim_{r\to
1^{-}}\frac{1}{2\pi}\sum_{m,p=1}^{\infty}\sum_{n=0}^{\infty}M_{\mu_{1}}(m,p)a_{m,n}\overline{a_{p,n}}r^{2n}R^{2(m\vee
p)}.$ (10)

On the other hand, using (8), note that

$\displaystyle\lim_{r\to
1^{-}}\sum_{k_{1}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\left|(T_{k_{1},0}f)(Re^{is_{1}},re^{it})\right|^{2}P_{\mu_{1}}(Re^{is_{1}})ds_{1}dt$
(11) $\displaystyle=$ $\displaystyle\lim_{r\to
1^{-}}\sum_{k_{1}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\left|\sum_{m=k_{1}}^{\infty}\sum_{n=0}^{\infty}a_{m,n}(Re^{is_{1}})^{m}(re^{it})^{n}\right|^{2}P_{\mu_{1}}(Re^{is_{1}})ds_{1}dt$
$\displaystyle=$ $\displaystyle\lim_{r\to
1^{-}}\sum_{k_{1}=1}^{\infty}\frac{1}{(2\pi)^{2}}\int_{0}^{2\pi}\sum_{m,p=k_{1}}^{\infty}\sum_{n,q=0}^{\infty}a_{m,n}\overline{a_{p,q}}R^{2(m\vee
p)}r^{n+q}e^{i(n-q)t}\widehat{\mu}_{1}(p-m)dt$ $\displaystyle=$
$\displaystyle\lim_{r\to
1^{-}}\sum_{k_{1}=1}^{\infty}\frac{1}{2\pi}\sum_{m,p=k_{1}}^{\infty}\sum_{n=0}^{\infty}a_{m,n}\overline{a_{p,n}}R^{2(m\vee
p)}r^{2n}\widehat{\mu}_{1}(p-m)$ $\displaystyle=$ $\displaystyle\lim_{r\to
1^{-}}\frac{1}{2\pi}\sum_{m,p=1}^{\infty}\sum_{n=0}^{\infty}(m\wedge
p)a_{m,n}\overline{a_{p,n}}r^{2n}R^{2(m\vee p)}\widehat{\mu}_{1}(p-m).$

Equations (10) and (11) put together completes the proof of lemma. ∎

We note down the following formula which is analogous to (10). In fact, we
also have Lemma 2.6. We omit the proof of the formula as well as Lemma 2.6,
since these are analogous to that of Lemma 2.5.

$\displaystyle D_{\boldsymbol{\mu},2}^{(2)}(f,R)=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\sum_{m=0}^{\infty}\sum_{n,q=1}^{\infty}M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{m,q}}~{}R^{2(n\vee
q)}r^{2m}.$ (12)

######  Lemma 2.6.

Let $0<R<1.$ If $f$ is analytic on $\mathbb{D}^{2}$, then

$\displaystyle D_{\boldsymbol{\mu},2}^{(2)}(f,R)=\lim_{r\to
1^{-}}\sum_{k_{2}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\left|(T_{0,k_{2}}f)(re^{it},Re^{is})\right|^{2}P_{\mu_{2}}(Re^{is})dsdt.$

The following lemma gives a relationship between
$D_{\boldsymbol{\mu},1}^{(2)}(z_{1}f,R)$ and
$D_{\boldsymbol{\mu},1}^{(2)}(f,R).$ In particular, it proves that, for any
$0<R<1,$ $D_{\boldsymbol{\mu},1}^{(2)}(z_{1}f,R)\geqslant
R^{2}D_{\boldsymbol{\mu},1}^{(2)}(f,R).$

######  Lemma 2.7.

Let $0<R<1$. If $f$ is analytic on $\mathbb{D}^{2},$ then

$\displaystyle
D_{\boldsymbol{\mu},1}^{(2)}(z_{1}f,R)-R^{2}D_{\boldsymbol{\mu},1}^{(2)}(f,R)=\lim_{r\to
1^{-}}\frac{R^{2}}{(2\pi)^{2}}\iint_{S}\left|f(Re^{is},re^{it})\right|^{2}P_{\mu_{1}}(Re^{is})dsdt.$
(13)

###### Proof.

Applying (10) to $z_{1}f,$ we get

$\displaystyle
D_{\boldsymbol{\mu},1}^{(2)}(z_{1}f,R)=R^{2}~{}\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\sum_{m,n,p=0}^{\infty}M_{\mu_{1}}(m+1,p+1)a_{m,n}\overline{a_{p,n}}~{}R^{2(m\vee
p)}~{}r^{2n}$ (14)

On the other hand, using (8), we observe that

$\displaystyle R.H.S.\mbox{ of }\eqref{first-difference-formula-
with-R}=\lim_{r\to
1^{-}}\frac{R^{2}}{2\pi}\sum_{m,n,p=0}^{\infty}a_{m,n}\overline{a_{p,n}}R^{2(m\vee
p)}r^{2n}\widehat{\mu}_{1}(p-m).$ (15)

Equations (10), (14), and (15) put together completes the proof of the lemma.
∎

For any $f\in\mathcal{D}^{2}(\boldsymbol{\mu}),$ taking limit $R$ tends to $1$
in (13), one gets

$\displaystyle
D_{\boldsymbol{\mu},1}^{(2)}(z_{1}f)-D_{\boldsymbol{\mu},1}^{(2)}(f)=\lim_{r\to
1^{-}}\frac{1}{(2\pi)^{2}}\iint_{S}\left|\tilde{f}(e^{is},re^{it})\right|^{2}P_{\mu_{1}}(e^{is})dsdt,$
(16)

where $\tilde{f}$ denotes the boundary value function corresponding to $f.$

######  Lemma 2.8.

There exists a constant $C=C(\boldsymbol{\mu})$ such that for every
$f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and $0<R<1,$

$\displaystyle\lim_{r\to
1^{-}}\frac{1}{(2\pi)^{2}}\iint_{S}\left|f(Re^{is},re^{it})\right|^{2}P_{\mu_{1}}(Re^{is})dsdt\leqslant
C\|f\|^{2}_{\boldsymbol{\mu}}$

###### Proof.

Let $Q:=\lim_{r\to
1^{-}}\frac{1}{(2\pi)^{2}}\iint_{S}\left|f(Re^{is},re^{it})\right|^{2}P_{\mu_{1}}(Re^{is})dsdt.$
Note that

$f(Re^{is},re^{it})=f(0,0)+(T_{1,0}f)(Re^{is},re^{it})+\sum_{n=1}^{\infty}a_{0,n}(re^{it})^{n}.$

Therefore Lemmas 2.5 and 2.1 imply that

$\displaystyle|f(Re^{is},re^{it})|^{2}$ $\displaystyle\leqslant
3\bigg{(}|f(0,0)|^{2}+|(T_{1,0}f)(Re^{is_{1}},re^{it})|^{2}+\Big{|}\sum_{n=1}^{\infty}a_{0,n}(re^{it})^{n}\Big{|}^{2}\bigg{)}$
$\displaystyle\leqslant
3\bigg{(}\|f\|^{2}_{H^{2}(\mathbb{D}^{2})}+\sum_{k_{1}=1}^{\infty}|(T_{k_{1},0}f)(Re^{is_{1}},re^{it})|^{2}+\Big{|}\sum_{n=1}^{\infty}a_{0,n}(re^{it})^{n}\Big{|}^{2}\bigg{)}.$

Using Lemma 2.5, we deduce that

$\displaystyle
Q\leqslant\frac{3}{2\pi}\|f\|^{2}_{H^{2}(\mathbb{D}^{2})}\widehat{\mu_{1}}(0)+3D_{\boldsymbol{\mu},1}^{(2)}(f,R)+\frac{3}{2\pi}\widehat{\mu_{1}}(0)\lim_{r\to
1^{-}}\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}|a_{m,n}|^{2}r^{2n}.$

Choosing $C=\max\big{\\{}\frac{3\widehat{\mu_{1}}(0)}{\pi},3\big{\\}}$
completes the proof of the lemma. ∎

In this lemma, it is observed that the multiplication operator $M_{z_{1}}$
behaves like an isometry with respect to semi-norm
$D_{\boldsymbol{\mu},2}^{(2)}(\cdot).$

######  Lemma 2.9.

If $f\in\mathcal{O}(\mathbb{D}^{2})$ and $0<R<1$, then
$D_{\boldsymbol{\mu},2}^{(2)}(z_{1}f,R)=D_{\boldsymbol{\mu},2}^{(2)}(f,R).$

###### Proof.

Proof of this lemma easily follows from (12), hence we skip the details. ∎

In the next two lemmas, we study the semi-norm
$D_{\boldsymbol{\mu},3}^{(2)}(\cdot).$ We observe that
$D_{\boldsymbol{\mu},3}^{(2)}(z_{1}f,R)\geqslant
D_{\boldsymbol{\mu},3}^{(2)}(f,R)$ for any $0<R<1.$

######  Lemma 2.10.

Let $0<R<1$. If $f$ is analytic on $\mathbb{D}^{2},$ then

$\displaystyle
D_{\boldsymbol{\mu},3}^{(2)}(f,R)=\sum_{k_{1}=1}^{\infty}\sum_{k_{2}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}|(T_{k_{1},k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}.$

###### Proof.

Let $f\in\mathcal{O}(\mathbb{D}^{2}).$ Using uniform convergence of
$\partial_{1}\partial_{2}f$ on $R\mathbb{D}^{2}$ and a bit of computation, we
get

$\displaystyle
D_{\boldsymbol{\mu},3}^{(2)}(f,R)=\frac{1}{4\pi^{2}}\sum_{m,n=1}^{\infty}\sum_{p,q=1}^{\infty}M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}R^{2(m\vee
p)}R^{2(n\vee q)}.$ (17)

Now, using (8), we get that

$\displaystyle\sum_{k_{1},k_{2}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}|(T_{k_{1},k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}$
$\displaystyle=\sum_{k_{1},k_{2}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\Big{|}\sum_{m=k_{1}}^{\infty}\sum_{n=k_{2}}^{\infty}a_{m,n}(Re^{is_{1}})^{m}(Re^{is_{2}})^{n}\Big{|}^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}$
$\displaystyle=\sum_{k_{1},k_{2}=1}^{\infty}\frac{1}{(2\pi)^{2}}\iint_{S}\sum_{m,p=k_{1}}^{\infty}\sum_{n,q=k_{2}}^{\infty}a_{m,n}\overline{a_{p,q}}R^{2(m\vee
p)}R^{2(n\vee
q)}e^{i(m-p)\theta_{1}}e^{i(n-q)\theta_{2}}d\mu_{1}(\theta_{1})d\mu_{2}(\theta_{2})$
$\displaystyle=\frac{1}{4\pi^{2}}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\sum_{p=1}^{\infty}\sum_{q=1}^{\infty}(m\wedge
p)(n\wedge q)a_{m,n}\overline{a_{p,q}}R^{2(m\vee p)}R^{2(n\vee
q)}\widehat{\mu}_{1}(p-m)\widehat{\mu}_{2}(q-n).$

This completes the proof of the lemma. ∎

######  Lemma 2.11.

Let $0<R<1.$ If $f$ is in $\mathcal{D}^{2}(\boldsymbol{\mu})$, then there
exists a constant $C=C(\boldsymbol{\mu})$ such that
$D_{\boldsymbol{\mu},3}^{(2)}(z_{1}f,R)\leqslant
C\|f\|^{2}_{\boldsymbol{\mu}}.$

###### Proof.

Since $T_{k_{1},k_{2}}f=T_{k_{1}+1,k_{2}}(z_{1}f)$ for all
$k_{1},k_{2}\in\mathbb{Z}_{\geqslant 0},$ it follows from Lemma 2.10 that

$\displaystyle
D_{\boldsymbol{\mu},3}^{(2)}(z_{1}f,R)=\sum_{k_{2}=1}^{\infty}\frac{R^{2}}{(2\pi)^{2}}\iint_{S}|(T_{0,k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}+D_{\boldsymbol{\mu},3}^{(2)}(f,R)$

Note that

$\displaystyle|(T_{0,k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}$
$\displaystyle=\Big{|}\sum_{m=0}^{\infty}\sum_{n=k_{2}}^{\infty}a_{m,n}(Re^{is_{1}})^{m}(Re^{is_{2}})^{n}\Big{|}^{2}$
$\displaystyle\leqslant
2\bigg{(}\Big{|}\sum_{n=k_{2}}^{\infty}a_{0,n}(Re^{is_{2}})^{n}\Big{|}^{2}+|(T_{1,k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\bigg{)}$
$\displaystyle\leqslant
2\bigg{(}\Big{|}\sum_{n=k_{2}}^{\infty}a_{0,n}(Re^{is_{2}})^{n}\Big{|}^{2}+\sum_{k_{1}=1}^{\infty}|(T_{k_{1},k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\bigg{)}.$

Therefore, using Lemma 2.10, we get

$\displaystyle\sum_{k_{2}=1}^{\infty}\frac{R^{2}}{(2\pi)^{2}}\iint_{S}|(T_{0,k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}$
$\displaystyle\leqslant\sum_{k_{2}=1}^{\infty}\frac{2R^{2}}{(2\pi)^{2}}\iint_{S}\Big{|}\sum_{n=k_{2}}^{\infty}a_{0,n}(Re^{is_{2}})^{n}\Big{|}^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}+2R^{2}D_{\boldsymbol{\mu},3}^{(2)}(f,R)$
(18)

Since

$\displaystyle\iint_{\mathbb{T}\times
R\mathbb{D}}|\partial_{2}f(0,z_{2})|^{2}$ $\displaystyle
P_{\mu_{2}}(z_{2})dA(z_{2})dt=\sum_{n,q=1}^{\infty}(m\wedge
q)a_{0,n}~{}\overline{a_{0,q}}\widehat{\mu_{2}}(q-n)R^{2(m\vee q)}$
$\displaystyle=\frac{1}{\widehat{\mu_{1}}(0)}\sum_{k_{2}=1}^{\infty}\iint_{S}\Big{|}\sum_{n=k_{2}}^{\infty}a_{0,n}(Re^{is_{2}})^{n}\Big{|}^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}.$

The inequality in (2) now gives

$\displaystyle\sum_{k_{2}=1}^{\infty}\frac{R^{2}}{(2\pi)^{2}}\iint_{S}|(T_{0,k_{2}}f)(Re^{is_{1}},Re^{is_{2}})|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(Re^{is_{1}},Re^{is_{2}})ds_{1}ds_{2}$
$\displaystyle\leqslant\frac{2R^{2}}{2\pi}\widehat{\mu_{1}}(0)D^{(2)}_{\boldsymbol{\mu},2}f(0,z_{2})+2R^{2}D_{\boldsymbol{\mu},3}^{(2)}(f,R)$

By Lemma 2.10, Lemma 2.4 and the above computation, we note that,

$\displaystyle\iint_{R\mathbb{D}^{2}}\left|\partial_{1}\partial_{2}z_{1}f(\boldsymbol{z})\right|^{2}\boldsymbol{P}_{\boldsymbol{\mu}}(\boldsymbol{z})d\boldsymbol{A}(\boldsymbol{z})$
$\displaystyle\leqslant\frac{R^{2}}{\pi}\widehat{\mu_{1}}(0)D^{(2)}_{\boldsymbol{\mu},2}(f,R)+2R^{2}D_{\boldsymbol{\mu},3}^{(2)}(f,R)+D_{\boldsymbol{\mu},3}^{(2)}(f,R)$
$\displaystyle\leqslant C\|f\|^{2}_{\boldsymbol{\mu}}$

where
$C=\max\big{\\{}\frac{R^{2}}{\pi}\widehat{\mu_{1}}(0),2R^{2},1\big{\\}}$. ∎

The proof of above lemma, in the limit $R\to 1,$ implies that

$\displaystyle D_{\boldsymbol{\mu},3}^{(2)}(z_{1}f)\geqslant
D_{\boldsymbol{\mu},3}^{(2)}(f).$ (19)

######  Theorem 2.12.

Suppose $f\in\mathcal{O}(\mathbb{D}^{2})$. Then the following statements are
equivalent

  1. (i)

$f\in\mathcal{D}^{2}(\boldsymbol{\mu})$

  2. (ii)

$z_{1}f\in\mathcal{D}^{2}(\boldsymbol{\mu})$

  3. (iii)

$z_{2}f\in\mathcal{D}^{2}(\boldsymbol{\mu})$.

Moreover, in this case,
$\|z_{i}f\|_{\boldsymbol{\mu}}\geqslant\|f\|_{\boldsymbol{\mu}}$ for $i=1,2.$

###### Proof.

It is enough to prove the equivalence of $(i)$ and $(ii)$ only. Combining
Lemma 2.7, Lemma 2.8, Lemma 2.9, and Lemma 2.11 proves that $(i)$ implies
$(ii)$. For the other way implication we apply equation (16) with Lemma 2.9
and equation (19). ∎

For an infinite matrix $X:(\\!(x_{mp})\\!),$ define $\sigma^{*}X$ (see [19,
page 151]) to be the matrix whose $(m,p)$-th entry is given by $x_{m+1,p+1}.$

######  Lemma 2.13.

For any polynomial $p\in\mathcal{D}^{2}(\boldsymbol{\mu}),$ and $i=1,2$ we
have the following relations;

$\|z_{i}^{2}p\|^{2}_{\boldsymbol{\mu}}-2\|z_{i}p\|^{2}_{\boldsymbol{\mu}}+\|p\|^{2}_{\boldsymbol{\mu}}=0.$

###### Proof.

On computing the norm of $p,z_{1}p,z_{1}^{2}p$ for
$p(z_{1},z_{2})=\sum_{m,n=0}^{k}a_{m,n}z_{1}^{m}z_{2}^{n}$ w.r.to the defined
norm on $\mathcal{D}^{2}(\boldsymbol{\mu})$, we have the following. Note that

$\|p\|_{H^{2}(\mathbb{D}^{2})}^{2}=\|z_{1}p\|_{H^{2}(\mathbb{D}^{2})}^{2}=\|z_{1}^{2}p\|_{H^{2}(\mathbb{D}^{2})}^{2}=\sum_{m,n=0}^{k}|a_{m,n}|^{2}.$

Therefore it is enough to check

$D_{\boldsymbol{\mu}}^{(2)}(z_{i}^{2}p)-2D_{\boldsymbol{\mu}}^{(2)}(z_{i}p)+D_{\boldsymbol{\mu}}^{(2)}(p)=0.$

Now, from (10), (12), (17), with $R=1,$ it follows that

$\displaystyle
D_{\boldsymbol{\mu},1}^{(2)}(p)=\frac{1}{2\pi}\sum_{n=0}^{k}\langle
M_{\mu_{1}}\boldsymbol{a}_{\cdot n},\boldsymbol{a}_{\cdot
n}\rangle\quad\mbox{and}\quad
D_{\boldsymbol{\mu},2}^{(2)}(p)=\frac{1}{2\pi}\sum_{m=0}^{k}\langle
M_{\mu_{2}}\boldsymbol{a}_{m\cdot},\boldsymbol{a}_{m\cdot}\rangle.$

In the same spirit, we also get the formulae:

$D_{\boldsymbol{\mu},3}^{(2)}(p)=\frac{1}{4\pi^{2}}\langle(M_{\mu_{1}}\otimes
M_{\mu_{2}})\boldsymbol{a},\boldsymbol{a}\rangle,\
D^{(2)}_{\boldsymbol{\mu},1}(z_{1}p)=\frac{1}{2\pi}\sum_{n=0}^{k}\langle\sigma^{*}M_{\mu_{1}}\boldsymbol{a}_{\cdot
n},\boldsymbol{a}_{\cdot n}\rangle,$
$D^{(2)}_{\boldsymbol{\mu},2}(z_{1}p)=D^{(2)}_{\boldsymbol{\mu},2}(z_{1}^{2}p)=\frac{1}{2\pi}\sum_{m=0}^{k}\langle
M_{\mu_{2}}\boldsymbol{a}_{m\cdot},\boldsymbol{a}_{m\cdot}\rangle,$
$D^{(2)}_{\boldsymbol{\mu},3}(z_{1}p)=\frac{1}{4\pi^{2}}\langle(\sigma^{*}M_{\mu_{1}}\otimes
M_{\mu_{2}})\boldsymbol{a},\boldsymbol{a}\rangle,\
D^{(2)}_{\boldsymbol{\mu},1}(z_{1}^{2}p)=\frac{1}{2\pi}\sum_{n=0}^{k}\langle(\sigma^{*})^{2}M_{\mu_{1}}\boldsymbol{a}_{\cdot
n},\boldsymbol{a}_{\cdot n}\rangle,$

and

$\displaystyle
D^{(2)}_{\boldsymbol{\mu},3}(z_{1}^{2}p)=\frac{1}{4\pi^{2}}\langle((\sigma^{*})^{2}M_{\mu_{1}}\otimes
M_{\mu_{2}})\boldsymbol{a},\boldsymbol{a}\rangle.$

The above computations put together prove that
$\|z_{1}^{2}p\|_{\boldsymbol{\mu}}^{2}-2\|z_{1}p\|_{\boldsymbol{\mu}}^{2}+\|p\|_{\boldsymbol{\mu}}^{2}=0.$
Similarly, we can prove that
$\|z_{2}^{2}p\|_{\boldsymbol{\mu}}^{2}-2\|z_{2}p\|_{\boldsymbol{\mu}}^{2}+\|p\|_{\boldsymbol{\mu}}^{2}=0.$
∎

For any analytic function $f$ in $\mathbb{D}^{2}$ and $0<r<1$, define
$f_{r}(z_{1},z_{2}):=f(rz_{1},rz_{2})$ for
$(z_{1},z_{2})\in\overline{\mathbb{D}}$.

######  Lemma 2.14.

For any $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$, the relation
$D^{(2)}_{\boldsymbol{\mu}}(f_{r})\leqslant D^{(2)}_{\boldsymbol{\mu}}(f)$
holds for any $0<r<1$.

###### Proof.

Let $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and $0<r<1.$ We shall prove the
relations $D_{\boldsymbol{\mu},i}^{(2)}(f_{r})\leqslant
D_{\boldsymbol{\mu},i}^{(2)}(f)$, for $i=1,2,3$. To this end, we claim that

$\displaystyle D_{\boldsymbol{\mu},1}^{(2)}(f_{r})\leqslant
D_{\boldsymbol{\mu},1}^{(2)}(f).$ (20)

We will first prove (20) for polynomials and then for any given
$f\in\mathcal{D}^{2}({\boldsymbol{\mu}})$. Let
$P(z_{1},z_{2})=\sum_{m,n=0}^{k}a_{m,n}z_{1}^{m}z_{2}^{n}$ be a polynomial of
degree at most $2k.$ Then, from (10) with $R=1$, we have

$\displaystyle
D^{(2)}_{\boldsymbol{\mu},1}(P)=\frac{1}{2\pi}\sum_{n=0}^{k}\langle
M_{\mu_{1}}\boldsymbol{a}_{\cdot n},\boldsymbol{a}_{\cdot n}\rangle.$

Since
$D^{(2)}_{\boldsymbol{\mu},1}(z_{1}^{2}P)-2D^{(2)}_{\boldsymbol{\mu},1}(z_{1}P)+D^{(2)}_{\boldsymbol{\mu},1}(P)=0$,
it then follows that $(\sigma^{*}-I)^{2}M_{\mu_{1}}=0$. Thus from [19, Theorem
3.11], we get that the matrix
$\big{(}\big{(}1-r^{m+p})M_{\mu_{1}}(m,p)\big{)}\big{)}_{m,p\geqslant 0}$ is
positive semidefinite. Now we have

$\displaystyle\sum_{n=0}^{k}\langle\big{(}(\boldsymbol{1}-\boldsymbol{r}^{*}\boldsymbol{r})\circ
M_{\mu_{1}}\big{)}\boldsymbol{a}_{\cdot n},\boldsymbol{a}_{\cdot
n}\rangle\geqslant 0,$

where $\boldsymbol{r}$ denotes the row vector $(1,r,r^{2},\ldots,r^{k}),$ and
$\boldsymbol{1}$ denotes the rank one matrix with all its entries equal to
$1.$ Here $X\circ Y$ denotes the Schur product of matrices $X$ and $Y.$
Therefore, it follows that

$\displaystyle
D^{(2)}_{\boldsymbol{\mu},1}(P)-D^{(2)}_{\boldsymbol{\mu},1}(P_{r})$
$\displaystyle=\sum_{m,n,p=0}^{k}(1-r^{m+p+2n})M_{\mu_{1}}(m,p)a_{m,n}\overline{a_{p,n}}$
$\displaystyle=\sum_{n=0}^{k}\langle\big{(}(\boldsymbol{1}-\boldsymbol{r}^{*}\boldsymbol{r})\circ
M_{\mu_{1}}\big{)}\boldsymbol{a}_{\cdot n},\boldsymbol{a}_{\cdot
n}\rangle+\sum_{n=0}^{k}(1-r^{2n})\langle\big{(}\boldsymbol{r}^{*}\boldsymbol{r}\circ
M_{\mu_{1}}\big{)}\boldsymbol{a}_{\cdot n},\boldsymbol{a}_{\cdot n}\rangle$
$\displaystyle\geqslant 0.$

The last inequality follows since the matrix $(r^{m+p})_{m,p\geqslant 0}$ is
positive definite. This proves the claim (20). In a similar fashion, we can
prove that $D^{(2)}_{\boldsymbol{\mu},2}(P_{r})\leqslant
D^{(2)}_{\boldsymbol{\mu},2}(P).$ Note that from (17) with $R=1$, we have

$\displaystyle
D^{(2)}_{\boldsymbol{\mu},3}(P)=\frac{1}{4\pi^{2}}\langle(M_{\mu_{1}}^{\tt
T}\otimes M_{\mu_{2}}^{\tt T})\boldsymbol{a},\boldsymbol{a}\rangle,$

where $\boldsymbol{a}$ denotes the $2$-tensor $((a_{mn})).$ As
$(\sigma^{*}-I)^{2}M_{\mu_{1}}=0$ and $(\sigma^{*}-I)^{2}M_{\mu_{2}}=0$,
therefore $((\sigma^{*}-I)^{2}M_{\mu_{1}})\otimes M_{\mu_{2}}=0$ and
$M_{\mu_{1}}\otimes((\sigma^{*}-I)^{2}M_{\mu_{2}})=0$ hold true. By [19,
Theorem 3.11], it turns out that for any $0\leqslant r<1$ the matrices
$\big{(}(\boldsymbol{1}-\boldsymbol{r}^{*}\boldsymbol{r})\circ
M_{\mu_{1}})\big{)}\otimes M_{\mu_{2}}$ and
$M_{\mu_{1}}\otimes\big{(}(\boldsymbol{1}-\boldsymbol{r}^{*}\boldsymbol{r})\circ
M_{\mu_{2}})\big{)}$ are positive semi-definite. This implies that

$\displaystyle\sum_{m,n,p,q=0}^{k}(1-r^{m+p})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}\geqslant
0$ (21)
$\displaystyle\sum_{m,n,p,q=0}^{k}(1-r^{n+q})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}\geqslant
0$ (22)

Note that

$\displaystyle
D^{(2)}_{\boldsymbol{\mu,3}}(P)-D^{(2)}_{\boldsymbol{\mu,3}}(P_{r})$
$\displaystyle=\sum_{m,n,p,q=0}^{k}(1-r^{m+n+p+q})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}$
$\displaystyle=\sum_{m,n,p,q=0}^{k}(1-r^{m+p}+r^{m+p}-r^{m+n+p+q})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}$
$\displaystyle=\sum_{m,n,p,q=0}^{k}(1-r^{m+p})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}~{}+$
$\displaystyle\quad\sum_{m,n,p,q=0}^{k}r^{m+p}(1-r^{n+q})M_{\mu_{1}}(m,p)M_{\mu_{2}}(n,q)a_{m,n}\overline{a_{p,q}}$
$\displaystyle\geqslant 0.$

The last inequality follows from (21), (22) and the fact that the matrix
$(\\!(r^{m+p})\\!)_{m,p=0}^{\infty}$ is positive semi-definite. To see that
the matrix $(\\!(r^{m+p})\\!)_{m,p=0}^{\infty}$ is positive semi-definite,
note that $(\\!(r^{m+p})\\!)=vv^{T}$, with $v$ given by
$v^{T}=\begin{bmatrix}1&r&r^{2}&r^{3}&\cdots\end{bmatrix}.$ This proves the
lemma if $f$ is a polynomial. Applying a standard uniform limit argument
proves the lemma for any function $f$ which is holomorphic in a neighborhood
of $\overline{\mathbb{D}}^{2}.$ Now for
$f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and $0<R<1$ we have
$f_{R}\in\mathcal{O}(\overline{\mathbb{D}}^{2}).$ Therefore this readily
implies that $D^{(2)}_{\boldsymbol{\mu}}(f_{R})_{r}\leqslant
D^{(2)}_{\boldsymbol{\mu}}(f_{R}),$ where $r$ being any number in $(0,1).$
Because $(f_{R})_{r}=(f_{r})_{R},$ we get that
$D^{(2)}_{\boldsymbol{\mu}}(f_{r})_{R}\leqslant
D^{(2)}_{\boldsymbol{\mu}}(f_{R}).$ This is true for any $0<R<1.$ Hence by
taking limit $R\rightarrow 1,$ we get that
$D^{(2)}_{\boldsymbol{\mu}}(f_{r})\leqslant D^{(2)}_{\boldsymbol{\mu}}(f).$ ∎

We skip the proof of the following theorem as it is standard and follows from
[3, Lemma 3.7] together with Fatou’s theorem applied to
$D_{\boldsymbol{\mu},3}^{(2)}(\cdot).$

######  Lemma 2.15.

For any $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$,
$\|f_{R}-f\|_{\boldsymbol{\mu}}\rightarrow 0$ as $R\rightarrow 1^{-}.$

In the following theorem we prove that the set of all polynomials is dense in
$\mathcal{D}^{2}(\boldsymbol{\mu}).$

######  Theorem 2.16.

The set of all polynomials is dense in $\mathcal{D}^{2}(\boldsymbol{\mu}).$

###### Proof.

Let $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and $\epsilon>0$. From Lemma 2.15,
it follows that there exists an $0<R<1$ such that
$\|f_{R}-f\|_{\boldsymbol{\mu}}<\epsilon/2.$ Let
$f_{R}(z_{1},z_{2})=\sum_{m,n=0}^{\infty}b_{m,n}z_{1}^{m}z_{2}^{n},$ where the
coefficient $b_{m,n}$ is a function of $R$. Consider the $k$-th partial sum
$S_{k}f_{R}(z_{1},z_{2})=\sum_{m,n=0}^{k}b_{m,n}z_{1}^{m}z_{2}^{n}$ of
$f_{R}(z_{1},z_{2})$. Note that $f_{R}(z_{1},z_{2})$ is holomorphic in any
neighbourhood of $\overline{\mathbb{D}}^{2}$. Therefore the sequences
$(S_{k}f_{R})_{k=0}^{\infty}$, $({\partial_{1}}(S_{k}f_{R}))_{k=0}^{\infty}$,
$({\partial_{2}}(S_{k}f_{R}))_{k=0}^{\infty},$ and
$({\partial_{1}\partial_{2}}(S_{k}f_{R}))_{k=0}^{\infty}$ will converge
uniformly on $\overline{\mathbb{D}}^{2}$ respectively to $f_{R}$,
${\partial_{1}}(f_{R})$, ${\partial_{2}}(f_{R}),$ and
${\partial_{1}\partial_{2}}(f_{R})$. Thus it follows that there exists
$k\in\mathbb{N}$ such that
$\|S_{k}f_{R}-f_{R}\|^{2}_{\mathcal{D}^{2}(\boldsymbol{\mu})}\leqslant{\epsilon/2}.$
We thus conclude that

$\displaystyle\|f-S_{k}f_{R}\|_{\boldsymbol{\mu}}\leqslant\|f-f_{R}\|_{\boldsymbol{\mu}}+\|S_{k}f_{R}-f_{R}\|_{\boldsymbol{\mu}}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$

This completes the proof of the theorem. ∎

The results due to Lemma 2.13 and Theorem 2.16 yield the following theorem.

######  Theorem 2.17.

The multiplication by the coordinate functions $M_{z_{1}}$ and $M_{z_{2}}$ are
$2$-isometries on ${\mathcal{D}^{2}(\boldsymbol{\mu})}$, i.e., for every
$f\in\mathcal{D}^{2}(\boldsymbol{\mu}),$ and $i=1,2,$

$\|z_{i}^{2}f\|_{\boldsymbol{\mu}}^{2}-2\|z_{i}f\|_{\boldsymbol{\mu}}^{2}+\|f\|_{\boldsymbol{\mu}}^{2}=0.$

Recall that for any $\nu\in\mathcal{M}_{+}(\mathbb{T}),$ one has

$\displaystyle\langle
z^{m},z^{n}\rangle_{\mathcal{D}(\nu)}=\begin{cases}\frac{m\wedge
n}{2\pi}\widehat{\nu}(n-m)&\quad ifm\neq n\\\
1+\frac{m}{2\pi}\widehat{\nu}(0)&\quad ifm=n.\end{cases}$ (23)

In the following lemma, we get a formula for inner product of monomials in the
space $\mathcal{D}^{2}(\boldsymbol{\mu})$, for any
$\boldsymbol{\mu}\in\mathcal{P}\mathcal{M}_{+}(\mathbb{T}^{2}).$

######  Lemma 2.18.

For $m,n,p,q\in\mathbb{Z}_{\geqslant 0},$ $\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\boldsymbol{\mu}}=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}(\mu_{1})}\langle
z_{2}^{n},z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})}.$

###### Proof.

Let $m,n,p,q\in\mathbb{Z}_{\geqslant 0},$ $f(z_{1},z_{2}):=z_{1}^{m}z_{2}^{n}$
and $g(z_{1},z_{2}):=z_{1}^{p}z_{2}^{q}.$ We divide the proof of the lemma
into the following five cases:

  * Case 1:

When either $f$ or $g$ is the constant function $1,$ i.e. $m=n=0$ or $p=q=0.$

  * Case 2:

When $f$ and $g$ both are function of $z_{1}$ only (similarly of $z_{2}$
only).

  * Case 3:

When $f$ is a function of $z_{1}$ only and $g$ is a function of $z_{2}$ only.

  * Case 4:

When $f$ is a function of $z_{1}$ only and $g$ is a function of both $z_{1}$
and $z_{2}$.

  * Case 5:

When $f$ and $g$ both are functions of both variables $z_{1}$ and $z_{2}$,
i.e. $m,n,p,q\in\mathbb{N}.$

Case 1: Suppose $f$ is the constant function $1$, then

$\langle 1,z_{1}^{p}z_{2}^{q}\rangle_{\boldsymbol{\mu}}=\langle
1,z_{1}^{p}z_{2}^{q}\rangle_{H^{2}(\mathbb{D}^{2})}=\langle
1,z_{1}^{p}\rangle_{H^{2}(\mathbb{D})}\langle
1,z_{2}^{q}\rangle_{H^{2}(\mathbb{D})}=\langle
1,z_{1}^{p}\rangle_{\mathcal{D}(\mu_{1})}\langle
1,z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})}.$

Case 2: From (8), we get that

$\displaystyle\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},1}}$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\iint_{\mathbb{D}\times\mathbb{T}}mz_{1}^{m-1}p\bar{{z}_{1}}^{p-1}P_{\mu_{1}}(z_{1})dA(z_{1})dt$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\frac{1}{\pi}\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{2\pi}mp~{}r_{1}^{m+p-1}r_{1}^{|m-p|}e^{i(m-p)\theta_{1}}dr_{1}dtd\mu_{1}(\theta_{1})$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\frac{1}{\pi}mp\int_{0}^{1}r_{1}^{m+p-1}r_{1}^{|m-p|}dr_{1}\int_{0}^{2\pi}dt\int_{0}^{2\pi}e^{i(m-p)\theta_{1}}d\mu_{1}(\theta_{1})$
$\displaystyle=\frac{1}{2\pi}(m\wedge p)\widehat{\mu}_{1}\left(p-m\right)$

On the other hand note that $\langle
z_{1}^{m},z_{1}^{p}\rangle_{H^{2}\left(\mathbb{D}^{2}\right)}=\delta_{mp}$ and
$0=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},2}}=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},3}}.$ Therefore,
we get

$\langle
z_{1}^{m},z_{1}^{p}\rangle_{\boldsymbol{\mu}}=\delta_{mp}+\frac{1}{2\pi}(m\wedge
p)\widehat{\mu}_{1}\left(p-m\right).$

Now, using (23) completes the proof in this case.

Case 3: Using the definitions, it follows that

$\langle z_{1}^{m},z_{2}^{q}\rangle_{H^{2}\left(\mathbb{D}^{2}\right)}=\langle
z_{1}^{m},z_{2}^{q}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},1}}=0=\langle
z_{1}^{m},z_{2}^{q}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},2}}=\langle
z_{1}^{m},z_{2}^{q}\rangle_{\mathcal{D}^{2}_{\boldsymbol{\mu},3}}.$

Thus, $z_{1}^{m}$ and $z_{2}^{q}$ are orthogonal in
$\mathcal{D}^{(2)}(\boldsymbol{\mu})$. For any $m,q\in\mathbb{N},$ it is
trivial to verify that $\langle
z_{1}^{m},1\rangle_{\mathcal{D}(\mu_{1})}\langle
1,z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})}=0.$

Case 4: From (8), note that

$\displaystyle\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},1}}$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\iint_{\mathbb{D}\times\mathbb{T}}mz_{1}^{m-1}p\bar{{z}_{1}}^{p-1}\bar{z_{2}}^{q}P_{\mu_{1}}(z_{1})dA(z_{1})dt$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\frac{1}{\pi}mp\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{2\pi}r_{1}^{m+p-1}r_{1}^{|m-p|}e^{i(m-p)\theta_{1}}r^{q}e^{-itq}d\mu_{1}(\theta_{1})dtdr_{1}.$

Since $q\geqslant 1,$ it easily follows that $\int_{0}^{2\pi}e^{-itq}dt=0.$
Therefore, we get that $\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},1}}=0.$
On the other hand, using definitions, note that

$\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{H^{2}\left(\mathbb{D}^{2}\right)}=0=\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},2}}=\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},3}}.$

Using (23), we observe that $\langle
1,z_{2}^{q}\rangle_{\mathcal{D}(\mu_{2})}=0.$ This establishes the equality in
the lemma in this case.

Case 5: Once again from (8), we get that

$\displaystyle\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},1}}$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\iint_{\mathbb{D}\times\mathbb{T}}mz_{1}^{m-1}z_{2}^{n}p\overline{z_{1}}^{p-1}\overline{z_{2}}^{q}P_{\mu_{1}}(z_{1})dtdA(z_{1})$
$\displaystyle=\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi}\frac{1}{\pi}mp\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{2\pi}r_{1}^{m+p-1}r^{n+q}e^{i(n-q)t}r_{1}^{|m-p|}e^{i(m-p)\theta_{1}}d\mu_{1}(\theta_{1})dtdr_{1}$
$\displaystyle=\delta_{nq}\lim_{r\rightarrow
1^{-}}\frac{1}{2\pi^{2}}mp\int_{0}^{1}\int_{0}^{2\pi}r_{1}^{m+p-1}r^{n+q}r_{1}^{|m-p|}e^{i(m-p)\theta_{1}}d\mu_{1}(\theta_{1})dr_{1}.$

Thus

$\displaystyle\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},1}}=\begin{cases}0&\mbox{
if }n\neq q\\\ \frac{1}{2\pi}(m\wedge p)\widehat{\mu_{1}}(p-m)&\mbox{ if
}n=q.\end{cases}$ (24)

Similarly, we can derive,

$\displaystyle\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},2}}=\begin{cases}0&\mbox{
if }m\neq p\\\ \frac{1}{2\pi}(n\wedge q)\widehat{\mu_{2}}(q-n)&\mbox{ if
}m=p.\end{cases}$ (25)

In the following, we use (8) twice to obtain that

$\displaystyle\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{(2)}_{\boldsymbol{\mu},3}}$
$\displaystyle=$
$\displaystyle\iint_{\mathbb{D}^{2}}mnz_{1}^{m-1}z_{2}^{n-1}~{}pq\overline{z_{1}}^{p-1}\overline{z_{2}}^{q-1}P_{\boldsymbol{\mu}}(z_{1},z_{2})dA(z_{1})dA(z_{2})$
(26) $\displaystyle=$ $\displaystyle\frac{1}{\pi^{2}}mnpq\frac{1}{2(m\vee
p)}\frac{1}{2(n\vee q)}\widehat{\mu}_{1}(p-m)\widehat{\mu}_{2}(q-n)$
$\displaystyle=$ $\displaystyle\frac{1}{4\pi^{2}}(m\wedge p)(n\wedge
q)\widehat{\mu}_{1}(p-m)\widehat{\mu}_{2}(q-n).$

Using (23), (24), (25), and (26), we get the result in this case. ∎

We note down the case when $m=p$ and $n=q$ of above lemma into the following
corollary.

######  Corollary 2.19.

For any $m,n\in\mathbb{Z}_{\geqslant 0},$
$\|z_{1}^{m}z_{2}^{n}\|_{\boldsymbol{\mu}}=\|z_{1}^{m}\|_{\boldsymbol{\mu}}\|z_{2}^{n}\|_{\boldsymbol{\mu}}.$

Lemma 2.18 suggests that there should be a relationship between
$\mathcal{D}^{2}(\boldsymbol{\mu})$ and
$\mathcal{D}(\mu_{1})\otimes\mathcal{D}(\mu_{2}).$ This is indeed the case, as
shown in the following theorem.

######  Theorem 2.20.

For any $\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T}),$ the pair of commuting
$2$-isometries $(M_{z_{1}},M_{z_{2}})$ is unitarily equivalent to the pair
$(M_{z}\otimes I,I\otimes M_{z})$ on
$(\mathcal{D}(\mu_{1})\otimes\mathcal{D}(\mu_{2}))\oplus(\mathcal{D}(\mu_{1})\otimes\mathcal{D}(\mu_{2})).$
In particular, the pair $(M_{z_{1}},M_{z_{2}})$ is doubly commuting.

###### Proof.

Define
$U:\mathcal{D}^{2}(\boldsymbol{\mu})\to\mathcal{D}(\mu_{1})\otimes\mathcal{D}(\mu_{2})$
by the rule $U(z_{1}^{m}z_{2}^{n}):=z^{m}\otimes z^{n},$ for
$m,n\in\mathbb{Z}_{\geqslant 0},$ and extend $U$ linearly. Since the set of
polynomials are dense in $\mathcal{D}^{2}(\boldsymbol{\mu}),$ $U$ extends to
whole space $\mathcal{D}^{2}(\boldsymbol{\mu})$ linearly. Using Lemma 2.18, it
follows that $U$ is an isometry. On the other hand, since $U$ has closed range
and $\mbox{span}\\{z^{m}\otimes z^{n}:m,n\in\mathbb{Z}_{\geqslant 0}\\}$ is
dense in $\mathcal{D}(\mu_{1})\otimes\mathcal{D}(\mu_{2}),$ it follows that
$U$ is a unitary map. Also note that

$UM_{z_{1}}(z_{1}^{m}z_{2}^{n})=z^{m+1}\otimes z^{n}=(M_{z}\otimes
I)U(z_{1}^{m}z_{2}^{n})$

and

$UM_{z_{2}}(z_{1}^{m}z_{2}^{n})=z^{m}\otimes z^{n+1}=(I\otimes
M_{z})U(z_{1}^{m}z_{2}^{n}).$

This proves that the pair $(M_{z_{1}},M_{z_{2}})$ is unitarily equivalent to
the pair $(M_{z}\otimes I,I\otimes M_{z}).$ This, in particular, implies that
$M_{z_{1}}$ and $M_{z_{2}}$ on $\mathcal{D}^{2}(\boldsymbol{\mu})$ is doubly
commuting. ∎

For the pair of commuting $2$-isometries $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\mu})$, denote the closed subspace
$(\mathcal{D}^{2}(\boldsymbol{\mu})\ominus
z_{1}\mathcal{D}^{2}(\boldsymbol{\mu}))\cap(\mathcal{D}^{2}(\boldsymbol{\mu})\ominus
z_{2}D^{2}(\boldsymbol{\mu}))$ by $\mathcal{W}$. The subspace $\mathcal{W}$
plays a major role in this article. With a simple computation, it is also easy
to see that $\mathcal{W}=\ker M_{z_{1}}^{*}\cap\ker M_{z_{2}}^{*}.$ Below, we
note down some of the important properties of the subspace $\mathcal{W}$.

######  Remark 2.21.

The proof of Theorem 2.20 tells us that the kernel function $K$ for
$\mathcal{D}^{2}(\boldsymbol{\mu})$ is given by

$\displaystyle
K((z_{1},z_{2}),(w_{1},w_{2}))=K^{\mu_{1}}(z_{1},w_{1})K^{\mu_{2}}(z_{2},w_{2}),\quad(z_{1},z_{2}),(w_{1},w_{2})\in\mathbb{D}^{2},$

where $K^{\mu_{j}}$ denotes the kernel function for $\mathcal{D}(\mu_{j}),$
$j=1,2.$

As corollaries of Theorem 2.20, we get the following results.

######  Corollary 2.22.

The closed subspace $\mathcal{W}$ is a wandering subspace of
$(M_{z_{1}},M_{z_{2}})$ on $\mathcal{D}^{2}(\boldsymbol{\mu})$ and the
dimension of $\mathcal{W}$ is $1$.

######  Corollary 2.23.

The subspace $\ker M_{z_{1}}^{*}\subseteq\mathcal{D}^{2}(\boldsymbol{\mu})$ is
a $M_{z_{2}}$-reducing subspace. Also, the subspace $\ker
M_{z_{2}}^{*}\subseteq\mathcal{D}^{2}(\boldsymbol{\mu})$ is a
$M_{z_{1}}$-reducing subspace.

Before we proceed further, for any $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and
$z_{1},z_{2}\in\mathbb{D},$ we define the operators $L_{1}$ and $L_{2}$ by the
rules

$\displaystyle(L_{1}f)(z_{1},z_{2}):=\frac{f(z_{1},z_{2})-f(0,z_{2})}{z_{1}}\quad\text{and}\quad(L_{2}f)(z_{1},z_{2}):=\frac{f(z_{1},z_{2})-f(z_{1},0)}{z_{2}}.$
(27)

Using the definition of norm, Lemma 2.4, and Theorem 2.12, it turns out that
$L_{1}$ and $L_{2}$ are indeed in
$\mathcal{B}(\mathcal{D}^{2}(\boldsymbol{\mu}))$. Here, we enlist some of the
key properties of these operators.

######  Theorem 2.24.

The operators $L_{1}$ and $L_{2}$ are the left inverses of $M_{z_{1}}$ and
$M_{z_{2}}$ respectively. Moreover, $L_{1}M_{z_{2}}=M_{z_{2}}L_{1}$ and
$L_{2}M_{z_{1}}=M_{z_{1}}L_{2}$.

###### Proof.

By using the definition of the operators $L_{1}$, defined above, we have

$L_{1}M_{z_{1}}f(z_{1},z_{2})=\frac{(z_{1}f)(z_{1},z_{2})-(z_{1}f)(0,z_{2})}{z_{1}}=f(z_{1},z_{2})$

for all $f\in\mathcal{D}^{2}(\boldsymbol{\mu})$ and for all
$(z_{1},z_{2})\in\mathbb{D}^{2}$. Similarly, we also have
$L_{2}M_{z_{2}}=M_{z_{2}}L_{2}$. On the other hand, for all
$(z_{1},z_{2})\in\mathbb{D}^{2}$, we have

$\displaystyle(M_{z_{2}}L_{1}f)(z_{1},z_{2})$
$\displaystyle=z_{2}\bigg{\\{}\frac{f(z_{1},z_{2})-f(0,z_{2})}{z_{1}}\bigg{\\}}$
$\displaystyle=z_{2}\Big{\\{}\sum_{m_{1}=1}^{\infty}\sum_{m_{2}=0}^{\infty}a_{m_{1},m_{2}}z_{1}^{m_{1}-1}z_{2}^{m_{2}}\Big{\\}}$
$\displaystyle=\sum_{m_{1}=1}^{\infty}\sum_{m_{2}=0}^{\infty}a_{m_{1},m_{2}}z_{1}^{m_{1}-1}z_{2}^{m_{2}+1}$
$\displaystyle=\frac{(z_{2}f)(z_{1},z_{2})-(z_{2}f)(0,z_{2})}{z_{1}}$
$\displaystyle=(L_{1}M_{z_{2}}f)(z_{1},z_{2})$

Analogously, one can also derive $L_{2}M_{z_{1}}=M_{z_{1}}L_{2}$. ∎

######  Remark 2.25.

Note that, by Theorem 2.12, the operators
$M_{z_{i}}\in\mathcal{B}(\mathcal{D}^{2}(\boldsymbol{\mu}))$ are bounded below
for $i=1,2$, hence $(M_{z_{i}}^{*}M_{z_{i}})^{-1}$ exists as well as
$M_{z_{i}}^{\prime}$s are left invertible. Moreover, for $i=1,2$,
$(M_{z_{i}}^{*}M_{z_{i}})^{-1}M_{z_{i}}^{*}$ and the above defined operators
$L_{i}$ are left-inverse of $M_{z_{i}}$ with $\ker
L_{i}=\ker(M_{z_{i}}^{*}M_{z_{i}})^{-1}M_{z_{i}}^{*}=\ker M_{z_{i}}^{*}$ and
therefore $L_{i}=(M_{z_{i}}^{*}M_{z_{i}})^{-1}M_{z_{i}}^{*}$.

##  3\. Wandering Subspace property for a pair of analytic 2-isometries

In this section we deal with a commuting pair of left invertible operators
acting on a Hilbert space and prove a multivariate version of S. Shimorin’s
result for a class of operators on a Hilbert space. We show that a class of
commuting pair of $2$-isometries on a Hilbert space always possess wandering
subspace property.

Let $\mathbf{T}=(T_{1},T_{2})$ be a pair of commuting $2$-isometries acting on
a Hilbert space $\mathcal{H}$ and we consider the adjoint of the Cauchy-dual
operators, that is,
$L_{i}:=(T_{i}^{*}T_{i})^{-1}T_{i}^{*}\in\mathcal{B}(\mathcal{H})$ which is
also a left-inverse of $T_{i}$ for $i=1,2$.

In the following lemma, we prove that joint kernel of adjoint is non-trivial
for any left-inverse commuting pair of analytic $2$-isometries.

######  Lemma 3.1.

Let $(T_{1},T_{2})$ be a left-inverse commuting pair of analytic
$2$-isometries on a Hilbert space $\mathcal{H}$. Then,

  1. (a)

the subspace $\ker T_{1}^{*}\bigcap\ker T_{2}^{*}$ is a non-trivial closed
subspace of $\mathcal{H}$ and

  2. (b)

the operator $(I-T_{1}L_{1})(I-T_{2}L_{2})$ is a projection on $\ker
T_{1}^{*}\bigcap\ker T_{2}^{*}$.

###### Proof.

(a) Let $x_{0}\in\ker T_{1}^{*}$. From the definition of $L_{1}$, it is easy
to see that $\ker L_{1}=\ker T_{1}^{*}$. So, we have $L_{1}x_{0}=0$ and
therefore $T_{2}L_{1}x_{0}=0$. Now, by using the hypothesis, we have
$L_{1}T_{2}x_{0}=0$, that is, $T_{2}x_{0}\in\ker T_{1}^{*}$. So, $\ker
T_{1}^{*}$ is invariant under $T_{2}$. On the other hand, using the
commutativity of $T_{1}$ and $T_{2}$, it is straightforward to see that $\ker
T_{1}^{*}$ is invariant under $T_{2}^{*}$. Hence, $\ker T_{1}^{*}$ is a
$T_{2}$-reducing subspace. Since $T_{2}\big{|}_{\ker T_{1}^{*}}:\ker
T_{1}^{*}\to\ker T_{1}^{*}$ is an analytic operator, that is,

$\bigcap_{n\geqslant 0}T_{2}^{n}\big{|}_{\ker T_{1}^{*}}(\ker
T_{1}^{*})=\\{0\\},$

therefore, $T_{2}\big{|}_{\ker T_{1}^{*}}$ cannot be onto. Moreover, since
$T_{2}\big{|}_{\ker T_{1}^{*}}$ is $2$-isometry, it’s range is closed.
Therefore, the subspace $\ker(T_{2}^{*}\big{|}_{\ker T_{1}^{*}})\subseteq\ker
T_{1}^{*}$ is non-trivial. Hence, $\ker T_{1}^{*}\bigcap\ker
T_{2}^{*}\neq\\{0\\}$.

(b) Using the definition of the operators $L_{1}$ and $L_{2}$, defined above,
it is easy to observe that the operators $I-T_{1}L_{1}$ and $I-T_{2}L_{2}$ are
projections. To show that the product of these operators
$(I-T_{1}L_{1})(I-T_{2}L_{2})$ is a projection, it is enough to show that

$(I-T_{1}L_{1})(I-T_{2}L_{2})=(I-T_{2}L_{2})(I-T_{1}L_{1}).$

Therefore, we only need to show that $L_{1}L_{2}=L_{2}L_{1}$. Since
$\mathcal{H}=\bigvee_{n\geqslant 0}T_{1}^{n}(\ker T_{1}^{*})$ and $L_{1}$ and
$L_{2}$ are bounded linear operators, it is enough to check the identity
$L_{1}L_{2}T_{1}^{k}y=L_{2}L_{1}T_{1}^{k}y$ for each $k\in\mathbb{Z}_{+}$ and
$y\in\ker T_{1}^{*}.$ Let $k\geqslant 1$ and $y\in\ker T_{1}^{*}.$ Note that

$\displaystyle
L_{1}L_{2}T_{1}^{k}y=L_{1}T_{1}^{k}L_{2}y=T_{1}^{k-1}L_{2}y=L_{2}T_{1}^{k-1}y=L_{2}L_{1}T_{1}^{k}y.$

Since $\ker T_{1}^{*}$ is a reducing subspace for $T_{2},$ it follows that
$L_{2}y$ is in $\ker T_{1}^{*}$ also. Therefore, we have

$\displaystyle L_{2}L_{1}y=0=L_{1}L_{2}y.$

This completes the proof of the lemma. ∎

Now, we proceed to prove the existence of the wandering subspace property for
the class of left-inverse commuting pair of analytic $2$-isometries. This
generelizes many existing results e.g. see [8, 9, 13, 5].

######  Theorem 3.2.

Let $T_{1}$ and $T_{2}$ be a pair of commuting, analytic 2-isometric operators
on $\mathcal{H}$ such that $L_{1}T_{2}=T_{2}L_{1}$ and
$L_{2}T_{1}=T_{1}L_{2}.$ Then there exists a wandering subspace
$\mathcal{W}\subseteq\mathcal{H}$ for the pair $(T_{1},T_{2})$. Moreover,
$\mathcal{W}=\ker T_{1}^{*}\bigcap\ker T_{2}^{*}$ is a wandering subspace for
the pair $(T_{1},T_{2})$ on $\mathcal{H}$.

###### Proof.

By Lemma 3.1, we know the subspace $\ker T_{1}^{*}\bigcap\ker T_{2}^{*}$ is
non-trivial. Also, for the subspace $\mathcal{W}=\ker T_{1}^{*}\bigcap\ker
T_{2}^{*}\subseteq\mathcal{H}$, one sided inclusion is trivial, that is,

$\bigvee_{n,m\geq 0}T_{1}^{n}T_{2}^{m}\mathcal{W}\subseteq\mathcal{H}.$

To prove the other inclusion, denote $P_{1}:=(I-T_{1}L_{1})$ and
$P_{2}:=(I-T_{2}L_{2}).$ Recall from Lemma 3.1 that $P=P_{1}P_{2}$ is a
projection of $\mathcal{H}$ onto $\mathcal{W}$. Using [10, pg. 209] and the
hypothesis that $(T_{1},T_{2})$ is a left-inverse commuting pair, we observe
that

$\displaystyle(I-T_{1}^{n}L_{1}^{n})(I-T_{2}^{m}L_{2}^{m})=\Big{(}\sum_{i=0}^{n-1}T_{1}^{i}P_{1}L_{1}^{i}\Big{)}\Big{(}\sum_{j=0}^{m-1}T_{2}^{j}P_{2}L_{2}^{j}\Big{)}=\sum_{i=0}^{n-1}\sum_{j=0}^{m-1}T_{1}^{i}T_{2}^{j}PL_{1}^{i}L_{2}^{j}.$

From the above computation, we conclude that for any $x\in\mathcal{H}$ and
$n,m\in\mathbb{N},$

$(I-T_{1}^{n}L_{1}^{n}-T_{2}^{m}L_{2}^{m}+T_{1}^{n}L_{1}^{n}T_{2}^{m}L_{2}^{m})x\in\bigvee_{n,m\geqslant
0}T_{1}^{n}T_{2}^{m}\mathcal{W}.$

To prove the claim, we first note that, by hypothesis, the operators $T_{1}$
and $T_{2}$ are 2-isometries. So, for any $x\in\mathcal{H},$ $i=1,2,$ and
$n\in\mathbb{N}$, we have

$\|T_{i}^{n}x\|^{2}-\|x\|^{2}=n\|D_{i}x\|^{2},$ (28)
$\|x\|^{2}=\sum_{k=0}^{n}\|P_{i}L_{i}^{k}x\|^{2}+\|L_{i}^{n+1}x\|^{2}+\sum_{k=1}^{n+1}\|D_{i}L_{i}^{k}x\|^{2},$
(29)

where $D_{i}$ denotes the positive square root of the operator
$T_{i}^{*}T_{i}-I$ and (29) follows from [10, equation (9)]. Along with that,
for each $i=1,2$, $T_{i}$ is also analytic and therefore, following an
analogous argument as in [10, Theorem 1], for each $x\in\mathcal{H}$ the
sequence $\\{T_{i}^{n_{i}}L_{i}^{n_{i}}x\\}$ has a subsequence which converge
to $0$ weakly. Hence, it is enough to show that the double sequence
$\\{T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\\}$ has a
subsequence which converges to 0 weakly. For more details we refer [10]. For
fixed $n_{2}\in\mathbb{N}$, applying the above equations (28) and (29) for
$T_{2}^{n_{2}}L_{2}^{n_{2}}x$ in place of $x$, we deduce

$\displaystyle\sum_{n_{1}=1}^{m_{1}}\frac{1}{n_{1}}~{}~{}\inf_{1\leqslant
n_{1}\leqslant
m_{1}}\big{(}\|T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}-\|L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}\big{)}$
$\displaystyle\leqslant\sum_{n_{1}=1}^{m_{1}}\frac{1}{n_{1}}\big{(}\|T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}-\|L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}\big{)}$
$\displaystyle\leqslant\sum_{n_{1}=1}^{m_{1}}\|D_{1}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}$
$\displaystyle\leqslant\|T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}.$

Therefore, we conclude that

$\displaystyle\liminf_{n_{1}\to\infty}\big{\\{}\|T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}-\|L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}\big{\\}}=0.$

From the equation (29) it is easy to observe that the sequence
$\\{\|L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}\\}$ is decreasing and
combining all these we get

$\liminf_{n_{1}\to\infty}\|T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}=\lim_{n_{1}\to\infty}\|L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}.$

By hypothesis, we have $L_{1}T_{2}=T_{2}L_{1}$ and by applying analogous
argument on the sequence $\\{T_{2}^{n_{2}}L_{1}^{n_{1}}L_{2}^{n_{2}}x\\}$ we
conclude

$\liminf_{n_{2}\to\infty}(\lim_{n_{1}\to\infty}\|T_{2}^{n_{2}}L_{2}^{n_{2}}L_{1}^{n_{1}}x\|^{2})=\lim_{n_{2}\to\infty}\lim_{n_{1}\to\infty}\|L_{2}^{n_{2}}L_{1}^{n_{1}}x\|^{2}$

and since the sequence $\\{\|L_{2}^{n_{2}}L_{1}^{n_{1}}x\|^{2}\\}$ is
decreasing, follows from the equation (29), therefore we have

$\liminf_{n_{2}\to\infty}\liminf_{n_{1}\to\infty}\|T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\|^{2}=\lim_{n_{2}\to\infty}\lim_{n_{1}\to\infty}\|L_{1}^{n_{1}}L_{2}^{n_{2}}x\|^{2}.$

Thus there exists a weakly convergent subsequence of
$\\{T_{1}^{n_{1}}L_{1}^{n_{1}}T_{2}^{n_{2}}L_{2}^{n_{2}}x\\}$ which converges
to $y$ weakly. Also, since $T_{1}$ and $T_{2}$ are $2$-isometries so, for each
$k_{1},k_{2}\in\mathbb{N}$, the subspaces $T_{i}^{k_{i}}\mathcal{H}$, are
weakly closed and $y\in T_{i}^{k_{i}}\mathcal{H}$ for each $i=1,2$. By using
the analyticity of each operator $T_{i}$, we conclude that $y=0$ and hence the
theorem follows. ∎

######  Remark 3.3.

We would like to remark here that the statement of Theorem 3.2 along with it’s
proof can be adapted easily for it to work for any analytic $2$-isometric
left-inverse commuting tuple $(T_{1},\ldots,T_{n}).$

##  4\. Model for a class of pair of cyclic analytic 2-isometries

In this section, we prove the following model theorem which shows that the
class of multiplication operators $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\mu})$ completely characterizes that of left-
inverse commuting pair of analytic $2$-isometries which satisfy certain
splitting properties. With the help of this theorem, we give an example of a
left-inverse commuting pair of analytic $2$-isometries which is not a doubly
commuting pair.

######  Theorem 4.1.

Let $(T_{1},T_{2})$ be a pair of cyclic, analytic, 2-isometries on a Hilbert
space $\mathcal{H}$ with unit cyclic vector $x_{0}\in\ker T_{1}^{*}\cap\ker
T_{2}^{*}$. Then $(T_{1},T_{2})$ on $\mathcal{H}$ is unitarily equivalent with
$(M_{z_{1}},M_{z_{2}})$ on $\mathcal{D}^{(2)}(\boldsymbol{\mu})$ if and only
if $L_{1}T_{2}=T_{2}L_{1}$, $L_{2}T_{1}=T_{1}L_{2}$ and for all
$m,n\in\mathbb{N}\setminus\\{0\\}$

$\displaystyle\langle T_{1}^{m}T_{2}^{n}x_{0},T_{1}T_{2}x_{0}\rangle=\langle
T_{1}^{m}x_{0},T_{1}x_{0}\rangle\langle T_{2}^{n}x_{0},T_{2}x_{0}\rangle,$
$\displaystyle\langle T_{1}^{m}T_{2}x_{0},T_{1}T_{2}^{n}x_{0}\rangle=\langle
T_{1}^{m}x_{0},T_{1}x_{0}\rangle\langle T_{2}x_{0},T_{2}^{n}x_{0}\rangle.$

###### Proof.

Since $(T_{1},T_{2})$ is an analytic left-inverse commuting pair, it follows
from Theorem 3.2 that $\ker T_{1}^{*}\cap\ker T_{2}^{*}$ is wandering subspace
for $(T_{1},T_{2}).$ Using the hypothesis that $(T_{1},T_{2})$ is a cyclic
pair, it follows that $\ker T_{1}^{*}\cap\ker T_{2}^{*}=span\\{x_{0}\\},$ for
some unit vector $x_{0}.$ Consider the operator $S_{1}:=T_{1}|_{\ker
T_{2}^{*}}$ and $S_{2}:=T_{2}|_{\ker T_{1}^{*}}.$ From Lemma 3.1, it follows
that $S_{1}$ and $S_{2}$ are bounded linear operators on $\ker T_{2}^{*}$ and
$\ker T_{1}^{*}$ respectively. Now it follows that $S_{1}$ and $S_{2}$ are
cyclic analytic $2$-isometries with cyclic subspace $\ker T_{1}^{*}\cap\ker
T_{2}^{*}.$ Using [12, Theorem 5.1], we have
$\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T})$ and unitary maps $U_{1}:\ker
T_{2}^{*}\rightarrow\mathcal{D}(\mu_{1})$ and $U_{2}:\ker
T_{1}^{*}\rightarrow\mathcal{D}(\mu_{2})$ such that

$U_{i}S_{i}=M_{z}U_{i}~{}~{}\quad~{}~{}\text{and}~{}~{}\quad U_{i}x_{0}=1.$

Let $\boldsymbol{\mu}:=\mu_{1}\times\mu_{2}$ be the product measure and
consider the corresponding Dirichlet-type spaces
$\mathcal{D}^{2}(\boldsymbol{\mu})$. We define a linear map
$U:\mathcal{H}\rightarrow\mathcal{D}^{2}(\boldsymbol{\mu})$, which is defined
by

$U(p(T_{1},T_{2})x_{0})=p(z_{1},z_{2}),$

where
$p(T_{1},T_{2})x_{0}=\sum_{k_{1},k_{2}=0}^{n_{1},n_{2}}a_{k_{1}k_{2}}T_{1}^{k_{1}}T_{2}^{k_{2}}x_{0}$
for any polynomial
$p(z_{1},z_{2})=\sum_{k_{1},k_{2}=0}^{n_{1},n_{2}}a_{k_{1}k_{2}}z_{1}^{k_{1}}z_{2}^{k_{2}}$.
We claim that the map defines a unitary map from $\mathcal{H}$ onto
$\mathcal{D}^{2}(\boldsymbol{\mu})$. So we need to show that
$\|P(T_{1},T_{2})x_{0}\|^{2}=\|P(z_{1},z_{2})\|^{2}_{\mathcal{D}^{2}(\mu)}$.
Therefore, it is enough to verify that

$\langle T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{2}(\mu)}\quad~{}~{}\text{for
all $m,n,p,q\in\mathbb{N}.$}$

First note that, in above, if $n,q=0$ then for any $m,p\in\mathbb{N}$,
considering the above unitary map $U_{1}$, we have

$\displaystyle\langle T_{1}^{m}x_{0},T_{1}^{p}x_{0}\rangle$
$\displaystyle=\langle
U_{1}T_{1}^{m}x_{0},U_{1}T_{1}^{p}x_{0}\rangle_{\mathcal{D}(\mu_{1})}$
$\displaystyle=\langle
M_{z}^{m}U_{1}x_{0},M_{z}^{p}U_{1}x_{0}\rangle_{\mathcal{D}(\mu_{1})}$
$\displaystyle=\langle z^{m},z^{p}\rangle_{\mathcal{D}(\mu_{1})}$
$\displaystyle=\langle
z_{1}^{m},z_{1}^{p}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}.$

Similarly, we also have $\langle T_{2}^{n}x_{0},T_{2}^{q}x_{0}\rangle=\langle
z_{2}^{n},z_{2}^{q}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}$ for any
$n,q\in\mathbb{N}$. On the other hand, by Lemma 3.1, we know that for any
$i,j=1,2$ with $i\neq j$ the subspace $KerT_{j}^{*}$ is $T_{i}$ reducing.
Without loss of generality, it is enough to prove for $i=1$ and $j=2$. By
using this fact it is obvious that

$\langle
T_{1}^{m}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=0\quad\quad~{}~{}\text{for any
$q\in\mathbb{N}\setminus\\{0\\}$ and $m,p\in\mathbb{N}.$}$

Now by combining the above equation and the Lemma 2.18 we conclude

$\langle T_{1}^{m}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=\langle
z_{1}^{m},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}=0\quad\quad~{}~{}\text{for
any $q\in\mathbb{N}\setminus\\{0\\}$ and $m,p\in\mathbb{N}.$}$

It remains to deal with the case where $m,n,p,q\in\mathbb{N}\setminus\\{0\\}$.
Again, without loss of generality, it is enough to establish the equation for
$m\geqslant p,n\geqslant q$ and $m\geqslant p,n\leqslant q$. First we consider
the case $m\geqslant p,n\geqslant q$. Since $T_{1}$ is a $2$-isometry
therefore we deduce

$\displaystyle\langle
T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=(m\wedge p)\langle
T_{1}^{m-p+1}T_{2}^{n}x_{0},T_{1}T_{2}^{q}x_{0}\rangle.$

Analogously, using the fact that $T_{2}$ is a $2$-isometry, we reduce the
above term further and it becomes

$\displaystyle(m\wedge p)\langle
T_{1}^{m-p+1}T_{2}^{n}x_{0},T_{1}T_{2}^{q}x_{0}\rangle$
$\displaystyle=(m\wedge p)(n\wedge q)\langle
T_{1}^{m-p+1}T_{2}^{n-q+1}x_{0},T_{1}T_{2}x_{0}\rangle.$

Combining all these we have

$\displaystyle\langle
T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle=(m\wedge p)(n\wedge
q)\langle T_{1}^{m-p+1}T_{2}^{n-q+1}x_{0},T_{1}T_{2}x_{0}\rangle.$

By the given hypothesis, the above identities on $T_{1}$ and $T_{2}$
individually and the Lemma 2.18 we conclude

$\displaystyle\langle T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle$
$\displaystyle=(m\wedge p)(n\wedge q)\langle
T_{1}^{m-p+1}T_{2}^{n-q+1}x_{0},T_{1}T_{2}x_{0}\rangle$
$\displaystyle=(m\wedge p)(n\wedge q)\langle
T_{1}^{m-p+1}x_{0},T_{1}x_{0}\rangle\langle
T_{2}^{n-q+1}x_{0},T_{2}x_{0}\rangle$ $\displaystyle=(m\wedge p)(n\wedge
q)\langle
z_{1}^{m-p+1},z_{1}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}\langle
z_{2}^{n-q+1},z_{2}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}$
$\displaystyle=\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}.$

Similarly, for the second case with $m\geqslant p,n\leqslant q$ by using the
given hypothesis and the Lemma 2.18 we deduce

$\displaystyle\langle T_{1}^{m}T_{2}^{n}x_{0},T_{1}^{p}T_{2}^{q}x_{0}\rangle$
$\displaystyle=(m\wedge p)(n\wedge q)\langle
T_{1}^{m-p+1}T_{2}x_{0},T_{1}T_{2}^{q-n+1}x_{0}\rangle$
$\displaystyle=(m\wedge p)(n\wedge q)\langle
T_{1}^{m-p+1}x_{0},T_{1}x_{0}\rangle\langle
T_{2}x_{0},T_{2}^{n-q+1}x_{0}\rangle$ $\displaystyle=(m\wedge p)(n\wedge
q)\langle
z_{1}^{m-p+1},z_{1}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}\langle
z_{2},z_{2}^{n-q+1}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}$
$\displaystyle=\langle
z_{1}^{m}z_{2}^{n},z_{1}^{p}z_{2}^{q}\rangle_{\mathcal{D}^{2}(\boldsymbol{\mu})}.$

The converse follows as an application of Lemma 2.18 and Proposition 2.24. ∎

######  Remark 4.2.

For any
$\boldsymbol{\mu},\boldsymbol{\nu}\in\mathcal{P}\mathcal{M}_{+}(\mathbb{T}^{2}),$
we content that $(M_{z_{1}},M_{z_{2}})$ on $\mathcal{D}^{2}(\boldsymbol{\mu})$
is unitarily equivalent to $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\nu})$ if and only if
$\boldsymbol{\mu}=\boldsymbol{\nu}.$ To prove this, note that $M_{z_{1}}$ and
$M_{z_{2}}$ are analytic $2$-isometries on $\mathcal{D}(\mu_{j})$ and
$\mathcal{D}(\nu_{j})$ for each $j=1,2.$ Therefore if $U$ is the unitary (as
in the proof of Theorem 4.1) intertwining pair $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\mu})$ and the pair $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}^{2}(\boldsymbol{\nu})$ then $M_{z_{1}}|_{\mathcal{D}(\mu_{1})}$
and $M_{z_{1}}|_{\mathcal{D}(\nu_{1})}$ is intertwined by the unitary map
$U|_{\mathcal{D}(\mu_{1})}.$ Therefore, from [12, Theorem 5.2], it follows
that $\mu_{1}=\nu_{1}.$ Similarly, one can prove $\mu_{2}=\nu_{2}.$

We conclude this section with the following example which exhibits the
importance of Theorem 3.2.

######  Example 4.3.

Let $\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T}).$ Consider the space
$\mathcal{D}(\mu_{1},\mu_{2})$ as introduced in [3]. We content that the pair
$(M_{z_{1}},M_{z_{2}})$ on $\mathcal{D}(\mu_{1},\mu_{2})$ is a left-inverse
commuting pair of analytic $2$-isometries.

To see this, we first note that from [3, Corollary 3.8] the pair
$(M_{z_{1}},M_{z_{2}})$ is a toral $2$-isometry and hence $2$-isometric pair.
Note that for any $f\in\mathcal{D}(\mu_{1},\mu_{2})$, the function
$f(0,z_{2})$ is in $\mathcal{D}(\mu_{1},\mu_{2}),$ see Lemma 2.4.

Define $g(z_{1},z_{2}):=f(z_{1},z_{2})-f(0,z_{2}).$ Note that
$g\in\mathcal{D}(\mu_{1},\mu_{2})$ and
$\\{(z_{1},z_{2})\in\mathbb{D}^{2}:z_{1}=0\\}$ is in the zero set of $g.$ Now
using [3, Theorem 2.2], we observe that $g/z_{1}$ (this is same as $L_{1}(f)$,
see (27)) is in $\mathcal{D}(\mu_{1},\mu_{2}).$ The operator $L_{1}$ serves as
a left inverse of $M_{z_{1}}$ on $\mathcal{D}(\mu_{1},\mu_{2}).$ Rerunning the
proof of Theorem 2.24 proves that $(M_{z_{1}},M_{z_{2}})$ is a left-inverse
commuting pair. By Theorem 3.2, the pair $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}(\mu_{1},\mu_{2})$ possess wandering subspace property.

Now we assume that each of $\mu_{1}$ and $\mu_{2}$ is normalized Lebesgue
measure $\sigma$ on the circle $\mathbb{T}$. With the help of the fact that
the set of monomials forms an orthogonal basis for
$\mathcal{D}(\sigma,\sigma)$, it follows from a simple computation that for
any $p\geqslant 1$ and $q\geqslant 0$

$M_{z_{1}}^{*}(z_{1}^{p}z_{2}^{q})=\frac{p+q+1}{p+q}z_{1}^{p-1}z_{2}^{q}.$

It helps us to obtain that

$M_{z_{2}}M_{z_{1}}^{*}(z_{1}^{p}z_{2}^{q})=\frac{p+q+1}{p+q}z_{1}^{p-1}z_{2}^{q+1}\neq\frac{p+q+2}{p+q+1}z_{1}^{p-1}z_{2}^{q+1}=M_{z_{1}}^{*}M_{z_{2}}(z_{1}^{p}z_{2}^{q}).$

Therefore, we can conclude that the pair $(M_{z_{1}},M_{z_{2}})$ on
$\mathcal{D}(\mu_{1},\mu_{2})$, where $\mu_{1}$ and $\mu_{2}$ are normalized
Lebesgue measures on the circle $\mathbb{T}$, is not a doubly commuting pair.

Acknowledgements: The first and second named author’s research are supported
by the DST-INSPIRE Faculty Grant with Fellowship No.
DST/INSPIRE/04/2020/001250 and DST/INSPIRE/04/2017/002367 respectively.

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