# A local Douglas formula for higher order weighted Dirichlet-type integrals

Soumitra Ghara  ,  Rajeev Gupta  and  Md. Ramiz Reza  Department of
Mathematics and Statistics
Laval University
Quebec City - G1V 0A6  School of Mathematics and Computer Science
Indian Institute of Technology Goa
Goa - 403401  School of Mathematics
Indian Institute of Science Education and Research Thiruvananthapuram
Kerala - 695551<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We prove a local Douglas formula for higher order weighted Dirichlet-type
integrals. With the help of this formula, we study the multiplier algebra of
the associated higher order weighted Dirichlet-type spaces
$\mathcal{H}_{\boldsymbol{\mu}},$ induced by an $m$-tuple
$\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ of finite non-negative Borel
measures on the unit circle. In particular, it is shown that any weighted
Dirichlet-type space of order $m,$ for $m\geqslant 3,$ forms an algebra under
pointwise product. We also prove that every non-zero closed $M_{z}$-invariant
subspace of $\mathcal{H}_{\boldsymbol{\mu}},$ has codimension $1$ property if
$m\geqslant 3$ or $\mu_{2}$ is finitely supported. As another application of
local Douglas formula obtained in this article, it is shown that for any
$m\geqslant 2,$ weighted Dirichlet-type space of order $m$ does not coincide
with any de Branges-Rovnyak space $\mathcal{H}(b)$ with equivalence of norms.

The first author was supported by the Fields Post-doctoral Fellowship of the
Fields Institute for Research in Mathematical Sciences, Canada and by a post-
doctoral fellowship of Laval University, Canada. The second author was
supported through the INSPIRE faculty grant (Ref. No.
DST/INSPIRE/04/2017/002367).

##  1\. Introduction

The symbols $\mathbb{Z},\mathbb{N}$ and $\mathbb{Z}_{\geqslant 0}$ will denote
the set of integers, positive integers and non-negative integers respectively.
We use the symbols $\mathbb{T}$ and $\mathbb{D}$ to denote the unit circle and
the open unit disc in the complex plane $\mathbb{C}$ respectively. The
notation $\mathcal{M}_{+}(\mathbb{T})$ stands for the set of all finite non-
negative Borel measures on $\mathbb{T}$. Let $\mathcal{O}(\mathbb{D})$ denote
the space of all complex valued holomorphic functions on $\mathbb{D}.$ For
each $\alpha\in\mathbb{R},$ let $\mathcal{D}_{\alpha}$ denote the subspace of
$\mathcal{O}(\mathbb{D})$ given by

$\mathcal{D}_{\alpha}:=\bigg{\\{}f=\sum_{k=0}^{\infty}a_{k}z^{k}\Big{|}\|f\|^{2}_{\alpha}:=\sum_{k=0}^{\infty}(k+1)^{\alpha}|a_{k}|^{2}<\infty\bigg{\\}},$

see [32] for a detailed study of the space $\mathcal{D}_{\alpha}.$ Many
classical function spaces arise as $\mathcal{D}_{\alpha}$ spaces. For examples
$\mathcal{D}_{-1}$ coincides with the Bergman space, the space
$\mathcal{D}_{0}$ coincides with the Hardy space $H^{2}$ and $\mathcal{D}_{1}$
coincides with the Dirichlet space $\mathcal{D}.$ Recall that the Dirichlet
space $\mathcal{D}$ consists of all functions $f$ in $\mathcal{O}(\mathbb{D})$
for which the Dirichlet integral $D(f)$, given by

$D(f):=\displaystyle\int_{\mathbb{D}}\big{|}f^{\prime}(z)\big{|}^{2}dA(z),$

is finite. Here $A(z)$ denotes the normalized Lebesgue area measure on
$\mathbb{D}.$ It turns out that $\mathcal{D}$ is contained in the Hardy space
$H^{2}.$ In his study of minimal surfaces, Douglas [13] derived and used the
following formula

$\displaystyle
D(f)=\int_{\mathbb{T}}\int_{\mathbb{T}}\frac{|f^{*}(\zeta)-f^{*}(\lambda)|^{2}}{|\zeta-\lambda|^{2}}d\sigma(\zeta)d\sigma(\lambda),\,\,f\in\mathcal{D},$

where $\sigma(\zeta)$ denotes the normalized arc length measure on
$\mathbb{T}$ and $f^{*}(\lambda)$ denotes the non-tangential limit of $f$ at
$\lambda\in\mathbb{T}$ (which exists a.e. on $\mathbb{T}$). This and similar
formulae turns out to be very useful in understanding the boundary behaviour
of functions in $\mathcal{D},$ see [6]. Later, Richter and Sundberg (see [19,
p. 356]) introduced the notion of the local Dirichlet-type integral
$D_{\lambda}(f)$ of a function $f$ in $H^{2}$ at a point
$\lambda\in\mathbb{T}$ in the following manner:

$\displaystyle
D_{\lambda}(f):=\int_{\mathbb{T}}\frac{|f^{*}(\zeta)-f^{*}(\lambda)|^{2}}{|\zeta-\lambda|^{2}}d\sigma(\zeta).$

They established various remarkable formulae for the local Dirichlet-type
integral in [19, Sec. 3]. One of the motivations for studying the local
Dirichlet-type integral was to understand the weighted Dirichlet-type integral
$D_{\mu}(f)$ for a function $f\in\mathcal{O}(\mathbb{D})$ and a measure
$\mu\in\mathcal{M}_{+}(\mathbb{T}),$ a notion introduced and studied by
Richter in order to describe all cyclic analytic $2$-isometries, see [27,
Theorem 5.1]. The weighted Dirichlet-type integral $D_{\mu}(f)$ turns out to
be a key ingredient in describing the structure of the invariant subspaces of
the Dirichlet shift, that is, the operator $M_{z}$ of multiplication by the
co-ordinate function $z$ on the Dirichlet space $\mathcal{D},$ see [20,
Theorem 3.2]. For any $m\in\mathbb{Z}_{\geqslant 0},$ the operator $M_{z}$ on
$\mathcal{D}_{m}$ turns out to be an analytic $(m+1)$ isometry, see [1], [2],
[3] for a detailed study of $m$-isometries. Thus if one desires to describe
the structure of the invariant subspaces of $M_{z}$ on $\mathcal{D}_{m},$ it
is natural to look for model for analytic $(m+1)$ isometries, see [28].
Following [28], for a measure $\mu$ in $\mathcal{M}_{+}(\mathbb{T})$,
$f\in\mathcal{O}(\mathbb{D})$ and for a positive integer $n,$ we consider the
weighted Dirichlet-type integral of $f$ of order $n$ by

$\displaystyle
D_{\mu,n}(f):=\frac{1}{n!(n-1)!}\displaystyle\int_{\mathbb{D}}\big{|}f^{(n)}(z)\big{|}^{2}P_{\\!\mu}(z)(1-|z|^{2})^{n-1}dA(z).$

Here $f^{(n)}(z)$ denotes the $n$-th order derivative of $f$ at $z,$ and
$P_{\\!\mu}(z)$ is the Poisson integral of the measure $\mu,$ that is,

$\displaystyle
P_{\\!\mu}(z):=\int_{\mathbb{T}}\frac{1-|z|^{2}}{|z-\zeta|^{2}}d\mu(\zeta),\,\,\,z\in\mathbb{D}.$

For a function $f\in\mathcal{O}(\mathbb{D}),$ it will be also useful for us to
consider the weighted Dirichlet-type integral of order $0$ defined by

$\displaystyle D_{\mu,0}(f):=\displaystyle\lim_{R\to
1^{-}}\displaystyle\int_{\mathbb{T}}\big{|}f(R\zeta)\big{|}^{2}P_{\\!\mu}(R\zeta)d\sigma(\zeta),$

provided the limit exists. Note that if $\mu=\sigma$ then the integral
$D_{\sigma,0}(f)$ coincides with the square of the Hardy norm of $f.$ In case
of $n=1,$ the integral $D_{\mu,1}(f)$ coincides with the weighted Dirichlet-
type integral $D_{\mu}(f)$ for every $\mu\in\mathcal{M}_{+}(\mathbb{T}).$ In
particular if $\mu=\sigma$ then $D_{\sigma,1}(f)$ coincides with $D(f).$ If
$\mu$ is $\delta_{\lambda}$, the Dirac measure at the point
$\lambda\in\mathbb{T}$, we adopt a simpler notation $D_{\lambda,n}(\cdot)$ in
place of $D_{\delta_{\lambda},n}(\cdot)$ and we refer it as a local Dirichlet-
type integral of order $n$ at $\lambda$. We would like to mention that a
similar notion, namely, local Dirichlet-type integral of order $n$ of a
function at a point $\lambda\in\mathbb{T}$ is introduced and studied recently
in [17, p. 3]. Nevertheless, we will see in Remark 5.4 that this notion
differs from that of local Dirichlet-type integral of order $n$ introduced in
this article.

For a measure $\mu\in\mathcal{M}_{+}(\mathbb{T})/\\{0\\}$ and for each
$n\in\mathbb{Z}_{\geqslant 0},$ we consider the semi-inner product space
$\mathcal{H}_{\mu,n}$ given by

$\mathcal{H}_{\mu,n}:=\big{\\{}f\in\mathcal{O}(\mathbb{D}):D_{\mu,n}(f)<\infty\big{\\}},$

associated to the semi-norm $\sqrt{D_{\mu,n}(\cdot)}.$ In case $\mu$ is
$\delta_{\lambda}$, $\lambda\in\mathbb{T},$ we will use a simpler notation
$\mathcal{H}_{\lambda,n}$ in place of $\mathcal{H}_{\delta_{\lambda},n}$ and
we refer it as a local Dirichlet space of order $n$ at $\lambda$. If $\mu=0,$
we set $\mathcal{H}_{\mu,n}=H^{2}$ for every $n\in\mathbb{Z}_{\geqslant 0}.$
If $\mu=\sigma,$ by a straightforward computation it follows that for a
holomorphic function $f=\sum_{k=0}^{\infty}a_{k}z^{k}$ in
$\mathcal{O}(\mathbb{D}),$ we have

$D_{\sigma,n}(f)=\sum_{k=n}^{\infty}\binom{k}{n}|a_{k}|^{2},\,\,\,\quad
n\in\mathbb{Z}_{\geqslant 0},$

where $\binom{k}{n}:=\frac{k!}{n!(k-n)!}$ for any $k\geqslant n.$ It follows
that for each $n\in\mathbb{Z}_{\geqslant 0},$ the space
$\mathcal{H}_{\sigma,n}$ coincides as a set with the space $\mathcal{D}_{n}.$
The reader is referred to [16] (see also [28]) for several properties of the
spaces $\mathcal{H}_{\mu,n}$ for an arbitrary non-negative measure $\mu$ and a
positive integer $n.$

Richter and Sundberg had shown that, for any $\lambda\in\mathbb{T},$
$D_{\lambda,1}(f)=D_{\lambda}(f),$ for every $f$ in $H^{2},$ see [19,
Proposition 2.2]. In these notations, we note that

$\displaystyle
D_{\lambda,1}(f)=D_{\sigma,0}\Big{(}\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)},$

for every $f\in\mathcal{H}_{\lambda,1},$ see also [29, Proposition 1] for an
alternative proof. This formula is often known as a local Douglas formula, see
[14, Section 7.2]. In this article, we establish a local Douglas formula for
higher order local Dirichlet-type integrals, namely, we prove the following
theorem.

######  Theorem 1.1.

Let $n$ be a positive integer, $\lambda\in\mathbb{T}$, and
$f\in\mathcal{O}(\mathbb{D})$. Then $f\in\mathcal{H}_{\lambda,n}$ if and only
if $f=\alpha+(z-\lambda)g$ for some $g$ in $\mathcal{H}_{\sigma,n-1}$ and
$\alpha\in\mathbb{C}.$ Moreover, in this case, the following statements hold:

  * (i)

$D_{\lambda,n}(f)=D_{\sigma,n-1}(g),$

  * (ii)

$f(z)\to\alpha$ as $z\to\lambda$ in each oricyclic approach region
$|z-\lambda|<\kappa(1-|z|^{2})^{\frac{1}{2}}$, $\kappa>0$. In particular,
$f^{*}(\lambda)$ exists and is equal to $\alpha.$

Thus one can rewrite the local Douglas formula for higher order Dirichlet-type
integrals in the following form:

$\displaystyle
D_{\lambda,n}(f)=D_{\sigma,n-1}\Big{(}\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)},\,\,\,\,f\in\mathcal{H}_{\lambda,n},\,\,\,n\in\mathbb{N}.$
(1)

We explore various applications of this generalized local Douglas formula (1)
in this article. For $0<r<1,$ let $f_{r}$ be the $r$-dilation of a function
$f\in\mathcal{O}(\mathbb{D})$ given by $f_{r}(z):=f(rz),\,z\in\mathbb{D}.$ As
a first application of the formula (1), we prove in Theorem 2.6 that for each
$n\geqslant 2$ and $\mu\in\mathcal{M}_{+}(\mathbb{T}),$

$D_{\mu,n}(f_{r})\leqslant\frac{4^{n-1}(2-r)r^{2n}}{(1+r)^{2n-2}}D_{\mu,n}(f)$
(2)

for every $f\in\mathcal{H}_{\mu,n}$ and $0<r<1$. Note that for any
$n\in\mathbb{N}$ and $0<r<1$, we have
$\frac{4^{n-1}(2-r)r^{2n}}{(1+r)^{2n-2}}\leqslant 1$. Thus the inequality (2)
is an improvement of [16, Theorem 1.1], where only the contractivity of the
map $f\mapsto f_{r}$ was shown. Analogous results are known in the case of
$n=1,$ see [19], [29], [14], [15], [18] and references therein.

In the study of function spaces, it is an important problem to determine the
associated multiplier algebra. Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $n$
be a non-negative integer. The multiplier algebra of $\mathcal{H}_{\mu,n}$ is
defined by

$\displaystyle\text{Mult\,}(\mathcal{H}_{\mu,n}):=\\{\varphi\in\mathcal{O}({\mathbb{D}}):\varphi
f\in\mathcal{H}_{\mu,n}\,\text{\,\,for
every\,}\,\,f\in\mathcal{H}_{\mu,n}\\}.$

Several studies have been done in characterizing the multipliers of the
Dirichlet space $\mathcal{D}$, and more generally for the multipliers of the
weighted Dirichlet-type space $\mathcal{H}_{\mu,1}$ for any
$\mu\in\mathcal{M}_{+}(\mathbb{T}),$ see [31],[20],[9] and references therein.
In Theorem 3.5, we characterize $\text{Mult\,}(\mathcal{H}_{\lambda,n})$ for
every $\lambda\in\mathbb{T}$ and $n\in\mathbb{N}.$ This characterization also
helps us to study the $\text{Mult\,}(\mathcal{H}_{\mu,n})$ for any
$\mu\in\mathcal{M}_{+}(\mathbb{T})$ and for every $n\in\mathbb{N}.$ It is well
known that the space $\mathcal{D}_{n}$ forms an algebra under pointwise
product for every $n\geqslant 2$ and consequently
$\text{Mult\,}(\mathcal{D}_{n})=\mathcal{D}_{n}$ for each $n\geqslant 2,$ see
[32, Remark, p. 239]. Even though $\mathcal{D}_{1},$ the classical Dirichlet
space, is not an algebra under pointwise product (see [14, Theorem 1.3.1]),
the subspace of bounded functions in it turns out to be an algebra, see [14,
Theorem 1.3.2]. More generally, it is well known that $\mathcal{H}_{\mu,1}$ is
never an algebra under pointwise product for any
$\mu\in\mathcal{M}_{+}(\mathbb{T}),$ see [14, Sec. 7.1, Exercise 4], but
$\mathcal{H}_{\mu,1}\cap H^{\infty}$ forms an algebra under pointwise product,
see [20, p. 170], where $H^{\infty}$ is the Banach algebra of bounded
holomorphic functions on $\mathbb{D}$. We extend this result in the following
theorem for higher order weighted Dirichlet-type spaces.

######  Theorem 1.2.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $n\in\mathbb{N}$. Then
$\mathcal{H}_{\mu,n}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1})$ is an
algebra.

As a consequence of this theorem, we find that $\mathcal{H}_{\mu,n}$ also
forms an algebra under pointwise product and
$\text{Mult\,}(\mathcal{H}_{\mu,n})=\mathcal{H}_{\mu,n}$ for every
$\mu\in\mathcal{M}_{+}(\mathbb{T})/\\{0\\}$ and for each $n\geqslant 3,$ see
Corollary 3.6. The answer to the question of whether $\mathcal{H}_{\mu,2}$ is
an algebra or not, is of mixed nature. We have found that when
$\mu=\delta_{\lambda}$ with $\lambda\in\mathbb{T},$ then
$\mathcal{H}_{\lambda,2}$ is not an algebra under pointwise product. On the
other hand we have $\mathcal{H}_{\sigma,2}=\mathcal{D}_{2}$ forms an algebra
under pointwise product, see [32, Remark, p. 239].

Let $m\in\mathbb{N}$ and $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ be an
$m$-tuple of non-negative measures in $\mathcal{M}_{+}(\mathbb{T}).$ Let
$\mathcal{H}_{\boldsymbol{\mu}}$ be the linear subspace of
$\mathcal{O}(\mathbb{D})$ given by
$\mathcal{H}_{\boldsymbol{\mu}}:=\bigcap_{j=1}^{m}\mathcal{H}_{\mu_{j},j}.$
Note that for each $j=1,\ldots,m,$ the space $\mathcal{H}_{\mu_{j},n}$ is
contained in the Hardy space $H^{2}$ for any $n\in\mathbb{N},$ see [16,
Corollary 2.5]. For $f\in\mathcal{H}_{\boldsymbol{\mu}},$ we associate the
norm $\|f\|_{\boldsymbol{\mu}}$ given by

$\displaystyle\|f\|^{2}_{\boldsymbol{\mu}}:=\|f\|^{2}_{\\!{}_{H^{2}}}+\sum\limits_{j=1}^{m}D_{\mu_{j},j}(f).$

The space $\mathcal{H}_{\boldsymbol{\mu}}$ with respect to the norm
$\|\cdot\|_{\boldsymbol{\mu}}$ turns out to be a reproducing kernel Hilbert
space, see [16, p. 13]. In particular, it follows that $\mathcal{H}_{\mu,n}$
is a Hilbert space with respect to the norm $\|\cdot\|_{\mu,n},$ given by

$\displaystyle\|f\|_{\mu,n}^{2}=\|f\|^{2}_{H^{2}}+D_{\mu,n}(f),\,\,\,\,f\in\mathcal{H}_{\mu,n}.$

The operator $M_{z}$ of multiplication by the co-ordinate function on
$\mathcal{H}_{\boldsymbol{\mu}}$ turns out to be a cyclic analytic
$(m+1)$-isometry, see [16, Theorem 4.1, Corollary 5.3]. Let
$Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ denote the lattice of all closed
$M_{z}$-invariant subspaces of $\mathcal{H}_{\boldsymbol{\mu}}.$ A subspace
$\mathcal{W}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ is said to have
codimension $k$ property if $dim(\mathcal{W}\ominus(z-\lambda)\mathcal{W})=k,$
for each $\lambda\in\mathbb{D},$ see [25, Definition 2.12]. In order to have a
Beurling-type description of $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ it
is expected that $\mathcal{W}$ has codimension $1$ property for every non-zero
$\mathcal{W}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ see [25,
Proposition 2.13]. In the case of $m=1,$ that is, when
$\boldsymbol{\mu}=\mu_{1}$ and $\mu_{1}\in\mathcal{M}_{+}(\mathbb{T}),$ it is
well known that any non-zero $\mathcal{W}$ in
$Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ has codimension $1$ property, see
[20, Theorem 3.2]. We find that for any $m\geqslant 3$ and any non-zero
$\mathcal{W}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ has codimension
$1$ property. For the remaining case $m=2,$ that is, when
$\boldsymbol{\mu}=(\mu_{1},\mu_{2}),$ we show that the result remains true
provided $\mu_{2}$ is finitely atomic, see Theorem 4.5.

There is an intimate connection between the local Dirichlet spaces
$\mathcal{H}_{\lambda}$ and the de Branges-Rovnyak spaces $\mathcal{H}(b)$
(see [8] for basic theory of $\mathcal{H}(b)$ spaces). For any $b$ in the unit
ball of $H^{\infty}$, the de Branges-Rovnyak space $\mathcal{H}(b)$ is the
reproducing kernel Hilbert space with the reproducing kernel
$\frac{1-b(z)\bar{b(w)}}{1-z\bar{w}}$, $z,w\in\mathbb{D}$. In [29], Sarason
showed that any local Dirichlet space $\mathcal{H}_{\lambda}$ coincides with a
de Branges-Ronyak space $\mathcal{H}(b),$ with equality of norms, where $b$
can be chosen to be
$b(z)=\frac{(1-r)\overline{\lambda}z}{1-r\overline{\lambda}z},$
$z\in\mathbb{D},$ with $r=\frac{3-\sqrt{5}}{2}.$ In [10, Theorem 3.1],
Chevrot, Guillot and Ransford showed that, in fact, if any $\mathcal{H}_{\mu}$
space is equal to some de Branges-Ronyak space $\mathcal{H}(b)$, with equality
of norms, then $\mu$ has to be a positive multiple of a point mass measure for
some $\lambda\in\mathbb{T}$ and further $b$ is identified explicitly.
Thereafter, in [12], the problem of characterizing $\mu$ and $b$ such that
$\mathcal{H}_{\mu}=\mathcal{H}(b)$ as a set was addressed by obtaining certain
necessary and some sufficient conditions for
$\mathcal{H}_{\mu}=\mathcal{H}(b).$ In this paper, we study the same question
for the spaces $\mathcal{H}_{\boldsymbol{\mu}}$ and $\mathcal{H}(b),$ where
$\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$, $m\geqslant 2$, is an arbitrary
$m$-tuple of positive measures in $\mathcal{M}_{+}(\mathbb{T})$.

This article is organized as follows. Section 2 is devoted to prove Theorem
1.1. Section 3 deals with a few applications of generalized local Douglas
Formula. In this section, we prove Theorem 1.2 and as a corollary it is
observed that $\mathcal{H}_{\mu,n}$ is an algebra for any
$\mu\in\mathcal{M}_{+}(\mathbb{T})/\\{0\\}$ and $n\geqslant 3,$ see Corollary
3.6. In Section 4, we discuss about the codimension $1$ property for any
closed $M_{z}$-invariant subspaces of $\mathcal{H}_{\mu,n}.$ In Section 5, it
is shown that, for $n\geqslant 2,$ any weighted Dirichlet-type space of order
$n$ does not coincide with de Branges-Rovnyak space $H(b)$, with equivalence
of norms, for any $b$ in the unit ball of $H^{\infty}$.

##  2\. A Generalized Local Douglas Formula

In this section, we shall prove a local Douglas formula for local Dirichlet
spaces of order $n$. For any $\lambda\in\mathbb{T}$ and $n\in\mathbb{N}$, let
$\nu_{\lambda,n}$ denote the positive weighted area measure on $\mathbb{D}$
given by

$\displaystyle
d\nu_{\lambda,n}(z)=\frac{1}{n!(n-1)!}P_{\delta_{\lambda}}(z)(1-|z|^{2})^{n-1}dA(z),$

where $P_{\delta_{\lambda}}(z)=\frac{1-|z|^{2}}{|z-\lambda|^{2}},$
$z\in\mathbb{D},$ is the Poisson integral of the Dirac measure
$\delta_{\lambda}.$ Let $A^{2}(d\nu_{\lambda,n})$ be the weighted Bergman
space corresponding to the measure $\nu_{\lambda,n}$, that is,

$A^{2}(d\nu_{\lambda,n}):=\\{f\in\mathcal{O}(\mathbb{D}):\|f\|^{2}=\int_{\mathbb{D}}|f(z)|^{2}d\nu_{\lambda,n}(z)<\infty\\}.$

######  Lemma 2.1.

For any $\lambda\in\mathbb{T}$ and $n\in\mathbb{N}$, $A^{2}(d\nu_{\lambda,n})$
is a reproducing kernel Hilbert space with the reproducing kernel

$(n+1)!(n-1)!\frac{(z-\lambda)(\overline{w}-\overline{\lambda})}{(1-z\overline{w})^{n+2}},~{}z,w\in\mathbb{D}.$

###### Proof.

Let $A^{2}\big{(}(1-|z|^{2})^{n}dA(z)\big{)}$ be the weighted Bergman space
corresponding to the weighted area measure $(1-|z|^{2})^{n}dA(z)$ on
$\mathbb{D},$ that is,

$\displaystyle
A^{2}\big{(}(1-|z|^{2})^{n}dA(z)\big{)}:=\big{\\{}f\in\mathcal{O}(\mathbb{D}):\|f\|^{2}=\int_{\mathbb{D}}|f(z)|^{2}(1-|z|^{2})^{n}dA(z)<\infty\big{\\}}.$

It is evident that the operator $\sqrt{n!(n-1)!}M_{z-\lambda}$ is a unitary
from $A^{2}\big{(}(1-|z|^{2})^{n}dA(z)\big{)}$ onto
$A^{2}\big{(}d\nu_{\lambda,n}\big{)}$, where $M_{z-\lambda}$ denotes the
operator of multiplication by the function $(z-\lambda)$. By a direct
computation, it is verified that

$\left\\{\sqrt{(n+1){n+k+1\choose k}}z^{k}:k\in\mathbb{Z}_{\geqslant
0}\right\\}$

forms an orthonormal basis of $A^{2}\big{(}(1-|z|^{2})^{n}dA(z)\big{)}$.
Therefore, by [24, Theorem 2.4], $A^{2}(d\nu_{\lambda,n})$ has the reproducing
kernel

$n!(n-1)!(n+1)(z-\lambda)(\overline{w}-\overline{\lambda})\sum_{k=0}^{\infty}\binom{n+k+1}{k}z^{k}\overline{w}^{k}=(n+1)!(n-1)!\frac{(z-\lambda)(\overline{w}-\overline{\lambda})}{(1-z\overline{w})^{n+2}},\quad
z,w\in\mathbb{D}.$

This completes the proof. ∎

######  Lemma 2.2.

Let $\lambda\in\mathbb{T}$, $n\in\mathbb{N}$, and
$f\in\mathcal{H}_{\sigma,n-1}.$ Then $((z-\lambda)f)^{(n)}(z)=0$ for all
$z\in\mathbb{D}$ if and only if $D_{\sigma,n-1}(f)=0$.

###### Proof.

Suppose that $f\in\mathcal{H}_{\sigma,n-1}$ and $((z-\lambda)f)^{(n)}(z)=0$
for all $z\in\mathbb{D}.$ Then $(z-\lambda)f$ is a polynomial of degree at
most $n-1$. Note that $\mathcal{H}_{\sigma,n-1}$ is contained in the Hardy
space $H^{2}$(see [16, Corollary 2.5], [32, Corollary 1, Theorem 7]).
Therefore, in view of [14, Lemma 1.5.4], we find that $\lim_{|z|\to
1^{-}}(1-|z|^{2})|f(z)|^{2}=0$. Hence it follows that $\lim_{r\to
1^{-}}(r\lambda-\lambda)f(r\lambda)=0.$ This gives us that the polynomial
$(z-\lambda)f$ must have a zero at $\lambda$ and consequently $f$ is a
polynomial of degree at most $n-2.$ Therefore, $D_{\sigma,n-1}(f)=0.$
Conversely, assume that $D_{\sigma,n-1}(f)=0$. This implies $f$ is a
polynomial of degree at most $n-2$, and therefore $((z-\lambda)f)^{(n)}(z)=0$
for all $z\in\mathbb{D}$. ∎

Let $j\in\mathbb{Z}_{\geqslant 0}.$ Note that, in general
$\sqrt{D_{\sigma,j}(\cdot)}$ represents a seminorm on $\mathcal{H}_{\sigma,j}$
(unless $j=0$). Consider, the subspace $\mathcal{U}_{\sigma,j}$ of
$\mathcal{H}_{\sigma,j}$ given by

$\mathcal{U}_{\sigma,j}=\big{\\{}f\in\mathcal{O}(\mathbb{D}):f(z)=\sum_{k=j}^{\infty}a_{k}z^{k},D_{\sigma,j}(f)<\infty\big{\\}}.$

It is straightforward to verify that $\sqrt{D_{\sigma,j}(\cdot)}$ represents a
norm on $\mathcal{U}_{\sigma,j}$ which is induced by an inner product.
Moreover $\mathcal{U}_{\sigma,j}$ is a reproducing kernel Hilbert space in
which the monomials given by $\\{{\binom{k}{j}}^{-\frac{1}{2}}z^{k}:k\geqslant
j\\}$ forms an orthonormal basis. The reproducing kernel $K_{\sigma,j}$ of
$\mathcal{U}_{\sigma,j}$ is given by

$K_{\sigma,j}(z,w)=\sum_{k=j}^{\infty}\binom{k}{j}^{-1}z^{k}\overline{w}^{k},\quad
z,w\in\mathbb{D}.$ (3)

For any $\alpha>0$ and $j\in\mathbb{Z}_{\geqslant 0}$, the Pochammer symbol
$(\alpha)_{j}$ is defined by

$(\alpha)_{j}=\begin{cases}\alpha(\alpha+1)\cdots(\alpha+j-1),~{}&j\geqslant
1\\\ 1,~{}&j=0.\end{cases}$

######  Proposition 2.3.

Let $n$ be a positive integer, $\lambda\in\mathbb{T}$, and
$T:\mathcal{O}(\mathbb{D})\to\mathcal{O}(\mathbb{D})$ be the linear map given
by

$(Tf)(z)=((z-\lambda)f)^{(n)}(z),~{}z\in\mathbb{D},~{}f\in\mathcal{O}(\mathbb{D}).$

Then $T$ maps the Hilbert space
$(\mathcal{U}_{\sigma,n-1},\sqrt{D_{\sigma,n-1}(\cdot)})$ isometrically onto
$A^{2}(d\nu_{\lambda,n}).$

###### Proof.

Consider the space $\mathcal{W}:=T(\mathcal{U}_{\sigma,n-1})$ endowed with the
norm

$\|T(f)\|_{\\!{}_{\mathcal{W}}}:=\sqrt{D_{\sigma,n-1}(f)},\quad
f\in\mathcal{H}_{\sigma,n-1}.$

By Lemma 2.2, $\|\cdot\|_{\\!{}_{\mathcal{W}}}$ is a well-defined norm on
$\mathcal{W}$. Clearly $T$ maps $\mathcal{U}_{\sigma,n-1}$ isometrically onto
$\mathcal{W}$. This gives us that the space $\mathcal{W}$ endowed with the
norm $\|\cdot\|_{\\!{}_{\mathcal{W}}}$ is a Hilbert space. Now we show that
$\mathcal{W}$ is a reproducing kernel Hilbert space. This is equivalent to
showing that for each $w\in\mathbb{D},$ the evaluation map
$E_{w}:\mathcal{W}\to\mathbb{C},$ defined by $E_{w}(g)=g(w),$ is continuous.
Let $w\in\mathbb{D}$, and $T(f_{n})$ converges to $T(f)$ in $\mathcal{W}$,
where $f_{n},f\in\mathcal{U}_{\sigma,n-1}$. Clearly $f_{n}$ converges to $f$
in $\mathcal{U}_{\sigma,n-1}$. Since $\mathcal{U}_{\sigma,n-1}$ is a
reproducing kernel Hilbert space, it follows that $f_{n}$ converges to $f$
uniformly on compact subsets of $\mathbb{D}$. Therefore $T(f_{n})$ also
converges to $T(f)$ uniformly on compact subsets of $\mathbb{D}$. This, in
particular, shows that $E_{w}$ is continuous on $\mathcal{W}$, proving that
$\mathcal{W}$ is a reproducing kernel Hilbert space.

The proposition will now be established by showing that the reproducing
kernels of $\mathcal{W}$ and $A^{2}(d\nu)$ coincide (see [24, Proposition
2.3]). As the operator $T$ is unitary, it maps the orthonormal basis
$\big{\\{}\binom{k}{n-1}^{-\frac{1}{2}}z^{k}:k\geqslant n-1\big{\\}}$ of
$\mathcal{U}_{\sigma,n-1}$ to an orthonormal basis of $\mathcal{W}.$ Thus the
reproducing kernel of $\mathcal{W}$ is given by

$K_{\\!{}_{\mathcal{W}}}(z,w)=\frac{\partial^{2n}}{\partial
z^{n}\partial\overline{w}^{n}}\Big{(}(z-\lambda)(\overline{w}-\overline{\lambda})K_{\sigma,n-1}(z,w)\Big{)},\quad
z,w\in\mathbb{D}.$ (4)

In view of (3), we see that the expression in (4) is equal to

$\displaystyle\frac{\partial^{2n}}{\partial
z^{n}\partial\overline{w}^{n}}\left(\sum_{k=n-1}^{\infty}{k\choose
n-1}^{-1}\Big{(}z^{k+1}\overline{w}^{k+1}+z^{k}\overline{w}^{k}-\lambda
z^{k}\overline{w}^{k+1}-\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}\right)$
$\displaystyle=$ $\displaystyle\sum_{k=n-1}^{\infty}\frac{1}{{k\choose
n-1}}\big{(}(k+2-n)_{n}\big{)}^{2}\,z^{k+1-n}\overline{w}^{k+1-n}+\sum_{k=n}^{\infty}\frac{1}{{k\choose
n-1}}\big{(}(k+1-n)_{n}\big{)}^{2}z^{k-n}\overline{w}^{k-n}$
$\displaystyle\hskip 34.14322pt-\sum_{k=n}^{\infty}\frac{1}{{k\choose
n-1}}(k+2-n)_{n}(k+1-n)_{n}\Big{(}\lambda
z^{k-n}\overline{w}^{k+1-n}+\overline{\lambda}z^{k+1-n}\overline{w}^{k-n}\Big{)}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{\infty}\big{(}(k+1)_{n}\big{)}^{2}\bigg{(}{k+n-1\choose
n-1}^{-1}+{k+n\choose n-1}^{-1}\bigg{)}z^{k}\overline{w}^{k}$
$\displaystyle\hskip 76.82234pt-\sum_{k=0}^{\infty}{k+n\choose
n-1}^{-1}(k+2)_{n}(k+1)_{n}\Big{(}\lambda
z^{k}\overline{w}^{k+1}+\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}$
$\displaystyle=$
$\displaystyle(n-1)!\Bigg{(}\sum_{k=0}^{\infty}(k+1)_{n}(2k+n+1)z^{k}\overline{w}^{k}-\sum_{k=0}^{\infty}(k+1)_{n+1}\Big{(}\lambda
z^{k}\overline{w}^{k+1}+\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}\Bigg{)}.$

Also, by a straightforward calculation, we see that

$\displaystyle~{}~{}~{}\frac{(z-\lambda)(\overline{w}-\overline{\lambda})}{(1-z\overline{w})^{n+2}}$
$\displaystyle=\sum_{k=0}^{\infty}{n+k+1\choose
k}\Big{(}z^{k+1}\overline{w}^{k+1}+z^{k}\overline{w}^{k}-\lambda
z^{k}\overline{w}^{k+1}-\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}$
$\displaystyle=1+\sum_{k=1}^{\infty}\Bigg{(}{n+k\choose k-1}+{n+k+1\choose
k}\Bigg{)}z^{k}\overline{w}^{k}-\sum_{k=0}^{\infty}{n+k+1\choose
k}\Big{(}\lambda
z^{k}\overline{w}^{k+1}+\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}$
$\displaystyle=1+\frac{1}{(n+1)!}\bigg{(}\sum_{k=1}^{\infty}(k+1)_{n}(2k+n+1)z^{k}\overline{w}^{k}-\sum_{k=0}^{\infty}(k+1)_{n+1}\Big{(}\lambda
z^{k}\overline{w}^{k+1}+\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}\bigg{)}$
$\displaystyle=\frac{1}{(n+1)!}\bigg{(}\sum_{k=0}^{\infty}(k+1)_{n}(2k+n+1)z^{k}\overline{w}^{k}-\sum_{k=0}^{\infty}(k+1)_{n+1}\Big{(}\lambda
z^{k}\overline{w}^{k+1}+\overline{\lambda}z^{k+1}\overline{w}^{k}\Big{)}\bigg{)}$
$\displaystyle=\frac{1}{(n+1)!(n-1)!}K_{\\!{}_{\mathcal{W}}}(z,w).$

Use of Lemma 2.1 now completes the proof. ∎

For any holomorphic function $f$ on the unit disc $\mathbb{D}$ and a complex
number $\lambda$ in $\mathbb{T}$, let $f^{*}(\lambda)$ denote the boundary
value $f$ at $\lambda$, defined by

$f^{*}(\lambda):=\lim_{r\to 1^{-}}f(r\lambda),$

whenever it exists. The following theorem (main theorem of this section) is
deduced as a consequence of Lemma 2.3. This theorem, in particular, tells us
that for any $f$ in $\mathcal{H}_{\lambda,n},$ the boundary value of $f$ at
$\lambda$ exists. As an another consequence of Theorem 1.1 and [16,
Proposition 2.6], we see that $\mathcal{H}_{\lambda,n}$ is contained in
$\mathcal{H}_{\sigma,n-1}.$

Proof of Theorem 1.1: Assume that $f\in\mathcal{H}_{\lambda,n}$. Thus we have
$f^{(n)}\in A^{2}(d\nu_{\lambda,n})$. Now, by Proposition 2.3, there exists a
$\tilde{g}$ in $\mathcal{U}_{\sigma,n-1}$ such that
$D_{\lambda,n}(f)=D_{\sigma,n-1}(\tilde{g})$ and
$((z-\lambda)\tilde{g})^{(n)}=f^{(n)}$. Therefore,

$\displaystyle f(z)=(z-\lambda)\tilde{g}(z)+p(z),\quad z\in\mathbb{D},$

for some polynomial $p$ of degree at most $n-1$. Thus, with
$q(z)=\frac{p(z)-p(\lambda)}{z-\lambda},$ we have

$\displaystyle f(z)=p(\lambda)+(z-\lambda)g(z),z\in\mathbb{D}$

where $g=\tilde{g}+q$. Clearly,
$D_{\sigma,n-1}(g)=D_{\sigma,n-1}(\tilde{g})=D_{\lambda,n}(f),$ which is
finite. Hence $g\in\mathcal{H}_{\sigma,n-1}$. Conversely, assume that
$f=\alpha+(z-\lambda)g$ for some $g$ with $g\in\mathcal{H}_{\sigma,n-1}.$ Then
it follows from Lemma 2.3 that $f^{(n)}\in A^{2}(d\nu_{\lambda,n})$, and hence
$f\in\mathcal{H}_{\lambda,n}$. This completes the proof of the first of the
theorem. Note that, for $f\in\mathcal{H}_{\lambda,n}$, the equality
$D_{\lambda,n}(f)=D_{\sigma,n-1}(g)$ is already established. To complete the
proof, suppose that $f=\alpha+(z-\lambda)g$ for some complex number $\alpha$
and $g$ in $\mathcal{H}_{\sigma,n-1}$. Since $\mathcal{H}_{\sigma,n-1}$ is
contained in the Hardy space $H^{2}$, by [14, Lemma 1.5.4] we have
$\lim_{|z|\to 1}(1-|z|^{2})^{\frac{1}{2}}|g(z)|=0$. If
$|z-\lambda|<\kappa(1-|z|^{2})^{\frac{1}{2}}$, $\kappa>0$, then

$|f(z)-\alpha|<\kappa(1-|z|^{2})^{\frac{1}{2}}|g(z)|.$

Hence it follows that $f(z)\to\alpha$ as $z\to\lambda$ in each oricyclic
approach region $|z-\lambda|<\kappa(1-|z|^{2})^{\frac{1}{2}}$. This completes
the proof of this Theorem.

######  Corollary 2.4.

Let $n$ be a positive integer and $\mu$ be a finite positive Borel measure on
the unit circle $\mathbb{T}$. Then for any $f$ in $\mathcal{H}_{\mu,n}$,
$f^{*}(\lambda)$ exists for $\mu$-almost every $\lambda\in\mathbb{T}$, and the
following holds:

$D_{\mu,n}(f)=\int_{\mathbb{T}}\int_{\mathbb{D}}\Big{|}\Big{(}\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)}^{(n-1)}(z)\Big{|}^{2}(1-|z|^{2})^{n-1}dA(z)d\mu(\lambda)$

###### Proof.

By Fubini’s theorem,

$\displaystyle
D_{\mu,n}(f)=\frac{1}{n!(n-1)!}\displaystyle\int_{\mathbb{D}}\big{|}f^{(n)}(z)\big{|}^{2}P_{\\!\mu}(z)(1-|z|^{2})^{n-1}dA(z)=\int_{\mathbb{\mathbb{missing}}T}D_{\lambda,n}(f)d\mu(\lambda).$

An application of Theorem 1.1 completes the proof of the corollary. ∎

We finish this section with a few applications of Theorem 1.1. The first
application shows that for any $n>1$ the map $f\mapsto f_{r}$, $0\leqslant
r<1$, is contractive on $\mathcal{H}_{\mu,n}$. Here for any
$f\in\mathcal{O}(\mathbb{D})$, the $r$th dilation $f_{r}$ of $f$ is defined by
$f_{r}(z):=f(rz)$, $z\in\mathbb{D}.$ We point out that this result is already
obtained in [29, Proposition 3] for $n=1$, and [16, Theorem 1.1] for every
$n\geqslant 1.$ One of the consequences of this result is that $\lim_{r\to
1^{-}}D_{\mu,n}(f_{r}-f)=0$ for any $f\in\mathcal{H}_{\mu,n}.$ We start with
the following lemma.

######  Lemma 2.5.

Let $\mathcal{H}$ be a semi-inner product space of holomorphic functions on
the unit disc $\mathbb{D}$ with the semi-norm $\|\cdot\|$. Suppose that
$zf\in\mathcal{H}$ whenever $f\in\mathcal{H}$. If $\|zf\|\geqslant\|f\|$ for
all $f\in\mathcal{H}$ then for $\lambda\in\mathbb{T}$, $0\leqslant r<1$, we
have

$\|(z-\lambda)f\|\leqslant\frac{2}{1+r}\|(z-r\lambda)f\|,~{}f\in\mathcal{H}.$

###### Proof.

Let $f\in\mathcal{H}$. Since $\|zf\|\geqslant\|f\|$, it follows that

$|\mbox{Re}~{}\lambda\langle f,zf\rangle|\leqslant|\lambda\langle
f,zf\rangle|\leqslant\|f\|\|zf\|\leqslant\|zf\|^{2},$

where for any complex number $w$, $\mbox{Re}\,w$ denotes the real part of $w$.
Hence

$\displaystyle\frac{4}{(1+r)^{2}}\|(z-r\lambda)f\|^{2}-\|(z-\lambda)f)\|^{2}$
$\displaystyle=\Big{(}\frac{4}{(1+r)^{2}}-1\Big{)}\|zf\|^{2}-\Big{(}1-\frac{4r^{2}}{(1+r)^{2}}\Big{)}\|f\|^{2}+2\Big{(}1-\frac{4r}{(1+r)^{2}}\Big{)}~{}\mbox{Re}~{}\lambda\langle
f,zf\rangle$
$\displaystyle\geqslant\Big{(}\big{(}\frac{4}{(1+r)^{2}}-1\big{)}-\big{(}1-\frac{4r^{2}}{(1+r)^{2}}\big{)}-2\big{(}1-\frac{4r}{(1+r)^{2}}\big{)}\Big{)}\|zf\|^{2}$
$\displaystyle=0.$

This completes the proof. ∎

######  Theorem 2.6.

Let $n\geqslant 1$ be an integer, and $\mu$ be in
$\mathcal{M}_{+}(\mathbb{T}).$ Then for any $f$ in $\mathcal{O}(\mathbb{D})$,
we have

$D_{\mu,n}(f_{r})\leqslant\frac{4^{n-1}(2-r)r^{2n}}{(1+r)^{2n-2}}D_{\mu,n}(f),~{}~{}0\leqslant
r<1.$ (5)

###### Proof.

We will establish the proof by induction on $n.$ The base case of $n=1$
follows from [18, Corollary 1.5]. Assume that the statement holds for every
$f\in\mathcal{O}(\mathbb{D}),$ $0\leqslant r<1,$ and $n=1,\ldots,k-1$ for
$k\geqslant 2.$ Let $f\in\mathcal{O}(\mathbb{D})$ and $0\leqslant r<1.$ Since
$D_{\mu,k}(f)=\int_{\mathbb{T}}D_{\lambda,k}(f)d\mu(\lambda)$, in order to
establish (5) for $n=k$, it suffices to show that the (5) holds for $n=k$ and
$\mu=\delta_{\lambda}$ for each $\lambda\in\mathbb{T}.$ Fix a $\lambda$ in
$\mathbb{T}$ and assume that $D_{\lambda,k}(f)<\infty.$ Then by Theorem 1.1 we
obtain that $f(z)=f^{*}(\lambda)+(z-\lambda)g(z)$ for some $g$ in
$\mathcal{H}_{\sigma,k-1}$ and $D_{\lambda,k}(f)=D_{\sigma,k-1}(g)$. By a
straightforward computation, we see that

$f_{r}(z)=f_{r}^{*}(\lambda)+(z-\lambda)h_{r}(z),$ (6)

where

$\displaystyle h(z)$
$\displaystyle=r\frac{(z-\lambda)g(z)-(r\lambda-\lambda)g(r\lambda)}{z-r\lambda}.$

Note that

$h(z)=r\frac{(z-\lambda)(g(z)-g(r\lambda))+(z-r\lambda)g(r\lambda)}{z-r\lambda}.$

Using this together with Lemma 2.5 we get that

$\displaystyle D_{\sigma,k-1}(h)$
$\displaystyle=r^{2}D_{\sigma,k-1}\left(\frac{(z-\lambda)(g(z)-g(r\lambda))}{z-r\lambda}\right)$
$\displaystyle\leqslant\frac{4r^{2}}{(1+r)^{2}}D_{\sigma,k-1}(g(z)-g(r\lambda))$
$\displaystyle=\frac{4r^{2}}{(1+r)^{2}}D_{\sigma,k-1}(g).$ (7)

By induction hypothesis, we have that

$D_{\sigma,k-1}(h_{r})\leqslant\frac{4^{k-2}(2-r)r^{2k-2}}{(1+r)^{2k-4}}D_{\sigma,k-1}(h).$
(8)

Using Theorem 1.1, together with (6), (2) and (8) we get

$\displaystyle D_{\lambda,k}(f_{r})$ $\displaystyle=D_{\sigma,k-1}(h_{r})$
$\displaystyle\leqslant\frac{4^{k-2}(2-r)r^{2k-2}}{(1+r)^{2k-4}}D_{\sigma,k-1}(h)$
$\displaystyle\leqslant\frac{4^{k-1}(2-r)r^{2k}}{(1+r)^{2k-2}}D_{\sigma,k-1}(g)=\frac{4^{k-1}(2-r)r^{2k}}{(1+r)^{2k-2}}D_{\lambda,k}(f).$

This completes the proof.

∎

The following theorem which is a generalization of [21, Theorem 3.1] (see also
[12, Lemma 4.2]) describes the weighted Dirichlet-type space
$\mathcal{H}_{\mu,n}$ when $\mu$ is finitely supported.

######  Theorem 2.7.

Let $n,k$ be two positive integers, and
$\mu=\sum_{j=1}^{k}c_{j}\delta_{\lambda_{j}}$ for some $c_{j}>0$ and
$\lambda_{j}\in\mathbb{T}$, $j=1,\ldots,k$. Then $f\in\mathcal{H}_{\mu,n}$ if
and only if there exists a polynomial $p$ of degree at most $k-1$ and a
function $g\in\mathcal{H}_{\sigma,n-1}$ such that

$f=p+g\prod_{j=1}^{k}(z-\lambda_{j}).$

###### Proof.

Note that $f\in\mathcal{H}_{\mu,n}$ if and only if
$f\in\mathcal{H}_{\lambda_{j},n}$ for each $j=1,\ldots,k.$ Suppose that
$f=p+g\prod_{j=1}^{k}(z-\lambda_{j})$ for some polynomial $p$ of degree at
most $k-1$ and a function $g\in\mathcal{H}_{\sigma,n-1}$. Since $z$ is a
multiplier of $\mathcal{H}_{\sigma,n-1}$, it follows that for any
$i=1,\ldots,k$, $g\prod_{j=1\\\ j\neq i}^{k}(z-\lambda_{j})$ is in
$\mathcal{H}_{\sigma,n-1}$. Therefore, by Theorem 1.1, the function
$g\prod_{j=1}^{k}(z-\lambda_{j})\in\mathcal{H}_{\lambda_{i},n}$ for each
$i=1,\ldots,k$. Hence $f\in\mathcal{H}_{\mu,n}.$
Conversely, suppose that $f\in\mathcal{H}_{\mu,n}$ and assume that, without
loss of generality, $\lambda_{1},\ldots,\lambda_{k}$ are distinct. Then
$f\in\mathcal{H}_{\lambda_{j},n},$ and applying Theorem 1.1 we see that
$f^{*}(\lambda_{j})$ exists for all $j=1,\ldots,k$. Let $p$ be the polynomial
of degree $k-1$ such that $p(\lambda_{j})=f^{*}(\lambda_{j})$, $j=1,\ldots,k$.
Again, by Theorem 1.1, it follows that $\frac{f-p}{z-\lambda_{j}}$ belongs to
$\mathcal{H}_{\sigma,n-1}$ for $j=1,\ldots,k.$ Since
$\lambda_{1},\ldots,\lambda_{k}$ are distinct, using partial fraction formula,
we see that

$\frac{f-p}{\prod_{j=1}^{k}(z-\lambda_{j})}=\sum_{j=1}^{k}\alpha_{j}\frac{f-p}{z-\lambda_{j}}$
(9)

for some $\alpha_{1},\ldots,\alpha_{k}\in\mathbb{C}.$ As the right hand side
in (9) belongs to $\mathcal{H}_{\sigma,n-1}$, this completes the proof. ∎

##  3\. Multipliers of $\mathcal{H}_{\mu,n}$

In the study of any function space, it is an important problem to characterize
the multipliers of the associated function space. Let $\mu$ be a non-zero
measure in $\mathcal{M}_{+}(\mathbb{T})$ and $n$ be a positive integer. The
set of multipliers of $\mathcal{H}_{\mu,n},$ is denoted by
$\text{Mult\,}(\mathcal{H}_{\mu,n})$ and is defined by

$\displaystyle\text{Mult\,}(\mathcal{H}_{\mu,n})=\\{\varphi\in\mathcal{O}(\mathbb{D}):\varphi
f\in\mathcal{H}_{\mu,n}\,\text{\,\,for
every\,}\,\,f\in\mathcal{H}_{\mu,n}\\}.$

In this section, we deduce many properties of a multiplier of
$\mathcal{H}_{\mu,n}$ and give some descriptions of the multiplier space
$\text{Mult\,}(\mathcal{H}_{\mu,n})$ in several different ways. Since the co-
ordinate function $z$ is a multiplier of $\mathcal{H}_{\mu,n}$ (see [16,
Proposition 2.6]), it follows that the set of all polynomials is contained in
$\text{Mult\,}(\mathcal{H}_{\mu,n}).$ Let $\mathcal{O}(\overline{\mathbb{D}})$
denote the set of functions which are holomorphic in some neighbourhood of the
closed unit disc $\overline{\mathbb{D}}$. As an application of the following
proposition we show that if $\varphi\in\mathcal{O}(\overline{\mathbb{D}})$
then $\varphi\in\text{Mult\,}(\mathcal{H}_{\mu,n}).$ Before proving the
proposition, we start with the following useful lemma.

######  Lemma 3.1.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})\setminus\\{0\\}$ and $n\in\mathbb{N}$.
Then for any $f\in\mathcal{H}_{\mu,n}$ and $j=0,\ldots,n,$ the integral

$\displaystyle
I_{\mu,j,n}(f):=\int_{\mathbb{D}}|f^{(j)}(z)|^{2}P_{\mu}(z)(1-|z|^{2})^{n-1}dA(z)$

is finite.

###### Proof.

Let $A^{2}(d\rho)$ be the weighted Bergman space with the weighted area
measure $\rho$ on the unit disc given by
$d\rho(z)=P_{\mu}(z)(1-|z|^{2})^{n-1}dA(z)$. Note that, a function $f$ in
$\mathcal{O}(\mathbb{D})$ satisfies $I_{\mu,j,n}(f)<\infty$ if and only if
$f^{(j)}\in A^{2}(d\rho)$. Let $f\in\mathcal{H}_{\mu,n}$. By definition it
follows that $f^{(n)}\in A^{2}(d\rho)$. Let $0\leqslant k\leqslant n-1$. The
proof will be established by showing that if $f^{(j)}\in A^{2}(d\rho)$ for all
$j=k+1,\ldots,n$, then $f^{(k)}\in A^{2}(d\rho)$. Note that, by the Leibniz’s
product rule for differentiation

$(z^{n-k}f)^{(n)}=z^{n-k}f^{(n)}+c_{0}z^{n-k-1}f^{(n-1)}+\cdots+c_{n-k}zf^{(k+1)}+{n\choose
n-k}(n-k)!f^{(k)}$ (10)

for some constants $c_{0},\ldots,c_{n-k}$. Since $z$ is a multiplier of
$\mathcal{H}_{\mu,n}$ (see [16, Proposition 2.6]), we see that
$(z^{n-k}f)^{(n)}\in A^{2}(d\rho)$. Since $f^{(j)}\in A^{2}(d\rho)$ for all
$j=k+1,\ldots,n$, it follows from (10) that $f^{(k)}\in A^{2}(d\rho).$

∎

######  Proposition 3.2.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $n\in\mathbb{N}$. Suppose
$\varphi\in\mathcal{O}(\mathbb{D})$ such that $\varphi^{(n)}\in H^{\infty}.$
Then $\varphi\in\text{Mult\,}(\mathcal{H}_{\mu,n}).$

###### Proof.

Let $\varphi\in\mathcal{O}(\mathbb{D})$ satisfying $\varphi^{(n)}\in
H^{\infty}.$ By repeated applications of the Mean Value Theorem, it follows
that $\varphi^{(j)}\in H^{\infty}$ for each $j=0,1,\ldots,n.$ By the Leibniz’s
product rule for differentiation and the Cauchy-Schwarz’s inequality, note
that, for any $f\in\mathcal{H}_{\mu,n},$

$\displaystyle|(\varphi
f)^{(n)}(z)|^{2}\leqslant\Big{(}\sum\limits_{j=0}^{n}|\varphi^{(n-j)}(z)|^{2}|f^{(j)}(z)|^{2}\Big{)}\Big{(}\sum\limits_{j=0}^{n}{\binom{n}{j}}^{2}\Big{)},\,\,\,z\in\mathbb{D}.$

Let $C_{n}=\sum_{j=0}^{n}{\binom{n}{j}}^{2}$ and
$\|g\|_{\infty}:=\sup\\{|g(z)|:z\in\mathbb{D}\\}$ for any $g\in H^{\infty}.$
It follows that for any $f\in\mathcal{H}_{\mu,n},$

$\displaystyle D_{\mu,n}(\varphi
f)\leqslant\frac{C_{n}}{n!(n-1)!}\sum\limits_{j=0}^{n}\|\varphi^{(n-j)}\|^{2}_{\infty}\int_{\mathbb{D}}|f^{(j)}(z)|^{2}P_{\mu}(z)(1-|z|^{2})^{n-1}dA(z).$

Applying Lemma 3.1, we obtain that $D_{\mu,n}(\varphi f)<\infty.$ Hence it
follows that $\varphi\in\text{Mult\,}(\mathcal{H}_{\mu,n}).$ ∎

######  Corollary 3.3.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and $n\in\mathbb{N}$. Then
$\mathcal{O}(\overline{\mathbb{D}})\subset\text{Mult\,}(\mathcal{H}_{\mu,n}).$

###### Proof.

Note that for any $\varphi\in\mathcal{O}(\overline{\mathbb{D}}),$ we have
$\varphi^{(j)}\in H^{\infty}$ for every $j\in\mathbb{N}.$ An application of
Proposition 3.2 now completes the proof. ∎

Let $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m}),$ where
$\mu_{j}\in\mathcal{M}_{+}(\mathbb{T})$ for every $j=1,\ldots,m.$ Recall that
the linear space
$\mathcal{H}_{\boldsymbol{\mu}}=\cap_{j=1}^{m}\mathcal{H}_{\mu_{j},j}$ is a
Hilbert space with respect to the norm $\|\cdot\|_{\boldsymbol{\mu}},$ given
by

$\displaystyle\|f\|_{\boldsymbol{\mu}}^{2}=\|f\|^{2}_{H^{2}}+\sum\limits_{j=1}^{m}D_{\mu_{j},j}(f),\quad
f\in\mathcal{H}_{\boldsymbol{\mu}}.$

Furthermore, the evaluation functional $ev_{z},$ defined by
$ev_{z}(f):=f(z),\,\,f\in\mathcal{H}_{\boldsymbol{\mu}},$ is bounded for every
$z\in\mathbb{D}.$ Thus $\mathcal{H}_{\boldsymbol{\mu}}$ is a reproducing
kernel Hilbert space of holomorphic functions on the unit disc, see [5], [24]
for definition and other basic properties of a reproducing kernel Hilbert
space. In particular, it follows that for any finite positive Borel measure
$\mu$ on $\mathbb{T}$ and a positive integer $n,$ the linear space
$\mathcal{H}_{\mu,n}$ is a reproducing kernel Hilbert space with respect to
the norm $\|\cdot\|_{\mu,n},$ given by

$\displaystyle\|f\|_{\mu,n}^{2}=\|f\|^{2}_{H^{2}}+D_{\mu,n}(f),\,\,\,\,f\in\mathcal{H}_{\mu,n}.$

Many properties of $\text{Mult\,}(\mathcal{H}_{\mu,n})$ can be derived using
the reproducing kernel Hilbert space structure of $\mathcal{H}_{\mu,n}.$ From
now onwards we will often use the notation
$(\mathcal{H}_{\mu,n},\|\cdot\|_{\mu,n})$ to denote the Hilbert space
$\mathcal{H}_{\mu,n}$ equipped with the norm $\|\cdot\|_{\mu,n}.$

In the following part of this section, we consider a general reproducing
kernel Hilbert space having some special properties and derive few properties
of the associated multiplier space.

Let $\mathscr{H}_{\kappa}$ be a reproducing kernel Hilbert space of
$\mathbb{C}$-valued holomorphic functions defined on the unit disc
$\mathbb{D}$ and let $\kappa:\mathbb{D}\times\mathbb{D}\rightarrow\mathbb{C}$
be the sesqui-analytic reproducing kernel for $\mathscr{H}_{\kappa},$ that is,
$\kappa(\cdot,w)\in\mathscr{H}_{\kappa}$ and

$\displaystyle\langle
f,\kappa(\cdot,w)\rangle_{\mathscr{H}_{\kappa}}=f(w),\quad
f\in\mathscr{H}_{\kappa},~{}w\in\mathbb{D}.$

In the remaining part of this section, we shall assume the following
properties on $\kappa$:

  1. (A1)

$\kappa$ is normalized at the origin, that is, $\kappa(w,0)=1$ for every
$w\in\mathbb{D},$

  2. (A2)

$z$ is a multiplier for $\mathscr{H}_{\kappa},$ that is,
$zf\in\mathscr{H}_{\kappa}$ for every $f\in\mathscr{H}_{\kappa},$

  3. (A3)

$\mathbb{C}[z],$ the set of all polynomials, is dense in
$\mathscr{H}_{\kappa}.$

We refer to the reproducing kernel Hilbert space $\mathscr{H}_{\kappa}$
satisfying (A1)-(A3) as the functional Hilbert space with property (A).

For example take $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m}),$ where
$\mu_{j}\in\mathcal{M}_{+}(\mathbb{T})$ for every $j=1,\ldots,m$ and consider
the associated reproducing kernel Hilbert space
$\mathcal{H}_{\boldsymbol{\mu}}.$ Since $\langle f,1\rangle=f(0)$ for every
$f\in\mathcal{H}_{\boldsymbol{\mu}},$ the kernel function associated to
$(\mathcal{H}_{\boldsymbol{\mu}},\|\cdot\|_{\boldsymbol{\mu}})$ is normalized
at the origin. This together with [16, Theorem 4.1, Proposition 4.2] gives us
that $(\mathcal{H}_{\boldsymbol{\mu}},\|\cdot\|_{\boldsymbol{\mu}})$ is a
functional Hilbert space with property (A). In particular, it follows that,
for any finite positive Borel measure $\mu$ on $\mathbb{T}$ and for any
$n\in\mathbb{N},$ the Hilbert space $(\mathcal{H}_{\mu,n},\|\cdot\|_{\mu,n})$
is also a functional Hilbert space with property (A).

For a reproducing kernel Hilbert space $\mathscr{H}_{\kappa},$ which consists
of holomorphic functions on the unit disc $\mathbb{D},$ the multipliers space
of $\mathscr{H}_{\kappa},$ is defined by

$\displaystyle\text{Mult\,}(\mathscr{H}_{\kappa}):=\\{\varphi\in\mathscr{H}_{\kappa}:\varphi
f\in\mathscr{H}_{\kappa}\,\text{\,\,for
every\,}\,\,f\in\mathscr{H}_{\kappa}\\}.$

Note that $\text{Mult\,}(\mathscr{H}_{\kappa})$ forms an algebra with respect
to the pointwise product, that is,
$\varphi\psi\in\text{Mult\,}(\mathscr{H}_{\kappa})$ for every
$\varphi,\psi\in\text{Mult\,}(\mathscr{H}_{\kappa}).$ Moreover one can
identify $\text{Mult\,}(\mathscr{H}_{\kappa})$ as a closed subalgebra of
$\mathscr{B}(\mathscr{H}_{\kappa})$ in the following manner. Given
$\varphi\in\text{Mult\,}(\mathscr{H}_{\kappa}),$ we write $M_{\varphi}$ for
the multiplication operator acting on $\mathscr{H}_{\kappa}$ given by

$\displaystyle M_{\varphi}(f)=\varphi f,\,\,f\in\mathscr{H}_{\kappa}.$

By an application of the Closed Graph Theorem, it follows that for any
$\varphi\in\text{Mult\,}(\mathscr{H}_{\kappa}),$ the operator $M_{\varphi}$ is
bounded from $\mathscr{H}_{\kappa}$ into itself. Consider the norm
$\|\cdot\|_{op}$ on $\text{Mult\,}(\mathscr{H}_{\kappa})$ defined by

$\displaystyle\|\varphi\|_{op}:=\|M_{\varphi}\|=\sup\\{\|\varphi
f\|_{\mathscr{H}_{\kappa}}:\|f\|_{\mathscr{H}_{\kappa}}=1\\},\,\,\,\varphi\in\text{Mult\,}(\mathscr{H}_{\kappa}).$

It follows that
$\big{(}\text{Mult\,}(\mathscr{H}_{\kappa}),\|\cdot\|_{op}\big{)}$ is a
commutative Banach algebra with identity. Let $\mathscr{H}_{\kappa}$ be a
functional Hilbert space with property (A). It follows that
$\text{Mult\,}(\mathscr{H}_{\kappa})$ can be identified as the commutant of
the operator $M_{z}$ of multiplication by the co-ordinate function on the
Hilbert space $\mathscr{H}_{\kappa}.$ Indeed if
$T\in\mathscr{B}(\mathscr{H}_{\kappa})$ satisfying $TM_{z}=M_{z}T,$ then a
routine argument using the polynomial density of the functional Hilbert space
$\mathscr{H}_{\kappa},$ it can be shown that $T$ is given by $T=M_{\varphi}$
for some $\varphi\in\text{Mult\,}(\mathscr{H}_{\kappa})$ where $\varphi=T(1)$.

Let $\mathscr{H}_{\kappa}$ be a functional Hilbert space with property (A). By
assumption (A1), we have
$\langle\kappa(\cdot,0),\kappa(\cdot,w)\rangle=\kappa(w,0)=1$ for each
$w\in\mathbb{D}.$ Therefore $\kappa(\cdot,w)\neq 0$ for each $w\in\mathbb{D}.$
Hence it follows that the multiplier algebra
$\text{Mult\,}(\mathscr{H}_{\kappa})$ is a subalgebra of $H^{\infty},$ the
algebra of bounded analytic functions on the unit disc, for every functional
Hilbert space $\mathscr{H}_{\kappa}$ with property (A), see [23, Problem 68],
[24, Corollary 5.22], [30, Proposition 3]. Thus, we obtain that
$\text{Mult\,}(\mathcal{H}_{\mu,n})\subseteq H^{\infty}$ for every measure
$\mu$ in $\mathcal{M}_{+}(\mathbb{T})$ and $n\in\mathbb{N}$.

Among the measures in $\mathcal{M}_{+}(\mathbb{T}),$ there are two special
measures. First one is the Lebesgue measure $\sigma$ and the second one is the
Dirac delta measure $\delta_{\lambda}$ for $\lambda\in\mathbb{T}.$ The space
$\mathcal{H}_{\sigma,n}$ coincides with the weighted Dirichlet-type space
$\mathcal{D}_{n}$ for each $n\in\mathbb{Z}_{\geqslant 0}.$ The set of
multipliers of $\mathcal{D}_{n}$ is well studied, see for example [32, 31] and
references therein. It is known that, for every $n\geqslant 2,$ the space
$\mathcal{D}_{n}$ forms an algebra, that is, $fg\in\mathcal{D}_{n},$ for every
$f,g\in\mathcal{D}_{n},$ see [32, Theorem 7]. Consequently it follows that
$\text{Mult\,}(\mathcal{H}_{\sigma,n})=\mathcal{H}_{\sigma,n}$ for every
$n\geqslant 2.$ Now we concentrate our study to characterize the multiplier of
$\mathcal{H}_{\lambda,n}$ for any positive integer $n.$ The following lemma
will be useful in describing the set $\text{Mult\,}(\mathcal{H}_{\lambda,n}).$
This lemma can be thought as a generalization of [19, Lemma 5.3].

######  Lemma 3.4.

Let $n$ be a positive integer, $\lambda\in\mathbb{T}$ and
$f\in\mathcal{H}_{\lambda,n}$ such that $D_{\lambda,n}(f)\neq 0.$ Suppose
$\varphi\in\text{Mult\,}(\mathcal{H}_{\sigma,n-1})\cap\mathcal{H}_{\lambda,n},$
then

$\displaystyle D_{\lambda,n}(\varphi f)$ $\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f)+2|f^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi)\,\,\,\,\text{and}$
$\displaystyle|f^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi)$
$\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f)+2D_{\lambda,n}(\varphi f),$

where

$\|\varphi\|_{s(\sigma,n-1)}:=\sup\Big{\\{}\sqrt{D_{\sigma,n-1}(\varphi
h)}:D_{\sigma,n-1}(h)=1\Big{\\}}.$

###### Proof.

For a function $g\in H^{2},$ if the non-tangential limit $g^{*}(\lambda)$ of
$g$ at $\lambda$ exists, then by the generalized local Douglas formula (see
Theorem 1.1), we have

$\displaystyle
D_{\lambda,n}(g)=D_{\sigma,n-1}\Big{(}\frac{g(z)-g(\lambda)}{z-\lambda}\Big{)}.$

Since $f,\varphi\in\mathcal{H}_{\lambda,n},$ the non-tangential limits
$f^{*}(\lambda)$ and $\varphi^{*}(\lambda),$ of $f$ and $\varphi$ at $\lambda$
respectively, exist, see Theorem 1.1. It follows that

$\displaystyle D_{\lambda,n}(\varphi f)=D_{\sigma,n-1}\Big{(}\frac{\varphi
f(z)-(\varphi f)^{*}(\lambda)}{z-\lambda}\Big{)}.$

Using the identity

$\displaystyle\frac{\varphi f(z)-\varphi
f(\lambda)}{z-\lambda}=\varphi(z)\frac{f(z)-f^{*}(\lambda)}{z-\lambda}+f^{*}(\lambda)\frac{\varphi(z)-\varphi(\lambda)}{z-\lambda},\,\,\,\,z\in\mathbb{D},$
(11)

it follows that

$\displaystyle D_{\lambda,n}(\varphi f)$ $\displaystyle\leqslant
2D_{\sigma,n-1}\Big{(}\varphi\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)}+2|f^{*}(\lambda)|^{2}D_{\sigma,n-1}\Big{(}\frac{\varphi(z)-\varphi(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\sigma,n-1}\Big{(}\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)}+2|f^{*}(\lambda)|^{2}D_{\sigma,n-1}\Big{(}\frac{\varphi(z)-\varphi^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f)+2|f^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi).$

Using the identity (11) one more time, we obtain that

$\displaystyle|f^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi)$
$\displaystyle=D_{\sigma,n-1}\Big{(}f^{*}(\lambda)\frac{\varphi(z)-\varphi(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle=D_{\sigma,n-1}\Big{(}\frac{\varphi f(z)-\varphi
f(\lambda)}{z-\lambda}-\varphi(z)\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle\leqslant 2D_{\sigma,n-1}\Big{(}\frac{\varphi f(z)-(\varphi
f)^{*}(\lambda)}{z-\lambda}\Big{)}+2D_{\sigma,n-1}\Big{(}\varphi(z)\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle\leqslant 2D_{\lambda,n}(\varphi
f)+2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\sigma,n-1}\Big{(}\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle=2D_{\lambda,n}(\varphi
f)+2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f).$

This completes the proof. ∎

The following proposition describes the multiplier algebra of
$\mathcal{H}_{\lambda,n}$ for any $n\in\mathbb{N}$ and $\lambda\in\mathbb{T}.$

######  Theorem 3.5.

Let $n\in\mathbb{N}$ and $\lambda\in\mathbb{T}.$ Then,
$\text{Mult\,}(\mathcal{H}_{\lambda,n})=\mathcal{H}_{\lambda,n}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1}).$

###### Proof.

By Lemma 3.4, it follows that
$\varphi\in\text{Mult\,}(\mathcal{H}_{\lambda,n})$ whenever
$\varphi\in\mathcal{H}_{\lambda,n}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1}).$
For the converse, let $\varphi\in\text{Mult\,}(\mathcal{H}_{\lambda,n}).$
Since the constant function $1\in\mathcal{H}_{\lambda,n},$ it follows that
$\varphi\in\mathcal{H}_{\lambda,n}.$ Consider the map
$L_{\lambda}:\mathcal{H}_{\lambda,n}\rightarrow\mathcal{H}_{\sigma,n-1}$
defined by

$\displaystyle
L_{\lambda}(f)(z):=\frac{f(z)-f^{*}(\lambda)}{z-\lambda},\,\,\,z\in\mathbb{D}.$

By the generalized local Douglas formula in Theorem 1.1, we have $L_{\lambda}$
is an isometry w.r.t the associated semi-norms, that is,
$D_{\sigma,n-1}(L_{\lambda}f)=D_{\lambda,n}(f)$ for every
$f\in\mathcal{H}_{\lambda,n}.$ Furthermore it also follows that $L_{\lambda}$
is onto. Let $f\in\mathcal{H}_{\lambda,n}.$ An application of the identity
(11) gives us

$\displaystyle
D_{\sigma,n-1}\Big{(}\varphi(z)\frac{f(z)-f^{*}(\lambda)}{z-\lambda}\Big{)}$
$\displaystyle=D_{\sigma,n-1}\Big{(}\frac{\varphi f(z)-(\varphi
f)^{*}(\lambda)}{z-\lambda}-f^{*}(\lambda)\frac{\varphi(z)-\varphi^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle\leqslant 2D_{\sigma,n-1}\Big{(}\frac{\varphi f(z)-(\varphi
f)^{*}(\lambda)}{z-\lambda}\Big{)}+2|f^{*}(\lambda)|^{2}D_{\sigma,n-1}\Big{(}\frac{\varphi(z)-\varphi^{*}(\lambda)}{z-\lambda}\Big{)},$
$\displaystyle=2D_{\lambda,n}(\varphi
f)+2|f^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi),$

Thus we have $D_{\sigma,n-1}(\varphi L_{\lambda}f)<\infty$ for every
$f\in\mathcal{H}_{\lambda,n}.$ Since $L_{\lambda}$ is onto, we obtain that
$D_{\sigma,n-1}(\varphi g)<\infty$ for every $g\in\mathcal{H}_{\sigma,n-1}.$
Hence $\varphi\in\text{Mult\,}(\mathcal{H}_{\sigma,n-1}).$ This completes the
proof. ∎

Proof of Theorem 1.2 : Let
$\varphi,\psi\in\mathcal{H}_{\mu,n}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1})$.
Since $\text{Mult\,}(\mathcal{H}_{\sigma,n-1})$ is an algebra, it is
sufficient to show that $\varphi\psi\in\mathcal{H}_{\mu,n}.$ Note that
$D_{\mu,n}(g)=\int_{\mathbb{T}}D_{\lambda,n}(g)d\mu(\lambda),$ for any
$g\in\mathcal{H}_{\mu,n}.$ Hence we obtain that
$\varphi,\psi\in\mathcal{H}_{\lambda,n}$ for $[\mu]$ a.e. $\lambda$ in
$\mathbb{T}.$ By Lemma 3.4 we have

$\displaystyle D_{\lambda,n}(\varphi\psi)$ $\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(\psi)+2|\psi^{*}(\lambda)|^{2}D_{\lambda,n}(\varphi),$
$\displaystyle\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(\psi)+2\|\psi\|_{\infty}^{2}D_{\lambda,n}(\varphi),~{}\lambda\in\mathbb{T}/A.$

Here in the last line we used the fact that
$\psi\in\text{Mult\,}(\mathcal{H}_{\sigma,n-1})\subseteq H^{\infty}$.
Integrating both sides w.r.t $\mu(\lambda)$ we obtain

$D_{\mu,n}(\varphi\psi)\leqslant
2\|\varphi\|_{s(\sigma,n-1)}^{2}D_{\mu,n}(\psi)+2\|\psi\|_{\infty}^{2}D_{\mu,n}(\varphi),~{}\lambda\in\mathbb{T}.$

This completes the proof of the theorem.

The above theorem helps us to characterize the multipliers of
$\mathcal{H}_{\mu,n}$ for any $\mu\in\mathcal{M}_{+}(\mathbb{T})$ and for any
positive integer $n\geqslant 3.$ In this case, it turns out that
$\mathcal{H}_{\mu,n}$ forms an algebra and consequently the multipliers of
$\mathcal{H}_{\mu,n}$ is itself.

######  Corollary 3.6.

Let $n\geqslant 3$ be a positive integer and
$\mu\in\mathcal{M}_{+}(\mathbb{T})$. Then $\mathcal{H}_{\mu,n}$ is an algebra
and consequently $\text{Mult\,}(\mathcal{H}_{\mu,n})=\mathcal{H}_{\mu,n}.$

###### Proof.

We have $\text{Mult\,}(\mathcal{H}_{\sigma,n-1})=\mathcal{H}_{\sigma,n-1}$ for
any $n\geqslant 3.$ Thus applying Theorem 3.6, it follows that
$\mathcal{H}_{\mu,n}\cap\mathcal{H}_{\sigma,n-1}$ is an algebra. By [16, Lemma
2.4], we also have that
$\mathcal{H}_{\mu,n}\subseteq\mathcal{H}_{\sigma,n-1}.$ Hence
$\mathcal{H}_{\mu,n}\cap\mathcal{H}_{\sigma,n-1}=\mathcal{H}_{\mu,n}.$ Thus
$\mathcal{H}_{\mu,n}$ is an algebra and consequently
$\text{Mult\,}(\mathcal{H}_{\mu,n})=\mathcal{H}_{\mu,n}.$ ∎

######  Corollary 3.7.

Let $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ be an $m$-tuple of finite
positive Borel measures on $\mathbb{T}$ with $\mu_{m}\neq 0.$ Then
$\mathcal{H}_{\boldsymbol{\mu}}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1})$ is
an algebra. In particular, $\mathcal{H}_{\boldsymbol{\mu}}$ is an algebra for
$m\geqslant 3.$

###### Proof.

It is well known that for every $n\in\mathbb{N},$
$\text{Mult\,}(\mathcal{H}_{\sigma,n})\subset\text{Mult\,}(\mathcal{H}_{\sigma,n-1}),$
see [32, p. 233, Corollary 1]. So, we have

$\displaystyle\mathcal{H}_{\boldsymbol{\mu}}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1})$
$\displaystyle=$
$\displaystyle\mathcal{H}_{\mu_{1},1}\cap\cdots\cap\mathcal{H}_{\mu_{m},m}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1}),$
$\displaystyle=$
$\displaystyle\big{(}\mathcal{H}_{\mu_{1},1}\cap\text{Mult\,}(\mathcal{H}_{\sigma,0})\big{)}\cap\cdots\cap\big{(}\mathcal{H}_{\mu_{m},m}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1})\big{)}.$

By an application of Theorem 3.6,
$\mathcal{H}_{\mu_{j},j}\cap\text{Mult\,}(\mathcal{H}_{\sigma,j-1})$ is an
algebra for each $j=1,\ldots,m.$ Hence
$\mathcal{H}_{\boldsymbol{\mu}}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1})$ is
an algebra. It also follows from Theorem 3.6 that the space
$\mathcal{H}_{\mu,j}\cap\text{Mult\,}(\mathcal{H}_{\sigma,j-1})=\mathcal{H}_{\mu,j}$
for any $j\geqslant 3.$ Hence
$\mathcal{H}_{\boldsymbol{\mu}}\cap\text{Mult\,}(\mathcal{H}_{\sigma,m-1})=\mathcal{H}_{\boldsymbol{\mu}}$
for any $m\geqslant 3.$ This completes the proof. ∎

Although $\mathcal{H}_{\mu,n}$ is an algebra for every $n\geqslant 3$ and for
any $\mu\in\mathcal{M}_{+}(\mathbb{T})$, it is well known that
$\mathcal{H}_{\mu,1}$ is never an algebra, see [14, Exercise 7.1.4]. So it is
natural to ask the following question.

######  Question 3.8.

Characterize all $\mu\in\mathcal{M}_{+}(\mathbb{T})$ such that
$\mathcal{H}_{\mu,2}$ is an algebra?

######  Example 3.9.

We now point out few cases when $\mathcal{H}_{\mu,2}$ is an algebra and when
it is not.

  * (i)

The space $\mathcal{H}_{\lambda,2}$ is not an algebra for any
$\lambda\in\mathbb{T}$. If possible assume that $\mathcal{H}_{\lambda,2}$ is
an algebra for some $\lambda\in\mathbb{T}$. Then we must have
$\mathcal{H}_{\lambda,2}=\text{Mult\,}(\mathcal{H}_{\lambda,2})\subset
H^{\infty}$. We now show that there exists a function in
$\mathcal{H}_{\lambda,2}$ which is not bounded on $\mathbb{D}$. Choose a
$\tilde{\lambda}\neq\lambda\in\mathbb{T}$ and a function
$g\in\mathcal{H}_{\sigma,1},$ the classical Dirichlet space, such that
$\lim_{r\to 1^{-}}g(r\tilde{\lambda})=\infty$. Then by Theorem 1.1,
$(z-\lambda)g\in\mathcal{H}_{\lambda,2}$. But $\lim_{r\to
1^{-}}(r\tilde{\lambda}-\lambda)g(r\tilde{\lambda})=\infty$. Hence
$(z-\lambda)g\notin H^{\infty}.$

  * (ii)

Note that $\mathcal{H}_{\sigma,2}=\mathcal{D}_{2}$ and $\mathcal{D}_{2}$ forms
an algebra, see [32, Theorem 7]. In fact, if
$\boldsymbol{\mu}=(\mu_{1},\mu_{2}+\sigma)$, where
$\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T})$, then
$\mathcal{H}_{\boldsymbol{\mu}}$ is an algebra. To see this, note that
$\mathcal{H}_{\boldsymbol{\mu}}=\mathcal{H}_{\mu_{1}}\cap\mathcal{H}_{\sigma,2}\cap\mathcal{H}_{\mu,2}.$
Since $\mathcal{H}_{\sigma,2}\subset\text{Mult\,}(\mathcal{H}_{\sigma,1})$, we
obtain that
$\mathcal{H}_{\boldsymbol{\mu}}\subset\text{Mult\,}(\mathcal{H}_{\sigma,1}).$
The conclusion now follows from Theorem 1.2.

The following theorem is a generalization of [14, Theorem 8.3.2]. As a
corollary to this theorem we obtain a stronger version of Theorem 1.2 of this
section.

######  Theorem 3.10.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})\setminus\\{0\\},$ $n\in\mathbb{N}.$ Let
$\mathcal{W}$ be a closed $M_{z}$-invariant subspace of $\mathcal{H}_{\mu,n}$
and $\phi\in\text{Mult\,}(\mathcal{H}_{\sigma,n-1}).$ Suppose
$f\in\mathcal{W}$ be such that $\phi f\in\mathcal{H}_{\mu,n},$ then $\phi
f\in\mathcal{W}.$

###### Proof.

Let $\phi$ be a multiplier of $\mathcal{H}_{\sigma,n-1}.$ For every $0<r<1,$
define $\phi_{r}(z):=\phi(rz).$ Note that for every $0<r<1,$
$\phi_{r}\in\mathcal{O}(\overline{\mathbb{D}})$ and hence
$\phi_{r}\in\text{Mult\,}(\mathcal{H}_{\mu,n}).$ Fix a $r$ satisfying $0<r<1.$
Choose a sequence of polynomials $\\{p_{j}\\}_{j\in\mathbb{N}}$ such that
$D_{\mu,n}(p_{j}-\phi_{r})\to 0$ as $j\to\infty.$ Let $f\in\mathcal{W}.$ Since
$\mathcal{W}$ is a $M_{z}$-invariant subspace of $\mathcal{H}_{\mu,n},$ we
have that $p_{j}f\in\mathcal{W}$ for all $j\in\mathbb{N}.$ As
$p_{n}f\to\phi_{r}f$ in $\mathcal{H}_{\mu,n},$ we have that $\phi_{r}f\in M.$
Since $r$ is an arbitrary number between $0$ and $1,$ we conclude that
$\phi_{r}f\in M$ for every $0<r<1.$ To show $\phi f\in M,$ it is enough to
show that

$\sup\\{D_{\mu,n}(\phi_{r}f):0<r<1\\}<\infty.$

Since if this is the case then $\\{\phi_{r}f\\}$ turns out to be a norm
bounded family of function in the Hilbert space $\mathcal{H}_{\mu,n}.$ And
therefore there exists a sequence $r_{n}$ converging to $1$ such that
$\phi_{r_{n}}f$ converges to $\phi f$ weakly and hence $\phi f\in M.$ Using
Lemma 3.4, we find that

$\displaystyle D_{\lambda,n}(\phi_{r}f)$ $\displaystyle\leqslant$
$\displaystyle
2\|\phi_{r}\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f)+2|f(\lambda)|^{2}D_{\lambda,n}(\phi_{r})$
$\displaystyle\leqslant$ $\displaystyle
6\|\phi\|_{s(\sigma,n-1)}^{2}D_{\lambda,n}(f)+4D_{\lambda,n}(\phi f)$

Integrating with respect to $\mu$ on both the sides of the above inequality,
we obtain that

$\sup\,\\{D_{\mu,n}(\phi_{r}f):0<r<1\\}<\infty.$

This completes the proof. ∎

######  Corollary 3.11.

Let $\mu\in\mathcal{M}_{+}(\mathbb{T})\setminus\\{0\\}$ and $n\in\mathbb{N}$.
If $\mathcal{W}$ is a closed $M_{z}$-invariant subspace of
$\mathcal{H}_{\mu,n}$ then
$\mathcal{W}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1})$ is an algebra.

###### Proof.

Let $\varphi,f\in\mathcal{W}\cap\text{Mult\,}(\mathcal{H}_{\sigma,n-1}).$ From
Theorem 1.2, we know that $\varphi f\in\mathcal{H}_{\mu,n}.$ Now by Theorem
3.10, we deduce that $\varphi f\in\mathcal{W}.$ This completes the proof. ∎

We have already seen that $\mathcal{H}_{\mu,n}$ is an algebra for every
$n\geqslant 3$ and for any $\mu\in\mathcal{M}_{+}(\mathbb{T}).$ There are many
interesting consequences of a reproducing kernel Hilbert space
$\mathscr{H}_{\kappa}$ being an algebra under pointwise product, see for
example [30, Proposition 31, pp. 94–95 ]. In this part of the section, we note
down some of these consequences of $\mathscr{H}_{\kappa}$ being an algebra.
Let $\mathscr{C}(\overline{\mathbb{D}})$ denote the algebra of continuous
functions on the closed unit disc $\overline{\mathbb{D}}.$ In the following
proposition, we first derive that under certain conditions, an abelian Banach
algebra contained in $\mathcal{O}(\mathbb{D})$ must lie in the disc algebra
$\mathcal{O}(\mathbb{D})\cap\mathscr{C}(\overline{\mathbb{D}}).$

######  Proposition 3.12.

Let $\mathcal{A}\subseteq\mathcal{O}(\mathbb{D})$ be an abelian Banach algebra
under pointwise product and the set of all polynomials is dense in
$\mathcal{A}.$ Suppose $\sigma_{\mathcal{A}}(z),$ the spectrum of the co-
ordinate function $z$ in $\mathcal{A},$ is equal to the closed unit disc
$\overline{\mathbb{D}}.$ Then it follows that
$\mathcal{A}\subseteq\mathscr{C}(\overline{\mathbb{D}})$ and for every
$\lambda\in\overline{\mathbb{D}},$ the evaluation map
$ev_{\lambda}:\mathcal{A}\to\mathbb{C}$ defined by
$ev_{\lambda}(f)=f(\lambda),\,f\in\mathcal{A},$ is continuous. Furthermore
$\sigma_{\mathcal{A}}(f)=f(\overline{\mathbb{D}})$ holds for every
$f\in\mathcal{A}.$

###### Proof.

Let $\Sigma$ be the maximal ideal space of $\mathcal{A},$ that is, the
collection of all non-zero homomorphisms of $\mathcal{A}$ into $\mathbb{C}.$
Consider $\Sigma$ with the weak* topology that it inherits as a subset of
$\mathcal{A}^{*}.$ It is well known that $\Sigma$ is a compact Hausdorff
space, see [11, Ch. VII, Theorem 8.6]. Since
$\mathcal{A}\subseteq\mathcal{O}(\mathbb{D}),$ it follows that $ev_{\lambda}$
belong to $\Sigma$ for every $\lambda\in\mathbb{D}.$ By our assumption the co-
ordinate function $z$ is a generator for the algebra $\mathcal{A}.$ By [11,
Ch. VII, Proposition 8.10], we have that there is a homeomorphism
$\tau:\Sigma\to\overline{\mathbb{D}}$ such that $\tau(ev_{\lambda})=\lambda.$
Furthermore, it follows that the Gelfand transform
$\gamma:\mathcal{A}\to\mathscr{C}(\overline{\mathbb{D}}),$ satisfies the
relation $\gamma(p)=p$ for every polynomial $p$ in $\mathbb{C}[z].$ Now let
$f\in\mathcal{A}.$ For each $\lambda\in\mathbb{D},$ we have that

$\displaystyle
f(\lambda)=ev_{\lambda}(f)=\big{(}\tau^{-1}(\lambda)\big{)}(f).$

So for each $\lambda\in\mathbb{T},$ it is natural to define
$f(\lambda):=\big{(}\tau^{-1}(\lambda)\big{)}(f).$ Since $\tau$ is a
homeomorphism, it follows that $f$ extends as a function in
$\mathscr{C}(\overline{\mathbb{D}}).$ Note that
$ev_{\lambda}=\tau^{-1}(\lambda)\in\Sigma$ for every
$\lambda\in\overline{\mathbb{D}}.$ Hence the continuity of $ev_{\lambda},$ for
each $\lambda\in\overline{\mathbb{D}},$ follows from [11, Ch. VII, Proposition
8.4]. Finally, in view of [11, Ch. VII, Theorem 8.6], it follows that

$\displaystyle\sigma_{\mathcal{A}}(f)=\\{h(f):h\in\Sigma\\}=\\{\tau^{-1}(\lambda)(f):\lambda\in\overline{\mathbb{D}}\\}=\\{f(\lambda):\lambda\in\overline{\mathbb{D}}\\}.$

This completes the proof. ∎

As an application of the aforementioned proposition, we obtain that every
functional Hilbert space $\mathscr{H}_{\kappa}$ with property (A), which is
also an algebra under pointwise product, must lie in the disc algebra.
Furthermore, for such instances, one can describe all the cyclic vectors in
$\mathscr{H}_{\kappa}.$ Recall that a vector $f$ in $\mathscr{H}_{\kappa}$ is
said to be cyclic if $\mathscr{H}_{\kappa}$ is the smallest closed
$M_{z}$-invariant subspace containing $f.$ The following results can be
thought as a straightforward generalization of the results in [30, Corollary
2, Corollary 3, p. 95].

######  Theorem 3.13.

Let $\mathscr{H}_{\kappa}$ be a functional Hilbert space with property (A).
Suppose $\mathscr{H}_{\kappa}$ is an algebra under pointwise product and the
spectrum $\sigma(M_{z})=\overline{\mathbb{D}}.$ Then
$\mathscr{H}_{\kappa}\subseteq\mathscr{C}(\overline{\mathbb{D}})$ and every
closed $M_{z}$-invariant subspace of $\mathscr{H}_{\kappa}$ is an ideal.
Furthermore, for a vector $f\in\mathscr{H}_{\kappa},$ the following are
equivalent:

  * (a)

$f$ is cyclic.

  * (b)

$f$ is invertible in $\text{Mult\,}(\mathscr{H}_{\kappa}).$

  * (c)

$f$ has no zero in $\overline{\mathbb{D}}.$

###### Proof.

Note that $\mathscr{H}_{\kappa}$ is an algebra under pointwise product if and
only if $\mathscr{H}_{\kappa}=\text{Mult\,}(\mathscr{H}_{\kappa}).$ Since
$\|1\|_{\mathscr{H}_{\kappa}}=\kappa(0,0)=1,$ it follows that
$\|f\|_{\mathscr{H}_{\kappa}}\leqslant\|M_{f}\|=\|f\|_{op}$ for every
$f\in\mathscr{H}_{\kappa}.$ This shows that the inclusion map from
$\big{(}\text{Mult\,}(\mathscr{H}_{\kappa}),\|\cdot\|_{op}\|\big{)}$ onto
$\big{(}\mathscr{H}_{\kappa},\|\cdot\|_{\mathscr{H}_{\kappa}}\big{)}$ is
continuous. By an application of open mapping theorem it follows that the two
norms $\|\cdot\|_{op}$ and $\|\cdot\|_{\mathscr{H}_{\kappa}}$ on
$\mathscr{H}_{\kappa}$ are equivalent. By assumption (A3), it now follows that
the set of all polynomials is dense in
$\big{(}\text{Mult\,}(\mathscr{H}_{\kappa}),\|\cdot\|_{op}\|\big{)}$ as well.
Since $\text{Mult\,}(\mathscr{H}_{\kappa})$ coincides with the commutant of
$M_{z}$ in $\mathcal{B}(\mathcal{H}_{\kappa}),$ it follows that the spectrum
of the co-ordinate function $z$ in the multiplier algebra
$\text{Mult\,}(\mathscr{H}_{\kappa})$ is equal to the spectrum of $M_{z}$ in
$\mathscr{B}(\mathscr{H}_{\kappa}).$ Now applying Proposition 3.12 to the
abelian Banach algebra
$\mathcal{A}=\big{(}\text{Mult\,}(\mathscr{H}_{\kappa}),\|\cdot\|_{op}\big{)},$
we obtain that
$\mathscr{H}_{\kappa}=\text{Mult\,}(\mathscr{H}_{\kappa})\subseteq\mathscr{C}(\overline{\mathbb{D}}).$
For the proof of the second part, consider an element
$f\in\mathscr{H}_{\kappa}.$ Let $\mathscr{I}(f)$ denote the smallest closed
$M_{z}$-invariant subspace of $\mathscr{H}_{\kappa},$ containing $f,$ that is,

$\displaystyle\mathscr{I}(f)=\bigvee\\{z^{n}f:n\in\mathbb{Z}_{\geqslant
0}\\}.$

To show every closed $M_{z}$-invariant subspace is an ideal, it is sufficient
to show that $\mathscr{I}(f)$ is an ideal. As the two norms $\|\cdot\|_{op}$
and $\|\cdot\|_{\mathscr{H}_{\kappa}}$ on $\mathscr{H}_{\kappa}$ are
equivalent and the set of all polynomials is dense in $\mathscr{H}_{\kappa},$
it follows that $\mathscr{I}(f)$ is an ideal. Since $\mathscr{I}(f)$ is a
closed ideal in the Banach algebra
$\mathcal{A}=\big{(}\text{Mult\,}(\mathscr{H}_{\kappa}),\|\cdot\|_{op}\|\big{)}$,
then either one of the following two mutually exclusive event must holds.

  * (i)

$\mathscr{I}(f)=\mathscr{H}_{\kappa},$

  * (ii)

$\mathscr{I}(f)$ is contained in a maximal ideal, that is,
$\mathscr{I}(f)\subseteq\ker(ev_{\lambda})$ for some
$\lambda\in\overline{\mathbb{D}}.$ In particular $f(\lambda)=0$ for some
$\lambda\in\overline{\mathbb{D}}.$

Hence we obtain that an element $f\in\mathscr{H}_{\kappa},$ is cyclic , that
is, $\mathscr{I}(f)=\mathscr{H}_{\kappa}$ if and only if $f$ has no zeros in
$\overline{\mathbb{D}}.$ Applying Proposition 3.12, we also have that
$\sigma_{\mathcal{A}}(f)=\\{f(\lambda):\lambda\in\overline{\mathbb{D}}\\}.$
Thus $f$ has no zeros in $\overline{\mathbb{D}}$ if and only if $f$ is
invertible in $\text{Mult\,}(\mathscr{H}_{\kappa}).$ This completes the proof.
∎

##  4\. Codimension $k$ property of Invariant subspaces

This section is an attempt to study the codimension $k$ property of non-zero
closed $M_{z}$-invariant subspaces of $\mathcal{H}_{\boldsymbol{\mu}}$ for any
$\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ with
$\mu_{j}\in\mathcal{M}_{+}(\mathbb{T})$ for $j=1,\ldots,m.$ Since
$M_{z}|_{{}_{\mathcal{M}}}$ is an analytic, $(m+1)$-isometry for every non-
zero $\mathcal{M}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ it turns
out that the dimension of
$\ker\big{(}(M_{z}|_{{}_{\mathcal{M}}})^{*}-\bar{\lambda}\big{)},$ that is,
the dimension of $\mathcal{M}\ominus(z-\lambda)\mathcal{M},$ is a positive
integer independent of $\lambda\in\mathbb{D},$ see [16, Proposition 6.4]. Thus
any $\mathcal{M}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ has
codimension $k$ property if and only if $dim(\mathcal{M}\ominus
z\mathcal{M})=k.$ In case of $m=1,$ that is, $\boldsymbol{\mu}=\mu_{1}$ and
$\mu_{1}\in\mathcal{M}_{+}(\mathbb{T}),$ it is well known that for any non-
zero $\mathcal{M}$ in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$
$\mathcal{M}$ has codimension $1$ property, see [20, Theorem 3.2]. Here,
firstly we show that for any $m\geqslant 3$ and any non-zero $\mathcal{M}$ in
$Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}})$ has codimension $1$ property.

######  Theorem 4.1.

Let $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ be an $m$-tuple of finite non-
negative Borel measure on $\mathbb{T}.$ Suppose
$\mathcal{H}_{\boldsymbol{\mu}}$ is an algebra under pointwise product, that
is, $\mathcal{H}_{\boldsymbol{\mu}}=Mult(\mathcal{H}_{\boldsymbol{\mu}}).$
Then for every non-zero $\mathcal{M}$ in
$Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ we have $\dim(\mathcal{M}\ominus
z\mathcal{M})=1.$ In particular if $m\geqslant 3$ and $\mu_{m}\neq 0,$ then
$\dim(\mathcal{M}\ominus z\mathcal{M})=1$ for any $\mathcal{M}$ in
$Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}).$

###### Proof.

Suppose $\mathcal{H}_{\boldsymbol{\mu}}$ is an algebra under pointwise
product, that is,
$\mathcal{H}_{\boldsymbol{\mu}}=Mult(\mathcal{H}_{\boldsymbol{\mu}}).$ By an
application of Closed Graph Theorem the multipliers norm $\|\cdot\|_{op}$ and
the Hilbertian norm $\|\cdot\|_{\boldsymbol{\mu}}$ on
$\mathcal{H}_{\boldsymbol{\mu}}$ are equivalent. In view of [16, Corollary
5.3], we obtain that the set of all polynomials is dense in
$\mathcal{H}_{\boldsymbol{\mu}}$ w.r.t the multiplier norm $\|\cdot\|_{op}.$
For any $f\in\mathcal{H}_{\boldsymbol{\mu}},$ let $[f]$ denotes the smallest
closed $M_{z}$-invariant subspace containing $f$ in
$\mathcal{H}_{\boldsymbol{\mu}}.$ Hence it follows that
$\varphi[f]\subseteq[f]$ for every
$\varphi,f\in\mathcal{H}_{\boldsymbol{\mu}}.$ Applying [7, Proposition 1] to
$\mathcal{H}_{\boldsymbol{\mu}}$, we get that $\mathcal{H}_{\boldsymbol{\mu}}$
is a cellular-indecomposable space. Now the first part of the theorem follows
immediately from [7, Theorem 1]. The last part follows using the first part
together with Corollary 3.7. ∎

For the remaining case $m=2,$ that is, $\boldsymbol{\mu}=(\mu_{1},\mu_{2})$
with $\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T}),$ we answer this question
partially in Theorem 4.5. We start with a basic lemma which will be useful in
the proof of Theorem 4.5.

######  Lemma 4.2.

Let $H$ be a Hilbert space and $V$ be a closed subspace of $H$ with $\dim
V^{\perp}=n.$ Let $W$ be another subspace of $H$, then $\dim(W\ominus(V\cap
W))\leqslant n.$

###### Proof.

Let $P$ denote the orthogonal projection of $H$ onto $V^{\perp}$. Suppose
$X=W\ominus(V\cap W).$ We claim that $P_{|_{X}}$ is an injective operator. To
verify the claim, let $x\in X$ be such that $P_{|_{X}}(x)=0$. Since $\ker
P=V,$ $x\in V.$ This implies $x=0.$ Thus claims stands verified and therefore
$\dim(W\ominus(V\cap W))\leqslant\dim V^{\perp}=n$. ∎

######  Lemma 4.3.

Let $\mu$ be a finite positive Borel measure on the unit circle $\mathbb{T}$.
Let $n\in\mathbb{N},$ $\lambda_{1},\ldots,\lambda_{n}\in\mathbb{T}$ be
distinct points and $c_{1},\ldots,c_{n}$ be positive real numbers. Let
$\boldsymbol{\mu}$ be the tuple
$(\mu,\sum_{j=1}^{n}c_{j}\delta_{\lambda_{j}})$. Then the subspace
$\ker({M_{z}^{*}}^{2}M_{z}^{2}-2M_{z}^{*}M_{z}+I)=\mathcal{V}$ (say) is a
closed $M_{z}$-invariant subspace of $\mathcal{H}_{\boldsymbol{\mu}}$ and the
operator $M_{z}|_{\mathcal{V}}$ is a cyclic analytic $2$-isometry. Moreover,
$\dim\mathcal{V}^{\perp}=n.$

###### Proof.

From [16, Theorem 4.1], we know that $M_{z}$ on
$\mathcal{H}_{\boldsymbol{\mu}}$ is a cyclic analytic $3$-isometry. Thus, by
[1, Proposition 1.6], it follows that $\mathcal{V}$ is a closed
$M_{z}$-invariant subspace of $\mathcal{H}_{\boldsymbol{\mu}}$ and the
operator $M_{z}|_{\mathcal{V}}$ is a $2$-isometry. Since $M_{z}$ on
$\mathcal{H}_{\boldsymbol{\mu}}$ is analytic, $M_{z}|_{\mathcal{V}}$ is also
analytic. In view of [26, Theorem 1], to show $M_{z}|_{\mathcal{V}}$ is
cyclic, it is sufficient to show that $dim(\mathcal{V}\ominus
z\mathcal{V})=1.$

Before proving that $dim(\mathcal{V}\ominus z\mathcal{V})=1,$ first we show
that $\dim\mathcal{V}^{\perp}=n$. In this regard, note that
$\mathcal{H}_{\boldsymbol{\mu}}\subseteq\mathcal{H}_{\lambda_{j},2}$ for each
$j=1,\ldots,n.$ Thus, by Theorem 1.1, $f^{*}(\lambda_{j})$ exists for all
$f\in\mathcal{H}_{\boldsymbol{\mu}}$ and for every $j=1,\ldots,n$. Since
$\mathcal{H}_{\boldsymbol{\mu}}$ is a reproducing kernel Hilbert space on
$\mathbb{D},$ the linear functional $f\mapsto f(r\lambda_{j})$ is bounded on
$\mathcal{H}_{\boldsymbol{\mu}}$ for any $0<r<1$ and for each $j=1,\ldots,n.$
An application of uniform boundedness principle gives that $f\mapsto
f^{*}(\lambda_{j})$ is a bounded linear functional on
$\mathcal{H}_{\boldsymbol{\mu}}$ for each $j=1,\ldots,n$. Therefore, by Riesz
representation theorem, there exist
$h_{1},\ldots,h_{n}\in\mathcal{H}_{\boldsymbol{\mu}}$ such that

$f^{*}(\lambda_{j})=\langle
f,h_{j}\rangle,~{}~{}f\in\mathcal{H}_{\boldsymbol{\mu}},\,\,\,j=1,\ldots,n.$
(12)

Since $M_{z}$ on $\mathcal{H}_{\boldsymbol{\mu}}$ is a $3$-isometry, by [1,
Proposition 1.5], the operator ${M_{z}^{*}}^{2}M_{z}^{2}-2M_{z}^{*}M_{z}+I$ is
positive. Thus a function $f$ of $\mathcal{H}_{\boldsymbol{\mu}}$ is in
$\mathcal{V}$ if and only if
$\|z^{2}f\|_{\boldsymbol{\mu}}^{2}-2\|zf\|_{\boldsymbol{\mu}}^{2}+\|f\|_{\boldsymbol{\mu}}^{2}=0.$
By [16, Proposition 2.7], this is equivalent to $D_{\tau,0}(f)=0,$ where
$\tau=\sum_{j=1}^{n}c_{j}\delta_{\lambda_{j}}.$ Since
$D_{\tau,0}(f)=\sum_{j=1}^{n}c_{j}|f^{*}(\lambda_{j})|^{2}$, we see that

$\mathcal{V}=\\{f\in\mathcal{H}_{\boldsymbol{\mu}}:f^{*}(\lambda_{j})=0,1\leqslant
j\leqslant n\\}.$

Hence, by (12), it follows that
$\mathcal{V}^{\perp}=\mbox{span}~{}\\{h_{1},\ldots,h_{n}\\}$. Since
$\lambda_{j}^{\prime}$s are distinct points on $\mathbb{T}$, it follows that
$h_{1},\ldots,h_{n}$ are linearly independent. This shows that
$\dim\mathcal{V}^{\perp}=n$. To finish the proof, note that $M_{z}$ on
$\mathcal{H}_{\boldsymbol{\mu}}$ is in the Cowen-Douglas class
$B_{1}(\mathbb{D})$, see [16, Corollary 6.1]. Since $\mathcal{V}$ is a closed
$M_{z}$-invariant subspace of finite co-dimension, it follows from [22, p. 71]
that $\dim(\mathcal{V}\ominus z\mathcal{V})=1.$ This completes the proof. ∎

######  Remark 4.4.

By Remark 3.9(ii), the space $\mathcal{H}_{\boldsymbol{\mu}}$ is an algebra,
where $\boldsymbol{\mu}=(\mu_{1},\mu_{2}+\sigma)$,
$\mu_{1},\mu_{2}\in\mathcal{M}_{+}(\mathbb{T}).$ Hence in view of Theorem 4.1
we obtain that for any $\mathcal{M}\in
Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ $\dim(\mathcal{M}\ominus
z\mathcal{M})=1.$

######  Theorem 4.5.

Let $\mu$ be a finite non-negative Borel measure on the unit circle
$\mathbb{T}$. Let $n\in\mathbb{N},$
$\lambda_{1},\ldots,\lambda_{n}\in\mathbb{T}$ be distinct points and
$c_{1},\ldots,c_{n}$ be positive real numbers. Suppose $\mathcal{M}$ is a non-
zero subspace in $Lat(M_{z},\mathcal{H}_{\boldsymbol{\mu}}),$ where
$\boldsymbol{\mu}=(\mu,\sum_{j=1}^{n}c_{j}\delta_{\lambda_{j}})$. Then
$\dim(\mathcal{M}\ominus z\mathcal{M})=1.$

###### Proof.

Let $\mathcal{V}$ be the subspace
$\ker({M_{z}^{*}}^{2}M_{z}^{2}-2M_{z}^{*}M_{z}+I)$. By Lemma 4.3, the operator
$T:=M_{z}|_{{}_{\mathcal{V}}}$ is a cyclic analytic $2$-isometry. As
$\mathcal{V}\cap\mathcal{M}=\mathcal{N}$ (say) is an invariant subspace for
$T$, from [20, Theorem 3.2], it follows that

$\displaystyle\dim(\mathcal{N}\ominus z\mathcal{N})\leqslant 1.$ (13)

Let $\mathcal{L}$ be a subspace of $\mathcal{M}$ such that
$\mathcal{M}=\mathcal{N}\oplus\mathcal{L}.$ Then from Lemma 4.2, it follows
that $\dim\mathcal{L}\leqslant n.$ Since $M_{z}$ is an injective map,
$z\mathcal{N}\cap z\mathcal{L}=\\{0\\}$ and $\dim\mathcal{L}=\dim
z\mathcal{L}$ as $\mathcal{L}$ is finite dimensional. From (13), we now have

$\displaystyle\dim(\mathcal{M}\ominus z\mathcal{M})$ $\displaystyle=$
$\displaystyle\dim((\mathcal{N}\oplus\mathcal{L})\ominus
z(\mathcal{N}\oplus\mathcal{L}))$ $\displaystyle=$
$\displaystyle\dim((\mathcal{N}\oplus\mathcal{L})\ominus(z\mathcal{N}+z\mathcal{L}))$
$\displaystyle\leqslant$ $\displaystyle 1.$

This completes the proof of the theorem. ∎

##  5\. Relationship between $\mathcal{H}_{\boldsymbol{\mu}}$ and
$\mathcal{H}(b)$

In what follows, for two reproducing kernel Hilbert spaces $\mathcal{H}_{1}$
and $\mathcal{H}_{2}$ consisting of holomorphic functions on the unit disc
$\mathbb{D}$, we write $\mathcal{H}_{1}=\mathcal{H}_{2}$ to mean that
$\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ are equal as a set but the norms are
not necessarily same. In case $\mathcal{H}_{1}=\mathcal{H}_{2},$ using closed
graph theorem, it can be shown that the norms on $\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ must be equivalent.

For a positive measure $\mu$ in $\mathcal{M}_{+}(\mathbb{T})$, we set the
notation (as in [12])

$V_{\mu}(z):=\int_{\mathbb{T}}\frac{d\mu(\lambda)}{|1-\lambda\overline{z}|^{2}},\quad
z\in\mathbb{D}.$

With the help of generalized local Douglas formula, for a fixed
$w\in\mathbb{D},$ the following lemma computes the semi-norm of Szegö kernel
$S(z,w):=\frac{1}{1-z\overline{w}}$ in $\mathcal{H}_{\mu,n},$ in terms of
$V_{\mu}(w).$

######  Lemma 5.1.

Let $\mu$ be a finite positive Borel measure on $\mathbb{T}$ and $n$ be a
positive integer. Then

$D_{\mu,n}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}=\frac{|w|^{2n}}{(1-|w|^{2})^{n}}V_{\mu}(w),\quad
w\in\mathbb{D}.$

###### Proof.

Let $\lambda\in\mathbb{T}$. Then by Theorem 1.1, we have

$\displaystyle D_{\lambda,n}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}$
$\displaystyle=D_{\sigma,n-1}\Big{(}\frac{\frac{1}{1-z\overline{w}}-\frac{1}{1-\lambda\overline{w}}}{z-\lambda}\Big{)}$
$\displaystyle=D_{\sigma,n-1}\Big{(}\frac{\overline{w}}{(1-z\overline{w})(1-\lambda\overline{w})}\Big{)}$
$\displaystyle=\frac{|w|^{2}}{|1-\lambda\overline{w}|^{2}}D_{\sigma,n-1}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}.$

To complete the proof for the case $\mu$ is point mass measure
$\delta_{\lambda}$ at the point $\lambda,$ we claim that

$\displaystyle
D_{\sigma,n-1}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}=\frac{|w|^{2n-2}}{(1-|w|^{2})^{n}}.$
(14)

Since $\frac{1}{1-z\overline{w}}$ is the reproducing kernel for the Hardy
space $H^{2},$ it is easy to verify (14) for the case $n=1.$ Suppose $n>1$,
then by definition of $D_{\sigma,n-1}(\cdot)$ and the fact that
$\frac{1}{(1-z\overline{w})^{n}}$ is the reproducing kernel of the weighted
Bergman space $A^{2}\big{(}(n-1)(1-|z|^{2})^{n-2}dA(z)\big{)}$, we get

$\displaystyle D_{\sigma,n-1}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}$
$\displaystyle=(n-1)|w|^{2n-2}\int_{\mathbb{D}}\frac{(1-|z|^{2})^{n-2}}{|1-z\overline{w}|^{2n}}dA(z)$
$\displaystyle=\frac{|w|^{2n-2}}{(1-|w|^{2})^{n}}.$

Hence equation(14) stands verified and consequently we have

$D_{\lambda,n}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}=\frac{|w|^{2n}}{|1-\lambda\overline{w}|^{2}(1-|w|^{2})^{n}}.$

Integrating both the sides with respect to $\mu,$ we get

$D_{\mu,n}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}=\int_{\mathbb{T}}D_{\lambda,n}\Big{(}\frac{1}{1-z\overline{w}}\Big{)}d\mu(\lambda)=\frac{|w|^{2n}}{(1-|w|^{2})^{n}}V_{\mu}(w).$

This completes the proof of the lemma. ∎

In the following main result of this section, we prove that the spaces
$\mathcal{H}_{\boldsymbol{\mu}}$ ($m\geqslant 2)$ and $\mathcal{H}(b)$ can not
be same even as sets.

######  Theorem 5.2.

Let $m\geqslant 2$ and $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$ be an
$m$-tuple of positive measures in $\mathcal{M}_{+}(\mathbb{T})$. If
$\mathcal{H}_{\boldsymbol{\mu}}=\mathcal{H}(b)$ for some $b\in H^{\infty}$
with $\|b\|_{\infty}\leqslant 1$ then

$\mu_{2}=\cdots=\mu_{m}=0.$

###### Proof.

Suppose that there exists a $b\in H^{\infty}$ with $\|b\|_{\infty}\leqslant 1$
such that $\mathcal{H}(b)=\mathcal{H}_{\boldsymbol{\mu}}$. Then, by an
application of closed graph theorem, there exist constants $\alpha>0$ and
$\beta>0$ such that

$\beta\|f\|_{\mathcal{H}(b)}\leqslant\|f\|_{\boldsymbol{\mu}}\leqslant\alpha\|f\|_{\mathcal{H}(b)},~{}f\in\mathcal{H}(b).$
(15)

Since $\mathcal{H}_{\boldsymbol{\mu}}$ contains all functions holomorphic on a
neighbourhood of $\overline{\mathbb{D}}$, it follows from [12, Theorem 2.1]
that $b$ is not an extreme point of the unit ball of $H^{\infty}$. Hence there
exists a outer function $a$ such that $(b,a)$ is a pair. Note that the
function $\frac{1}{1-z\overline{w}}$ is holomorphic on a neighbourhood of
$\overline{\mathbb{D}}$ for all $w\in\mathbb{D}$. By [12, Lemma 3.4], we have

$\Big{\|}\frac{1}{1-z\overline{w}}\Big{\|}^{2}_{\mathcal{H}(b)}=\frac{1+\big{|}\frac{b(w)}{a(w)}\big{|}^{2}}{1-|w|^{2}},~{}w\in\mathbb{D}.$

Also by Lemma 5.1 we have

$\Big{\|}\frac{1}{1-z\overline{w}}\Big{\|}^{2}_{\boldsymbol{\mu}}=\frac{1}{1-|w|^{2}}+\sum_{j=1}^{m-1}\frac{|w|^{2j}}{(1-|w|^{2})^{j}}V_{\mu_{j}}(w),~{}w\in\mathbb{D}.$

Hence by (15) we obtain

$\frac{1}{1-|w|^{2}}+\sum_{j=1}^{m-1}\frac{|w|^{2j}}{(1-|w|^{2})^{j}}V_{\mu_{j}}(w)\leqslant\alpha~{}\frac{1+\big{|}\frac{b(w)}{a(w)}\big{|}^{2}}{1-|w|^{2}},~{}~{}w\in\mathbb{D}.$

Thus for $2\leqslant j\leqslant m-1$, we have

$\frac{|w|^{2j}}{(1-|w|^{2})^{j}}V_{\mu_{j}}(w)\leqslant\alpha~{}\frac{1+\big{|}\frac{b(w)}{a(w)}\big{|}^{2}}{1-|w|^{2}}.$

Since $V_{\mu_{j}}(w)\geqslant\frac{\mu_{j}(\mathbb{T})}{4}$ for all
$w\in\mathbb{D}$, we get that

$\mu_{j}(\mathbb{T})\leqslant
4\alpha(1-|w|^{2})^{j-1}\frac{1+\big{|}\frac{b(w)}{a(w)}\big{|}^{2}}{|w|^{2j}}.$

Let $\zeta$ be any point in $\mathbb{T}$ such that
$\frac{b^{*}(\zeta)}{a^{*}(\zeta)}$ exists. Since $j\geqslant 2$, putting
$w=r\zeta$ in the above inequality and taking limit $r\to 1^{-}$, we get that
$\mu_{j}(\mathbb{T})=0$, completing the proof of the theorem. ∎

######  Remark 5.3.

Let $\boldsymbol{B}:\mathbb{D}\to(\ell^{2})_{1}$ be an analytic map with
$\boldsymbol{B}(z)=(b_{i}(z))_{i=0}^{\infty}$, where $(\ell^{2})_{1}$ is the
closed unit ball in $\ell^{2}$. The space $\mathcal{H}(\boldsymbol{B})$
corresponding to $\boldsymbol{B}$ is defined to be the reproducing kernel
Hilbert space with the kernel function
$\frac{1-\sum_{i=0}^{\infty}b_{i}(z)\overline{b_{i}(w)}}{1-z\overline{w}}$,
$z,w\in\mathbb{D}$. The space $\mathcal{H}(\boldsymbol{B})$ is said to be of
finite rank if there exists a $N\in\mathbb{N}$ and an analytic map
$\boldsymbol{C}:\mathbb{D}\to(\ell^{2})_{1}$,
$\boldsymbol{C}(z)=(c_{i}(z))_{i=0}^{\infty}$ such that $c_{i}=0$ for all
$i>N$ and $\mathcal{H}(\boldsymbol{B})=\mathcal{H}(\boldsymbol{C}),$ with
equality of norms. Since the backward shift operator $L$, defined by
$(Lf)(z)=\frac{f(z)-f(0)}{z}$, $z\in\mathbb{D}$, is contractive on
$\mathcal{H}_{\boldsymbol{\mu}}$ (see [16, Lemma 2.9]), it follows from [4,
Proposition 2.1] that $\mathcal{H}_{\boldsymbol{\mu}}$ coincides with the
space $\mathcal{H}(\boldsymbol{B})$ for some $\boldsymbol{B}$ with
$\boldsymbol{B}(0)=0$. In the case of finite rank
$\mathcal{H}(\boldsymbol{B})$ spaces with $\boldsymbol{B}(0)=0$, if the
operator $M_{z}$ is bounded, then it turns out that
$\overline{\mbox{ran}}(M_{z}^{*}M_{z}-I)$ is finite dimensional, (see [17,
Lemma 5.1]). On the contrary, for the space $\mathcal{H}_{\boldsymbol{\mu}}$,
where $\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$, $m\geq 2$, it is worth
noting that the space $\ker(M_{z}^{*}M_{z}-I)$ is finite dimensional. This
follows easily from [16, Proposition 2.7]. Therefore, in this case,
$\overline{\mbox{ran}}(M_{z}^{*}M_{z}-I)$ is infinite dimensional.
Consequently, we conclude that if $m\geqslant 2$, then for any
$\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m})$, $\mu_{m}\neq 0$,
$\mathcal{H}_{\boldsymbol{\mu}}$ can not be equal to any
$\mathcal{H}(\boldsymbol{B})$ space of finite rank with equality of norms. It
will be interesting to know whether for a fixed
$\boldsymbol{\mu}=(\mu_{1},\ldots,\mu_{m}),$ does there exists a
$\mathcal{H}(\boldsymbol{B})$ space of finite rank such that
$\mathcal{H}_{\boldsymbol{\mu}}=\mathcal{H}(\boldsymbol{B})$ with just
equivalence of norms.

######  Remark 5.4.

The last part of [17, Theorem 1.1] says that if $M_{z}$ on $\mathcal{H}(b)$ is
a strict $2m$-isometry for some $m\in\mathbb{N}$ then there exists a
$\zeta\in\mathbb{T}$ such that $\mathcal{H}(b)$ is equal to
$\mathcal{D}_{\zeta}^{m}$-space defined in [17, page 3] with equivalence of
norms. Theorem 5.2 indicates that the local Dirichlet space of order $m$ in
the present article differs from that in [17].

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