#  Toeplitz operators on weighted harmonic Bergman spaces

Zipeng Wang  College of Mathematics and Information Science, Xi’an, Shaanxi
Normal University, 710062, PR China<EMAIL_ADDRESS>and  Xianfeng
Zhao  College of Mathematics and Statistics, Chongqing University, Chongqing,
401331, PR China and Shanghai Center for Mathematical Sciences, Shanghai,
Fudan University, 200433, PR China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we study Toeplitz operators on the weighted harmonic Bergman
spaces with nonnegative symbols, the weights we choose here are Muckenhoupt
$\mathcal{A}_{2}$ weights. Results obtained include characterizations of
bounded Toeplitz operators, compact Toeplitz operators, invertible Toeplitz
operators and Toeplitz operators in the Schatten classes.

###### Key words and phrases:

Toeplitz operator, weighted harmonic Bergman space, boundedness, compactness,
invertibility

MR(2010) Subject Classification 47B 35, 47B 65

This work was partially supported by a NSFC grant(11271387) and Chongqing
Natural Sience Foundation (cstc2013jjB0050).

##  1\. Introduction

Let $1\leqslant p<\infty$ and $\omega$ be a nonnegative integrable function on
the unit disk $\mathbb{D}$. $L^{p}(\omega)$ denotes the Banach space with norm

$\|f\|_{L^{p}(\omega)}:=\bigg{(}\int_{\mathbb{D}}|f(z)|^{p}\omega(z)dA(z)\bigg{)}^{\frac{1}{p}}.$

The weighted harmonic (analytic) Bergman space $L^{p}_{h}(\omega)$
($L^{p}_{a}(\omega)$) is the subspace of $L^{p}(\omega)$ which is consisting
of harmonic (analytic) functions on $\mathbb{D}$. The goal of this paper is to
provide a framework to study operator properties (boundedness, compactness,
Schatten classes and inveritibility) of Toeplitz operators on
$L_{h}^{2}(\omega)$ with nonnegative symbols.

Weighted analytic function spaces and their Toeplitz operators have captured
people’s attentions for a long time. It is now well known ([24]) that several
results on unweighted Bergman space can be extended to the standard weighted
Bergman space $L_{a}^{2}(\omega_{\alpha})$, where
$\omega_{\alpha}(z)=(1+\alpha)(1-|z|^{2})^{\alpha}$ and $-1<\alpha<\infty$. In
recent papers [19] and [20], Peláez and Rättyä characterized the bounded and
Schatten class Toeplitz operators (induced by a positive Borel measure) on a
weighted Bergman space, here the weight is a radial function satisfying the
doubling property $\int_{r}^{1}\omega(s)ds\leqslant
C\int_{\frac{1+r}{2}}^{1}\omega(s)ds.$ The first results of non-radial
weighted Bergman space seems to be due to Luecking ([13]) who investigated the
structure of weighted Bergman space with Békollé-Bonami weight. Based on
Luecking’s representation and duality theorems in [13], Chacón ([2]) and
Constantin ([5], [6]) studied the boundedness and compactness of Toeplitz
operators on certain weighted Bergman spaces. In [17], Mitkovski and Wick
established a reproducing kernel thesis for operators on Bergman type space,
and their definitions include weighted versions of Bergman spaces on more
complicated domains.

We will be primarily interested in weighted harmonic Bergman space
$L_{h}^{2}(\omega)$. Our choice of the weight $\omega$ is motivated by the
characterization of boundedness of $P_{h}$ acting on $L^{2}(\omega)$, where
$P_{h}$ is the unweghted harmonic Bergman (orthogonal) projection from
$L^{2}(dA)$ to $L_{h}^{2}(dA)$. It is well known that $L_{h}^{2}(dA)$ is a
reproducing kernel Hilbert space and

$P_{h}f(z)=\int_{\mathbb{D}}f(\lambda)\bigg{[}\frac{1}{(1-\overline{\lambda}z)^{2}}+\frac{1}{(1-\overline{z}\lambda)^{2}}-1\bigg{]}dA(\lambda).$

Clearly, $P_{h}$ is a Calderón-Zygmund operator on the homogeneous space
$(\mathbb{D},d,dA)$, where $d$ is the Euclidean distance and $dA$ is the
Lebesgue area measure on $\mathbb{D}$, normalized so that the measure of
$\mathbb{D}$ is $1$. For the definitions of Calderón-Zygmund operator and
homogeneous space, we refer to [1].

The most successful understanding of the (one) weight theory of Calderón-
Zgumund operator was spurred by Munckenhoupt’s work in 1970s ([18]), which led
to the introduction of the class of $\mathcal{A}_{p}$ weight and developments
of weighted inequality. We will restrict our attentions to $\mathcal{A}_{2}$
weight on $(\mathbb{D},d,dA)$.

Let $0<\omega\in L^{1}(\mathbb{D},dA)$, it is called a Muckenhoupt
$\mathcal{A}_{2}$ weight if

$[\omega]_{\mathcal{A}_{2}}:=\sup_{a\in\mathbb{D},\
0<r<1}\frac{|B(a,r)|_{\omega}|B(a,r)|_{\omega^{-1}}}{|B(a,r)|^{2}}<+\infty,$

where

$B(a,r)=\\{z\in\mathbb{D}:d(a,z)=|z-a|<r\\},$
$|B(a,r)|_{\omega}=\int_{B(a,r)}\omega(z)dA(z)$

and $|\cdot|$ is the normalized Lebesgue measure on $\mathbb{D}$.

It follows form the remarkable $\mathcal{A}_{2}$ theorem ([1],[9]) that
_$P_{h}$ is bounded from $L^{2}(\omega)$ to $L_{h}^{2}(\omega)$ provided
$\omega$ is a Muckenhoupt $\mathcal{A}_{2}$ weight._ As mentioned above, we
will focus on the weighted harmonic Bergman space $L_{h}^{2}(\omega)$ with
$\omega\in\mathcal{A}_{2}$. Little is known about this nature function space.
However, we will see in Section 2 that $L_{h}^{2}(\omega)$ is a reproducing
kernel Hilbert space with the reproducing kernel $K^{\omega}_{z}(\lambda)$,
i.e., $f(z)=\langle f,K^{\omega}_{z}\rangle_{L^{2}(\omega)}$ for all $f$ in
$L_{h}^{2}(\omega)$.

For a positive finite Borel measure $\nu$ on $\mathbb{D}$, we densely define
the Toeplitz operator $T_{\nu}$ on $L_{h}^{2}(\omega)$ by

$T_{\nu}f(z)=\big{\langle}T_{\nu}f,K^{\omega}_{z}\big{\rangle}_{L^{2}(\omega)}=\int_{\mathbb{D}}f(\lambda)\overline{K^{\omega}_{z}(\lambda)}d\nu(\lambda)\
\ \ \big{(}z\in\mathbb{D}\big{)}.$

For a bounded function $\varphi$, using the integral representation for the
projection operator (from $L^{2}(\omega)$ to $L_{h}^{2}(\omega)$), we can
express the Toeplitz operator $T_{\varphi}$ (on $L_{h}^{2}(\omega)$) as the
following:

$T_{\varphi}f(z)=\int_{\mathbb{D}}f(\lambda)\overline{K^{\omega}_{z}(\lambda)}\varphi(\lambda)\omega(\lambda)dA(\lambda)\
\ \ \big{(}z\in\mathbb{D}\big{)}.$

Although we follow Luecking’s methods in [13] and [14] for the weighted
Bergman spaces, some new difficulties arise in the study of the space
$L_{h}^{2}(\omega)$ and the corresponding operators. For instance, harmonic
functions do not share many powerful tools with analytic functions. One can
use the Cauchy formula to estimate the local values of analytic functions
easily. However, because of the tedious remainder, the harmonic version Cauchy
formula (known as Cauchy-Pompeiu formula) is not valid now. We instead must
rely on some known estimations on harmonic functions. In addition, just as the
weighted Bergman space, one cannot write down an explicit formula for the
reproducing kernel for $L_{h}^{2}(\omega)$. To overcome this obstacle, we will
use the reproducing kernel for the unweighted space $L_{h}^{2}$ to help us
study the representation theory of $L_{h}^{2}(\omega)$. However, the
properties of reproducing kernel for $L_{h}^{2}$ are much more complicated
than that of Bergman space $L_{a}^{2}$.

Using some properties of harmonic functions and $\mathcal{A}_{2}$ weights, we
establish two different atomic decompositions for functions in
$L_{h}^{2}(\omega)$ (Theorems 2.4 and 2.10), which extend the representation
theorems in [13] to the harmonic case.

In Section 3, we characterize the boundedness, compactness and Schatten $p$
class of Toeplitz operators $T_{\nu}$ on $L_{h}^{2}(\omega)$ by means of
Berezin transform and Carleson measure. We are pleasure to mention here that
Miao ([15]) have obtained characterizations for the Toeplitz operators with
nonnegative symbols to be bounded, compact and in Schatten classes on
unweighted harmonic Bergman space $L_{h}^{2}$.

Section 4 of this paper is devoted to studying the invertibility of Toeplitz
operators on the standard weighted harmonic Bergman space
$L^{2}_{h}(\omega_{\alpha})$. A little surprising to us, the results
illustrate that the invertibility of Toeplitz operators on Bergman space
$L_{a}^{2}$ can imply a reverse Carleson inequality for
$L_{h}^{2}(\omega_{\alpha})$, see Theorems 4.2 and 4.4. Based on this
inequality, we generalize the result on the invertibility of Bergman Toeplitz
operators with nonnegative symbols ([23]) to the case of
$L_{h}^{2}(\omega_{\alpha})$. As a consequence, we obtain a relationship of
the invertibility between Toeplitz operators with bounded nonnegative symbols
on $L_{a}^{2}(\omega_{\alpha})$ and $L_{h}^{2}(\omega_{\alpha})$, see
Corollary 4.5.

In the final section, we establish a reverse Carleson type inequality for
$L_{h}^{2}(\omega)$ with $\omega\in\mathcal{A}_{2}$. Indeed, we obtain a
sufficient condition for $\chi_{G}dA$ to be a reverse Carleson measure for
$L_{h}^{2}(\omega)$, where $G$ is a measurable set in $\mathbb{D}$, see
Theorem 5.1, which extends Theorem 3.9 for the weighted (analytic) Bergman
space in [13] to the harmonic setting.

Throughout the paper, positive constants will be explicitly denoted by
$C,C_{0},C_{1},\cdots$, which may depend on some fixed numbers and change at
each occurrence.

##  2\. The Space $L_{h}^{2}(\omega)$ and its Representation

In this section, we present some elementary structures of $L_{h}^{2}(\omega)$
with $\omega\in\mathcal{A}_{2}$. To study the harmonic Bergman spaces, we need
the following important properties of harmonic functions.

######  Lemma 2.1.

([10]) Suppose that $f$ is a harmonic function on the disk $\mathbb{D}$ and
$0<p<\infty$. There exists a positive constant $C=C(p)$ such that for every
ball $B(a,r)=\\{z\in\mathbb{D}:|z-a|<r\\}$ in $\mathbb{D}$,

$|f(a)|^{p}\leqslant\frac{C}{|B(a,r)|}\int_{B(a,r)}|f(z)|^{p}dA(z).$

In particular, if $p\geqslant 1$, then the constant $C\equiv 1$. Using this
result one can get the following useful inequalities easily: given
$0<p<\infty$ and $0<r<1$, there exist positive constants $C_{1}=C_{1}(p)$ and
$C_{2}=C_{2}(p)$ such that

$|f(a)|^{p}\leqslant\frac{C_{1}}{(1-r)^{2}}\cdot\frac{1}{|D(a,r)|}\int_{D(a,r)}|f(z)|^{p}dA(z)\
\ \ \big{(}a\in\mathbb{D}\big{)}$

and

$|\partial
f(a)|^{p}\leqslant\frac{C_{2}}{(1-r)^{2+p}}\cdot\frac{1}{|D(a,r)|^{\frac{p+2}{2}}}\int_{D(a,r)}|f(z)|^{p}dA(z)\
\ \ \big{(}a\in\mathbb{D}\big{)}$

for all $f$ harmonic on $\mathbb{D}$, where $\partial f=\frac{\partial
f}{\partial z}$.

_Remark 1._ From the above inequalities, it is easy to show that point
evaluations are bounded linear functionals on $L_{h}^{2}(\omega)$ with
$\omega\in\mathcal{A}_{2}$. As a consequence, $L_{h}^{2}(\omega)$ is a
reproducing kernel Hilbert space.

_Remark 2._ It is not clear whether $L^{2}_{h}(\omega)$ is complete for a
general weight. However, if $p$ is an analytic polynomial on $\mathbb{D}$ and
$\omega(z)=|p(z)|^{2}$, Douglas and Wang [7] showed that $L_{a}^{2}(\omega)$
is complete, whose proof heavily depends on some particular properties of
polynomials.

It is clear that $L_{h}^{2}(\omega)$ coincides with its dual space with
respect to the $L^{2}(\omega)$ inner product. The next result illustrates that
the dual space of $L_{h}^{2}(\omega)$ can be identified with
$L^{2}_{h}(\omega^{-1})$ via the unweighted inner product, which generalizes
Luecking’s result for $L_{a}^{2}(\omega)$ (see Theorem 2.1 in [13]) to the
setting of $L_{h}^{2}(\omega)$.

######  Lemma 2.2.

Suppose that $\omega$ is an $\mathcal{A}_{2}$ weight. Then the dual space of
$L_{h}^{2}(\omega)$ can be identified with $L_{h}^{2}(\omega^{-1})$. The
pairing is given by

$\langle f,g\rangle=\int_{\mathbb{D}}f(z)\overline{g(z)}dA(z).$

Consequently, there exists a bounded, bijective and linear operator
$\mathscr{F}:L_{h}^{2}(\omega^{-1})\rightarrow(L_{h}^{2}(\omega))^{*}$ such
that

$\mathscr{F}(f)(g)=\langle g,\overline{f}\rangle_{L^{2}(dA)}$

for $f\in L_{h}^{2}(\omega^{-1})$ and $g\in L_{h}^{2}(\omega)$.

###### Proof.

Let $\omega$ be an $\mathcal{A}_{2}$ weight. Recall that orthogonal projection
$P_{h}:L^{2}(dA)\to L^{2}_{h}(dA)$ is a Calderón-Zygmund operator on
$(\mathbb{D},d,dA)$. Then $P_{h}$ is bounded on both $L^{2}(\omega)$ and
$L^{2}(\omega^{-1})$. Thus for each $f\in L^{2}(\omega)$ and $g\in
L^{2}(\omega^{-1})$, we have

$\langle P_{h}f,g\rangle=\langle f,P_{h}g\rangle.$

Note that each $f\in L_{h}^{2}(\omega)$ (or $f\in L_{h}^{2}(\omega^{-1})$) has
the following form:

$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}+\sum_{n=1}^{\infty}b_{n}\overline{z^{n}},$

thus we obtain

$\displaystyle P_{h}f(z)$
$\displaystyle=\int_{\mathbb{D}}f(\lambda)R_{z}(\lambda)dA(\lambda)$
$\displaystyle=\lim_{s\rightarrow
1^{-}}\int_{s\mathbb{D}}f(\lambda)\Big{[}\frac{1}{(1-\overline{z}\lambda)^{2}}+\frac{1}{(1-\overline{\lambda}z)^{2}}-1\Big{]}dA(\lambda)$
$\displaystyle=f(z)$

for $z\in\mathbb{D}$. Then the rest of this proof are exactly the same as the
one of Theorem 2.1 in [13], we omit the details. ∎

Let $a\in\mathbb{D}$ and $0<r<1$. A pseudohyperbolic disk $D(a,r)$ is defined
by

$D(a,r)=\bigg{\\{}z\in\mathbb{D}:\rho(z,a)=\Big{|}\frac{z-a}{1-\overline{a}z}\Big{|}<r\bigg{\\}}.$

We will frequently use the following property of $\mathcal{A}_{2}$ weights on
pseudohyperbolic disks. For the sake of complete, we include a proof of this
fact as follows.

######  Lemma 2.3.

Let $0<r\leqslant\frac{1}{4}$ and $z\in\mathbb{D}$. If $\xi\in D(z,r)$, then
we have

$|D(z,r)|_{\omega}<8[\omega]_{\mathcal{A}_{2}}|D(\xi,r)|_{\omega}.$

###### Proof.

Observe that $D(z,r)\subset D(\xi,2r)$. Now it suffices to show the following
doubling inequality:

$|D(\xi,2r)|_{\omega}<8[\omega]_{\mathcal{A}_{2}}|D(\xi,r)|_{\omega}\ \ \
\big{(}\xi\in\mathbb{D}\big{)}.$

Since $\omega$ is an $\mathcal{A}_{2}$weight, we have

$\frac{|D(\xi,2r)|_{\omega}|D(\xi,2r)|_{\omega^{-1}}}{|D(\xi,2r)|^{2}}\leqslant[\omega]_{\mathcal{A}_{2}}.$

Recall that a pseudohyperbolic disk $D(z,r)$ is a Euclidean disk with center
and radius given by

$\mathfrak{C}=\frac{1-r^{2}}{1-r^{2}|z|^{2}}z,\quad\mathfrak{R}=\frac{1-|z|^{2}}{1-r^{2}|z|^{2}}r.$

Combining the above with Cauchy-Schwarz’s inequality gives

$\displaystyle|D(\xi,2r)|_{\omega}$
$\displaystyle\leqslant[\omega]_{\mathcal{A}_{2}}\frac{|D(\xi,2r)|^{2}}{|D(\xi,2r)|_{\omega^{-1}}}$
$\displaystyle\leqslant[\omega]_{\mathcal{A}_{2}}\frac{|D(\xi,r)|^{2}}{|D(\xi,r)|_{\omega^{-1}}}\cdot\frac{|D(\xi,2r)|^{2}}{|D(\xi,r)|^{2}}$
$\displaystyle\leqslant
4[\omega]_{\mathcal{A}_{2}}|D(\xi,r)|_{\omega}\cdot\bigg{(}\frac{1-r^{2}|\xi|^{2}}{1-4r^{2}|\xi|^{2}}\bigg{)}^{2}$
$\displaystyle<8[\omega]_{\mathcal{A}_{2}}|D(\xi,r)|_{\omega},$

where the last inequality is due to $r\leqslant\frac{1}{4}$. This completes
the proof of Lemma 2.3. ∎

We now turn to the representation theory of the space $L_{h}^{2}(\omega)$.
These results and their proof strategies are motivated by Luecking’s works on
weighted Bergman space ([13], [14]).

Before studying the representation theory of $L_{h}^{2}(\omega)$, we need to
recall the concept of $\epsilon$-lattice in the unit disk. Let
$\epsilon\in(0,1)$, a sequence $\\{a_{n}\\}_{n=1}^{\infty}$ in the unit disk
is called an $\epsilon$-lattice in the pseudo-hyperbolic metric if

  * •

$\mathbb{D}=\bigcup_{n=1}^{\infty}D(a_{n},\epsilon)$ and

  * •

$\inf\limits_{n\neq
m}\Big{|}\frac{a_{n}-a_{m}}{1-\overline{a_{n}}a_{m}}\Big{|}\geqslant\frac{\epsilon}{2}.$

Now, we are ready to state the atomic decomposition for $L^{2}_{h}(\omega)$.

######  Theorem 2.4.

Let $\omega$ be an $\mathcal{A}_{2}$ weight. Then there is an
$\epsilon$-lattice $\\{a_{n}\\}_{n=1}^{\infty}$ such that for each $f\in
L_{h}^{2}(\omega)$ we have

$f(z)=\sum_{n=1}^{\infty}c_{n}(1-|a_{n}|^{2})^{2}|D(a_{n},\epsilon)|_{\omega}^{-\frac{1}{2}}R_{a_{n}}(z)$

for some sequence $\\{c_{n}\\}$ in $\ell^{2}(\mathbb{N})$, where

$R_{\lambda}(z)=\frac{1}{(1-\overline{z}\lambda)^{2}}+\frac{1}{(1-\overline{\lambda}z)^{2}}-1$

is the reproducing kernel for $L_{h}^{2}$ at $\lambda\in\mathbb{D}$.

_Remarks._ We have the following estimate of the module of $R_{\lambda}$:
there exists an $r_{0}\in(0,\frac{1}{4}]$ such that if $0<r\leqslant r_{0}$,
then

$\frac{\frac{1}{2}}{(1-|\lambda|)^{2}}\leqslant|R_{\lambda}(z)|\leqslant\frac{3}{(1-|\lambda|)^{2}}$

for all $z\in D(\lambda,r)$. For the proof of this fact, we refer to Lemma 2.2
in [4]. In what follows, we will use $r_{0}$ to denote the constant provided
in this remarks.

To prove Theorem 2.4, we need to establish a harmonic version of Luecking’s
theorems in [13] and [14].

######  Theorem 2.5.

Let $\omega$ be an $\mathcal{A}_{2}$ weight. Then there exists an
$\epsilon$-lattice $\\{a_{n}\\}_{n=1}^{\infty}$ for some
$0<\epsilon<\frac{1}{16}$ such that

$\sum_{n=1}^{\infty}|f(a_{n})|^{2}|D(a_{n},\epsilon)|_{\omega}\approx\|f\|_{L^{2}(\omega)}^{2}$

for all $f$ in $L_{h}^{2}(\omega)$. That is, there exist two positive constant
$C_{1}$ and $C_{2}$ such that

$C_{1}\|f\|_{L^{2}(\omega)}^{2}\leqslant\sum_{n=1}^{\infty}|f(a_{n})|^{2}|D(a_{n},\epsilon)|_{\omega}\leqslant
C_{2}\|f\|_{L^{2}(\omega)}^{2}.$

for all $f$ in $L_{h}^{2}(\omega)$.

Once Theorem 2.5 is established, we can quickly present a proof of Theorem 2.4
as follows.

###### Proof of Theorem 2.4.

Note that both $\omega$ and $\omega^{-1}$ are $\mathcal{A}_{2}$ weights, it
follows from Theorem 2.5 that we can choose $\epsilon\in(0,\frac{1}{16})$ and
an $\epsilon$-lattice $\\{a_{n}\\}_{n=1}^{\infty}$ such that

$\|g\|_{L^{2}(\omega^{-1})}^{2}\approx\sum_{n=1}^{\infty}|g(a_{n})|^{2}|D(a_{n},\epsilon)|_{\omega^{-1}}.$

By Cauchy-Schawrz inequality and the definition of $\mathcal{A}_{2}$ weight,

$|D(a_{n},\epsilon)|^{2}\leqslant|D(a_{n},\epsilon)|_{\omega}\cdot|D(a_{n},\epsilon)|_{\omega^{-1}}\leqslant[\omega]_{\mathcal{A}_{2}}|D(a_{n},\epsilon)|^{2}.$

Therefore,

$\|g\|_{L^{2}(\omega^{-1})}^{2}\approx\sum_{n=1}^{\infty}|g(a_{n})|^{2}(1-|a_{n}|^{2})^{4}|D(a_{n},\epsilon)|_{\omega}^{-1}.$

This implies that the linear operator
$\mathscr{L}:L_{h}^{2}(\omega^{-1})\rightarrow\ell^{2}(\mathbb{Z})$ defined by

$\mathscr{L}(g):=\bigg{\\{}g(a_{n})(1-|a_{n}|^{2})^{2}|D(a_{n},\epsilon)|_{\omega}^{-\frac{1}{2}}\bigg{\\}}_{n=1}^{\infty}$

is bounded below and so its range is closed. It follows from the closed range
theorem that $\mathscr{L}^{*}$ is surjective.

From the proof of Lemma 2.2, we have

$g(a_{n})=\langle g,R_{a_{n}}\rangle_{L^{2}(dA)}\ \ \ \ (*)$

for each $g\in L_{h}^{2}(\omega^{-1})$ and every $n\geqslant 1$. Let
$\\{c_{n}\\}_{n=1}^{\infty}\in\ell^{2}(\mathbb{N})$. Using $(*)$ to obtain
that

$\mathscr{L}^{*}(\\{c_{n}\\})(z)=\sum_{n=1}^{\infty}c_{n}R_{a_{n}}(z)(1-|a_{n}|^{2})^{2}|D(a_{n},\epsilon)|_{\omega}^{-\frac{1}{2}},$

which gives the desired result. This completes the proof of Theorem 2.4. ∎

We now turn to the proof of Theorem 2.5. Let $0<\epsilon<\frac{1}{16}$ and
$\\{a_{n}\\}_{n=1}^{\infty}\subset\mathbb{D}$ be an $\epsilon$-lattice. Define
a measure $\mu=\mu_{\epsilon}$ on $\mathbb{D}$ by

$\mu(z)=\sum_{n=1}^{\infty}\delta_{a_{n}}(z)\big{|}D(a_{n},\frac{\epsilon}{4})\big{|}_{\omega},$

where $\delta_{a_{n}}$ is the Dirac measure concentrated at $a_{n}$. Indeed,
the conclusion of Theorem 2.4 tells us that $\mu$ is a Carleson and reverse
Carleson measure for $L_{h}^{2}(\omega)$. Firstly, we establish a sufficient
condition for a general (positive) measure to be the $L_{h}^{p}(\omega)$
Carleson measure, where $0<p<\infty$.

######  Proposition 2.6.

Suppose that $\nu$ is a positive Borel measure on $\mathbb{D}$. If there exist
an $0<r\leqslant r_{0}$ and a constant $C>0$ independent of $z\in\mathbb{D}$
such that

$\nu(D(z,r))\leqslant C|D(z,r)|_{\omega}$

for all $z\in\mathbb{D}$, then $\nu$ is a Carlseon measure for
$L_{h}^{p}(\omega)$ ($0<p<\infty$), i.e., there is a positive constant $C_{p}$
such that

$\int_{\mathbb{D}}|f(z)|^{p}d\nu(z)\leqslant C_{p}\|f\|_{L^{p}(\omega)}^{p}$

for $f\in L_{h}^{p}(\omega)$. Consequently, $\mu$ is a Carleson measure for
$L_{h}^{2}(\omega)$, i.e., there is an absolute constant $C>0$ such that

$\int_{\mathbb{D}}|f(z)|^{2}d\mu(z)=\sum_{n=1}^{\infty}|f(a_{n})|^{2}\big{|}D(a_{n},\frac{\epsilon}{4})\big{|}_{\omega}\leqslant
C\|f\|_{L^{2}(\omega)}^{2}$

for all $f$ in $L_{h}^{2}(\omega)$.

###### Proof.

Fix an $r\leqslant r_{0}$. By Lemma 2.1, we obtain

$|f(z)|^{\frac{p}{2}}\leqslant\frac{C}{|D(z,r)|}\int_{D(z,r)}|f(\xi)|^{\frac{p}{2}}dA(\xi)\
\ \ \big{(}z\in\mathbb{D}\big{)},$

where $0<p<\infty$ and $C=C(p,r)$. Cauchy-Schwartz inequality and
$\mathcal{A}_{2}$ condition give us that

$\displaystyle|f(z)|^{p}$
$\displaystyle\leqslant\frac{C^{2}}{|D(z,r)|^{2}}\bigg{(}\int_{D(z,r)}|f(\xi)|^{\frac{p}{2}}dA(\xi)\bigg{)}^{2}$
$\displaystyle\leqslant\frac{C^{2}}{|D(z,r)|^{2}}\bigg{(}\int_{D(z,r)}|f(\xi)|^{p}\omega(\xi)dA(\xi)\bigg{)}\cdot\bigg{(}\int_{D(z,r)}\frac{1}{\omega(\xi)}dA(\xi)\bigg{)}$
$\displaystyle\leqslant
C^{2}[\omega]_{\mathcal{A}_{2}}\frac{\int_{D(z,r)}|f(\xi)|^{p}\omega(\xi)dA(\xi)}{|D(z,r)|_{\omega}}.$

Integrating the above over the unit disk to obtain

$\displaystyle\int_{\mathbb{D}}|f(z)|^{p}d\nu(z)$ $\displaystyle\leqslant
C\int_{\mathbb{D}}|D(z,r)|_{\omega}^{-1}\int_{D(z,r)}|f(\xi)|^{p}\omega(\xi)dA(\xi)d\nu(z)$
$\displaystyle=C\int_{\mathbb{D}}|D(z,r)|_{\omega}^{-1}\int_{\mathbb{D}}|f(\xi)|^{p}\chi_{D(z,r)}(\xi)\omega(\xi)dA(\xi)d\nu(z)$
$\displaystyle=C\int_{\mathbb{D}}\int_{\mathbb{D}}|D(z,r)|_{\omega}^{-1}|f(\xi)|^{p}\omega(\xi)\chi_{D(\xi,r)}(z)dA(\xi)d\nu(z),$

here the constant $C$ depends only on $p$ and $r$. Note that $\xi\in D(z,r)$,
we have by Lemma 2.3 that

$|D(z,r)|_{\omega}\geqslant C|D(\xi,r)|_{\omega}$

for some absolute constant $C$. Therefore,

$\displaystyle\int_{\mathbb{D}}|f(z)|^{2}d\nu(z)$ $\displaystyle\leqslant
C\int_{\mathbb{D}}\int_{\mathbb{D}}|D(z,r)|_{\omega}^{-1}|f(\xi)|^{p}\omega(\xi)\chi_{D(\xi,r)}(z)dA(\xi)d\nu(z)$
$\displaystyle\leqslant
C\int_{\mathbb{D}}\int_{\mathbb{D}}|D(\xi,r)|_{\omega}^{-1}|f(\xi)|^{p}\omega(\xi)\chi_{D(\xi,r)}(z)d\nu(z)dA(\xi).$

Now using our hypothesis on $\nu$ to get

$\displaystyle\int_{\mathbb{D}}|f(z)|^{p}d\nu(z)$ $\displaystyle\leqslant
C\int_{\mathbb{D}}|D(\xi,r)|_{\omega}^{-1}\nu(D(\xi,r))|f(\xi)|^{p}\omega(\xi)dA(\xi)$
$\displaystyle\leqslant C_{1}\int_{\mathbb{D}}|f(\xi)|^{p}\omega(\xi)dA(\xi),$

the constant $C_{1}>0$ comes from the assumption and $C_{1}$ is independent of
$f\in L_{h}^{p}(\omega)$.

For the second conclusion of this proposition, it sufficient for us to show
the following inequality:

$\mu(D(a,\frac{1}{4}))\leqslant C\int_{D(a,\frac{1}{4})}\omega(z)dA(z)\ \ \ \
\ \big{(}a\in\mathbb{D}\big{)}$

for some absolute constant $C>0$. Indeed, by the definition of $\mu$, we have

$\mu(D(a,\frac{1}{4}))=\sum_{\rho(a_{n},a)<\frac{1}{4}}\int_{D(a_{n},\frac{\epsilon}{4})}\omega
dA=\sum_{\rho(a_{n},a)<\frac{1}{4}}\big{|}D(a_{n},\frac{\epsilon}{4})\big{|}_{\omega}.$

If $\rho(a,a_{n})<\frac{1}{4}$, then for each $z\in
D(a_{n},\frac{\epsilon}{4})$ we get

$\rho(z,a)\leqslant\rho(z,a_{n})+\rho(a_{n},a)<\frac{\epsilon}{4}+\frac{1}{4}<\frac{1}{2}.$

Which shows that

$D(a_{n},\frac{\epsilon}{4})\subset D(a,\frac{1}{2})$

for every $n\geqslant 1$ provided $\rho(a,a_{n})<\frac{1}{4}$.

Since $D(a_{n},\frac{\epsilon}{4})\cap
D(a_{m},\frac{\epsilon}{4})=\varnothing$ for $n\neq m$, we obtain

$\bigcup_{\rho(a_{n},a)<\frac{1}{4}}D(a_{n},\frac{\epsilon}{4})\subset
D(a,\frac{1}{2})$

and

$\mu(D(a,\frac{1}{4}))\leqslant\big{|}D(a,\frac{1}{2})\big{|}_{\omega}\leqslant
C\big{|}D(a,\frac{1}{4})\big{|}_{\omega}$

for every $a\in\mathbb{D}$, the constant $C>0$ (independent of $a$) comes from
Lemma 2.3. This completes the proof of Proposition 2.6. ∎

In order to finish the proof of Theorem 2.5, we need to show that there is an
$\epsilon\in(0,\frac{1}{16})$ such that $\mu=\mu_{\epsilon}$ is a reverse
$L_{h}^{2}(\omega)$ Carleson measure. More precisely, we will prove the
following proposition.

######  Proposition 2.7.

There exists an $\epsilon\in(0,\frac{1}{16})$ such that

$\int_{\mathbb{D}}|f(z)|^{2}d\mu(z)=\sum_{n=1}^{\infty}|f(a_{n})|^{2}|D(a_{n},\epsilon)|_{\omega}\geqslant
C\|f\|_{L^{2}(\omega)}^{2}$

for all $f$ in $L_{h}^{2}(\omega)$, where $C>0$ is an absolute constant.

The rest of this section is devoted to the proof of the above _reverse
Carleson inequality_. To do so, we need to prove the following two lemmas
related to harmonic functions, which extend the results in [13] and [14] for
weighted Bergman spaces to the present situation.

######  Lemma 2.8.

Let $f$ be a harmonic function on $\mathbb{D}$ and
$\epsilon\in(0,\frac{1}{16})$. Then there exists a constant $C_{1}>0$
(independent of $z,\epsilon$ and $f$) such that

$|f(z)-f(0)|\leqslant
C_{1}\epsilon\int_{D(0,\frac{1}{4})}|f(\lambda)|dA(\lambda)$

when $|z|<\epsilon.$ As a consequence, there exists a constant $C_{2}>0$
(independent of $z,\epsilon$ and $f$) such that

$|f(w)-f(\xi)|^{2}\leqslant\frac{C_{2}\epsilon^{2}}{|D(\xi,\frac{1}{4})|_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)$

when $\xi\in D(w,\epsilon)$.

###### Proof.

Observe that

$f(z)-f(0)=\int_{0}^{1}\frac{\partial}{\partial
t}\big{[}f(tz)\big{]}dt=\int_{0}^{1}\big{[}\nabla f(tz)\cdot z\big{]}dt$

for $|z|<\epsilon<\frac{1}{16}$. Thus we have

$|f(z)-f(0)|\leqslant\sup_{|\xi|\leqslant\epsilon}|\nabla f(\xi)|\cdot|z|.$

Recall that

$\displaystyle|\nabla f|^{2}$ $\displaystyle=2\big{(}|\partial
f|^{2}+|\overline{\partial}f|^{2}\big{)}=2\big{(}|\partial
f|^{2}+|\partial\overline{f}|^{2}\big{)}$
$\displaystyle\leqslant\big{[}\sqrt{2}(|\partial
f|+|\partial\overline{f}|)\big{]}^{2},$

where $\overline{\partial}f=\frac{\partial f}{\partial\overline{z}}.$ By Lemma
2.1, there is an absolute constant $C>0$ such that

$\displaystyle|\nabla f(\xi)|$
$\displaystyle\leqslant\frac{2\sqrt{2}C}{(1-\frac{1}{16})^{3}}\cdot\frac{1}{|D(\xi,\frac{1}{16})|^{\frac{3}{2}}}\int_{D(\xi,\frac{1}{16})}|f(\lambda)|dA(\lambda)$
$\displaystyle=\frac{2\sqrt{2}C}{\big{(}1-\frac{1}{16}\big{)}^{3}}\cdot\bigg{(}\frac{1-(\frac{1}{16})^{2}|\xi|^{2}}{\frac{1}{16}(1-|\xi|^{2})}\bigg{)}^{3}\int_{D(\xi,\frac{1}{16})}|f(\lambda)|dA(\lambda).$

Note that $|\xi|\leqslant\epsilon<\frac{1}{16}$ and if $\lambda\in
D(\xi,\frac{1}{16})$, then

$|\lambda|<|\xi|+\frac{1}{16}|1-\overline{\lambda}\xi|<\frac{1}{4},$

so we have $D(\xi,\frac{1}{16})\subset D(0,\frac{1}{4})$ and

$|\nabla f(\xi)|\leqslant C_{1}\int_{D(0,\frac{1}{4})}|f(\lambda)|dA(\lambda)$

for all $|\xi|\leqslant\epsilon$, where the constant $C_{1}$ is independent of
$z,\xi$ and $\epsilon$. Therefore,

$|f(z)-f(0)|\leqslant\sup_{|\xi|\leqslant\epsilon}|\nabla
f(\xi)|\cdot|z|\leqslant
C_{1}|z|\int_{D(0,\frac{1}{4})}|f(\lambda)|dA(\lambda)$

for $|z|<\epsilon$ with $\epsilon\in(0,\frac{1}{16})$, this proves the first
part of the lemma.

Let $\varphi_{\xi}$ be the Möbius map, then $f\circ\varphi_{\xi}$ is harmonic
on $\mathbb{D}$. Changing of variable gives that

$\displaystyle\big{|}f(\varphi_{\xi}(z))-f(\xi)\big{|}\leqslant\frac{C_{3}\epsilon}{|D(\xi,\frac{1}{4})|}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|dA(\lambda)$

for some absolute constant $C_{3}>0$. By Cauchy-Schwartz inequality we obtain

$\displaystyle\big{|}f(\varphi_{\xi}(z))-f(\xi)|^{2}$
$\displaystyle\leqslant\frac{C_{3}^{2}\epsilon^{2}}{|D(\xi,\frac{1}{4})|^{2}}\bigg{(}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)\bigg{)}\cdot\bigg{(}\int_{D(\xi,\frac{1}{4})}\frac{1}{\omega(\lambda)}dA(\lambda)\bigg{)}$
$\displaystyle\leqslant\frac{C_{3}^{2}[\omega]_{\mathcal{A}_{2}}\epsilon^{2}}{\big{|}D(\xi,\frac{1}{4})\big{|}_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda).$

Let $w=\varphi_{\xi}(z)$, then $|\varphi_{\xi}(w)|=|z|<\epsilon$, which gives
that

$|f(w)-f(\xi)|^{2}\leqslant\frac{C_{2}\epsilon^{2}}{\big{|}D(\xi,\frac{1}{4})\big{|}_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)$

if $\xi\in D(w,\epsilon)$, as desired. ∎

######  Lemma 2.9.

Let $f$ be a harmonic function and $\epsilon\in(0,\frac{1}{16})$. Let $\mu$ be
the measure defined in Section 2. Then there exists a constant $C>0$
(independent of $\epsilon$) such that

$\int_{\mathbb{D}}\int_{\mathbb{D}}\chi_{\epsilon}(z,\xi)|D(\xi,\epsilon)|_{\omega}^{-1}|f(z)-f(\xi)|^{2}\omega(z)dA(z)d\mu(\xi)\leqslant
C\epsilon^{2}\|f\|_{L^{2}(\omega)}^{2}$

for all $f\in L_{h}^{2}(\omega)$, where

$\chi_{\epsilon}(z,\xi)=\begin{cases}1,&\mathrm{if}\ z\in D(\xi,\epsilon),\\\
0,&\mathrm{otherwise}.\end{cases}$

###### Proof.

By Lemma 2.8, we have

$\chi_{\epsilon}(z,\xi)|f(z)-f(\xi)|^{2}\frac{\omega(z)}{|D(\xi,\epsilon)|_{\omega}}\leqslant\bigg{(}\frac{C\epsilon^{2}}{\big{|}{D(\xi,\frac{1}{4})}\big{|}_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)\bigg{)}\frac{\chi_{\epsilon}(z,\xi)\omega(z)}{|D(\xi,\epsilon)|_{\omega}}.$

Integrating over $z\in\mathbb{D}$ on both sides gives

$\int_{\mathbb{D}}\chi_{\epsilon}(z,\xi)|f(z)-f(\xi)|^{2}\frac{\omega(z)}{|D(\xi,\epsilon)|_{\omega}}dA(z)\leqslant\frac{C\epsilon^{2}}{\big{|}D(\xi,\frac{1}{4})\big{|}_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda).$

Now integrate with respect to $d\mu(\xi)$ to obtain

$\int_{\mathbb{D}}\int_{\mathbb{D}}|f(z)-f(\xi)|^{2}\frac{\chi_{\epsilon}(z,\xi)\omega(z)}{|D(\xi,\epsilon)|_{\omega}}dA(z)d\mu(\xi)\leqslant\int_{\mathbb{D}}\frac{C\epsilon^{2}}{\big{|}D(\xi,\frac{1}{4})\big{|}_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)d\mu(\xi).$

Using Fubini’s Theorem on the right side and noting
$\chi_{D(\xi,\frac{1}{4})}(\lambda)=\chi_{D(\lambda,\frac{1}{4})}(\xi)$, we
have

$\displaystyle\int_{\mathbb{D}}\frac{C\epsilon^{2}}{|D(\xi,\frac{1}{4})|_{\omega}}\int_{D(\xi,\frac{1}{4})}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)d\mu(\xi)=C\epsilon^{2}\int_{\mathbb{D}}\bigg{(}\int_{\mathbb{D}}\frac{\chi_{D(\lambda,\frac{1}{4})}(\xi)}{|D(\xi,\frac{1}{4})|_{\omega}}d\mu(\xi)\bigg{)}|f(\lambda)|^{2}\omega(\lambda)dA(\lambda)$

Since $\lambda\in D(\xi,\frac{1}{4})$, Lemma 2.3 tells us that there is an
absolute constant $C>0$ such that

$\big{|}D(\xi,\frac{1}{4})\big{|}_{\omega}\geqslant
C\big{|}D(\lambda,\frac{1}{4})\big{|}_{\omega}.$

Thus we obatin

$\int_{\mathbb{D}}\frac{\chi_{D(\lambda,\frac{1}{4})}(\xi)}{|D(\xi,\frac{1}{4})|_{\omega}}d\mu(\xi)\leqslant
C^{-1}\frac{\mu(D(\lambda,\frac{1}{4}))}{|D(\lambda,\frac{1}{4})|_{\omega}}.$

By Lemma 2.6, we have

$\mu(D(\lambda,\frac{1}{4}))\cdot\big{|}D(\lambda,\frac{1}{4})\big{|}_{\omega}^{-1}\leqslant
C_{1}$

for some constant $C_{1}>0$ (independent of $\epsilon$), completing the proof.
∎

Now we are ready to prove the reverse Carleson inequality in Proposition 2.7.

###### Proof of Proposition 2.7.

Recall that $\mu$ satisfies the condition:

$\mu(D(a,\frac{1}{4}))\leqslant C\big{|}D(a,\frac{1}{4})\big{|}_{\omega}$

for all $a\in\mathbb{D}$. Applying Lemma 2.9 to $\epsilon\in(0,\frac{1}{16})$
we have

$\bigg{[}\int_{\mathbb{D}}\int_{\mathbb{D}}\frac{\chi_{\epsilon}(z,\xi)}{|D(\xi,\epsilon)|_{\omega}}|f(z)-f(\xi)|^{2}\omega(z)dA(z)d\mu(\xi)\bigg{]}^{\frac{1}{2}}\leqslant
C\epsilon\|f\|_{L^{2}(\omega)}.$

The triangle inequality gives

$I-J\leqslant C\epsilon\|f\|_{L^{2}(\omega)},$

where

$I:=\bigg{[}\int_{\mathbb{D}}\int_{\mathbb{D}}\frac{\chi_{\epsilon}(z,\xi)}{|D(\xi,\epsilon)|_{\omega}}|f(z)|^{2}\omega(z)dA(z)d\mu(\xi)\bigg{]}^{\frac{1}{2}}$

and

$J:=\bigg{[}\int_{\mathbb{D}}\int_{\mathbb{D}}\frac{\chi_{\epsilon}(z,\xi)}{|D(\xi,\epsilon)|_{\omega}}|f(\xi)|^{2}\omega(z)dA(z)d\mu(\xi)\bigg{]}^{\frac{1}{2}}.$

For the first integral $I$, we note that for each $z\in\mathbb{D}$,

$\displaystyle\int_{\mathbb{D}}\frac{\chi_{\epsilon}(z,\xi)}{|D(\xi,\epsilon)|_{\omega}}d\mu(\xi)$
$\displaystyle=\int_{D(z,\epsilon)}\frac{d\mu(\xi)}{|D(\xi,\epsilon)|_{\omega}}\geqslant
C_{1}\frac{\mu(D(z,\epsilon))}{|D(z,\epsilon)|_{\omega}},$

where the $``\geqslant"$ follows from Lemma 2.3 and $C_{1}$ is an absolute
constant.

Since $\mathbb{D}=\bigcup_{n=1}^{\infty}D(a_{n},\epsilon)$, we can select a
disk $D(a_{j},\epsilon)$ such that $z\in D(a_{j},\epsilon)$ for each
$z\in\mathbb{D}$. Applying Lemma 2.3 to get

$\displaystyle\mu(D(z,\epsilon))$
$\displaystyle=\sum_{n=1}^{\infty}\delta_{a_{n}}(D(z,\epsilon))\int_{D(a_{n},\frac{\epsilon}{4})}\omega
dA$
$\displaystyle\geqslant\big{|}D(a_{j},\frac{\epsilon}{4})\big{|}_{\omega}\geqslant
C_{2}|D(a_{j},\epsilon)|_{\omega}\geqslant C_{3}|D(z,\epsilon)|_{\omega},$

where $C_{2}$ and $C_{3}$ are absolute positive constants. Therefore, we have

$\int_{\mathbb{D}}\frac{\chi_{\epsilon}(z,\xi)}{|D(\xi,\epsilon)|_{\omega}}d\mu(\xi)\geqslant\widetilde{C}$

for some absolute constant $\widetilde{C}>0$. These give us that

$I\geqslant\widetilde{C}\|f\|_{L^{2}(\omega)}.$

For the second integral, we observe that

$\int_{\mathbb{D}}\chi_{\epsilon}(z,\xi)\omega(z)dA(z)=|D(\xi,\epsilon)|_{\omega},$

to get

$J=\bigg{(}\int_{\mathbb{D}}|f(\xi)|^{2}d\mu(\xi)\bigg{)}^{\frac{1}{2}}.$

Thus we have

$\widetilde{C}\|f\|_{L^{2}(\omega)}-\bigg{(}\int_{\mathbb{D}}|f(\xi)|^{2}d\mu(\xi)\bigg{)}^{\frac{1}{2}}\leqslant
I-J\leqslant C\epsilon\|f\|_{L^{2}(\omega)}.$

Equivalently,

$(\widetilde{C}-C\epsilon)\|f\|_{L^{2}(\omega)}\leqslant\bigg{(}\int_{\mathbb{D}}|f(\xi)|^{2}d\mu(\xi)\bigg{)}^{\frac{1}{2}}$

for each $0<\epsilon<\frac{1}{16}$.

Since $C,\widetilde{C}$ are both independent of $\epsilon$, we can choose

$0<\epsilon<\min\bigg{\\{}\frac{1}{16},\frac{\widetilde{C}}{2C},r_{0}\bigg{\\}}$

such that

$\displaystyle\|f\|_{L^{2}(\omega)}^{2}\leqslant\frac{1}{(\widetilde{C}-C\epsilon)^{2}}\int_{\mathbb{D}}|f(\xi)|^{2}d\mu(\xi).$

Recall the definition of $\mu$, we conclude that

$\displaystyle\|f\|_{L^{2}(\omega)}^{2}$
$\displaystyle\leqslant\frac{4}{\widetilde{C}^{2}}\sum_{n=1}^{\infty}|f(a_{n})|^{2}|D(a_{n},\frac{\epsilon}{4})|_{\omega}$
$\displaystyle\leqslant\frac{4}{\widetilde{C}^{2}}\sum_{n=1}^{\infty}|f(a_{n})|^{2}|D(a_{n},\epsilon)|_{\omega}$

This completes the proof. ∎

The proof of Theorem 2.4 implies the following result immediately.

######  Theorem 2.10.

Suppose that $\omega$ satisfies the $\mathcal{A}_{2}$ condition. Then there is
an $\epsilon$-lattice $\\{a_{n}\\}_{n=1}^{\infty}\subset\mathbb{D}$ such that
any $f\in L_{h}^{2}(\omega)$ has the following form:

$f(z)=\sum_{n=1}^{\infty}c_{n}K^{\omega}_{a_{n}}(z)|D(a_{n},\epsilon)|_{\omega}^{\frac{1}{2}}$

for some sequence $\\{c_{n}\\}$ in $\ell^{2}(\mathbb{N})$, where
$K^{\omega}_{a_{n}}$ is the reproducing kernel for $L_{h}^{2}(\omega)$.

###### Proof.

We consider the linear map
$\mathscr{S}:L_{h}^{2}(\omega)\rightarrow\ell^{2}(\mathbb{N})$:

$\mathscr{S}f=\bigg{\\{}f(a_{n})|D(a_{n},\epsilon)|_{\omega}^{\frac{1}{2}}\bigg{\\}}_{n=1}^{\infty}.$

Since $L_{h}^{2}(\omega)$ is a Hilbert space, we apply Propositions 2.6 and
2.7 to deduce that $\mathscr{S}^{*}:\ell^{2}(\mathbb{N})\rightarrow
L_{h}^{2}(\omega)$ is surjective and

$\big{\langle}\mathscr{S}^{*}(\\{c_{n}\\}),\
f\big{\rangle}_{L^{2}(\omega)}=\bigg{\langle}\sum_{n=1}^{\infty}c_{n}K^{\omega}_{a_{n}}(z)|D(a_{n},\epsilon)|_{\omega}^{\frac{1}{2}},\
f\bigg{\rangle}_{L^{2}(\omega)}$

for $\\{c_{n}\\}\in\ell^{2}(\mathbb{N})$ and $f\in L^{2}_{h}(\omega)$.
Therefore,

$\mathscr{S}^{*}(\\{c_{n}\\})(z)=\sum_{n=1}^{\infty}c_{n}K^{\omega}_{a_{n}}(z)|D(a_{n},\epsilon)|_{\omega}^{\frac{1}{2}}.$

This completes the proof of this theorem. ∎

##  3\. Boundedness and Compactness of $T_{\nu}$ on $L^{2}_{h}(\omega)$

Recall Toeplitz operator $T_{\nu}$ initially defined on a dense subspace of
$L^{2}_{h}(\omega)$ is given by

$T_{\nu}f(z)=\int_{\mathbb{D}}f(\lambda)\overline{K^{\omega}_{z}(\lambda)}d\nu(\lambda)\
\ \ \big{(}z\in\mathbb{D}\big{)}.$

In this section, we will characterize the boundedness and compactness of
$T_{\nu}$ on $L_{h}^{2}(\omega)$ via Berezin transform and Carleson measure
for $L_{h}^{2}(\omega)$. Firstly, we define the Berezin transform
$\widetilde{\nu}$ of $\nu$ as follows:

$\widetilde{\nu}(z)=\frac{1}{\|R_{z}\|_{L^{2}(\omega)}^{2}}\int_{\mathbb{D}}|R_{z}(\lambda)|^{2}d\nu(\lambda),$

where

$R_{z}(\lambda)=\frac{1}{(1-\overline{\lambda}z)^{2}}+\frac{1}{(1-\overline{z}\lambda)^{2}}-1$

is the reproducing kernel for $L_{h}^{2}$. The first main result of this
section is Theorem 3.1.

######  Theorem 3.1.

Let $\nu$ be a positive finite Borel measure on $\mathbb{D}$ and $\omega$ be
an $\mathcal{A}_{2}$ weight. The following conditions are equivalent:
(1) $T_{\nu}$ extends to a bounded linear operator on $L_{h}^{2}(\omega)$;
(2) $\nu$ is a Carlseon measure for $L_{h}^{2}(\omega)$;
(3) There exist an $0<r\leqslant r_{0}$ and a constant $C>0$ independent of
$z\in\mathbb{D}$ such that

$\nu(D(z,r))\leqslant C|D(z,r)|_{\omega}$

for all $z\in\mathbb{D}$;
(4) Berezin transform $\widetilde{\nu}$ is bounded.

To prove Theorem 3.1, we need the following useful lemma.

######  Lemma 3.2.

Let $\omega\in\mathcal{A}_{2}$. If $0<r\leqslant r_{0}$, then there is a
constant $C=C(r)$ such that

$\frac{|D(\lambda,r)|_{\omega}}{2(1-|\lambda|)^{4}}\leqslant\|R_{\lambda}\|_{L^{2}(\omega)}^{2}\leqslant
C\frac{|D(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}$

for all $\lambda\in\mathbb{D}$.

###### Proof.

Let $\lambda\in\mathbb{D}$. By the remarks below Theorem 2.4, there exists an
$r_{0}\in(0,\frac{1}{4}]$ such that if $0<r\leqslant r_{0}$, then

$\frac{\frac{1}{2}}{(1-|\lambda|)^{2}}\leqslant|R_{\lambda}(z)|\leqslant\frac{3}{(1-|\lambda|)^{2}}$

for all $z\in D(\lambda,r)$. It follows that for each $r\in(0,r_{0}]$ we have

$\displaystyle\|R_{\lambda}\|_{L^{2}(\omega)}^{2}$
$\displaystyle=\int_{\mathbb{D}}|R_{\lambda}(z)|^{2}\omega(z)dA(z)$
$\displaystyle\geqslant\int_{D(\lambda,r)}|R_{\lambda}(z)|^{2}\omega(z)dA(z)$
$\displaystyle\geqslant\frac{1}{4}\int_{D(\lambda,r)}\frac{\omega(z)}{(1-|\lambda|)^{4}}dA(z)$
$\displaystyle=\frac{|D(\lambda,r)|_{\omega}}{4(1-|\lambda|)^{4}}.$

To show the other inequality, observe that

$|z-\lambda|<r(1-|\lambda|)<r|1-\overline{z}\lambda|\ \ \ \
\big{(}z,\lambda\in\mathbb{D}\big{)},$

we have

$S(\lambda,r):=\\{z\in\mathbb{D}:|z-\lambda|<r(1-|\lambda|)\\}\subset
D(\lambda,r).$

Thus, we have by Lemma 2.1 in [6] that

$C_{1}\frac{|S(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}\leqslant\|K_{\lambda}\|_{L^{2}(\omega)}^{2}\leqslant
C_{2}\frac{|S(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}\leqslant
C_{2}\frac{|D(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}$

for some positive constants $C_{1}=C_{1}(r)$ and $C_{2}=C_{2}(r)$, here
$K_{\lambda}(z)=\frac{1}{(1-\overline{\lambda}z)^{2}}$ is the reproducing
kernel for $L_{a}^{2}$ at $\lambda$. Recall that

$R_{\lambda}(z)=2\mathrm{Re}(K_{\lambda}(z))-1,$

we have

$\displaystyle\|R_{\lambda}\|_{L^{2}(\omega)}^{2}\leqslant\frac{4C|D(\lambda,r)|\omega}{(1-|\lambda|)^{4}}+2\|\omega\|_{L^{1}}.$

Consequently, to complete the proof we need only to show that there is a
constant $C_{3}$ depending only on $r$ such that

$\|\omega\|_{L^{1}}\leqslant
C_{3}\frac{|D(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}$

for every $\lambda\in\mathbb{D}.$ Indeed, we may assume that
$\|\omega\|_{L^{1}}=1$. Then it is easy to see that

$C(1-|\lambda|)^{2}\leqslant|D(\lambda,r)|$

for some constant $C=C(r)$. Thus we have

$\displaystyle C(1-|\lambda|)^{2}$
$\displaystyle\leqslant|D(\lambda,r)|=\int_{D(\lambda,r)}\omega(z)^{\frac{1}{2}}\omega(z)^{-\frac{1}{2}}dA(z)$
$\displaystyle\leqslant|D(\lambda,r)|_{\omega}^{\frac{1}{2}}\cdot|D(\lambda,r)|_{\omega^{-1}}^{\frac{1}{2}}$
$\displaystyle\leqslant|D(\lambda,r)|_{\omega}^{\frac{1}{2}}\cdot\|\omega^{-1}\|_{L^{1}}^{\frac{1}{2}}.$

This shows that

$\frac{|D(\lambda,r)|_{\omega}}{(1-|\lambda|)^{4}}\geqslant C_{3},$

to complete the proof of Lemma 3.2. ∎

Now we are ready to prove Theorem 3.1.

###### Proof of Theorem 3.1.

Our strategy is $(4)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1)\Rightarrow(4)$.

$(4)\Rightarrow(3)$: Note that there exists an $r\in(0,r_{0}]$ such that

$\frac{\frac{1}{2}}{(1-|z|)^{2}}\leqslant|R_{z}(\lambda)|\leqslant\frac{3}{(1-|z|)^{2}}$

for $\lambda\in D(z,r)$. From the definition of $\widetilde{\nu}$, there is a
constant $C>0$ such that

$\frac{1}{\|R_{z}\|_{L^{2}(\omega)}^{2}}\int_{D(z,r)}|R_{z}(\lambda)|^{2}d\nu(\lambda)\leqslant\widetilde{\nu}(z)\leqslant
C$

for all $z\in\mathbb{D}.$ Combining these with Lemma 3.2 gives us that

$\nu(D(z,r))\leqslant C|D(z,r)|_{\omega}\ \ \ \big{(}z\in\mathbb{D}\big{)}$

for some positive constant $C=C(r)$.

$(3)\Rightarrow(2)$: This was proved in Proposition 2.6.

$(2)\Rightarrow(1)$: Assume that $\nu$ is a Carleson measure. By Condition
$(2)$ and Lemma 3.2, we obtain Condition $(3)$. Now Proposition 2.6 implies
that $\nu$ is also a Carleson measure for $L_{h}^{1}(\omega)$. Then for $f,g$
are harmonic on a neighborhood of $\overline{\mathbb{D}}$ (by Theorem 2.4,
these functions are dense in $L_{h}^{2}(\omega)$), we have

$\big{|}\langle
T_{\nu}f,g\rangle\big{|}\leqslant\int_{\mathbb{D}}|f(z)g(z)|d\nu(z)\leqslant
C\|fg\|_{L^{1}(\omega)}\leqslant C\|f\|_{L^{2}(\omega)}\|g\|_{L^{2}(\omega)},$

which means that $T_{\nu}$ is bounded.

$(1)\Rightarrow(4)$: Suppose that $T_{\nu}$ is bounded on $L_{h}^{2}(\omega)$.
We consider the partial sum
$\sigma_{N}=\sum_{n=1}^{N}t_{n}K^{\omega}_{a_{n}}$, where $N\geqslant 1$,
$\\{t_{n}\\}$ are complex numbers and $\\{a_{n}\\}\subset\mathbb{D}$. Direct
calculation shows that

$\|\sigma_{N}\|_{L^{2}(\nu)}\leqslant C\|\sigma_{N}\|_{L^{2}(\omega)}$

for some constant $C>0$. This implies that if
$\lim\limits_{N\rightarrow\infty}\|\sigma_{N}-g\|_{L^{2}(\omega)}=0$ for some
$g\in L_{h}^{2}(\omega)$, then

$\lim_{N\rightarrow\infty}\langle f,\sigma_{N}\rangle_{L^{2}(\nu)}=\langle
f,g\rangle_{L^{2}(\nu)}$

for every $f\in L^{2}_{h}(\omega)$. Applying Theorem 2.10 to obtain

$\langle T_{\nu}f,g\rangle_{L^{2}(\omega)}=\langle f,g\rangle_{L^{2}(\nu)}$

for every $f,g\in L_{h}^{2}(\omega)$. In particular, we have

$\langle T_{\nu}R_{z},R_{z}\rangle_{L^{2}(\omega)}=\langle
R_{z},R_{z}\rangle_{L^{2}(\nu)}=\widetilde{\nu}(z)\|R_{z}\|_{L^{2}(\omega)}^{2},$

to get $\widetilde{\nu}(z)\leqslant\|T_{\nu}\|$ for all $z\in\mathbb{D}$. The
proof of Theorem 3.1 is complete now. ∎

From the above theorem, it is natural to characterize the compactness of
$T_{\nu}$ via vanishing Carleson measure. In fact, we will characterize the
compact Toeplitz operators with positive measures as the symbols via not only
vanishing Carlson measure (for the $\mathcal{A}_{2}$ weighted harmonic Bergman
space) but also Berezin transform.

######  Theorem 3.3.

Let $\nu$ be a positive finite Borel measure on $\mathbb{D}$ and
$\omega\in\mathcal{A}_{2}$. The following conditions are equivalent:
(1) $T_{\nu}$ is compact on $L_{h}^{2}(\omega)$;
(2) $\nu$ is a vanishing Carlseon measure for $L_{h}^{2}(\omega)$, i.e.,

$\lim_{|z|\rightarrow 1^{-}}\frac{\nu(D(z,r))}{|D(z,r)|_{\omega}}=0$

for some $r\in(0,1)$;
(3) The Berezin transform $\lim\limits_{|z|\rightarrow
1^{-}}\widetilde{\nu}(z)=0.$

###### Proof.

We will show that $(2)\Rightarrow(1)\Rightarrow(3)\Rightarrow(2)$.

$(2)\Rightarrow(1):$ To prove (1), we need only to show that the inclusion
operator $i:L_{h}^{2}(\omega)\rightarrow L^{2}(\nu)$ is compact, i.e.,

$\lim_{n\rightarrow\infty}\int_{\mathbb{D}}|f_{n}(z)|^{2}d\nu(z)=0$

whenever $\|f_{n}\|_{L^{2}(\omega)}\rightarrow 0\ (n\rightarrow\infty)$, where
$\\{f_{n}\\}_{n=1}^{\infty}$ is a bounded sequence in $L_{h}^{2}(\omega)$
which converges to zero uniformly on each compact subset of $\mathbb{D}$.

From the proof of Proposition 2.6, there exists a positive constant $C=C(r)$
such that

$\displaystyle\int_{\mathbb{D}}|f_{n}(z)|^{2}d\nu(z)$ $\displaystyle\leqslant
C\int_{\mathbb{D}}|D(\xi,r)|_{\omega}^{-1}\nu(D(\xi,r))|f_{n}(\xi)|^{2}\omega(\xi)dA(\xi)$
$\displaystyle=C\bigg{(}\int_{|\xi|\leqslant
s}+\int_{|\xi|>s}\bigg{)}\frac{\nu(D(\xi,r))}{|D(\xi,r)|_{\omega}}|f_{n}(\xi)|^{2}\omega(\xi)dA(\xi),$

where $s\in(0,1)$. Under the assumption in (2), we can choose an
$s_{0}\in(0,1)$ to make the second integral as small as we like; Fix $s_{0}$,
it is easy to show the first integral converges to zero, since
$f_{n}\rightarrow 0\ (n\rightarrow\infty)$ uniformly on compact subsets. This
proves $(2)\Rightarrow(1)$.

$(1)\Rightarrow(3):$ Observe that

$\widetilde{\nu}(z)=\bigg{\langle}T_{\nu}\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}},\
\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}}\bigg{\rangle}_{L^{2}(\omega)}\leqslant\bigg{\|}T_{\nu}\bigg{(}\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}}\bigg{)}\bigg{\|}_{L^{2}(\omega)}.$

So, it sufficient for us to show that
$\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}}$ converges to zero weakly in
$L_{h}^{2}(\omega)$ as $|z|\rightarrow 1^{-}.$ Note that
$\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}}$ is an unit vector in
$L_{h}^{2}(\omega)$, we need only to show it converges to zero uniformly on
compact subsets of $\mathbb{D}$ as $|z|\rightarrow 1^{-}$. Observe that Lemma
3.2 implies that there exists a positive constants $C=C(r_{0})$ such that

$\bigg{|}\frac{R_{z}(\lambda)}{\|R_{z}\|_{L^{2}(\omega)}}\bigg{|}^{2}\leqslant
C|R_{z}(\lambda)|^{2}\int_{D(z,r_{0})}{\omega^{-1}}dA.$

It is clear that $\frac{R_{z}}{\|R_{z}\|_{L^{2}(\omega)}}$ converges to zero
uniformly (as $|z|\rightarrow 1^{-}$) on each disk $|\lambda|\leqslant s<1$,
since $|D(z,r_{0})|\rightarrow 0$ as $|z|\rightarrow 1^{-}$ and
$\omega^{-1}\in L^{1}(dA)$.

$(3)\Rightarrow(2):$ By the definition of $\widetilde{\nu}$ and Lemma 3.2,
there exists a constant $C=C(r_{0})>0$ such that

$\displaystyle\widetilde{\nu}(z)$
$\displaystyle=\frac{1}{\|R_{z}\|_{L^{2}(\omega)}^{2}}\int_{\mathbb{D}}|R_{z}(\lambda)|^{2}d\nu(\lambda)$
$\displaystyle\geqslant
C\frac{(1-|z|)^{4}}{|D(z,r_{0})|_{\omega}}\int_{D(z,r_{0})}|R_{z}(\lambda)|^{2}d\nu(\lambda)$
$\displaystyle\geqslant\frac{C}{4}\frac{\nu(D(z,r_{0}))}{|D(z,r_{0})|_{\omega}}.$

Then the desired result follows, to complete the proof of Theorem 3.3. ∎

In the rest of this section, we will consider the special class of compact
Toeplitz operators to give a characterization of $\nu$ for $T_{\nu}$ to be in
the Schatten class $\mathcal{S}^{p}\ (p\geqslant 1)$. The following theorem is
the third main result in Section 3.

######  Theorem 3.4.

Let $\nu$ be a positive finite Borel measure on $\mathbb{D}$ and
$\omega\in\mathcal{A}_{2}$. Then for $1\leqslant p<\infty$,
$T_{\nu}\in\mathcal{S}^{p}$ if and only if

$\sum_{n=1}^{\infty}\bigg{(}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}\bigg{)}^{p}<+\infty,$

where $\big{(}\\{a_{n}\\}_{n=1}^{\infty},\ \epsilon\big{)}$ is the
$\epsilon$-lattice obtained by Theorem 2.4.

In order to prove the above result, we need one more lemma, which is a
straightforward consequence of Lemmas 2.1 and 3.2.

######  Lemma 3.5.

Let $\omega\in\mathcal{A}_{2}$ and $0<r\leqslant r_{0}$. There exists a
constant $C=C(r)>0$ such that

$C^{-1}\leqslant K^{\omega}_{z}(z)\cdot|D(z,r)|_{\omega}\leqslant C$

for $z\in\mathbb{D}$, where $K^{\omega}_{z}$ is the reproducing kernel of
$L_{h}^{2}(\omega)$.

###### Proof.

Note that
$K^{\omega}_{\lambda}(\lambda)=\|K^{\omega}_{\lambda}\|_{L^{2}(\omega)}^{2}$
for each $\lambda\in\mathbb{D}$. Applying Lemma 2.1 to the function
$K^{\omega}_{\lambda}(z)$, we get a constant $C$ depends only on $r$ such that

$|K^{\omega}_{\lambda}(z)|^{2}\leqslant\frac{C\|K^{\omega}_{\lambda}\|_{L^{2}(\omega)}^{2}}{|D(z,r)|_{\omega}}=\frac{CK^{\omega}_{\lambda}(\lambda)}{|D(z,r)|_{\omega}}.$

Taking $\lambda=z$ to get the inequality on the right hand side in Lemma 3.5.

For the reverse inequality, note that for each $z\in\mathbb{D}$ we have

$\displaystyle\frac{1}{(1-|z|)^{2}}$
$\displaystyle\leqslant\frac{2}{(1-|z|^{2})^{2}}-1$
$\displaystyle=R_{z}(z)=\big{\langle}R_{z},K^{\omega}_{z}\big{\rangle}_{L^{2}(\omega)}$
$\displaystyle\leqslant\|R_{z}\|_{L^{2}(\omega)}\cdot\|K^{\omega}_{z}\|_{L^{2}(\omega)}$
$\displaystyle\leqslant
C\frac{|D(z,r)|_{\omega}^{\frac{1}{2}}}{(1-|z|)^{2}}\cdot\sqrt{K^{\omega}_{z}(z)},$

where the constant $C$ comes from Lemma 3.2. This finishes the proof. ∎

We are ready to prove Theorem 3.4. The method of its proof is quite standard.

###### Proof of Theorem 3.4.

Suppose that the series converges. We consider the $\epsilon$-lattice
$\\{a_{n}\\}_{n=1}^{\infty}$ given by Theorem 2.4, recall that
$\epsilon<r_{0}$ (see the proof of Proposition 2.7). For an arbitrary
orthonormal basis $\\{e_{n}\\}_{n=1}^{\infty}$ of $L_{h}^{2}(\omega)$, we have

$\displaystyle\sum_{n=1}^{\infty}\langle T_{\nu}e_{n},e_{n}\rangle$
$\displaystyle=\sum_{n=1}^{\infty}\int_{\mathbb{D}}|e_{n}(z)|^{2}d\nu(z)=\int_{\mathbb{D}}K^{\omega}_{z}(z)d\nu(z)$
$\displaystyle\leqslant\sum_{n=1}^{\infty}\int_{D(a_{n},\epsilon)}K^{\omega}_{z}(z)d\nu(z)$
$\displaystyle\leqslant
C\sum_{n=1}^{\infty}\int_{D(a_{n},\epsilon)}|D(z,\epsilon)|_{\omega}^{-1}d\nu(z),$

the constant $C$ comes from Lemma 3.5, which depends only on $\epsilon$. Note
that $\rho(z,a_{n})<\epsilon$ for every $n\geqslant 1$, by Lemma 2.3 and its
proof we can choose a constant $C_{1}=C_{1}(\epsilon)$ such that

$\sum_{n=1}^{\infty}\langle T_{\nu}e_{n},e_{n}\rangle\leqslant
C_{1}\sum_{n=1}^{\infty}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}.$

This shows that $T_{\nu}\in\mathcal{S}^{1}$ provided
$\sum_{n=1}^{\infty}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}<+\infty$.

On the other hand, if $\sup\limits_{n\geqslant
1}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}<+\infty$, then
by the proof Theorem 7.4 in [24] (or the proof of $(3)\Rightarrow(2)$ in
Theorem 3.1), we deduce that $T_{\nu}$ is bounded on $L_{h}^{2}(\omega)$,
i.e., $T_{\nu}\in\mathcal{S}^{\infty}$. Now applying the interpolation theorem
for the Schatten classes (see Theorem 2.6 in [24] if needed), we obtain that
$T_{\nu}\in\mathcal{S}^{p}$ for each $p\in(1,+\infty)$ if

$\sum_{n=1}^{\infty}\bigg{(}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}\bigg{)}^{p}<+\infty.$

Conversely, we assume that $T_{\nu}\in\mathcal{S}^{p}$ for $1\leqslant
p<\infty$. We recall by Theorem 2.4 that each $f\in L_{h}^{2}(\omega)$ has the
following form:

$\displaystyle f(z)$
$\displaystyle=\sum_{n=1}^{\infty}c_{n}(1-|a_{n}|)^{2}R_{a_{n}}(z)|D(a_{n},\epsilon)|_{\omega}^{-\frac{1}{2}}$
$\displaystyle:=\sum_{n=1}^{\infty}c_{n}h_{n}(z),$

where $\\{c_{n}\\}_{n=1}^{\infty}\in\ell^{2}(\mathbb{N})$. Cauchy-Schwartz
inequality shows that

$\|f\|_{L^{2}(\omega)}^{2}\leqslant\bigg{(}\sum_{n=1}^{\infty}|c_{n}|^{2}\bigg{)}\cdot\bigg{(}\sum_{n=1}^{\infty}\frac{(1-|a_{n}|)^{4}}{|D(a_{n},\epsilon)|_{\omega}}\|R_{a_{n}}\|_{L^{2}(\omega)}^{2}\bigg{)}.$

Since $\epsilon<r_{0}$, we have by Lemma 3.2 that

$\|f\|_{L^{2}(\omega)}^{2}\leqslant C\sum_{n=1}^{\infty}|c_{n}|^{2}$

where the constant $C>0$ depends only on $\epsilon$.

Fix an orthonormal basis $\\{e_{n}\\}_{n=1}^{\infty}$ for $L_{h}^{2}(\omega)$
and define a linear operator $\mathscr{A}$ on $L_{h}^{2}(\omega)$ by

$\mathscr{A}\bigg{(}\sum_{n=1}^{\infty}c_{n}e_{n}\bigg{)}=\sum_{n=1}^{\infty}c_{n}h_{n}.$

Then $\mathscr{A}$ is a bounded surjective linear operator on
$L_{h}^{2}(\omega)$. Thus the $\mathscr{A}^{*}$ is well-defined on
$L_{h}^{2}(\omega)$ and $\mathscr{A}^{*}T_{\nu}\mathscr{A}\in\mathcal{S}^{p}$,
so that

$\sum_{n=1}^{\infty}\big{|}\langle\mathscr{A}^{*}T_{\nu}\mathscr{A}e_{n},e_{n}\rangle_{L^{2}(\omega)}\big{|}^{p}<+\infty.$

On the other hand,

$\displaystyle\sum_{n=1}^{\infty}\big{|}\langle\mathscr{A}^{*}T_{\nu}\mathscr{A}e_{n},e_{n}\rangle_{L^{2}(\omega)}\big{|}^{p}$
$\displaystyle=\sum_{n=1}^{\infty}\big{|}\langle
T_{\nu}\mathscr{A}e_{n},\mathscr{A}e_{n}\rangle_{L^{2}(\omega)}\big{|}^{p}$
$\displaystyle=\sum_{n=1}^{\infty}\big{|}\langle
T_{\nu}h_{n},h_{n}\rangle_{L^{2}(\omega)}\big{|}^{p}\ \ \ \big{(}\mathrm{by\
the\ definition\ of}\ \mathscr{A}\big{)}$
$\displaystyle=\sum_{n=1}^{\infty}\bigg{(}\int_{\mathbb{D}}|h_{n}(z)|^{2}d\nu(z)\bigg{)}^{p}$
$\displaystyle\geqslant\sum_{n=1}^{\infty}\bigg{(}\int_{D(a_{n},\epsilon)}|h_{n}(z)|^{2}d\nu(z)\bigg{)}^{p}.$

Recall

$|h_{n}(z)|^{2}=\frac{(1-|a_{n}|)^{4}|R_{a_{n}}(z)|^{2}}{|D(a_{n},\epsilon)|_{\omega}},$

we have by Lemma 3.2 that

$|h_{n}(z)|^{2}\geqslant\frac{1}{4|D(a_{n},\epsilon)|_{\omega}}$

if $\rho(z,a_{n})<\epsilon<r_{0}$ for each $n\geqslant 1$. Therefore,

$\displaystyle\sum_{n=1}^{\infty}\big{|}\langle\mathscr{A}^{*}T_{\nu}\mathscr{A}e_{n},e_{n}\rangle_{L^{2}(\omega)}\big{|}^{p}\geqslant
4^{-p}\sum_{n=1}^{\infty}\bigg{(}\frac{\nu(D(a_{n},\epsilon))}{|D(a_{n},\epsilon)|_{\omega}}\bigg{)}^{p}.$

This completes the proof of Theorem 3.4. ∎

##  4\. Invertibility of Toeplitz operators on $L_{h}^{2}(\omega_{\alpha})$

A fundamental and interesting problem is to determine when a Toeplitz operator
is invertible on the Hardy or Bergman space. In this section, we study the
invertibility problem of Toeplitz operators on the standard weighted harmonic
Bergman space $L_{h}^{2}(\omega_{\alpha})$ with
$\omega_{\alpha}=(1+\alpha)(1-|z|^{2})^{\alpha},\alpha>-1$. Recall that the
reproducing kernel for $L_{h}^{2}(\omega_{\alpha})$ is given by

$R^{\alpha}_{z}(\lambda)=K^{\alpha}_{z}(\lambda)+\overline{K^{\alpha}_{z}(\lambda)}-1\
\ \ \big{(}z,\lambda\in\mathbb{D}\big{)},$

where

$K^{\alpha}_{z}(\lambda)=\frac{1}{(1-\overline{z}\lambda)^{2+\alpha}}$

is the reproducing kernel for the weighted Bergman space
$L_{a}^{2}(\omega_{\alpha})$, see [21] if needed.

For the (unweighted) Bergman space setting, the second author and Zheng
provided a necessary and sufficient condition for the Toeplitz operators with
nonnegative symbols to be invertible on $L_{a}^{2}$ ([23]). The main tool used
in [23] is Luecking’s result on the reverse Carleson measure for Bergman space
([11]), which also holds for the harmonic Bergman space. More precisely,
Luecking established the following result.

######  Lemma 4.1.

([12]) Suppose that $G$ is a measurable subset of $\mathbb{D}$. Then the
following are equivalent:
(i) There exists a $\delta\in(0,1)$ such that

$|G\cap K|\geqslant\delta|\mathbb{D}\cap K|$

for every ball $K$ whose center lies on $\partial\mathbb{D}$;
(ii) $\chi_{G}dA$ is a reverse Carleson measure for
$L_{h}^{2}(\omega_{\alpha})$. That is, there is a constant $C>0$ such that

$\int_{\mathbb{D}}|f(z)|^{2}\omega_{\alpha}(z)dA(z)\leqslant
C\int_{G}|f(z)|^{2}\omega_{\alpha}(z)dA(z)$

for all $f\in L_{h}^{2}(\omega_{\alpha})$.

Motivated by the ideas and techniques used in [23], we are able to
characterize the invertiblility of Toeplitz operator $T_{\varphi}$
($\varphi\geqslant 0$) on $L_{h}^{2}(\omega_{\alpha})$ in terms of the reverse
Carleson measure and Berezin transform.

######  Theorem 4.2.

Let $\varphi\geqslant 0$ be in $L^{\infty}(\omega_{\alpha})$. The following
conditions are equivalent:
$(1)$ The Toeplitz operator $T_{\varphi}$ is invertible on
$L_{h}^{2}(\omega_{\alpha})$;
$(2)$ The Berezin transform $\widetilde{\varphi}$ is invertible in
$L^{\infty}(\omega_{\alpha})$, where

$\widetilde{\varphi}(z):=\frac{1}{\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\int_{\mathbb{D}}\varphi(\lambda)|R_{\alpha}(z,\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$

and

$\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}=R^{\alpha}_{z}(z)=\frac{2}{(1-|z|^{2})^{2+\alpha}}-1;$

$(3)$ There exists $r>0$ such that

$G:=\\{z\in\mathbb{D}:\varphi(z)>r\\},$

$\chi_{G}dA$ is a reverse Carleson measure for $L_{h}^{2}(\omega_{\alpha})$;
$(4)$ There exists a constant $C>0$ such that

$\int_{\mathbb{D}}|f(z)|^{2}\varphi(z)\omega_{\alpha}(z)dA(z)\geqslant
C\int_{\mathbb{D}}|f(z)|^{2}\omega_{\alpha}(z)dA(z)$

for $f\in L_{h}^{2}(\omega_{\alpha})$.

Before giving the proof the main theorem of this section, we need another
lemma, which was proved in [12], [16] and [22].

######  Lemma 4.3.

Suppose that the ball $K$ has radius $0<t<1$ and center
$u=(1,0)\in\mathbb{R}^{2}$. Let $f$ be the harmonic function

$f(\lambda)=f_{s}(\lambda):=\sqrt{1+\alpha}R^{\alpha}_{z_{0}}(\lambda)(1-|z_{0}|^{2})^{\frac{2+\alpha}{2}},$

where $z_{0}=(1-st)u$, $0<s<1$. Then for each $\epsilon>0$, there exist
$s=s(\epsilon)$ and a positive constant $C=C(\epsilon)$ (independent of $K$)
such that

$\int_{B\setminus
K}|f(\lambda)|^{2}(1-|\lambda|)^{\alpha}dA(\lambda)<\epsilon$

and

$\int_{G\cap K}|f(\lambda)|^{2}(1-|\lambda|)^{\alpha}dA(\lambda)\leqslant
C\bigg{(}\frac{|G\cap K|}{|\mathbb{D}\cap K|}\bigg{)}^{\beta},$

where

$\beta=\begin{cases}1,&\mathrm{if}\ 0\leqslant\alpha<\infty,\\\
1-\frac{1}{\gamma},&\mathrm{if}\ -1<\alpha<0,\end{cases}$

$\gamma$ is a number in $(1,-\frac{1}{\alpha})$ if $-1<\alpha<0$.

Now we are ready to prove Theorem 4.2.

###### Proof of Theorem 4.2.

We will show that
$(1)\Rightarrow(2)\Rightarrow(3)\Rightarrow(4)\Rightarrow(1)$. Without loss of
generality, we may assume that $0\leqslant\varphi\leqslant 1$.

$(1)\Rightarrow(2)$: This is trivial.

$(2)\Rightarrow(3)$: Suppose that $\widetilde{\varphi}$ is bounded below by
some positive constant $\delta$. By Lemma 4.1, it sufficient to show that
there exists a $\delta^{\prime}\in(0,1)$ such that

$|G\cap K|\geqslant\delta^{\prime}|\mathbb{D}\cap K|$

for all balls $K$ whose centers lie on $\partial\mathbb{D}$.

Since $\omega_{\alpha}dA$ is a rotation invariant measure, it is no loss of
generality to assume that $K$ has its center at the point $(1,0)$. It is also
clear that we need only to prove the inequality for sufficient small radius
$t$, say $t<1$.

Now we consider the subset
$G=\\{\lambda\in\mathbb{D}:\varphi(\lambda)>\frac{\delta}{4}\\}$. For each
$z\in\mathbb{D}$,

$\displaystyle\delta$
$\displaystyle\leqslant\widetilde{\varphi}(z)=\frac{1}{\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\int_{\mathbb{D}}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle=\frac{1}{\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\bigg{(}\int_{G}+\int_{\mathbb{D}\setminus
G}\bigg{)}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle\leqslant(1-|z|^{2})^{2+\alpha}\int_{G}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)+\frac{\delta}{4}.$

Let $L_{z}$ be the following integral:

$\displaystyle
L_{z}=\frac{1}{\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\int_{G\cap
K}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda).$

Then for each $z\in\mathbb{D}$, we have

$\displaystyle L_{z}$
$\displaystyle=\frac{1}{\|R^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\bigg{(}\int_{G}-\int_{G\backslash
K}\bigg{)}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle\geqslant\frac{1}{2}(1-|z|^{2})^{2+\alpha}\bigg{(}\int_{G}-\int_{G\backslash
K}\bigg{)}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle\geqslant\frac{3\delta}{8}-\frac{1}{2}(1-|z|^{2})^{2+\alpha}\int_{G\backslash
K}\varphi(\lambda)|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle\geqslant\frac{3\delta}{8}-\frac{1}{2}(1-|z|^{2})^{2+\alpha}\int_{G\backslash
K}|R^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)\ \ \
\big{(}\mathrm{using}\ 0\leqslant\varphi\leqslant 1\big{)}.$

For the $\delta$ above, Lemma 4.3 guarantees that we can select
$z_{0}\in\mathbb{D}$ to define a function $f$ (as the one in Lemma 4.3)
satisfies the following three inequalities:

$\displaystyle L_{z_{0}}$
$\displaystyle\geqslant\frac{3\delta}{8}-\frac{1}{2}(1-|z_{0}|^{2})^{2+\alpha}\int_{G\backslash
K}|R^{\alpha}_{z_{0}}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle=\frac{3\delta}{8}-\frac{1}{2}\int_{G\backslash
K}|f(\lambda)|^{2}(1-|\lambda|^{2})^{\alpha}dA(\lambda),$ $\int_{G\backslash
K}|f(\lambda)|^{2}(1-|\lambda|^{2})^{\alpha}dA(\lambda)<\frac{\delta}{4}$

and

$\int_{G\cap K}|f(\lambda)|^{2}(1-|\lambda|)^{\alpha}dA(\lambda)\leqslant
C\bigg{(}\frac{|G\cap K|}{|\mathbb{D}\cap K|}\bigg{)}^{\beta},$

where the constant $C$ depends only on $\delta$ and $\alpha$. Therefore,

$\displaystyle\frac{\delta}{4}$ $\displaystyle\leqslant
L_{z_{0}}\leqslant(1-|z_{0}|^{2})^{2+\alpha}\int_{K\cap
G}|R^{\alpha}_{z_{0}}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda)$
$\displaystyle=\int_{G\cap
K}|f(\lambda)|^{2}(1-|\lambda|^{2})^{\alpha}dA(\lambda)$
$\displaystyle\leqslant C\bigg{(}\frac{|G\cap K|}{|\mathbb{D}\cap
K|}\bigg{)}^{\beta}.$

Now we get $(3)$ by Lemma 4.1.

$(3)\Rightarrow(4)$: Observer that

$\displaystyle\int_{\mathbb{D}}|f(z)|^{2}\varphi(z)\omega_{\alpha}(z)dA(z)$
$\displaystyle\geqslant\int_{G}\varphi(z)|f(z)|^{2}\omega_{\alpha}(z)dA(z)$
$\displaystyle>r\int_{G}|f(z)|^{2}\omega_{\alpha}(z)dA(z)$
$\displaystyle\geqslant\frac{r}{C}\int_{\mathbb{D}}|f(z)|^{2}\omega_{\alpha}(z)dA(z),$

the last inequality follows from the definition of the reverse Carleson
measure.

$(4)\Rightarrow(1)$: Use the same arguments as the proof of Corollary 3 in
[11], we obtain that $\|I-T_{\varphi}\|<1$, which implies that $T_{\varphi}$
is invertible on $L_{h}^{2}(\omega_{\alpha})$. This completes the whole proof
of the theorem. ∎

Let $\mathbb{T}_{\varphi}$ denote the Toeplitz operator with symbol $\varphi$
on the weighted Bergman space $L_{a}^{2}(\omega_{\alpha})$. Combining the main
result in [11] and the techniques used in the proof of Theorem 4.2, we can
generalize Theorem 3.2 in [23] to the case of standard weighted Bergman space.

######  Theorem 4.4.

Let $\varphi\geqslant 0$ be in $L^{\infty}(\omega_{\alpha})$. Then the
following are equivalent:
$(1)$ The Toeplitz operator $\mathbb{T}_{\varphi}$ is invertible on
$L_{a}^{2}(\omega_{\alpha})$;
$(2)$ The Berezin transform $\widehat{\varphi}$ is invertible in
$L^{\infty}(\omega_{\alpha})$, where

$\widehat{\varphi}(z):=\frac{1}{\|K^{\alpha}_{z}\|^{2}_{L^{2}(\omega_{\alpha})}}\int_{\mathbb{D}}\varphi(\lambda)|K^{\alpha}_{z}(\lambda)|^{2}\omega_{\alpha}(\lambda)dA(\lambda);$

$(3)$ There exists $r>0$ such that

$G:=\\{z\in\mathbb{D}:\varphi(z)>r\\},$

$\chi_{G}dA$ is a reverse Carleson measure for $L_{a}^{2}(\omega_{\alpha})$;
$(4)$ There exists a $\delta\in(0,1)$ such that

$|G\cap K|\geqslant\delta|\mathbb{D}\cap K|$

for every ball $K$ whose center lies on $\partial\mathbb{D}$;
$(5)$ There exists a constant $C>0$ such that

$\int_{\mathbb{D}}|f(z)|^{2}\varphi(z)\omega_{\alpha}(z)dA(z)\geqslant
C\int_{\mathbb{D}}|f(z)|^{2}\omega_{\alpha}(z)dA(z)$

for $f\in L_{a}^{2}(\omega_{\alpha})$.

###### Proof.

From the proof of Theorem 4.2, it sufficient to show that one can replace the
harmonic function $f$ defined in Lemma 4.3 by a suitable analytic function
$g$. Indeed, we construct the desired function $g$ as follows. Suppose that
$K$ has radius $0<t<1$ and center $u=(1,0)\in\mathbb{R}^{2}$. Define

$g(\lambda)=\sqrt{\alpha+1}K^{\alpha}_{z_{0}}(\lambda)(1-|z_{0}|^{2})^{\frac{2+\alpha}{2}},$

where $z_{0}=(1-st)u$, $0<s<1$. Then it is not hard to check that the two
inequalities in Lemma 4.3 both hold for $g$. Now the rest of the proof
parallels exactly one given in Theorem 4.2. ∎

Based on the theorem above, we can establish a relationship of the
invertibility between Toeplitz operators with nonnegative symbols on
$L_{a}^{2}(\omega_{\alpha})$ and $L_{h}^{2}(\omega_{\alpha})$.

######  Corollary 4.5.

Let $\varphi$ be a nonnegative bounded measurable function on $\mathbb{D}$.
The following four conditions are equivalent:
$(1)$ $\mathbb{T}_{\varphi}$ is invertible on $L_{a}^{2}(\omega_{\alpha})$;
$(2)$ $T_{\varphi}$ is invertible on $L_{h}^{2}(\omega_{\alpha})$;
$(3)$ $\widehat{\varphi}$ is invertible in $L^{\infty}(\omega_{\alpha})$;
$(4)$ $\widetilde{\varphi}$ is invertible in $L^{\infty}(\omega_{\alpha})$.

To end this section, we study Toeplitz operators with bounded analytic symbols
on $L_{h}^{2}(\omega_{\alpha})$. Unlike the Bergman space
$L_{a}^{2}(\omega_{\alpha})$ setting, the spectrum of analytic Toeplitz
operator $T_{\varphi}$ on $L_{h}^{2}(\omega_{\alpha})$ is not equal to the
closure of $\varphi(\mathbb{D})$. Let $H^{\infty}$ denote space of analytic
functions in $L^{\infty}(\omega_{\alpha})$, we have the following result.

######  Theorem 4.6.

Suppose that $\varphi\in H^{\infty}$. If $\varphi$ is invertible, then the
Toeplitz operator $T_{\varphi}$ is invertible on $L_{h}^{2}(\omega_{\alpha})$.
However, the converse is not true.

Let $\varphi\in H^{\infty}$ be invertible. However,
$T_{\varphi}T_{1/\varphi}=T_{1/\varphi}T_{\varphi}$ does not hold on the
harmonic Bergman space in the general case (see Theorem 5 in [3]). For this
reason, we need to find some relationships between $T_{\varphi}$ and
$\mathbb{T}_{\varphi}$ to study the invertibility problem. To do so, let us
introduce some notations first.

For $\varphi\in L^{\infty}(\omega_{\alpha})$, we define

$\varphi^{\star}(z)=\varphi(\overline{z})\ \ \mathrm{and}\ \
\varphi^{*}(z)=\overline{\varphi(\overline{z})}.$

Next, we define the unitary operator
$W:L_{h}^{2}(\omega_{\alpha})=zL_{a}^{2}(\omega_{\alpha})\oplus\overline{L_{a}^{2}(\omega_{\alpha})}\rightarrow
zL_{a}^{2}(\omega_{\alpha})\oplus L_{a}^{2}(\omega_{\alpha})$ by

$W=\left({\begin{array}[]{*{20}{c}}{{I}}&{{O}}\\\ {{O}}&{{U}}\\\
\end{array}}\right),$

where $U$ is a unitary operator on $L^{2}(\omega_{\alpha})$, defined by
$Uf(z)=f(\overline{z})$.

It is clear that $W^{*}$ maps $zL_{a}^{2}(\omega_{\alpha})\oplus
L_{a}^{2}(\omega_{\alpha})$ to $L_{h}^{2}(\omega_{\alpha})$ and

$W^{*}=\left({\begin{array}[]{*{20}{c}}{{I}}&{{O}}\\\ {{O}}&{{U}}\\\
\end{array}}\right).$

For $f$ and $g$ in $L^{2}(\omega_{\alpha})$, let $f\otimes g$ be the rank one
operator defined by

$(f\otimes g)h=\langle h,g\rangle_{L^{2}(\omega_{\alpha})}f$

for $h\in L^{2}(\omega_{\alpha})$.

Let $P$ denote the orthogonal projection from $L^{2}(\omega_{\alpha})$ to
$L_{a}^{2}(\omega_{\alpha})$, we define the Hankel operator with symbol
$\varphi$ acting on $L_{a}^{2}(\omega_{\alpha})$ by
$\mathbb{H}_{\varphi}f=P(\varphi Uf)$. Using the notations above, we obtain
the following matrix representation of $WT_{\varphi}W^{*}$, which will be used
to study the invertibility problem.

######  Lemma 4.7.

Let $\varphi\in L^{\infty}(\omega_{\alpha})$. On the space
$zL_{a}^{2}(\omega_{\alpha})\oplus L_{a}^{2}(\omega_{\alpha})$ we have

$WT_{\varphi}W^{*}=\left({\begin{array}[]{*{20}{c}}{{\mathbb{T}_{\varphi}-(1\otimes\overline{\varphi})}}&{{\mathbb{H}_{\varphi}-(1\otimes\varphi^{*})}}\\\
{{\mathbb{H}_{\varphi^{\star}}}}&{{\mathbb{T}_{\varphi^{\star}}}}\\\
\end{array}}\right).$

###### Proof.

The proof is exactly the same as the proof of Theorem 2.1 in [8], we omit the
details here. ∎

We are now ready to prove Theorem 4.6.

###### Proof of Theorem 4.6..

For $\varphi\in H^{\infty}$ and $f\in zL_{a}^{2}(\omega_{\alpha})$ we have

$\mathbb{H}_{\varphi^{\star}}f=P(\varphi^{\star}Uf)=P(\varphi(\overline{z})f(\overline{z}))=0$

and

$(1\otimes\overline{\varphi})f=\langle
f,\overline{\varphi}\rangle_{L^{2}(\omega_{\alpha})}1=0,$

since $f(0)=0$.

Consequently, the matrix representation of $WT_{\varphi}W^{*}$ with
$\varphi\in H^{\infty}$ is given by

$WT_{\varphi}W^{*}=\left({\begin{array}[]{*{20}{c}}{{\mathbb{T}_{\varphi}}}&{{\mathbb{H}_{\varphi}-(1\otimes\varphi^{*})}}\\\
{{O}}&{{\mathbb{T}_{\varphi^{\star}}}}\\\ \end{array}}\right).$

Note that $\varphi\in H^{\infty}$ is invertible implies that $\varphi^{\star}$
is also invertible, so that $\mathbb{T}_{\varphi^{\star}}$ is invertible on
$L_{a}^{2}(\omega_{\alpha})$.

On the other hand, $\mathbb{T}_{\varphi}$ is invertible on
$zL_{a}^{2}(\omega_{\alpha})$ follows from $\varphi$ is invertible. The above
matrix representation tells us that $WT_{\varphi}W^{*}$ is invertible on
$zL_{a}^{2}(\omega_{\alpha})\oplus L_{a}^{2}(\omega_{\alpha})$. Thus
$T_{\varphi}$ is invertible on $L_{h}^{2}(\omega_{\alpha})$.

To show there exists a function $\varphi$ in $H^{\infty}$ such that $\varphi$
is not invertible in $L^{\infty}(\omega_{\alpha})$ but $T_{\varphi}$ is
invertible on $L_{h}^{2}(\omega_{\alpha})$, we consider the function
$\varphi(z)=z.$ It is easy to show that

$\mathrm{Ker}(T_{z})=\mathrm{Ker}(T^{*}_{z})=\mathrm{Ker}(T_{\overline{z}})=\\{0\\}.$

Observe that $T_{z}$ is Fredholm, we conclude that $T_{z}$ is invertible on
$L_{h}^{2}(\omega_{\alpha})$. This finishes the proof. ∎

##  5\. A Reverse Carleson type Inequality for $L_{h}^{2}(\omega)$

In the preceding section, we study the invertibility problem of Toeplitz
operators via the reverse Carleson measures for standard weighted harmonic
Bergman spaces. In this section, we establish a sufficient condition for
$\chi_{G}dA$ to be a reverse Carleson measure for the space
$L_{h}^{2}(\omega)$, where $\omega\in\mathcal{A}_{2}$ and $G$ is a measurable
set in $\mathbb{D}$.

For $a\in\mathbb{D}$, $0<r<1$, recall that

$S(a,r)=\\{z\in\mathbb{D}:|z-a|<r(1-|a|)\\}.$

The main result in this section is Theorem 5.1, which is a harmonic version of
Theorem 3.9 in [13].

######  Theorem 5.1.

Suppose that $G\subset\mathbb{D}$ and $\omega$ satisfies $\mathcal{A}_{2}$
condition. If there exist $\delta\in(0,1)$ and $r\in(0,1)$ such that for all
$a\in\mathbb{D}$

$|G\cap S(a,r)|\geqslant\delta|S(a,r)|,$

then there exists a positive constant $C=C(r,\delta)$ such that

$\int_{\mathbb{D}}|f(z)|^{2}\omega(z)dA(z)\leqslant
C\int_{G}|f(z)|^{2}\omega(z)dA(z)$

for all $f\in L^{2}_{h}(\omega)$.

To prove the above theorem, we will adopt some ideas and techniques in [13].
Firstly, we need to introduce a new (weight) function $\omega^{*}$ and discuss
some properties of $\omega^{*}$. In the rest of this section, we use “$r$” and
“$\delta$” to denote the numbers provided in Theorem 5.1.

Now we define a positive function $\omega^{*}$ on the open unit disk as
follows:

$\omega^{*}(z)=\omega_{r}^{*}(z):=\frac{|S(z,r)|_{\omega}}{|S(z,r)|}.$

It is clear that $\omega^{*}\in L^{1}(dA)$, and so $\omega^{*}$ is a weight.
Moreover, $\omega^{*}$ has the following important property.

######  Lemma 5.2.

Let $z\in\mathbb{D}$, there exist constants $C_{1}$ and $C_{2}$ depending only
on $r$ such that

$C_{1}\omega^{*}(a)\leqslant\omega^{*}(z)\leqslant C_{2}\omega^{*}(a)$

for all $a\in S(z,r)$. Consequently, we have

$\int_{\mathbb{D}}\frac{\omega^{*}(a)}{|S(a,r)|}\chi_{S(a,r)}(z)dA(a)\leqslant
C_{3}\omega^{*}(z)\ \ \ \big{(}z\in\mathbb{D}\big{)},$

where $C_{3}=C_{3}(r)$ is a constant.

###### Proof.

By Lemma 2.2 in [6], there exists a positive constant $C$ depending only on
$r$ such that

$C^{-1}|S(a,r)|_{\omega}\leqslant|S(z,r)|_{\omega}\leqslant
C|S(a,r)|_{\omega}$

Moreover, it is well known that $|S(z,r)|$ is equivalent to $|S(a,r)|$ (with
constants independent of $a$ and $z$) if $a\in S(z,r)$. This gives the first
conclusion of the lemma. Based on this result, we have

$\displaystyle\int_{\mathbb{D}}\frac{\omega^{*}(a)}{|S(a,r)|}\chi_{S(a,r)}(z)dA(a)$
$\displaystyle\leqslant
C\int_{\mathbb{D}}\frac{\omega^{*}(z)}{|S(a,r)|}\chi_{S(a,r)}(z)dA(a)\ \ \
\big{(}\mathrm{using}\ z\in S(a,r)\big{)}$ $\displaystyle\leqslant
C\int_{\mathbb{D}}\frac{\omega^{*}(z)}{|S(a,r)|}\chi_{D(a,r)}(z)dA(a)\ \ \
\big{(}\mathrm{since}\ S(a,r)\subset D(a,r)\big{)}$
$\displaystyle=C\int_{\mathbb{D}}\frac{\omega^{*}(z)}{|S(a,r)|}\chi_{D(z,r)}(a)dA(a)$
$\displaystyle=C\int_{D(z,r)}\frac{\omega^{*}(z)}{|S(a,r)|}dA(a)$
$\displaystyle\leqslant C_{3}\omega^{*}(z),$

where $C_{3}$ is a positive constant depending only on $r$, as desired. ∎

Another property of $\omega^{*}$ is given by the following inequality, which
will be used to estimate the integral of $|f|^{2}\omega$ over the subset $G$.

######  Lemma 5.3.

Let $\omega$ be an $\mathcal{A}_{2}$ weight. Then there exists a constant
$C>0$ depending only on $r$ such that

$\|f\|_{L^{2}(\omega)}^{2}\leqslant C\|f\|_{L^{2}(\omega^{*})}^{2}$

for all $f\in L_{h}^{2}(\omega)$.

###### Proof.

By definitions, we have

$\displaystyle\|f\|_{L^{2}(\omega^{*})}^{2}$
$\displaystyle=\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)$
$\displaystyle=\int_{\mathbb{D}}\omega(\xi)\bigg{(}\int_{\mathbb{D}}|f(z)|^{2}\frac{\chi_{S(z,r)}(\xi)}{|S(z,r)|}dA(z)\bigg{)}dA(\xi),$

We next deal with the bracketed expression. Observe

$S\Big{(}\xi,\frac{r}{2(1+r)}\Big{)}\subset\big{\\{}z\in\mathbb{D}:|z-\xi|<r(1-|z|)\big{\\}},$

we obtain

$\displaystyle\int_{\mathbb{D}}|f(z)|^{2}\frac{\chi_{S(z,r)}(\xi)}{|S(z,r)|}dA(z)$
$\displaystyle=\int_{\\{z\in\mathbb{D}:|z-\xi|<r(1-|z|\\}}\frac{|f(z)|^{2}}{|S(z,r)|}dA(z)$
$\displaystyle\geqslant\int_{S(\xi,\frac{r}{2(1+r)})}\frac{|f(z)|^{2}}{|S(z,r)|}dA(z)$
$\displaystyle\geqslant\frac{C}{|S(\xi,r)|}\int_{S(\xi,\frac{r}{2(1+r)})}|f(z)|^{2}dA(z)\
\ \ \ \bigg{(}\mathrm{since}\ z\in
S\big{(}\xi,\frac{r}{2(1+r)}\big{)}\bigg{)}$
$\displaystyle=C\bigg{[}\frac{1}{|S(\xi,\frac{r}{2(1+r)})|}\int_{S(\xi,\frac{r}{2(1+r)})}|f(z)|^{2}dA(z)\bigg{]}\cdot\frac{|S(\xi,\frac{r}{2(1+r)})|}{|S(\xi,r)|}$
$\displaystyle\geqslant\frac{C}{16}|f(\xi)|^{2},$

the last inequality follows form the subharmonicity of $|f|^{2}$ (see Lemma
2.1) and $C$ depends only on $r$. Thus we get

$\|f\|_{L^{2}(\omega^{*})}^{2}\geqslant\frac{C}{16}\int_{\mathbb{D}}|f(\xi)|^{2}\omega(\xi)dA(\xi)=\frac{C}{16}\|f\|^{2}_{L^{2}(\omega)},$

which finishes the proof of Lemma 5.3. ∎

In order to complete the proof the main theorem in this section, the following
two key lemmas are also needed.

######  Proposition 5.4.

Let $G$ be the subset which satisfies the assumption in Theorem 5.1. For
$\eta\in(0,1)$, we define a subset $F$ as the following:

$F:=\\{z\in\mathbb{D}:\omega(z)\geqslant\eta\omega^{*}(z)\\}.$

Then one can choose $\eta$ (depending only on $\delta$ and $r$) sufficiently
small such that

$|F\cap S(a,r)|\geqslant(1-\frac{\delta}{2})|S(a,r)|$

and

$\Big{|}G\cap S(a,r)\cap F\Big{|}\geqslant\frac{\delta}{2}|S(a,r)|$

for all $a\in\mathbb{D}$,

###### Proof.

We first claim that for any $\delta^{\prime}\in(0,1)$, there exists
$\eta^{\prime}=\eta^{\prime}(\delta^{\prime})>0$ such that

$\big{|}\\{z\in
S(a,r):\omega(z)<\eta^{\prime}\omega^{*}(a)\\}\big{|}<\delta^{\prime}|S(a,r)|$

for all $a\in\mathbb{D}$.

Indeed, for each $\kappa\in(0,1)$ and $a\in\mathbb{D}$, we have

$\displaystyle\big{|}\\{z\in
S(a,r):\omega(z)<\kappa\omega^{*}(a)\\}\big{|}\cdot\frac{1}{\kappa\omega^{*}(a)}$
$\displaystyle<\int_{\\{z\in
S(a,r):\omega(z)<\kappa\omega^{*}(a)\\}}\frac{1}{\omega(z)}dA(z)$
$\displaystyle\leqslant|S(a,r)|_{\omega^{-1}}$
$\displaystyle\leqslant[\omega]_{\mathcal{A}_{2}}|S(a,r)|^{2}\cdot|S(a,r)|_{\omega}^{-1}.$

So, we obtain

$\big{|}\\{z\in
S(a,r):\omega(z)<\kappa\omega^{*}(a)\\}\big{|}<([\omega]_{\mathcal{A}_{2}}\kappa)|S(a,r)|$

for all $a\in\mathbb{D}$ and $\kappa\in(0,1)$. By this inequality, we can
choose any
$0<\eta^{\prime}\leqslant[\omega]_{\mathcal{A}_{2}}^{-1}\delta^{\prime}$ to
finish the proof of the claim.

Lemma 5.2 guarantees that there is a constant $C=C(r)$ such that

$\\{z\in S(a,r):\omega(z)<C\tau\omega^{*}(z)\\}\subset\\{z\in
S(a,r):\omega(z)<\tau\omega^{*}(a)\\}\ \ \ \big{(}a\in\mathbb{D}\big{)}$

for every $\tau\in(0,1)$. By the claim and its proof, there exists a
$\tau=\tau(\delta)<C^{-1}$ such that

$\big{|}\\{z\in
S(a,r):\omega(z)<\tau\omega^{*}(a)\\}\big{|}<\frac{\delta}{2}|S(a,r)|\ \ \
\big{(}a\in\mathbb{D}\big{)}.$

Therefore, we can take $\eta=\eta(\delta,r)=C\tau<1$ such that

$\big{|}\\{z\in
S(a,r):\omega(z)<\eta\omega^{*}(z)\\}\big{|}<\frac{\delta}{2}|S(a,r)|\ \ \
\big{(}a\in\mathbb{D}\big{)}.$

Using this $\eta$ to define the corresponding $F$, so that

$|F\cap S(a,r)|=\big{|}\\{z\in
S(a,r):\omega(z)\geqslant\eta\omega^{*}(z)\\}\big{|}\geqslant(1-\frac{\delta}{2})|S(a,r)|$

for all $a\in\mathbb{D}$.

By assumption

$|G\cap S(a,r)|\geqslant\delta|S(a,r)|,$

we have

$\displaystyle\delta|S(a,r)|$ $\displaystyle\leqslant|G\cap S(a,r)|$
$\displaystyle=\Big{|}\big{[}G\cap S(a,r)\cap F\big{]}\cup\big{[}G\cap
S(a,r)\cap(\mathbb{D}\setminus F)\big{]}\Big{|}$
$\displaystyle\leqslant\Big{|}G\cap S(a,r)\cap
F\Big{|}+|S(a,r)\cap(\mathbb{D}\setminus F)|$ $\displaystyle=\Big{|}G\cap
S(a,r)\cap F\Big{|}+|S(a,r)|-|S(a,r)\cap F|$
$\displaystyle\leqslant\Big{|}G\cap S(a,r)\cap
F\Big{|}+|S(a,r)|-(1-\frac{\delta}{2})|S(a,r)|.$

This gives that

$\Big{|}G\cap S(a,r)\cap F\Big{|}\geqslant\frac{\delta}{2}|S(a,r)|$

for all $a\in\mathbb{D}$, as desired. ∎

######  Lemma 5.5.

If $G_{0}$ is a measurable subset of $\mathbb{D}$ that satisfies

$|G_{0}\cap S(a,r)|\geqslant\delta_{0}|S(a,r)|\ \ \
\big{(}a\in\mathbb{D}\big{)}$

for some $\delta_{0}>0$, then there exists constant $C=C(\delta_{0},r)>0$ such
that

$\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant
C\int_{G_{0}}|f(z)|^{2}\omega^{*}(z)dA(z)$

for all $f\in L_{h}^{2}(\omega^{*})$.

The proof of the preceding lemma is some what long and it requires a number of
technical lemmas. Let us assume this result for the moment and we will prove
it at the end of this section. Now we give the proof of Theorem 5.1.

###### Proof of Theorem 5.1.

By Proposition 5.4 and Lemma 5.5, we have

$\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant C_{1}\int_{G\cap
F}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant C_{1}\eta^{-1}\int_{G}|f|^{2}\omega
dA$

for all $f\in L_{h}^{2}(\omega^{*})$, where $C_{1}$ is a constant depending
only on $r$ and $\eta=\eta(\delta,r)<1$ is chosen by Proposition 5.4.

From Lemma 5.3, it is clear that

$\int_{\mathbb{D}}|f(z)|^{2}\omega(z)dA(z)\leqslant
C_{1}\eta^{-1}\int_{G}|f(z)|^{2}\omega(z)dA(z)$

for all $f\in L_{h}^{2}(\omega)$, which gives the desired inequality in
Theorem 5.1. ∎

Now we turn to the proof of Lemma 5.5. Before giving the proof, we need to
introduce some notations and prove three technical lemmas.

Let $0<\theta<\frac{1}{2}$, we define the subset

$E_{\theta}(a)=E_{\theta}(f,a):=\Big{\\{}z\in
S(a,r):|f(z)|>\theta|f(a)|\Big{\\}}$

and the following operator:

$B_{\theta}f(a):=\frac{1}{|E_{\theta}(a)|}\int_{E_{\theta}(a)}|f(z)|^{2}dA(z)\
\ \ \big{(}a\in\mathbb{D}\big{)}.$

It is clear that

$B_{\theta}f(a)\geqslant\frac{1}{|S(a,r)|}\int_{S(a,r)}|f(z)|^{2}dA(z)\ \ \
\big{(}a\in\mathbb{D}\big{)}.$

For $\epsilon\in(0,1)$, we consider the following two subsets, which are very
useful to establish our main result. Define

$A=A_{\epsilon}:=\bigg{\\{}a\in\mathbb{D}:|f(a)|^{2}\leqslant\frac{\epsilon}{|S(a,r)|}\int_{S(a,r)}|f(z)|^{2}dA(z)\bigg{\\}}$

and

$B=B_{\epsilon}:=\bigg{\\{}a\in\mathbb{D}:|f(a)|^{2}\leqslant\epsilon^{2}B_{\theta}f(a)\bigg{\\}}.$

A useful estimation for the Lebesgue measure of $\big{\\{}z\in
S(a,r):|f(z)|>\theta|f(a)|\big{\\}}$ with $f$ harmonic is the following
inequality.

######  Lemma 5.6.

Fix $\epsilon\in(0,1)$. For any $\delta^{\prime}\in(0,1)$, there exists
$\theta\in(0,\frac{1}{2})$ such that

$\bigg{|}\Big{\\{}z\in
S(a,r):|f(z)|>\theta|f(a)|\Big{\\}}\bigg{|}>\big{(}1-\frac{\delta^{\prime}}{2}\big{)}|S(a,r)|$

for every $f$ harmonic on $\mathbb{D}$ and satisfies

$|f(a)|^{2}>\frac{\epsilon^{2}}{|S(a,r)|}\int_{S(a,r)}|f(z)|^{2}dA(z)\ \ \
(a\in\mathbb{D}).$

###### Proof.

See the proof of Lemma 2 in [12]. ∎

The next lemma provides an estimation for the integral of $|f|^{2}\omega^{*}$
over the set $A$.

######  Lemma 5.7.

Let $\epsilon\in(0,1)$, then there exists a constant $C$ (independent of
$\epsilon$) such that

$\int_{A}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant
C\epsilon\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)$

for all $f\in L_{h}^{2}(\omega^{*})$.

###### Proof.

For $a\in A$, we have

$|f(a)|^{2}\leqslant\frac{\epsilon}{|S(a,r)|}\int_{\mathbb{D}}\chi_{S(a,r)}(z)|f(z)|^{2}dA(z).$

Multiplying the above inequality by $\omega^{*}(a)$ and integrating over $A$
to obtain

$\displaystyle\int_{A}|f(a)|^{2}\omega^{*}(a)dA(a)$
$\displaystyle\leqslant\int_{A}\frac{\epsilon}{|S(a,r)|}\int_{\mathbb{D}}\chi_{S(a,r)}(z)|f(z)|^{2}dA(z)\omega^{*}(a)dA(a)$
$\displaystyle=\epsilon\int_{\mathbb{D}}|f(z)|^{2}\bigg{(}\int_{A}\frac{\chi_{S(a,r)}(z)}{|S(a,r)|}\omega^{*}(a)dA(a)\bigg{)}dA(z).$

By the second conclusion of Lemma 5.2, we have

$\displaystyle\int_{A}\frac{\chi_{S(a,r)}(z)}{|S(a,r)|}\omega^{*}(a)dA(a)$
$\displaystyle\leqslant C\omega^{*}(z),$

where $C$ depends only on $r$. This completes the proof. ∎

The proof of Lemma 5.5 requires the following inequality, which can be proved
by the preceding lemma.

######  Lemma 5.8.

Let $\epsilon\in(0,1)$, then there exists a constant $C=C(r)$ such that

$\int_{B}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant
C\epsilon\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)$

for all $f\in L_{h}^{2}(\omega^{*})$.

###### Proof.

Observe that

$\displaystyle\int_{B}|f(z)|^{2}\omega^{*}(z)dA(z)$ $\displaystyle=\int_{B\cap
A}|f(z)|^{2}\omega^{*}(z)dA(z)+\int_{B\setminus
A}|f(z)|^{2}\omega^{*}(z)dA(z)$
$\displaystyle\leqslant\int_{A}|f(z)|^{2}\omega^{*}(z)dA(z)+\int_{B\setminus
A}|f(z)|^{2}\omega^{*}(z)dA(z).$

Based on Lemma 5.7, it sufficient for us to show the following inequality
holds for some constant $C=C(r)$:

$J:=\int_{B\setminus A}|f(z)|^{2}\omega^{*}(z)dA(z)\leqslant
C\epsilon\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z).$

Recall that for each $a\in B$ we have

$|f(a)|^{2}\leqslant\frac{\epsilon^{2}}{|E_{\theta}(a)|}\int_{E_{\theta}(a)}|f(z)|^{2}dA(z).$

From the above inequality we have

$\displaystyle J$ $\displaystyle=\int_{B\setminus
A}|f(a)|^{2}\omega^{*}(a)dA(a)$
$\displaystyle\leqslant\epsilon^{2}\int_{B\setminus
A}\bigg{(}\frac{1}{|E_{\theta}(a)|}\int_{E_{\theta}(a)}|f(z)|^{2}dA(z)\bigg{)}\omega^{*}(a)dA(a)$
$\displaystyle=\epsilon^{2}\int_{\mathbb{D}}\bigg{(}\int_{B\setminus
A}\frac{\omega^{*}(a)}{|E_{\theta}(a)|}\chi_{E_{\theta}(a)}(z)dA(a)\bigg{)}|f(z)|^{2}dA(z)$
$\displaystyle\leqslant\epsilon^{2}\int_{\mathbb{D}}|f(z)|^{2}\bigg{(}\int_{B\setminus
A}\omega^{*}(a)\frac{\chi_{S(a,r)}(z)}{|E_{\theta}(a)|}dA(a)\bigg{)}dA(z).$

The last “$\leqslant$” is due to $E_{\theta}(a)\subset S(a,r)$.

To finish the proof, we need the following claim.
Claim. There is a positive constant $C=C(r)$ such that

$|E_{\theta}(a)|\geqslant C\epsilon|S(a,r)|\ \ \ \mathrm{or}\ \ \
|E_{\theta}(a)|\geqslant C|S(a,r)|$

for each $a\notin A.$

If the above claim is true, then we get

$\displaystyle\int_{B\setminus
A}\omega^{*}(a)\frac{\chi_{S(a,r)}(z)}{|E_{\theta}(a)|}dA(a)$
$\displaystyle\leqslant C^{-1}\epsilon\int_{B\setminus
A}\omega^{*}(a)\frac{\chi_{S(a,r)}(z)}{|S(a,r)|}dA(a).$

Use Lemma 5.2 again, we have

$\displaystyle\int_{B\setminus
A}\omega^{*}(a)\frac{\chi_{S(a,r)}(z)}{|S(a,r)|}dA(a)\leqslant
C_{1}\omega^{*}(z),$

the constant $C_{1}$ depends only on $r$. From the definition of $J$, we
obtain

$J\leqslant C\epsilon\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)$

for some positive constant $C=C(r)$.

Now we turn to prove the claim. For each $a\notin A$, we have

$\displaystyle|f(a)|^{2}$
$\displaystyle>\frac{\epsilon}{|S(a,r)|}\int_{S(a,r)}|f(z)|^{2}dA(z)$
$\displaystyle=\frac{\epsilon}{r^{2}(1-|a|)^{2}}\int_{S(a,r)}|f(z)|^{2}dA(z).$

Using the change of variables $z=a+r(1-|a|)\lambda$ to get

$\displaystyle|f(a)|^{2}$
$\displaystyle>\epsilon\int_{\mathbb{D}}|f(a+r(1-|a|)\lambda)|^{2}dA(\lambda).$

Let $g(\lambda):=f(a+r(1-|a|)\lambda)$, then $g$ is also harmonic on
$\mathbb{D}$ and

$|g(0)|^{2}>\epsilon\int_{\mathbb{D}}|g(\lambda)|^{2}dA(\lambda).$

Applying Lemma 2.8 to the function $g$ to get a constant $C_{0}=C_{0}(r)$ such
that

$\displaystyle|g(z)-g(0)|$ $\displaystyle\leqslant
C_{0}|z|\int_{D(0,\frac{r}{4})}|g(\lambda)|dA(\lambda)\leqslant
C_{0}|z|\int_{\mathbb{D}}|g|dA$

whenever $|z|\leqslant\frac{r}{16}.$ Cauchy-Schwartz inequality gives that

$|g(z)-g(0)|\leqslant
C_{0}|z|\bigg{(}\int_{\mathbb{D}}|g(\lambda)|^{2}dA(\lambda)\bigg{)}^{\frac{1}{2}}\leqslant
C_{0}\epsilon^{-\frac{1}{2}}|g(0)|\cdot|z|.$

provided $|z|\leqslant\frac{r}{16}$.

If

$|z|<\min\Big{\\{}\frac{r}{16},\frac{\epsilon^{\frac{1}{2}}}{2C_{0}}\Big{\\}},$

then

$|g(z)|\geqslant|g(0)|-|g(z)-g(0)|\geqslant\frac{|g(0)|}{2}.$

Recall that $0<\theta<\frac{1}{2}$, we have $|g(z)|>\theta|g(0)|$ for
$|z|<\min\big{\\{}\frac{r}{16},\frac{\epsilon^{\frac{1}{2}}}{2C_{0}}\big{\\}}.$
This means that

$B\big{(}0,\frac{r}{16}\big{)}\subset\Big{\\{}z\in\mathbb{D}:|g(z)|>\theta|g(0)|\Big{\\}}$

or

$B\big{(}0,\frac{\epsilon^{\frac{1}{2}}}{2C_{0}}\big{)}\subset\Big{\\{}z\in\mathbb{D}:|g(z)|>\theta|g(0)|\Big{\\}}.$

On the other hand, observe that

$\displaystyle|E_{\theta}(a)|$ $\displaystyle=\int_{\\{z\in
S(a,r):|f(z)|>\theta|f(a)|\\}}dA(z)$
$\displaystyle=\int_{\big{\\{}|\frac{z-a}{r(1-|a|)}|<1:|f(z)|>\theta|f(a)|\big{\\}}}dA(z)$
$\displaystyle=|S(a,r)|\int_{\big{\\{}|\lambda|<1:|f(a+r(1-|a|)\lambda)|>\theta|f(a)|\big{\\}}}dA(\lambda)$
$\displaystyle=|S(a,r)|\int_{\\{|\lambda|<1:|g(\lambda)|>\theta|g(0)|\\}}dA(\lambda)$
$\displaystyle=|S(a,r)|\cdot\Big{|}\\{\lambda\in\mathbb{D}:|g(\lambda)|>\theta|g(0)|\\}\Big{|}.$

Therefore, we obtain

$|E_{\theta}(a)|\geqslant\Big{(}\frac{r}{16}\Big{)}^{2}|S(a,r)|$

or

$|E_{\theta}(a)|\geqslant\frac{\epsilon}{4C_{0}^{2}}|S(a,r)|.$

This gives the proof of the claim, and so we complete the proof of Lemma 5.8.
∎

We are now ready to prove Lemma 5.5.

###### Proof of Lemma 5.5.

Suppose that $|G_{0}\cap S(a,r)|\geqslant\delta_{0}|S(a,r)|$. From Lemmas 5.7
and 5.8, we can choose $\epsilon$ small enough so that

$\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)<2\int_{\mathbb{D}\setminus
B}|f(z)|^{2}\omega^{*}(z)dA(z).$

On the other hand, if $a\notin B$, then

$\displaystyle|f(a)|^{2}$
$\displaystyle>\epsilon^{2}B_{\theta}f(a)\geqslant\frac{\epsilon^{2}}{|S(a,r)|}\int_{S(a,r)}|f(z)|^{2}dA(z).$

For the $\delta_{0}$ above, we apply Lemma 5.6 to choose a
$\theta\in(0,\frac{1}{2})$ such that

$\bigg{|}\Big{\\{}z\in
S(a,r):|f(z)|>\theta|f(a)|\Big{\\}}\bigg{|}>(1-\frac{\delta_{0}}{2})|S(a,r)|.$

Since $|G_{0}\cap S(a,r)|\geqslant\delta_{0}|S(a,r)|$, we have

$\bigg{|}\Big{\\{}z\in S(a,r)\cap
G_{0}:|f(z)|>\theta|f(a)|\Big{\\}}\bigg{|}>\frac{\delta_{0}}{2}|S(a,r)|,$

and so

$\frac{1}{|S(a,r)|}\int_{S(a,r)\cap
G_{0}}|f(z)|^{2}dA(z)>\frac{\theta^{2}\delta_{0}}{2}|f(a)|^{2}\ \ \
\big{(}a\notin B\big{)}.$

Multiplying the above inequality by $\omega^{*}(a)$ and integrating over
$\mathbb{D}\backslash B$ to get

$\displaystyle\frac{\theta^{2}\delta_{0}}{2}\int_{\mathbb{D}\backslash
B}\omega^{*}(a)|f(a)|^{2}dA(a)$ $\displaystyle<\int_{\mathbb{D}\backslash
B}\frac{\omega^{*}(a)}{|S(a,r)|}\int_{S(a,r)\cap G_{0}}|f(z)|^{2}dA(z)dA(a)$
$\displaystyle=\int_{G_{0}}|f(z)|^{2}\bigg{(}\int_{\mathbb{D}\backslash
B}\frac{\omega^{*}(a)}{|S(a,r)|}\chi_{S(a,r)}(z)dA(a)\bigg{)}dA(z)$
$\displaystyle\leqslant C\int_{G_{0}}|f(z)|^{2}\omega^{*}(z)dA(z),$

the last “$\leqslant$” follows from Lemma 5.2 and $C$ depends only on $r$.

Therefore,

$\int_{G_{0}}|f(z)|^{2}\omega^{*}(z)dA(z)>\frac{\delta_{0}\theta^{2}}{4C}\int_{\mathbb{D}}|f(z)|^{2}\omega^{*}(z)dA(z)$

for all $f\in L_{h}^{2}(\omega^{*})$. This completes the proof of Lemma 5.5. ∎

Acknowledgments The authors are grateful to Professor Kunyu Guo for helpful
suggestions. Both authors would like to express their deepest gratitude to
Professor Dechao Zheng, X. Zhao’s Ph.D thesis adviser, for his valuable
guidance, encouragement and support in many years.

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