# On weighted Compactness of Commutators of square function and semi-group
maximal function associated to Schrödinger operators

Shifen Wang  Shifen Wang: School of Mathematical Sciences
Beijing Normal University
Beijing 100875
People’s Republic of China<EMAIL_ADDRESS>,  Qingying Xue∗ Qingying
Xue: School of Mathematical Sciences
Beijing Normal University
Laboratory of Mathematics and Complex Systems
Ministry of Education
Beijing 100875
People’s Republic of China<EMAIL_ADDRESS>and  Chunmei Zhang  Chunmei
Zhang: School of Mathematical Sciences
Beijing Normal University
Beijing 100875
People’s Republic of China<EMAIL_ADDRESS>
###### Abstract.

In this paper, the object of our investigation is the following Littlewood-
Paley square function $g$ associated with the Schrödinger operator
$L=-\Delta+V$ defined by:

$g(f)(x)=\Big{(}\int_{0}^{\infty}\Big{|}\frac{d}{dt}e^{-tL}(f)(x)\Big{|}^{2}tdt\Big{)}^{1/2}.$

where $\Delta$ is the laplacian operator on $\mathbb{R}^{n}$ and $V$is a
nonnegative potential. We show that the commutators of $g$ are compact
operators from $L^{p}(w)$ to $L^{p}(w)$ for $1<p<\infty$ if $b\in\rm
CMO_{\theta}(\rho)$ and $w\in A_{p}^{\rho,\theta}$, where $\rm
CMO_{\theta}(\rho)$ is the closure of
$\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$ in the $\rm BMO_{\theta}(\rho)$
topology which is more larger than the classical $\rm CMO$ space and
$A_{p}^{\rho,\theta}$ is a weights class which is more larger than Muckenhoupt
$A_{p}$ weight class. An extra weight condition in a privious weighted
compactness result is removed for the commutators of the semi-group maximal
function defined by $\mathcal{T}^{*}(f)(x)=\sup_{t>0}|e^{-tL}f(x)|.$

###### Key words and phrases:

Schrödinger operator, Littlewood-Paley function, semi-group maximal operator,
commutator, compactness.
2010 Mathematics Subject Classification. Primary 42B25, Secondary 35J10.

The second author was supported partly by the National Key Research and
Development Program of China (Grant No. 2020YFA0712900) and NSFC (No.
11871101).

∗ Corresponding author, e-mail address<EMAIL_ADDRESS>
##  1\. Introduction

This paper is devoted to study the weighted compactness of the commutators of
the Littlewood-Paley square function $g$ associated with the Schrödinger
operator $L=-\Delta+V$, where $\Delta$ is the well-known Laplacian operator
and the function $V$ is a nonnegative potential enjoying certain reverse
Hölder inequality in $\mathbb{R}^{n}(n\geq 3$).

It is well-known that the Schrödinger operator $L$ plays a fundamental role in
Harmonic analysis and PDE [26]. In the past few decades, Great achievements
have been made in the study of the operator $L$ and its associated operators
such as Riesz transforms $\nabla L^{-1/2}$. It is worthy to pointing out that
the estimates of the solutions of $L$ and Riesz transforms $\nabla L^{-1/2}$
were given in [15, 27, 39]. Great interests have been shown to investigate the
properties of Hardy type spaces associated with $L$, such like, the Hardy
space $H^{p}_{L}$ [14, 12] for $0<p\leq 1$; the local Hardy spaces $h_{L}$
[38]. The semigroup maximal function $e^{-tL}(t>0)$ and the Riesz transforms
$\nabla L^{-1/2}$ on Heisenberg groups were further considered in [23].

In order to state more known results, we need to introduce some notations and
definitions. Let $V$ be a nonnnegative function defined on ${\mathbb{R}^{n}}.$
For every ball $B(x,r)\subset\mathbb{R}^{n}$, we say $V$ is in the reverse
Hölder class $RH_{q}$ ($q>n/2$), if there exists a constant $C>0$ such that
$V$ enjoys the following reverse Hölder inequality

$\Big{(}\frac{1}{|B(x,r)|}\int_{B(x,r)}V^{q}(y)dy\Big{)}^{\frac{1}{q}}\leq
C\Big{(}\frac{1}{|B(x,r)|}\int_{B(x,r)}V(y)dy\Big{)}.$

We need to give the definition of a class of more generalized $\rm BMO$ space
associated with $\rho.$

######  Definition 1.1.

($\rm BMO_{\theta}(\rho)({\mathbb{R}^{n}})$ space, [1]). For $\theta>0$, we
defined the class $\rm BMO_{\theta}(\rho)({\mathbb{R}^{n}})$ of locally
integrable functions $f$ such that

(1.1) $\frac{1}{|B(x,r)|}\int_{B(x,r)}|f(y)-f_{B}|dy\leq
C\Big{(}1+\frac{r}{\rho(x)}\Big{)}^{\theta},$

for all $x\in\mathbb{R}^{n}$ and $r>0$, where $f_{B}=\frac{1}{|B|}\int_{B}f$
and the function $\rho$ [27] associated with $V(x)\in RH_{q}$ is defined by

(1.2) $\rho(x)=\sup_{r>0}\Big{\\{}r:\frac{1}{r^{n-2}}\int_{B(x,r)}V(y)dy\leq
1\Big{\\}}.$

For a function $f\in\rm BMO_{\theta}(\rho)({\mathbb{R}^{n}})$, the infimum of
the constants in (1.1) is denoted by $\|f\|_{\rm BMO_{\theta}(\rho)}$.

We now need tointroduce a more larger weights class than the classical $A_{p}$
weights class as follows.

######  Definition 1.2.

($A_{p}^{\rho,\theta}$ weights class, [2]). Let $w$ be a nonnegative, locally
integrable function on $\mathbb{R}^{n}$. For $1<p<\infty$, a weight $w$ is
said in the class $A_{p}^{\rho,\theta}$ if there exists a positive constant
$C$ such that for all balls $B=B(x,r)$, it holds that

(1.3)
$\Big{(}\frac{1}{|B|}\int_{B}w(y)dy\Big{)}\Big{(}\frac{1}{|B|}\int_{B}w(y)^{-1/(p-1)}dy\Big{)}^{p-1}\leq
C\Big{(}1+\frac{r}{\rho(x)}\Big{)}^{\theta p}.$

$w\in A_{1}^{\rho,\theta}$ means that there exists a constant $C$ such that
for all balls $B$, the following inequality holds

$M^{\theta}_{V}(w)(x)\leq Cw(x),\ a.e.\ x\in\mathbb{R}^{n},$

where $M^{\theta}_{V}(f)(x)=\sup_{x\in
B}\frac{1}{\Psi_{\theta}(B)|B|}\int_{B}|f(y)|dy$ and
$\Psi_{\theta}(B)=(1+r/\rho(x))^{\theta}$ for $B=B(x,r)$.

######  Remark 1.3.

It is easy to see that $\rm BMO({\mathbb{R}^{n}})\subset\rm
BMO_{\theta}(\rho)({\mathbb{R}^{n}})\subset\rm
BMO_{\theta^{\prime}}(\rho)({\mathbb{R}^{n}})$ for $0<\theta<\theta^{\prime}$,
$A_{p}\subsetneq A_{p}^{\rho,\theta}$ and the classes $A_{p}^{\rho,\theta}$
are increasing with $\theta$ for $1\leq p<\infty$.

The Littlewood-Paley $g$ function and the semi-group maximal function
associated to Schrödinger operator $L$ are defined by

(1.4)
$g(f)(x)=\Big{(}\int_{0}^{\infty}\Big{|}\frac{d}{dt}e^{-tL}(f)(x)\Big{|}^{2}tdt\Big{)}^{1/2},$

and

(1.5)
$\mathcal{T}^{*}(f)(x)=\sup_{t>0}|e^{-tL}f(x)|=\sup_{t>0}\Big{|}\int_{\mathbb{R}^{n}}k_{t}(x,y)f(y)dy\Big{|},$

where $k_{t}$ is the kernel of the operator $e^{-tL}$. The commutators of $g$
and $\mathcal{T}^{*}$ with $b\in\rm BMO_{\theta}(\rho)({\mathbb{R}^{n}})$ are
defined by

(1.6)
$g_{b}(f)(x)=\Big{(}\int_{0}^{\infty}\Big{|}\frac{d}{dt}e^{-tL}(b(x)-b(\cdot))(f)(x)\Big{|}^{2}tdt\Big{)}^{1/2};$

and

(1.7)
$\mathcal{T}_{b}^{*}(f)(x)=\sup_{t>0}\Big{|}\int_{\mathbb{R}^{n}}k_{t}(x,y)(b(x)-b(y))f(y)dy\Big{|}.$

Recall that, in 2011, Bongioanni, Harboure and Salinas [1] first considered
the strong boundedness of the commutators of Riesz transforms related to $L$
with $\rm BMO_{\theta}(\rho)({\mathbb{R}^{n}})$ functions. Weighted estimates
were demonstrated in [2] for $g$ function and $\mathcal{T}^{*}$ related to $L$
with $A_{p}^{\rho,\theta}$ weights as follows:

Theorem A ([2]). For $1<p<\infty$, the operator $g$ and $\mathcal{T}^{*}$ are
bounded on $L^{p}(w)$ when $w\in A_{p}^{\rho,\theta}$.

In another paper, Tang [28] considered the weighted norm inequalities for
commutators of $g$ function and the maximal function $\mathcal{T}^{*}$
associated with Schrödinger operators $L.$

Theorem B ([28]). Let $1<p<\infty$, $w\in A_{p}^{\rho,\theta}$ and $b\in\rm
BMO_{\theta}(\rho)$, then there exists a constant $C$ such that

$\|g_{b}(f)\|_{L^{p}(w)}\leq C\|b\|_{\rm
BMO_{\theta}(\rho)}\|f\|_{L^{p}(w)},\quad\|{T}_{b}^{*}(f)\|_{L^{p}(w)}\leq
C\|b\|_{\rm BMO_{\theta}(\rho)}\|f\|_{L^{p}(w)}.$

As was said before, this paper is devoted to studying the weighted compactness
for commutators of Littlewood-Paley function related to Schrödinger operator
$L$. Recall that, a linear or sublinear operator $T$ is said to be a compact
operator from a Banach space $X$ to a Banach space $Y$, if the image under $T$
of any bounded subset in $X$ is a relatively compact subset in $Y.$ Given a
locally integrable function $b$, the commutator $[b,T]$ of $T$ is given by
$[b,T](f)(x)=bTf(x)-T(bf)(x).$

We now present some known result about the compactness of some classical
operators. In 1975, Corders [11] first showed that the commutators of the
Fourier multipliers and pseudodifferential operators are compact on
$L^{p}({\mathbb{R}^{n}})$ provided $b\in{\rm CMO}(\mathbb{R}^{n})$, where
${\rm CMO}(\mathbb{R}^{n})$ is the closure of $C_{c}^{\infty}$ in the $\rm
BMO(\mathbb{R}^{n})$ topology. In 1978, Uchiyama [32] surprisingly proved that
the commutator of singular intergral opertors with ${\rm Lip}_{1}({\rm
S}^{n-1})$ kernels is compact on $L^{p}(\mathbb{R}^{n})$ for $1<p<\infty$ if
and only if $b\in{\rm CMO}(\mathbb{R}^{n})$. Since then, the study on the
compactness of commutators of different operators has attracted much more
attention. We refer the reader to [21, 22, 33, 7, 8, 9, 25] for the
compactness of commutators of linear operators, [3, 4, 37, 30, 31] for
multilinear operators and [6, 17, 18, 19] for extrapolation theory of
compactness. In particular, in a recent paper , the result of Corders [11] was
extended to the bilinear case and the the condition $b\in\rm
CMO({\mathbb{R}^{n}})$ was improved to a more bigger space which is called
$\rm{XMO}({\mathbb{R}^{n}})$.

Note that these compactness results are all concerned with the space
$\text{CMO}(\mathbb{R}^{n})$. However, Theorem B shows that the $L^{p}$
boundedness holds for more larger space
$\text{BMO}_{\theta}(\rho)(\mathbb{R}^{n})$, rather than
$\text{BMO}(\mathbb{R}^{n})$ and the weights class $A_{p}^{\rho,\theta}$ is
more larger than $A_{p}$ weights class. Let $\text{
CMO}_{\theta}(\rho)(\mathbb{R}^{n})$ be the closure of
$\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$ in the $\rm
BMO_{\theta}(\rho)(\mathbb{R}^{n})$ topology. Then, it is quite natural to ask
the following question:

######  Question 1.4.

Whether the operator $g_{b}$ is compact from $L^{p}(w)$ to $L^{p}(w)$ when
$w\in A_{p}^{\rho,\theta}$ and $b$ belongs to the space $\text{
CMO}_{\theta}(\rho)(\mathbb{R}^{n})$?

One may ask the same question for the operator $\mathcal{T}^{*}_{b}$. Very
recently, we obtained the following weighted compactness for
$\mathcal{T}^{*}_{b}$ in [34]. However, an extra weights condition needs to be
assumed. Theorem C ([34]). Let $1<p<\infty$, $w\in A_{p}^{\rho,\theta}$ and
$b\in\rm CMO_{\theta}(\rho)({\mathbb{R}^{n}})$. If $w$ satisfies the following
condition

(1.8)
$\lim_{A\rightarrow+\infty}A^{-np+n}\int_{|x|>1}\frac{w(Ax)}{{|x|}^{np}}dx=0,$

then the operator $\mathcal{T}^{*}_{b}$ is a compact operator from $L^{p}(w)$
to $L^{p}(w)$.

Therefore, another question arises naturally:

######  Question 1.5.

Is it possible to remove condition (1.8) from Theorem C ?

The main purpose of this paper is to give firm answers to the above questions.
Our results are as follows:

######  Theorem 1.1.

Let $1<p<\infty$. If $w\in A_{p}^{\rho,\theta}$ and $b\in\rm
CMO_{\theta}(\rho)(\mathbb{R}^{n})$, then the operator $g_{b}$ is a compact
operator from $L^{p}(w)$ to $L^{p}(w)$.

######  Theorem 1.2.

Let $1<p<\infty$. If $w\in A_{p}^{\rho,\theta}$ and $b\in\rm
CMO_{\theta}(\rho)(\mathbb{R}^{n})$, then the operator $\mathcal{T}^{*}_{b}$
defined by (1.7) is a compact operator from $L^{p}(w)$ to $L^{p}(w)$.

The organization of this paper is as follows. In section 2, we prepare some
definitions and preliminary lemmas, which play a fundamental role in our
analysis. Section 3 is devoted to give the proof of Theorem 1.1 via smoothness
truncated techniques, which is different from what have been used in [34]. The
domain of integration will be divided into several cases, some cases are
combinable, but various subcases also arise, which increases the difficulty we
need to deal with.

Throughout the paper, the letter $C$ or $c$, sometimes with certain
parameters, will stand for positive constants not necessarily the same one at
each occurrence, but are independent of the essential variables. $A\sim B$
means that there exists constants $C_{1}>0$ and $C_{2}>0$ such that
$C_{2}B\leq A\leq C_{1}B$.

##  2\. Preliminaries

We first recall some notation and lemmas which will be used later. Given a
Lebesgue measurable set $E\subset\mathbb{R}^{n}$, $|E|$ will denote the
Lebesgue measure of $E$. If $B=B(x,r)$ is a ball in $\mathbb{R}^{n}$ and
$\lambda$ is a real number, then $\lambda B$ shall stand for the ball with the
same center as $B$ and radiu $\lambda$ times that of $B$. A weight $w$ is a
non-negative measurable function on $\mathbb{R}^{n}$. The measure associated
with $w$ is the set function given by $w(E)=\int_{E}wdx$. For $0<p<\infty$ we
denote by $L^{p}(w)$ the space of all Lebesgue measurable function $f(x)$ such
that

$\|f\|_{L^{p}(w)}=\Big{(}\int_{\mathbb{R}^{n}}|f(x)|^{p}w(x)dx|\Big{)}^{1/p}.$

The auxiliary function $\rho$ defined in (1.2) enjoys the following property.

######  Lemma 2.1.

([27]). There exist $k_{0}\geq 1$ and $C>0$ such that for all $x,\
y\in{\mathbb{R}^{n}}$,

$C^{-1}\rho(x)\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-k_{0}}\leq\rho(y)\leq
C\rho(x)(1+\frac{|x-y|}{\rho(x)}\Big{)}^{\frac{k_{0}}{k_{0}+1}}.$

In particular, $\rho(x)\sim\rho(y)$ if $|x-y|<C\rho(x)$.

It is known that $A_{p}^{\rho,\theta}$ weights class has some properties
analogy to $A_{p}$ weights class.

######  Lemma 2.2.

([2, 29]). Let $1<p<\infty$ and $w\in A_{p}^{\rho,\infty}=\bigcup_{\theta\geq
0}A_{p}^{\rho,\theta}$. Then

  1. (i)

If $1\leq p_{1}<p_{2}<\infty$, then $A_{p_{1}}^{\rho,\theta}\subset
A_{p_{2}}^{\rho,\theta}.$

  2. (ii)

$w\in A_{p}^{\rho,\theta}$ if and only if $w^{-\frac{1}{p-1}}\in
A_{p^{\prime}}^{\rho,\theta}$, where $1/p+1/p^{\prime}=1$.

  3. (iii)

If $w\in A_{p}^{\rho,\infty}$, $1<p<\infty$, then there exists $\epsilon>0$
such that $w\in A_{p-\epsilon}^{\rho,\infty}$.

  4. (iv)

if $w\in A_{p}^{\rho,\theta}$, $1\leq p<\infty$, then for any cube $Q$, we
have

$\frac{1}{\Psi_{\theta}(Q)|Q|}\int_{Q}|f(y)|dy\leq
C\Big{(}\frac{1}{w(5Q)}\int_{Q}|f|^{p}w(y)dy\Big{)}^{1/p},$

where $\Psi_{\theta}$ is the same function appeared in Definition 1.2.

It should be pointed out that (iii) of Lemma 2.2 was proved by Bongioanni et
al. [2] and (iv) of Lemma 2.2 was proved by Tang [29].

For maximal operator $M_{V}^{\theta}$ in Definition 1.2, the following result
holds:

######  Lemma 2.3.

([29]). Let $1<p<\infty$ and suppose that $w\in A_{p}^{\rho,\theta}$. If
$p<p_{1}<\infty$, then

$\int_{\mathbb{R}^{n}}|M_{V}^{\theta}f(x)|^{p_{1}}w(x)dx\leq
C_{p}\int_{\mathbb{R}^{n}}|f(x)|^{p_{1}}w(x)dx.$

By Lemma 2.3, $M_{V}^{\theta}$ may not be bounded on $L^{p}(w)$ for all $w\in
A_{p}^{\rho,\theta}$ and $1<p<\infty$. A variant maximal operator $M_{V,\eta}$
was introduced by Tang ([29]) as follows:

$M_{V,\eta}f(x)=\sup_{x\in
B}\frac{1}{(\Psi_{\theta}(B))^{\eta}|B|}\int_{B}|f(y)|dy,\ \ \ 0<\eta<\infty.$

This variant maximal function $M_{V,\eta}$ then enjoys the following $L^{p}$
boundedness.

######  Lemma 2.4.

([29]). Let $1<p<\infty$, $p^{\prime}=p/(p-1)$ and suppose that $w\in
A_{p}^{\rho,\theta}$. Then there exists a constant $C>0$ such that

$\|M_{V,p^{\prime}}f\|_{L^{p}(w)}\leq C\|f\|_{L^{p}(w)}.$

Let $q_{t}$ be the kernel of $\frac{d}{dt}e^{-tL}$. Now, we list some
pointwise estimates of $q_{t}$.

######  Lemma 2.5.

([10]) For every $N$, there are constants $c$ and $C_{N}$ such that

$|q_{t}(x,y)|\leq\frac{C_{N}}{t^{n/2+1}}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}.$

######  Lemma 2.6.

([10]) There exist $0<\delta<1$ and $c>0$ such that for every $N>0$ there is a
constant $C_{N}>0$ so that, for all $|h|\leq\sqrt{t}$,

$|q_{t}(x+h)-q_{t}(x,y)|\leq\frac{C_{N}}{t^{n/2+1}}\Big{(}\frac{|h|}{\sqrt{t}}\Big{)}^{\delta}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}.$

Let $X$ and $Y$ be Banach spaces and suppose $T$ and $\\{T_{n}\\}$ be
operators from $X$ to $Y$. Then, we need the limiting property of compact
operators.

######  Lemma 2.7.

([36, p. 278, Theorem(iii)]) Let a sequence $\\{T_{n}\\}$ of compact operators
converge to an operator T in sense of the uniform operator toplogy, i.e.,
$\lim\limits_{n\rightarrow\infty}\|T-T_{n}\|=0$. Then $T$ is also compact.

We end this section by introducing the general weighted version of Frechet-
Kolmogorov theorems, which was given by Xue, Yabuta and Yan in [37].

######  Lemma 2.8.

([37]). Let $w$ be a weight on $\mathbb{R}^{n}$. Assume that
$w^{-1/(p_{0}-1)}$ is also a weight on $\mathbb{R}^{n}$ for some $p_{0}>1$.
Let $0<p<\infty$ and $\mathcal{F}$ be a subset in $L^{p}(w)$, then
$\mathcal{F}$ is sequentially compact in $L^{p}(w)$ if the following three
conditions are satisfied:

  1. (i)

$\mathcal{F}$ is bounded, i.e.,
$\sup\limits_{f\in\mathcal{F}}\|f\|_{L^{p}(w)}<\infty$;

  2. (ii)

$\mathcal{F}$ uniformly vanishes at infinity, i.e.,

$\lim\limits_{N\rightarrow\infty}\sup\limits_{f\in\mathcal{F}}\int_{|x|>N}|f(x)|^{p}w(x)dx=0;$

  3. (iii)

$\mathcal{F}$ is uniformly equicontinuous, i.e.,

$\lim\limits_{|h|\rightarrow
0}\sup\limits_{f\in\mathcal{F}}\int_{\mathbb{R}^{n}}|f(\cdot+h)-f(\cdot)|^{p}w(x)dx=0.$

##  3\. Proof of Theorem 1.1 and 1.2

###### Proof of Theorem 1.1.

We shall prove Theorem 1.1 via smoothness truncated techniques, which is
different from that used in [34]. Let $\varphi\in C^{\infty}([0,\infty))$
satisfy

(3.1) $\displaystyle 0\leq\varphi\leq 1\ \ \ and\ \ \
\varphi(t)=\begin{cases}1,&t\in[0,1],\\\ 0,&t\in[2,\infty).\end{cases}$

For any $\gamma>0$, let

(3.2) $q_{t,\gamma}(x,y)=q_{t}(x,y)(1-\varphi(\frac{\sqrt{t}}{\gamma})),$

where $q_{t}$ is the kernel of $\frac{d}{dt}e^{-tL}$. Note that this truncated
function essentially just truncates $t$ and does not contribute anything to
$x$ and $y$. With the kernel $q_{t,\gamma}$, we define

(3.3)
$g_{\gamma}f(x)=\Big{(}\int_{0}^{\infty}\Big{|}\int_{{\mathbb{R}^{n}}}q_{t,\gamma}(x,y)f(y)dy\Big{|}^{2}tdt\Big{)}^{1/2}$

and

(3.4)
$g_{b,\gamma}f(x)=\Big{(}\int_{0}^{\infty}\Big{|}\int_{{\mathbb{R}^{n}}}q_{t,\gamma}(x,y)(b(x)-b(y))f(y)dy\Big{|}^{2}tdt\Big{)}^{1/2}.$

Then, for any $|h|\leq\sqrt{t}$, by Lemma 2.5 and Lemma 2.6, we have

(3.5)
$\displaystyle\begin{split}|q_{t,\gamma}(x,y)|\leq|q_{t}(x,y)|\leq\frac{C_{N}}{t^{n/2+1}}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}\end{split}$

and

(3.6)
$\displaystyle\begin{split}|q_{t,\gamma}(x+h,y)-q_{t,\gamma}(x,y)|&\leq|q_{t}(x+h,y)-q_{t}(x,y)|\\\
&\leq\frac{C_{N}}{t^{n/2+1}}\Big{(}\frac{|h|}{\sqrt{t}}\Big{)}^{\delta}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}.\end{split}$

For any $b\in\mathcal{C}_{c}^{\infty}({\mathbb{R}^{n}})$ and $\gamma,\theta,\
\eta>0$, by (3.2), (3.4) and Lemma 2.5 with $N=\theta\eta$, one has

(3.7) $\displaystyle\begin{split}&|g_{b,\gamma}f(x)-g_{b}f(x)|\\\ &\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}\int_{\mathbb{R}^{n}}t^{-\frac{n}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}|b(x)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}\int_{|x-y|<\sqrt{t}}t^{-\frac{n}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}|b(x)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}\int_{|x-y|\geq\sqrt{t}}t^{-\frac{n}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}|b(x)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&=:I_{1}+I_{2}.\end{split}$

For $I_{1}$, we have

(3.8) $\displaystyle\begin{split}I_{1}&\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}t^{-\frac{n}{2}-1}\sqrt{t}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}\int_{|x-y|<\sqrt{t}}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}t^{-\frac{n}{2}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}\int_{|x-y|<\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq C\Big{(}\int_{0}^{4\gamma^{2}}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\ &\leq
C\gamma M_{V,\eta}f(x).\end{split}$

For $I_{2}$, using the estimate $e^{-s}\leq\frac{C}{s^{M/2}}$with
$M>n+1+\theta\eta$ and splitting to annuli, it follows that

(3.9) $\displaystyle\begin{split}I_{2}&\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}t^{\frac{M-n}{2}-1}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}\int_{|x-y|\geq\sqrt{t}}\frac{|b(x+h)-b(y)||f(y)|}{|x-y|^{M}}dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}t^{\frac{M-n}{2}-1}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}\sum_{k=1}^{\infty}\frac{2^{k}\sqrt{t}}{(2^{k}\sqrt{t})^{M}}\int_{|x-y|\sim
2^{k}\sqrt{t}}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\ &\leq
C\displaystyle\Big{(}\int_{0}^{4\gamma^{2}}\Big{(}t^{-\frac{n}{2}}\Big{(}1+\frac{2^{k}\sqrt{t}}{\rho(x)}\Big{)}^{-\theta\eta}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-1-\theta\eta)}}{(2^{k}\sqrt{t})^{M}}\int_{|x-y|<2^{k}\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq C\gamma M_{V,\eta}f(x).\end{split}$

Combing (3.9) with (3.7) and (3.8) may lead to

$|g_{b,\gamma}(f)(x)-g_{b}(f)(x)|\leq C\gamma M_{V,\eta}f(x).$

Then, Lemma 2.4 with $p^{\prime}\leq\eta<\infty$ gives that

(3.10) $\|g_{b,\gamma}(f)-g_{b}(f)\|_{L^{p}(w)}\leq C\gamma\|f\|_{L^{p}(w)},$

which implies that

(3.11) $\lim_{\gamma\rightarrow
0}\|g_{b,\gamma}(f)-g_{b}(f)\|_{L^{p}(w)\rightarrow L^{p}(w)}=0.$

On the other hand, if $b\in\rm CMO_{\theta}(\rho)({\mathbb{R}^{n}})$, then for
any $\epsilon>0$, there exists
$b_{\epsilon}\in\mathcal{C}_{c}^{\infty}({\mathbb{R}^{n}})$ such that
$\|b-b_{\epsilon}\|_{\rm BMO_{\theta}(\rho)}<\epsilon$. Therefore

$\|g_{b}(f)-g_{b_{\epsilon}}(f)\|_{L^{p}(w)}\leq\|g_{b-b_{\epsilon}}(f)\|_{L^{p}(w)}\leq
C\|b-b_{\epsilon}\|_{\rm BMO_{\theta}(\rho)}\|f\|_{L^{p}(w)}\leq C\epsilon.$

Thus, to prove $g_{b}$ is compact on $L^{p}(w)$ for any $b\in\rm
CMO_{\theta}(\rho)$, it suffices to prove that $g_{b}$ is compact on
$L^{p}(w)$ for any $b\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$. By (3.11)
and Lemma 2.7, it suffices to show that $g_{b,\gamma}$ is compact for any
$b\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$ when $\gamma>0$ is small
enough. To this end, for arbitrary bounded set $F$ in $L^{p}(w)$, let

$\mathcal{F}=\\{g_{b,\gamma}f:f\in F\\}.$

Then, we need to show that for $b\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$,
$\mathcal{F}$ satisfies the conditions$\rm(i)$-$\rm(iii)$ of Lemma 2.8.

From the definition of $q_{t,\gamma}$, we know that $g_{\gamma}f(x)\leq gf(x)$
and $g_{b,\gamma}f(x)\leq g_{b}f(x)$. Hence, Theorem A and Theorem B still
hold for $g_{\gamma}$ and $g_{b,\gamma}$ respectively. Then

$\sup\limits_{f\in F}\|g_{b,\gamma}f\|_{L^{p}(w)}\leq C\sup\limits_{f\in
F}\|f\|_{L^{p}(w)}\leq C,$

which yields the fact that the set $\mathcal{F}$ is bounded.

To prove that for any $w\in A_{p}^{\rho,\theta}$, (ii) of Lemma 2.8

(3.12) $\lim_{A\rightarrow\infty}\int_{|x|>A}|g_{b,\gamma}f(x)|^{p}w(x)dx=0,$

holds for $f\in F$, we need the following claim

Claim. For any $N>0$, there is a constant $C_{N}>0$ so that

(3.13)
$\Big{(}\int_{0}^{\infty}|q_{t,\gamma}(x,y)|^{2}tdt\Big{)}^{\frac{1}{2}}\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.$

In fact, by (3.5), we may decompose the left side of (3.13) as

$\displaystyle\displaystyle\Big{(}\int_{0}^{\infty}|q_{t,\gamma}(x,y)|^{2}tdt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq\displaystyle
C_{N}\Big{(}\int_{0}^{|x-y|^{2}}t^{-n-2}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2N}tdt\Big{)}^{\frac{1}{2}}$
$\displaystyle\qquad+\displaystyle
C_{N}\Big{(}\int_{|x-y|^{2}}^{\infty}t^{-n-2}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2N}tdt\Big{)}^{\frac{1}{2}}$
$\displaystyle=:J_{1}+J_{2}.$

In order to estimate $J_{1}$ and $J_{2}$, we need the following inequality:
for any $M>0$, there exists a constant $C>0$, such that

(3.14) $e^{-\frac{c|x-y|^{2}}{t}}\leq C\frac{t^{\frac{M}{2}}}{|x-y|^{M}}.$

For $J_{1}$, using (3.14) with $M=n+N+1$, we obtain

$\displaystyle J_{1}$ $\displaystyle\leq\displaystyle
C_{N}\Big{(}\int_{0}^{|x-y|^{2}}t^{-n-2}\Big{(}\frac{t}{|x-y|^{2}}\Big{)}^{n+N+1}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2N}tdt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq
C_{N}\Big{(}\int_{0}^{|x-y|^{2}}\frac{1}{|x-y|^{2(n+1)}}\Big{(}\frac{\sqrt{t}}{|x-y|}\Big{)}^{2N}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2N}dt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq\frac{C_{N}}{|x-y|^{n+1}}\Big{(}\int_{0}^{|x-y|^{2}}\Big{(}\frac{\sqrt{t}+\rho(x)}{|x-y|+\rho(x)}\Big{)}^{2N}\Big{(}\frac{\sqrt{t}+\rho(x)}{\rho(x)}\Big{)}^{-2N}dt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.$

For $J_{2}$, using (3.14) with $M=2(n-1)$, we get

$\displaystyle J_{2}$ $\displaystyle\leq\displaystyle
C_{N}\Big{(}\int_{|x-y|^{2}}^{\infty}t^{-n-1}\Big{(}\frac{t}{|x-y|^{2}}\Big{)}^{n-1}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-2N}dt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq\frac{C_{N}}{|x-y|^{n-1}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}\Big{(}\int_{|x-y|^{2}}^{\infty}t^{-2}dt\Big{)}^{\frac{1}{2}}$
$\displaystyle\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.$

Therefore, (3.13) holds, and this finishes the proof of this claim.

Now we are in the position to use the above claim to prove (3.12). Assume
$b\in\mathcal{C}_{c}^{\infty}(\mathbb{R}^{n})$ and
$\operatorname{supp}(b)\subset B(0,R)$, where $B(0,R)$ is the ball of radius
$R$ center at origin in $\mathbb{R}^{n}$. For any $|x|>A>2R$, $w\in
A_{p}^{\rho,\theta}$, $1<p<\infty$ and $f\in F$. Then, by the inequality
(3.13), we have

$\begin{split}|g_{b,\gamma}(f)(x)|&\leq\displaystyle\int_{|y|<R}\Big{(}\int_{0}^{\infty}|q_{t,\gamma}(x,y)|^{2}tdt\Big{)}^{\frac{1}{2}}|b(y)f(y)|dy\\\
&\leq
C_{N}\|b\|_{\infty}\displaystyle\int_{|y|<R}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}\frac{|f(y)|}{|x-y|^{n}}dy\\\
&\leq\frac{C_{N}\|b\|_{\infty}}{|x|^{n}}\frac{\|f\|_{L^{p}(w)}}{(1+1/2|x|m_{V}(x))^{N}}\displaystyle\Big{(}\int_{|y|<R}w^{-p^{\prime}/p}(y)dy\Big{)}^{1/p^{\prime}}.\end{split}$

By Lemma 2.1, there exist $k_{0}\geq 1$ and $C_{0}>0$ such that

$m_{V}(x)\geq C_{0}m_{V}(0)(1+|x|m_{V}(0))^{-\frac{k_{0}}{k_{0}+1}}.$

Therefore, it holds that

$\begin{split}\frac{1}{1+1/2|x|m_{V}(x)}&\leq\frac{C}{1+|x|m_{V}(0)(1+|x|m_{V}(0))^{-\frac{k_{0}}{k_{0}+1}}}\\\
&\leq\frac{C}{(1+|x|m_{V}(0))(1+|x|m_{V}(0))^{-\frac{k_{0}}{k_{0}+1}}}\\\
&=\frac{C}{(1+|x|/\rho(0))^{\frac{1}{k_{0}+1}}}\end{split}$

which leads to

$\begin{array}[]{ll}{}&\displaystyle\Big{(}\int_{|x|>A}|g_{b,\gamma}f(x)|^{p}w(x)dx\Big{)}^{1/p}\\\
&\leq
C\displaystyle\sum_{j=0}^{\infty}\frac{1}{(2^{j}A)^{n}}\frac{1}{(1+2^{j}A/\rho(0))^{\frac{N}{k_{0}+1}}}\Big{(}\int_{2^{j}A<|x|<2^{j+1}A}w(x)dx\Big{)}^{1/p}\displaystyle\Big{(}\int_{|y|<R}w^{-p^{\prime}/p}(y)dy\Big{)}^{1/p^{\prime}}\\\
&\leq
C\displaystyle\sum_{j=0}^{\infty}\frac{1}{(2^{j}A)^{n}}\frac{1}{(1+2^{j}A/\rho(0))^{\frac{N}{k_{0}+1}}}\Big{(}\int_{B(0,2^{j+1}A)}w(x)dx\Big{)}^{1/p}\displaystyle\Big{(}\int_{B(0,R)}w^{-p^{\prime}/p}(y)dy\Big{)}^{1/p^{\prime}}.\end{array}$

Let $Q=B(0,2^{j+1}A)$, $E=B(0,R)$. Since $w\in A_{p}^{\rho,\theta}$, then, by
Lemma 2.2, we can get

$\begin{array}[]{ll}w(5Q)&\leq
C\displaystyle\Big{(}\frac{\Psi_{\theta}(Q)|Q|}{|E|}\Big{)}^{p}w(E)\\\ &\leq
C\displaystyle
w(B(0,R))\Big{(}\frac{(1+2^{j+1}A/\rho(0))^{\theta}(2^{j+1}A)^{n}}{R^{n}}\Big{)}^{p}\\\
&\leq C\displaystyle w(B(0,R))(1+2^{j+1}A/\rho(0))^{\theta
p}\frac{(2^{j+1}A)^{np}}{R^{np}}\end{array}$

Taking $N>(k_{0}+1)\theta$, we may continue to estimate
$\Big{(}\int_{|x|>A}|g_{b,\gamma}f(x)|^{p}w(x)dx\Big{)}^{1/p}$ in the way that

$\begin{array}[]{ll}&\quad\displaystyle\Big{(}\int_{|x|>A}|g_{b,\gamma}f(x)f(x)|^{p}w(x)dx\Big{)}^{1/p}\\\
&\quad\leq
C\displaystyle\sum_{j=0}^{\infty}\frac{(2^{j+1}A)^{n}}{(2^{j}A)^{n}}\frac{(1+2^{j+1}A/\rho(0))^{\theta}}{(1+2^{j}A/\rho(0))^{\frac{N}{k_{0}+1}}}\Big{(}\frac{1}{(1+\frac{R}{\rho(0)})^{\theta}R^{n}}\\\
&\qquad\displaystyle\times\int_{B(0,R)}w(x)dx\Big{)}^{1/p}\Big{(}\frac{1}{(1+\frac{R}{\rho(0)})^{\theta}R^{n}}\int_{B(0,R)}w^{-p^{\prime}/p}(y)dy\Big{)}^{1/p^{\prime}}\\\
&\quad\leq
C\displaystyle\sum_{j=0}^{\infty}\frac{(1+2^{j}A/\rho(0))^{\theta}}{(1+2^{j}A/\rho(0))^{\frac{N}{k_{0}+1}}}\\\
&\quad\leq CA^{-(\frac{N}{k_{0}+1}-\theta)}.\end{array}$

Therefore, we obtain

$\lim_{A\rightarrow\infty}\int_{|x|>A}|g_{b,\gamma}f(x)|^{p}w(x)dx=0,$

holds uniformly for $f\in F$.

It remains to show that $\rm(iii)$ of Lemma 2.8 holds, namely, the set
$\mathcal{F}$ is uniformly equicontinuous. It suffices to verify that

$\lim\limits_{|h|\rightarrow
0}\|g_{b,\gamma}(f)(h+\cdot)-g_{b,\gamma}(f)(\cdot)\|_{L^{p}(w)}=0,$

holds uniformly for $f\in F$.

Taking $\gamma=|h|$ and $|h|\in(0,1)$. When $\sqrt{t}<|h|$, it holds that

$\varphi(\frac{\sqrt{t}}{\gamma})=\varphi(\frac{\sqrt{t}}{|h|})=1.$

This implies $q_{t,\gamma}(x+h,y)=q_{t,\gamma}(x,y)=0.$ Therefore, we have

(3.15) $\displaystyle\begin{split}&|g_{b}(f)(x+h)-g_{b}(f)(x)|\\\
&\leq\displaystyle\Big{(}\int_{|h|^{2}}^{\infty}\Big{|}\int_{\mathbb{R}^{n}}q_{t,\gamma}(x+h,y)(b(x+h)-b(y))f(y)dy\\\
&\quad-\int_{\mathbb{R}^{n}}q_{t,\gamma}(x,y)(b(x)-b(y))f(y)dy\Big{|}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\infty}\Big{|}\int_{\mathbb{R}^{n}}(q_{t,\gamma}(x+h,y)-q_{t,\gamma}(x,y))(b(x+h)-b(y))f(y)dy\Big{|}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{|h|^{2}}^{\infty}\Big{|}\int_{\mathbb{R}^{n}}q_{t,\gamma}(x,y)(b(x+h)-b(x))f(y)dy\Big{|}^{2}tdt\Big{)}^{1/2}\\\
&=:L_{1}+L_{2}.\end{split}$

For $L_{2}$, it holds that

(3.16) $L_{2}(x)=C\displaystyle|b(x+h)-b(x)|g_{\gamma}(f)(x)\leq
C\displaystyle|h|g(f)(x).$

By the $L^{p}(w)$-bounds of $g$-function, we have

(3.17) $\|L_{2}\|_{L^{p}(w)}\leq C|h|\|f\|_{L^{p}(w)}.$

For ${L}_{1}$, by (3.6), we get

(3.18) $\displaystyle\begin{split}L_{1}&\leq
C\displaystyle|h|^{\delta}\Big{(}\int_{|h|^{2}}^{\infty}\Big{|}\int_{\mathbb{R}^{n}}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{|}^{2}tdt\Big{)}^{1/2}\\\ &\leq
C\displaystyle|h|^{\delta}\Big{\\{}\Big{(}\int_{|h|^{2}}^{1}\Big{|}\int_{|x-y|\geq\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{|}^{2}\Big{)}^{1/2}\\\
&\quad+\displaystyle\Big{(}\int_{1}^{\infty}\Big{|}\int_{|x-y|\geq\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{|}^{2}\Big{)}^{1/2}\\\
&\quad+\displaystyle\Big{(}\int_{|h|^{2}}^{1}\Big{|}\int_{|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{|}^{2}\Big{)}^{1/2}\\\
&\quad+\displaystyle\Big{(}\int_{1}^{\infty}\Big{|}\int_{|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{|}^{2}\Big{)}^{1/2}\Big{\\}}\\\
&=:C|h|^{\delta}\Big{\\{}L_{11}+L_{12}+L_{13}+L_{14}\Big{\\}}.\end{split}$

For $L_{11}$, if $t<1$, then $t^{-\delta/2}<t^{-1/2}$. If
$|x-y|<2^{k}\rho(x)$, $k=1,2,\cdots$, and $|h|\leq\sqrt{t}$, then

$|b(x+h)-b(x)|\leq C(2^{k}\rho(x)+\sqrt{t}).$

Using (3.14) and splitting the integrand domain into annuli, we obtain

$\displaystyle\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\int_{|x-y|\geq\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}|b(x+h)-b(y)||f(y)|dy$
$\displaystyle\leq
C\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}t^{\frac{M-n-1}{2}-1}\sum_{k=1}^{\infty}\int_{2^{k-1}\rho(x)\leq|x-y|<2^{k}\rho(x)}\frac{|f(y)|}{|x-y|^{M}}$
$\displaystyle\quad\times|b(x+h)-b(y)|dy$ $\displaystyle\leq
C\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}t^{\frac{M-n-1}{2}-1}\sum_{k=1}^{\infty}\frac{2^{k}\rho(x)+\sqrt{t}}{(2^{k}\rho(x))^{M}}\int_{|x-y|<2^{k}\rho(x)}|f(y)|dy$
$\displaystyle\leq
C\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}t^{-1}\Big{\\{}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{M-n-1}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-1-\theta\eta)}}{2^{k\theta\eta}(2^{k}\rho(x))^{n}}$
$\displaystyle\quad+\displaystyle\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{M-n}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-\theta\eta)}}{2^{k\theta\eta}(2^{k}\rho(x))^{n}}\Big{\\}}\int_{|x-y|<2^{k}\rho(x)}|f(y)|dy$
$\displaystyle\leq C\displaystyle
t^{-1}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N+1}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{M-n-1}M_{V,\eta}f(x).$

Choosing $M$ and $N$ such that $N>M>n+1+\theta\eta$, then it yields that

(3.19) $\displaystyle\begin{split}L_{11}&\leq
C\Big{(}\int_{0}^{1}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2(N-1)}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{2(M-n-1)}\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

For $L_{12}$, if $t>1$, then $t^{-\delta/2}<1$. Hence

$\displaystyle\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\int_{|x-y|\geq\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}|b(x+h)-b(y)||f(y)|dy$
$\displaystyle\leq
C\displaystyle\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}t^{\frac{M-n}{2}-1}\sum_{k=1}^{\infty}\frac{1}{(2^{k}\rho(x))^{M}}\int_{|x-y|<2^{k}\rho(x)}|f(y)|dy$
$\displaystyle\leq C\displaystyle
t^{-1}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{M-n}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-\theta\eta)}}{2^{k\theta\eta}(2^{k}\rho(x))^{n}}$
$\displaystyle\quad\times\int_{|x-y|<2^{k}\rho(x)}|f(y)|dy$ $\displaystyle\leq
C\displaystyle
t^{-1}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{M-n}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}M_{V,\eta}f(x).$

Choosing $M$ and $N$ such that $N>M>n+\theta\eta$, we have

(3.20)
$\displaystyle\begin{split}L_{12}\leq&C\Big{(}\int_{1}^{\infty}\Big{(}\frac{\sqrt{t}}{\rho(x)}\Big{)}^{2(M-n)}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-2N}\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

For $L_{13}$, decompose $L_{13}$ into the following three parts

(3.21) $\displaystyle\begin{split}L_{13}&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}\Big{(}\int_{|x-y|<\sqrt{t}}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}\Big{(}\int_{\sqrt{t}\leq|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{1}\Big{(}\int_{|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&=:L^{1}_{13}+L^{2}_{13}+L^{3}_{13}.\end{split}$

If $|x-y|<\sqrt{t}$, $t<\rho(x)^{2}<1$ and $|h|\leq\sqrt{t}$, then
$|b(x+h)-b(x)|\leq C\sqrt{t}.$ This leads to

(3.22) $\displaystyle\begin{split}L^{1}_{13}&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}\Big{(}\int_{|x-y|<\sqrt{t}}t^{-\frac{n+\delta+1}{2}}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}t^{-\delta}\Big{(}t^{-\frac{n}{2}}\int_{|x-y|<\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{\rho(x)^{2}}t^{-\delta}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

If $|x-y|<2^{k}\sqrt{t}$, $k=1,\cdots,n$, $t<\rho(x)^{2}<1$ and
$|h|\leq\sqrt{t}$, then

$|b(x+h)-b(x)|\leq C2^{k}\sqrt{t},\ \ \ \ \ \frac{\sqrt{t}}{\rho(x)}<1.$

Choosing $M,N>0$ such that $N>M>n+1+\theta\eta$, and applying (3.14) again, we
obtain

(3.23) $\displaystyle\begin{split}L^{2}_{13}&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}\Big{(}t^{\frac{M-n-\delta}{2}-1}\sum_{k=1}^{\infty}\frac{2^{k}\sqrt{t}}{(2^{k}\sqrt{t)^{M}}}\int_{2^{k-1}\sqrt{t}\leq|x-y|<2^{k}\sqrt{t}}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{|h|^{2}}^{\rho(x)^{2}}t^{-\delta}\Big{(}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-1-\theta\eta)}}{2^{k\theta\eta}(2^{k}\sqrt{t})^{n}}\int_{|x-y|<2^{k}\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{\rho(x)^{2}}t^{-\delta}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

If $|x-y|<\rho(x)$, $\rho(x)^{2}<t<1$ and $|h|\leq\sqrt{t}$, then
$|b(x+h)-b(x)|\leq C\sqrt{t}$ and $\frac{\sqrt{t}}{\rho(x)}<1.$ Therefore

(3.24) $\displaystyle\begin{split}L^{3}_{13}&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{1}\Big{(}\int_{|x-y|<\rho(x)}t^{-\frac{n+\delta+1}{2}}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{1}t^{-\delta}\Big{(}t^{-\frac{n}{2}}\int_{|x-y|<\rho(x)}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{1}t^{-\delta}\Big{(}\rho(x)^{-n}\int_{|x-y|<\rho(x)}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{0}^{1}t^{-\delta}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

This inequality, combining with (3.21), (3.22) and (3.23) yields that

(3.25) $L_{13}\leq CM_{V,\eta}f(x).$

For $L_{14}$, decompose it into the following three parts

(3.26) $\displaystyle\begin{split}L_{14}&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}\Big{(}\int_{|x-y|<\sqrt{t}}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}\Big{(}\int_{\sqrt{t}\leq|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\quad+C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{\infty}\Big{(}\int_{|x-y|<\rho(x)}t^{-\frac{n+\delta}{2}-1}e^{-\frac{c|x-y|^{2}}{t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad\times|b(x+h)-b(y)||f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&=:L^{1}_{14}+L^{2}_{14}+L^{3}_{14}.\end{split}$

It is easy to see that

(3.27) $\displaystyle\begin{split}L^{1}_{14}&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}\Big{(}\int_{|x-y|<\sqrt{t}}t^{-\frac{n+\delta}{2}-1}|f(y)|dy\Big{)}^{2}tdt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}t^{-(1+\delta)}\Big{(}t^{-\frac{n}{2}}\int_{|x-y|<\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}t^{-(1+\delta)}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

Using (3.14) with $M>n+\theta\eta$, we get

(3.28) $\displaystyle\begin{split}L^{2}_{14}&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}t^{-(1+\delta)}\Big{(}t^{\frac{M-n}{2}}\sum_{k=1}^{\infty}\int_{2^{k-1}\sqrt{t}\leq|x-y|<2^{k}\sqrt{t}}\frac{|f(y)|}{|x-y|^{M}}dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}t^{-(1+\delta)}\Big{(}\sum_{k=1}^{\infty}\frac{2^{-k(M-n-\theta\eta)}}{2^{k\theta\eta}(2^{k}\sqrt{t})^{n}}\int_{|x-y|<2^{k}\sqrt{t}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{1}^{\rho(x)^{2}}t^{-(1+\delta)}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

If $t\geq 1$, then $t^{-\delta/2}\leq 1$. Taking $N=1$, we have
$(\sqrt{t}/\rho(x))^{-2N}=\rho(x)^{2}/t$, therefore

(3.29) $\displaystyle\begin{split}L^{3}_{14}&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{\infty}\Big{(}\frac{\rho(x)}{t}\Big{)}^{2}\Big{(}\int_{|x-y|<\rho(x)}t^{-\frac{n}{2}}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{\infty}\Big{(}\frac{\rho(x)}{t}\Big{)}^{2}\Big{(}\rho(x)^{-n}\int_{|x-y|<\rho(x)}|f(y)|dy\Big{)}^{2}dt\Big{)}^{1/2}\\\
&\leq
C\displaystyle\Big{(}\int_{\rho(x)^{2}}^{\infty}\Big{(}\frac{\rho(x)}{t}\Big{)}^{2}dt\Big{)}^{1/2}M_{V,\eta}f(x)\\\
&\leq CM_{V,\eta}f(x).\end{split}$

Sum up (3.26)-(3.29) in all, we have

(3.30) $L_{14}\leq CM_{V,\eta}f(x).$

Inequality (3.30) together with (3.25), (3.20), (3.19) and (3.18) gives that

(3.31) $L_{1}\leq C|h|^{\delta}M_{V,\eta}f(x).$

By Lemma 2.4 for any $p^{\prime}\leq\eta<\infty$, we get

(3.32) $\|L_{1}\|_{L^{p}(w)}\leq C|h|^{\delta}\|M_{V,\eta}f\|_{L^{p}(w)}\leq
C|h|^{\delta}\|f\|_{L^{p}(w)}.$

which together with (3.15) and (3.16) yields that

$\|g_{b}f(h+\cdot)-g_{b}f(\cdot)\|_{L^{p}(w)}\leq
C(|h|+|h|^{\delta})\|f\|_{L^{p}(w)}.$

This completes the proof of Theorem 1.1. ∎

###### Proof of Theorem 1.2.

We will complete the proof of Theorem 1.2 by slightly modifying the proof in
[34]. We only give the proof of the modified part here. Since the condition
(1.8) was used only in the step to prove for $w\in A_{p}^{\rho,\theta}$,

(3.33)
$\lim_{A\rightarrow\infty}\int_{|x|>A}|\mathcal{T}^{*}_{b,\gamma}f(x)|^{p}w(x)dx=0,$

whenever $f\in F$. Therefore it suffices to prove (3.33) without the condition
(1.8). As the proof of Theorem 1.1, it suffices to show that for any $N>0$,
there is a constant $C_{N}>0$ so that

(3.34)
$\sup_{t>0}k_{t}(x,y)\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.$

In fact, we may control the left side of (3.34) by

$\displaystyle\begin{split}\sup_{t>0}k_{t}(x,y)&\leq
C_{N}\sup_{t>0}t^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{5t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)}\Big{)}^{-N}\\\
&\leq
C_{N}\sup_{\sqrt{t}\geq|x-y|}t^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{5t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\quad+C_{N}\sup_{\sqrt{t}<|x-y|}t^{-\frac{n}{2}}e^{-\frac{|x-y|^{2}}{5t}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&=I+J.\end{split}$

We need to consider the contributions of $I$ and $J$. For $I$, using the
estimate $e^{-s}\leq\frac{C}{s^{n/2}}$, we have

$I\leq\sup_{\sqrt{t}\geq|x-y|}\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.\\\
$

For $J$, using the estimate $e^{-s}\leq\frac{C}{s^{n/2+N/2}}$, it follows that

$\displaystyle\begin{split}J&\leq
C_{N}\sup_{\sqrt{t}<|x-y|}t^{-\frac{n}{2}}\Big{(}\frac{t}{|x-y|^{2}}\Big{)}^{\frac{n}{2}+\frac{N}{2}}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&=\sup_{\sqrt{t}<|x-y|}\frac{C_{N}}{|x-y|^{n}}\Big{(}\frac{\sqrt{t}}{|x-y|}\Big{)}^{N}\Big{(}1+\frac{\sqrt{t}}{\rho(x)}\Big{)}^{-N}\\\
&\leq\sup_{\sqrt{t}<|x-y|}\frac{C_{N}}{|x-y|^{n}}\Big{(}\frac{\sqrt{t}+\rho(x)}{|x-y|+\rho(x)}\Big{)}^{N}\Big{(}\frac{\sqrt{t}+\rho(x)}{\rho(x)}\Big{)}^{-N}\\\
&=\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N}.\end{split}$

Therefore

$\sup_{t>0}k_{t}(x,y)\leq\frac{C_{N}}{|x-y|^{n}}\Big{(}1+\frac{|x-y|}{\rho(x)}\Big{)}^{-N},$

holds for any $N>0$. Then, similar to the proof of Theorem 1.1, we may show
that (3.33) holds. This finishes the proof of Theorem 1.2.

∎

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