1 Introduction and results

Let \(\mathcal{L}=-\Delta +V\) be the Schrödinger operator, where the nonnegative potential V belongs to the reverse Hölder class \(\mathit{RH}_{s}\) with \(s> n/2\), \(n\geq 3\). Many papers related to Schrödinger operator have appeared (see [1,2,3,4,5]). In recent years, some researchers have studied the boundedness of the commutators generated by the operators associated with \(\mathcal{L}\) and the BMO type space (see [6,7,8,9]). In this paper, we investigated the boundedness of the commutator \([b,T_{\beta}]\), where \(T_{\beta }=V^{\alpha}\nabla\mathcal{L}^{-\beta} \) and the function \(b\in \operatorname{BMO}_{\theta }(\rho )\). We note that the space \(\operatorname{BMO}_{\theta }(\rho )\) is larger than the space \(\operatorname{BMO}(\mathbb{R}^{n})\).

For \(s>1\), a nonnegative locally \(L^{s}\)-integrable function V is said to belong to \(\mathit{RH}_{s}\) if there exists a constant \(C>0\) such that the reverse Hölder inequality

$$ \biggl(\frac{1}{ \vert B \vert } \int _{B}V(y)^{s}\,dy \biggr)^{1/s}\leq \frac{C}{ \vert B \vert } \int _{B}V(y)\,dy $$

holds for every ball \(B\subset \mathbb{R}^{n}\). It is obvious that \(\mathit{RH}_{s_{1}}\subseteq \mathit{RH}_{s_{2}}\) for \(s_{1}\geq s_{2}\).

As in [2], for a given potential \(V\in \mathit{RH}_{s}\) with \(s>n/2\), we will use the auxiliary function \(\rho (x)\) defined as

$$ \rho (x)=\sup \biggl\{ r>0: \frac{1}{r^{n-2}} \int _{B(x,r)}V(y)\,dy\leq 1 \biggr\} , \quad x\in \mathbb{R}^{n}. $$

It is well known that \(0<\rho (x)<\infty \) for any \(x\in \mathbb{R} ^{n}\).

Let \(\mathcal{L}=-\Delta +V\) be the Schrödinger operator on \(\mathbb{R}^{n}\), where \(V\in \mathit{RH}_{s}\) with \(s>n/2\) and \(n\geq 3\). We know \(\mathcal{L}\) generates a \((C_{0})\) semigroup \(\{e^{-t \mathcal{L}}\}_{t>0}\). The maximal function with respect to the semigroup \(\{e^{-t\mathcal{L}}\}_{t>0}\) is defined by \(M^{\mathcal{L}}f(x)= \sup_{t>0}|e^{-t\mathcal{L}}f(x)|\). The Hardy space associated with \(\mathcal{L}\) is defined as follows (see [3, 4]).

Definition 1

We say that f is an element of \(H_{\mathcal{L}}^{1}(\mathbb{R}^{n})\) if the maximal function \(M^{\mathcal{L}}f\) belongs to \(L^{1}( \mathbb{R}^{n})\). The quasi-norm of f is defined by

$$ \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}= \bigl\Vert M^{\mathcal{L}}f \bigr\Vert _{L^{1}(\mathbb{R}^{n})}. $$

Definition 2

Let \(1< q\leq \infty \). A measurable function a is called an \((1,q)_{\rho }\)-atom related to the ball \(B(x_{0},r)\) if \(r<\rho (x _{0})\) and the following conditions hold:

  1. (1)

    \(\operatorname{supp} a\subset B(x_{0},r)\);

  2. (2)

    \(\|a\|_{L^{q}(\mathbb{R}^{n})}\leq |B(x_{0},r)|^{1/q-1}\);

  3. (3)

    \(\int _{B(x_{0},r)}a(x)\,dx=0\) if \(r<\rho (x_{0})/4\).

The space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) admits the following atomic decomposition (see [3, 4]).

Proposition 1

Let \(f\in L^{1}(\mathbb{R}^{n})\). Then \(f\in H_{\mathcal{L}}^{1}( \mathbb{R}^{n})\) if and only if f can be written as \(f=\sum_{j} \lambda _{j}a_{j}\), where \(a_{j}\) are \((1,q)_{\rho }\)- atoms, \(\sum_{j}|\lambda _{j}|<\infty \), and the sum converges in the \(H_{\mathcal{L}}^{1}(\mathbb{R}^{n})\) quasi-norm. Moreover

$$ \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}\sim \inf \biggl\{ \sum _{j} \vert \lambda _{j} \vert \biggr\} , $$

where the infimum is taken over all atomic decompositions of f into \((1,q)_{\rho }\)- atoms.

Following [10], the space \(\operatorname{BMO}_{\theta }(\rho )\) with \(\theta \geq 0\) is defined as the set of all locally integrable functions b such that

$$ \frac{1}{ \vert B(x,r) \vert } \int _{B(x,r)} \bigl\vert b(y)-b_{B} \bigr\vert \,dy \leq C \biggl(1+\frac{r}{ \rho (x)} \biggr)^{\theta } $$

for all \(x\in \mathbb{R}^{n}\) and \(r>0\), where \(b_{B}=\frac{1}{|B|} \int _{B}b(y)\,dy\). A norm for \(b\in \operatorname{BMO}_{\theta }(\rho )\), denoted by \([b]_{\theta }\), is given by the infimum of the constants in the inequalities above. Clearly, \(\operatorname{BMO}\subset \operatorname{BMO}_{\theta }(\rho )\).

We consider the operator

$$ T_{\beta }=V^{\alpha }\nabla \mathcal{L}^{-\beta },\quad \frac{1}{2}\leq \beta \leq 1, \beta -\alpha =\frac{1}{2}. $$

The boundedness of operator \(T_{1/2}\) and its commutator have been researched under the condition \(V\in \mathit{RH}_{s}\) for \(n/2< s< n\). In [2], Shen showed that \(T_{1/2}\) is bounded on \(L^{p}(\mathbb{R} ^{n}) \) for \(1< p< p_{0}\), \(\frac{1}{p_{0}}=\frac{1}{s}-\frac{1}{n}\). For \(b\in \operatorname{BMO}(\mathbb{R}^{n})\), Guo, Li and Peng [11] investigated the \(L^{p}\)-boundedness of commutator \([b,T_{1/2}]\) for \(1< p< p_{0}\); Li and Peng [12] studied the boundedness of \([b, T_{1/2}]\) from \(H_{\mathcal{L}}^{1}(\mathbb{R}^{n})\) into weak \(L^{1}(\mathbb{R}^{n})\). When \(b\in \operatorname{BMO}_{\theta }(\rho ) \), Bongioanni, Harboure and Salinas [10] obtained the \(L^{p}\)-boundedness of \([b,T_{1/2}]\) and Liu, Sheng and Wang [13] proved that \([b,T_{1/2}]\) is bounded from \(H_{\mathcal{L}}^{1}(\mathbb{R}^{n})\) to weak \(L^{1}(\mathbb{R}^{n})\). More boundedness of commutator \([b,T_{1/2}]\) can be found in [14] and [15].

For \(1/2<\beta \leq 1\), \(\beta -\alpha =1/2\), \(n/2< s< n\), Sugano [5] established the estimate for \(T^{*}_{\beta }\) (the adjoint operator of \(T_{\beta }\)), and proved that there exists a constant C such that

$$ \bigl\vert T^{*}_{\beta }f(x) \bigr\vert \leq C M\bigl( \vert f \vert ^{{p}'_{\alpha }}\bigr) (x)^{1/{p}'_{\alpha }} $$

for all \(f\in C_{0}^{\infty }(\mathbb{R}^{n})\), where \(\frac{1}{{p} _{\alpha }}=\frac{\alpha +1}{s}-\frac{1}{n}\), and \(\frac{1}{{p}_{ \alpha }}+ \frac{1}{{p}'_{\alpha }}=1\). Then, by the boundedness of maximal function, we get

Theorem 1

Suppose \(V\in \mathit{RH}_{s}\) with \(n/2< s< n\). Let \(1/2< \beta \leq 1\), \(\frac{1}{p _{\alpha }}=\frac{\alpha +1}{s}-\frac{1}{n}\). Then

$$ \bigl\Vert T^{*}_{\beta }f \bigr\Vert _{L^{p}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{L^{p}( \mathbb{R}^{n})} $$

for \(p'_{\alpha }< p\leq \infty \), and by duality we get

$$ \Vert T_{\beta }f \Vert _{L^{p}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{L^{p}(\mathbb{R} ^{n})} $$

for \(1\leq p< p_{\alpha }\).

Inspired by the above results, in the present work, we are interested in the boundedness of \([b,T_{\beta }]\). Our main results are as follows.

Theorem 2

Suppose \(V\in \mathit{RH}_{s}\) with \(n/2< s< n\). Let \(1/2< \beta \leq 1\), \(b \in \operatorname{BMO}_{\theta }(\rho )\). Then,

$$ \bigl\Vert \bigl[b,T^{*}_{\beta }\bigr](f) \bigr\Vert _{L^{p}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{L^{p}( \mathbb{R}^{n})} $$

for \(p'_{\alpha }< p< \infty \), and

$$ \bigl\Vert [b,T_{\beta }](f) \bigr\Vert _{L^{p}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{L^{p}( \mathbb{R}^{n})} $$

for \(1< p< p_{\alpha }\), where \(\frac{1}{p_{\alpha }}= \frac{\alpha +1}{s}-\frac{1}{n}\).

Theorem 3

Suppose \(V\in \mathit{RH}_{s}\) with \(n/2< s< n\). Let \(1/2<\beta \leq 1\), \(b\in \operatorname{BMO} _{\theta }(\rho )\). Then,

$$ \bigl\Vert [b,T_{\beta }](f) \bigr\Vert _{WL^{1}(\mathbb{R}^{n})}\leq C \Vert f \Vert _{H_{ \mathcal{L}}^{1}(\mathbb{R}^{n})}. $$

In this paper, we shall use the symbol \(A\lesssim B\) to indicate that there exists a universal positive constant c, independent of all important parameters, such that \(A\leq cB\). \(A\sim B\) means that \(A\lesssim B\) and \(B\lesssim A\).

2 Some preliminaries

We recall some important properties concerning the auxiliary function \(\rho (x)\) which have been proved by Shen [2]. Throughout this section we always assume \(V\in \mathit{RH}_{s}\) with \(n/2< s< n\).

Proposition 2

There exist constants C and \(k_{0}\geq 1\) such that

$$ C^{-1}\rho (x) \biggl(1+\frac{ \vert x-y \vert }{\rho (x)} \biggr)^{-k_{0}}\leq \rho (y)\leq C\rho (x) \biggl(1+\frac{ \vert x-y \vert }{\rho (x)} \biggr)^{\frac{k _{0}}{1+k_{0}}} $$

for all \(x,y \in \mathbb{R}^{n}\).

Assume that \(Q=B(x_{0},\rho (x_{0}))\), for any \(x\in Q \), then Proposition 2 tells us that \(\rho (x)\sim \rho (y)\), if \(|x-y|< C\rho (x)\). It is easy to get the following result from Proposition 2.

Lemma 1

Let \(k\in \mathbb{N}\) and \(x\in 2^{k+1}B(x_{0},r)\setminus 2^{k}B(x _{0},r)\). Then we have

$$ \frac{1}{ (1+\frac{2^{k}r}{\rho (x)} )^{N}}\lesssim \frac{1}{ (1+\frac{2^{k}r}{\rho (x_{0})} )^{N/(k_{0}+1)}}. $$

Lemma 2

There exists a constant \(l_{0}>0\) such that

$$ \frac{1}{r^{n-2}} \int _{B(x,r)}V(y)\,dy\lesssim \biggl(1+ \frac{r}{\rho (x)} \biggr)^{l_{0}}. $$

The following finite overlapping property was given by Dziubański and Zienkiewicz in [3].

Proposition 3

There exists a sequence of points \(\{x_{k}\}_{k=1}^{\infty }\) in \(\mathbb{R}^{n}\), so that the family of critical balls \(Q_{k}=B(x_{k}, \rho (x_{k}))\), \(k\geq 1\), satisfies

  1. (i)

    \(\bigcup_{k} Q_{k}=\mathbb{R}^{n}\).

  2. (ii)

    There exists \(N=N(\rho )\) such that for every \(k\in N\), \(\operatorname{card}\{j: 4Q_{j}\cap 4Q_{k}\}\leq N\).

For \(\alpha >0\), \(g\in L_{\mathrm{loc}}^{1}(\mathbb{R}^{n})\) and \(x\in \mathbb{R}^{n}\), we introduce the following maximal functions:

$$ M_{\rho ,\alpha }g(x)=\sup_{x\in B\in \mathcal{B}_{\rho ,\alpha }} \frac{1}{ \vert B \vert } \int _{B} \bigl\vert g(y) \bigr\vert \,dy, $$

and

$$ M^{\sharp }_{\rho ,\alpha }g(x)= \sup_{x\in B\in \mathcal{B}_{\rho ,\alpha }} \frac{1}{ \vert B \vert } \int _{B} \bigl\vert g(y)-g _{B} \bigr\vert \,dy, $$

where \(\mathcal{B}_{\rho ,\alpha }=\{B(z,r): z\in \mathbb{R}^{n} \text{ and } r\leq \alpha \rho (y)\}\).

The following Fefferman–Stein type inequality can be found in [10].

Proposition 4

For \(1< p<\infty \), then there exist δ and γ such that if \(\{Q_{k}\}_{k}\) is a sequence of balls as in Proposition 3 then

$$ \int _{\mathbb{R}^{n}} \bigl\vert M_{\rho ,\delta }g(x) \bigr\vert ^{p}\,dx\lesssim \int _{\mathbb{R}^{n}} \bigl\vert M^{\sharp }_{\rho ,\gamma }g(x) \bigr\vert ^{p}\,dx +\sum_{k} \vert Q_{k} \vert \biggl(\frac{1}{ \vert Q_{k} \vert } \int _{2Q_{k}} \vert g \vert \biggr)^{p} $$

for all \(g\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})\).

We have the following result for the function \(b\in \operatorname{BMO}_{\theta }( \rho )\).

Lemma 3

([10])

Let \(1\leq s<\infty \), \(b\in \operatorname{BMO}_{\theta }(\rho )\), and \(B=B(x,r)\). Then

$$ \biggl(\frac{1}{ \vert 2^{k}B \vert } \int _{2^{k}B} \bigl\vert b(y)-b_{B} \bigr\vert ^{s}\,dy \biggr)^{1/s} \lesssim [b]_{\theta }k \biggl(1+ \frac{2^{k}r}{\rho (x)} \biggr)^{ \theta '} $$

for all \(k\in \mathbb{N} \), with \(r>0\), where \(\theta '=(k_{0}+1) \theta \) and \(k_{0}\) is the constant appearing in Proposition 2.

We give an estimate of fundamental solutions; this result can be found in [2]. We denote by \(\varGamma (x,y,\lambda )\) the fundamental solution of \(-\Delta +(V(x)+i\lambda )\), and then \(\varGamma (x,y,\lambda )=\varGamma (y,x,-\lambda )\).

Lemma 4

Assume that \(-\Delta u+(V(x)+i\lambda )u=0\) in \(B(x_{0},2R)\) for some \(x_{0}\in \mathbb{R}^{n}\). Then, there exists a \(k'_{0}\) such that

$$ \biggl( \int _{B(x_{0}, R)} \vert \nabla u \vert ^{t}\,dx \biggr)^{1/t}\lesssim R^{n/s-2} \biggl(1+\frac{R}{\rho (x_{0})} \biggr)^{k'_{0}} \sup_{B(x_{0},2R)} \vert u \vert , $$

where \(1/t=1/s-1/n\).

Suppose \(\mathcal{W}_{\beta }= \nabla \mathcal{L}^{-\beta }\). Let \(\mathcal{W}_{\beta }^{*}\) be the adjoint operator of \(\mathcal{W} _{\beta }\), K and \(K^{*}\) be the kernels of \(\mathcal{W}_{\beta }\) and \(\mathcal{W}_{\beta }^{*}\) respectively, then \(K(x,z)=K^{*}(z,x)\), and we have the following estimates.

Lemma 5

Suppose \(1/2<\beta \leq 1\).

  1. (i)

    For every N there exists a constant \(C_{N}\) such that

    $$ \bigl\vert K^{*}(x,z) \bigr\vert \leq \frac{C_{N}}{ (1+\frac{ \vert x-z \vert }{\rho (x)} )^{N}} \frac{1}{ \vert x-z \vert ^{n-2\beta }} \biggl( \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert \xi -z \vert ^{n-1}}\,d\xi +\frac{1}{ \vert x-z \vert } \biggr). $$

    Moreover, the inequality above also holds with \(\rho (x)\) replaced by \(\rho (z)\).

  2. (ii)

    For every N and \(0<\delta <\min \{1,2-n/q_{0}\}\) there exists a constant \(C_{N}\) such that

    $$\begin{aligned} \bigl\vert K^{*}(x,z) -K^{*}(y,z) \bigr\vert &\leq \frac{C_{N}}{ (1+ \frac{ \vert x-z \vert }{\rho (x)} )^{N}} \\ &\quad {} \times \frac{ \vert x-y \vert ^{\delta }}{ \vert x-z \vert ^{n-2\beta +\delta }} \biggl( \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert \xi -z \vert ^{n-1}}\,d\xi +\frac{1}{ \vert x-z \vert } \biggr) \end{aligned}$$

    whenever \(|x-y|<\frac{1}{16}|x-z|\). Moreover, the inequality above also holds with \(\rho (x)\) replaced by \(\rho (z)\).

Proof

The proof of (i) can be found in [5], page 449. Let us prove (ii). By (6) of [5] we know

$$ K (x,z)= \textstyle\begin{cases} \frac{1}{2\pi }\int _{\mathbb{R}}(-i\tau )^{-\beta }\nabla _{x}\varGamma (x,z,\tau )\,d \tau , &\text{for } \frac{1}{2}< \beta < 1, \\ \nabla _{x}\varGamma (x,z,0),& \text{for } \beta =1. \end{cases} $$

Then

$$ \bigl\vert K^{*} (x,z)-K^{*} (y,z) \bigr\vert \lesssim \int _{-\infty }^{\infty } \vert \tau \vert ^{- \beta } \bigl\vert \nabla _{z}\varGamma (z,x,\tau )-\nabla _{z} \varGamma (z,y,\tau ) \bigr\vert \,d \tau $$

for \(\frac{1}{2}<\beta <1\) and

$$ \bigl\vert K^{*} (x,z)-K^{*} (y,z) \bigr\vert \lesssim \bigl\vert \nabla _{z}\varGamma (z,x,0)- \nabla _{z}\varGamma (z,y,0) \bigr\vert $$

for \(\beta =1\).

Fix \(x,z\in \mathbb{R}^{n}\) and let \(R=|x-z|/8\), \(1/t=1/s-1/n\), \(\delta =2-n/s>0\). For any \(|x-y|< R/2\), it follows from the Morrey embedding theorem (see [16]) and Lemma 4 that

$$\begin{aligned} & \bigl\vert \nabla _{z}\varGamma (z,x,\tau )-\nabla _{z} \varGamma (z,y,\tau ) \bigr\vert \\ &\quad \lesssim \vert x-y \vert ^{1-n/t} \biggl( \int _{B(x,R)} \bigl\vert \nabla _{u}\nabla _{z} \varGamma (z,u,\tau ) \bigr\vert ^{t}\,du \biggr)^{1/t} \\ &\quad \lesssim \vert x-y \vert ^{1-n/t}R^{(n/s)-2} \biggl(1+ \frac{R}{\rho (x)} \biggr)^{k _{0}}\sup_{u\in B(x,2R)} \bigl\vert \nabla _{z} \varGamma (z,u,\tau ) \bigr\vert . \end{aligned}$$

It follows from [11, p. 428] that

$$\begin{aligned} &\sup_{u\in B(x,2R)} \bigl\vert \nabla _{z} \varGamma (z,u, \tau ) \bigr\vert \\ &\quad \lesssim \frac{C_{k_{1}}}{(1+ \vert \tau \vert ^{1/2} \vert z-u \vert )^{k_{1}} (1+\frac{ \vert z-u \vert }{ \rho (z)} )^{k_{1}}} \frac{1}{ \vert z-u \vert ^{n-2}} \\ &\qquad {} \times \biggl( \int _{B(z, \vert z-u \vert /4)}\frac{V(\xi )}{ \vert z-\xi \vert ^{n-1}}\,d \xi +\frac{1}{ \vert z-u \vert } \biggr). \end{aligned}$$

Then, by the fact that \(6R\leq |z-u|\leq 10R\), we get

$$\begin{aligned} & \bigl\vert \nabla _{z}\varGamma (z,x,\tau )-\nabla _{z} \varGamma (z,y,\tau ) \bigr\vert \\ &\quad \lesssim \frac{ \vert x-y \vert ^{\delta }}{ \vert x-z \vert ^{n-2+\delta }} \frac{C_{N}}{(1+ \vert \tau \vert ^{1/2} \vert x-z \vert )^{N} (1+\frac{ \vert x-z \vert }{\rho (x)} )^{N}} \\ &\qquad {} \times \biggl( \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert z-\xi \vert ^{n-1}}\,d \xi +\frac{1}{ \vert x-z \vert } \biggr). \end{aligned}$$

Thus, for \(\beta =1\),

$$\begin{aligned} \bigl\vert K^{*} (x,z)-K^{*} (y,z) \bigr\vert &\lesssim \bigl\vert \nabla _{z}\varGamma (z,x,0)- \nabla _{z}\varGamma (z,y,0) \bigr\vert \\ &\lesssim \frac{ \vert x-y \vert ^{\delta }}{ \vert x-z \vert ^{n-2+\delta }} \frac{C_{N}}{ (1+\frac{ \vert x-z \vert }{\rho (x)} )^{N}} \biggl( \int _{B(z, \vert x-z \vert /4)}\frac{V( \xi )}{ \vert z-\xi \vert ^{n-1}}\,d\xi +\frac{1}{ \vert x-z \vert } \biggr). \end{aligned}$$

Note that

$$ \int _{-\infty }^{\infty } \frac{ \vert \tau \vert ^{-\beta }\,d \tau }{(1+ \vert \tau \vert ^{1/2} \vert x-z \vert )^{k}}\lesssim \vert x-z \vert ^{2 \beta -2}. $$

Then, for \(\frac{1}{2}<\beta <1\), we have

$$\begin{aligned} \bigl\vert K^{*} (x,z)- K^{*} (y,z) \bigr\vert &\lesssim \frac{ \vert x-y \vert ^{\delta }}{ \vert x-z \vert ^{n+ \delta -2\beta }} \\ &\quad {}\times \frac{C_{N}}{ (1+\frac{ \vert x-z \vert }{\rho (x)} )^{N}} \biggl( \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert \xi -z \vert ^{n-1}}\,d\xi +\frac{1}{ \vert x-z \vert } \biggr). \end{aligned}$$

By Lemma 2, we know that the inequality above also holds with \(\rho (x)\) replaced by \(\rho (z)\). □

3 Proof of main results

Before proving Theorem 2, we need to give some necessary lemmas.

Lemma 6

Let \(V\in \mathit{RH}_{s}\) with \(n/2< s< n\), \(\frac{1}{{p}_{\alpha }}=\frac{\alpha +1}{s}-\frac{1}{n}\), and \(b\in \operatorname{BMO}_{\theta }(\rho )\). Then, for any \({p}'_{\alpha }< t<\infty \), we have

$$ \frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert \bigl[b, T^{*}_{\beta } \bigr]f \bigr\vert \lesssim [b]_{\theta } \inf_{y\in Q}M_{t}f(y) $$

for all \(f\in L^{t}_{\mathrm{loc}}(\mathbb{R}^{n})\) and every ball \(Q=B(x_{0}, \rho (x_{0}))\).

Proof

Let \(f\in L^{t}_{\mathrm{loc}}(\mathbb{R}^{n})\) and \(Q=B(x_{0},\rho (x _{0}))\). We consider

$$ \bigl[b,T^{*}_{\beta }\bigr]f=(b-b_{Q})T^{*}_{\beta }f-T^{*}_{\beta } \bigl(f(b-b_{Q})\bigr). $$
(1)

By Hölder’s inequality with \(t>{p}'_{\alpha }\) and Lemma 3,

$$\begin{aligned} \frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert (b-b_{Q})T^{*}_{\beta }f \bigr\vert &\lesssim \biggl( \frac{1}{ \vert Q \vert } \int _{Q} \vert b-b_{Q} \vert ^{t'} \biggr)^{1/t'} \biggl(\frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert T^{*}_{\beta }f \bigr\vert ^{t} \biggr)^{1/t} \\ &\lesssim [b]_{\theta } \biggl(\frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert T^{*}_{\beta }f \bigr\vert ^{t} \biggr)^{1/t}. \end{aligned}$$

Write \(f=f_{1}+f_{2}\) with \(f_{1}=f\chi _{2Q}\). By Theorem 1, we know that \(T^{*}_{\beta }\) is bounded on \(L^{t}(\mathbb{R}^{n})\) with \(t> {p}'_{\alpha } \), and then

$$ \biggl(\frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert T^{*}_{\beta }f_{1} \bigr\vert ^{t} \biggr)^{1/t} \lesssim \biggl( \frac{1}{ \vert Q \vert } \int _{2Q} \vert f \vert ^{t} \biggr)^{1/t}\lesssim \inf_{y\in Q}M_{t}f(y). $$

For \(x\in Q\), using (i) in Lemma 5, we get

$$ \bigl\vert T^{*}_{\beta }f_{2}(x) \bigr\vert = \biggl\vert \int _{(2Q)^{c}}V(z)^{\alpha }K^{*}(x,z)f(z)\,dz \biggr\vert \lesssim I_{1}(x)+I_{2}(x), $$

where

$$ I_{1}(x)\lesssim \int _{(2Q)^{c}}\frac{ \vert f(z) \vert }{ (1+\frac{ \vert x-z \vert }{ \rho (x)} )^{N}} \frac{V(z)^{\alpha }}{ \vert x-z \vert ^{n-2\beta +1}}\,dz $$

and

$$ I_{2}(x)\lesssim \int _{(2Q)^{c}}\frac{ \vert f(z) \vert }{ (1+\frac{ \vert x-z \vert }{ \rho (x)} )^{N}} \frac{V(z)^{\alpha }}{ \vert x-z \vert ^{n-2\beta }} \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert \xi -z \vert ^{n-1}}\,d\xi \,dz. $$

To deal with \(I_{2}(x)\), note that \(\rho (x)\sim \rho (x_{0})\) and \(|x-z|\sim |x_{0}-z|\) for \(x\in Q\). We split \((2Q)^{c}\) into annuli to obtain

$$ I_{2}(x)\lesssim \sum_{k\geq 2} \frac{2^{-kN}(2^{k}\rho (x_{0}))^{2 \beta }}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q} \bigl\vert f(z) \bigr\vert V(z)^{\alpha } \mathcal{I}_{1}(V\chi _{2^{k}Q}) (z)\,dz. $$

Observe that \(\frac{1}{{p}'_{\alpha }}+\frac{\alpha }{s}+\frac{1}{q _{1}}=1\), \(\frac{1}{q_{1}}=\frac{1}{s}-\frac{1}{n}\), \(t> {p}'_{\alpha }\), and \(\beta -\alpha =1/2\). Then by Hölder’s inequality and the boundedness of fractional integral \(\mathcal{I}_{1}: L^{s}\rightarrow L^{q_{1}}\) with \(\frac{1}{q_{1}}=\frac{1}{s}-\frac{1}{n}\), we get

$$\begin{aligned} I_{2}(x)&\lesssim \sum_{k\geq 2}{2^{-kN}} {\bigl(2^{k}\rho (x_{0})\bigr)^{2 \beta }} \biggl( \frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q} \bigl\vert f(z) \bigr\vert ^{ {p}'_{\alpha }}\,dz \biggr)^{1/{{p}'_{\alpha }}} \\ &\quad {}\times \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q}V(z)^{s}\,dz \biggr) ^{\alpha /s} \biggl( \frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q} \bigl\vert \mathcal{I}_{1}(V\chi _{2^{k}Q}) (z) \bigr\vert ^{q_{1}}\,dz \biggr)^{1/{q_{1}}} \\ &\lesssim \sum_{k\geq 2}{2^{-kN}} { \bigl(2^{k}\rho (x_{0})\bigr)^{2\beta +n/s-n/ {q_{1}}}} \biggl( \frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q}V(z)^{s}\,dz \biggr) ^{\alpha /s} \\ &\quad {}\times \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q}V(z)^{s}\,dz \biggr) ^{1/s}\inf _{y\in Q}M_{t}f(y). \end{aligned}$$

Then, since \(V\in \mathit{RH}_{s}\), from Lemma 2 and \(2\beta +n(1/s-1/ {q_{1}})-2\alpha -2=0\), we get

$$\begin{aligned} I_{2}(x)&\lesssim \sum_{k\geq 2}2^{-kN} \bigl(2^{k}\rho (x_{0})\bigr)^{2\beta +n(1/s-1/ {q_{1}})-2\alpha -2} \bigl(1+2^{k}\bigr)^{(\alpha +1) l_{0}}\inf_{y\in Q}M_{t}f(y) \\ &\lesssim \inf_{y\in Q}M_{t}f(y). \end{aligned}$$
(2)

For \(I_{1}(x)\), we split \((2Q)^{c}\) into annuli to obtain

$$ I_{1}(x)\lesssim \sum_{k\geq 1} \frac{2^{-kN}(2^{k}\rho (x_{0}))^{2 \beta -1}}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q} \bigl\vert f(z) \bigr\vert V(z)^{\alpha }\,dz. $$

By Hölder’s inequality with \(\frac{1}{{p}'_{\alpha }}+\frac{ \alpha }{s}+\frac{1}{q_{1}}=1\), \(t> {p}'_{\alpha }\), \(\beta -\alpha =1/2\), and Lemma 2, we get

$$\begin{aligned} I_{1}(x)&\lesssim \sum_{k\geq 1}{2^{-kN}} {\bigl(2^{k}\rho (x_{0})\bigr)^{2 \beta -1}} \biggl( \frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q} \bigl\vert f(z) \bigr\vert ^{ {p}'_{\alpha }}\,dz \biggr)^{1/{{p}'_{\alpha }}} \\ &\quad {}\times \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q}V(z)^{s}\,dz \biggr) ^{\alpha /s} \\ &\lesssim \sum_{k\geq 1}\frac{2^{-kN}}{(2^{k}\rho (x_{0}))^{1-2 \beta }} \biggl( \frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q}V(z)\,dz \biggr) ^{\alpha }\inf _{y\in Q}M_{t}f(y) \\ &\lesssim \sum_{k\geq 1}2^{-kN} \bigl(1+2^{k}\bigr)^{\alpha l_{0}}\inf_{y \in Q}M_{t}f(y) \lesssim \inf_{y\in Q}M_{t}f(y). \end{aligned}$$
(3)

To deal with the second term of (1), we write again \(f=f_{1}+f_{2}\). Choosing \({p}'_{\alpha }<\bar{t}<t\) and denoting \(\nu =\frac{\bar{t} t}{t-\bar{t}}\), using the boundedness of \(T_{\beta }^{*}\) on \(L^{\bar{t}}(\mathbb{R}^{n})\) and applying Hölder’s inequality,

$$\begin{aligned} \frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert T_{\beta }^{*}f_{1}(b-b_{Q}) \bigr\vert &\lesssim \biggl(\frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert T_{\beta }^{*}f_{1}(b-b_{Q}) \bigr\vert ^{\bar{t}} \biggr)^{1/\bar{t}} \\ &\lesssim \biggl(\frac{1}{ \vert Q \vert } \int _{Q} \bigl\vert f_{1}(b-b_{Q}) \bigr\vert ^{\bar{t}} \biggr) ^{1/\bar{t}} \\ &\lesssim \biggl(\frac{1}{ \vert Q \vert } \int _{2Q} \vert f \vert ^{t} \biggr)^{1/t} \biggl(\frac{1}{ \vert Q \vert } \int _{2Q} \vert b-b_{q} \vert ^{\nu } \biggr)^{1/\nu } \\ &\lesssim [b]_{\theta }\inf_{y\in Q}M_{t}f(y). \end{aligned}$$

For the remaining term, we have

$$ {I}'_{1}(x)\lesssim \int _{(2Q)^{c}}\frac{ \vert f(z)(b-b_{Q}) \vert }{ (1+\frac{ \vert x-z \vert }{ \rho (x)} )^{N}} \frac{V(z)^{\alpha }}{ \vert x-z \vert ^{n-2\beta +1}}\,dz $$

and

$$ {I}'_{2}(x)\lesssim \int _{(2Q)^{c}}\frac{ \vert f(z)(b-b_{Q}) \vert }{ (1+\frac{ \vert x-z \vert }{ \rho (x)} )^{N}} \frac{V(z)^{\alpha }}{ \vert x-z \vert ^{n-2\beta }} \int _{B(z, \vert x-z \vert /4)}\frac{V(\xi )}{ \vert \xi -z \vert ^{n-1}}\,d\xi \,dz. $$

Since \(1\leq {p}'_{\alpha }< t\), we can choose t̄ such that \({p}'_{\alpha }<\bar{t} <t\). Let \(\nu =\frac{\bar{t} t}{t-\bar{t}} \), and then by Hölder’s inequality and Lemma 3, we get

$$\begin{aligned} & \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q} \bigl\vert f(z) \bigl(b(z)-b _{Q} \bigr) \bigr\vert ^{{p}'_{\alpha }}\,dz \biggr)^{1/{{p}'_{\alpha }}} \\ &\quad \lesssim \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k+1}Q} \bigl\vert f(z) \bigl(b(z)-b _{Q} \bigr) \bigr\vert ^{\bar{t}}\,dz \biggr)^{1/\bar{t}} \\ &\quad \lesssim \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q} \bigl\vert f(z) \bigr\vert ^{t}\,dz \biggr) ^{1/t} \\ &\qquad {}\times \biggl(\frac{1}{(2^{k}\rho (x_{0}))^{n}} \int _{2^{k}Q} \bigl\vert \bigl(b(z)-b _{Q}\bigr) \bigr\vert ^{\nu }\,dz \biggr)^{1/\nu } \\ &\quad \lesssim k2^{k\theta '}[b]_{\theta }\inf_{y\in Q}M_{t}f(y). \end{aligned}$$
(4)

Then, similar to the estimate of (3), we get

$$ {I}'_{1}(x) \lesssim \sum_{k\geq 1}2^{-kN} \bigl(1+2^{k}\bigr)^{\alpha l_{0}}k2^{k \theta '}[b]_{\theta } \inf_{y\in Q}M_{t}f(y) \lesssim [b]_{\theta } \inf_{y\in Q}M_{t}f(y). $$

By (4) and similar to the estimate of (2), we can get

$$ {I}'_{2}(x) \lesssim [b]_{\theta }\inf _{y\in Q}M_{t}f(y). $$

This completes the proof of Lemma 6. □

Lemma 7

Let \(V\in \mathit{RH}_{s}\) for \(n/2< s< n\), \(\frac{1}{{p}_{\alpha }}=\frac{\alpha +1}{s}-\frac{1}{n}\), and \(b\in \operatorname{BMO}_{\theta }(\rho )\). Then, for any \({p}'_{\alpha }< t<\infty \) and \(\gamma \geq 1\) we have

$$ \int _{(2B)^{c}} \bigl\vert K^{*}(x,z)-K^{*}(y,z) \bigr\vert V(z)^{\alpha } \bigl\vert b(z)-b_{B} \bigr\vert \bigl\vert f(z) \bigr\vert \,dz \lesssim [b]_{\theta }\inf _{u\in B}M_{t}f(u), $$

for all f and \(x,y\in B=B(x_{0},r)\) with \(r<\gamma \rho (x_{0})\).

Proof

Denote \(Q=B(x_{0},\gamma \rho (x_{0}))\). By Lemma 5 and since in our situation \(\rho (x)\sim \rho (x_{0})\) and \(|x-z|\sim |x _{0}-z|\), we need to estimate the following four terms:

$$\begin{aligned}& J_{1}=r^{\delta } \int _{Q\setminus 2B}\frac{ \vert f(z) \vert V(z)^{\alpha } \vert b(z)-b _{B} \vert }{ \vert x_{0}-z \vert ^{n-2\beta +\delta +1}}\,dz, \\& J_{2}=r^{\delta }\rho (x_{0})^{N} \int _{Q^{c}}\frac{ \vert f(z) \vert V(z)^{\alpha } \vert b(z)-b_{B} \vert }{ \vert x_{0}-z \vert ^{n-2\beta +\delta +1+N}}\,dz, \\& J_{3}=r^{\delta } \int _{Q\setminus 2B}\frac{ \vert f(z) \vert V(z)^{\alpha } \vert b(z)-b _{B} \vert }{ \vert x_{0}-z \vert ^{n-2\beta +\delta }} \int _{B(x_{0},4 \vert x_{0}-z \vert )} \frac{V(u)}{ \vert u-z \vert ^{n-1}}\,du\,dz, \end{aligned}$$

and

$$ J_{4}=r^{\delta }\rho (x_{0})^{N} \int _{Q^{c}}\frac{ \vert f(z) \vert V(z)^{\alpha } \vert b(z)-b_{B} \vert }{ \vert x_{0}-z \vert ^{n-2\beta +\delta +N}} \int _{B(x_{0},4 \vert x_{0}-z \vert )} \frac{V(u)}{ \vert u-z \vert ^{n-1}}\,du\,dz. $$

Splitting into annuli, we have

$$ J_{1}\lesssim \sum_{j=2}^{j_{0}}2^{-j\delta } \bigl(2^{j}r\bigr)^{2\beta -1} \frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha }\,dz, $$

where \(j_{0}\) is the least integer such that \(2^{j_{0}}\geq \gamma \rho (x_{0})/r\). By Hölder’s inequality with \(\frac{1}{{p}'_{ \alpha }}+\frac{\alpha }{s}+\frac{1}{q_{1}}=1\), \(t> {p}'_{\alpha }\), similar to the estimate of (4), we have

$$\begin{aligned} &\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha }\,dz \\ &\quad \lesssim \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl( \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b _{B} \bigr\vert \bigr)^{{p}'_{\alpha }} \,dz \biggr)^{1/{p}'_{\alpha }} \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B}V(z)^{s} \,dz \biggr)^{\alpha /s} \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\alpha }[b]_{\theta }\inf _{y\in B}M_{t}f(y) \biggl(1+\frac{2^{j}r}{\rho (x_{0})} \biggr)^{\theta '+l_{0}\alpha } \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{1-2\beta }[b]_{\theta }\inf _{u\in B}M_{t}f(u). \end{aligned}$$

Then, using \(\beta -\alpha =1/2\), we get

$$ J_{1} \lesssim [b]_{\theta }\inf_{u\in B}M_{t}f(u). $$

To deal with \(I_{2}\), we split into annuli and get

$$ J_{2}\lesssim \biggl(\frac{\rho (x_{0})}{r} \biggr)^{N}\sum _{j=j_{0}-1} ^{\infty }2^{-j(\delta +N)} \bigl(2^{j}r\bigr)^{2\beta -1}\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha }\,dz. $$

Notice that

$$\begin{aligned} &\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha }\,dz \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\alpha }[b]_{\theta }\inf _{y\in B}M_{t}f(y) \biggl(1+\frac{2^{j}r}{\rho (x_{0})} \biggr)^{\theta '+l_{0}\alpha } \\ &\quad \lesssim j2^{j(\theta '+l_{0}\alpha )} \biggl(\frac{\rho (x_{0})}{r} \biggr) ^{-(\theta '+l_{0}\alpha )} \bigl(2^{j}r\bigr)^{1-2\beta }[b]_{\theta }\inf _{u \in B}M_{t}f(u). \end{aligned}$$

Then, taking \(N>\theta '+l_{0}\alpha \), we get

$$ J_{2} \lesssim [b]_{\theta }\inf_{u\in B}M_{t}f(u). $$

For \(J_{3}\), splitting into annuli, we obtain

$$ J_{3}\lesssim \sum_{j=2}^{j_{0}}2^{-j\delta } \bigl(2^{j}r\bigr)^{2\beta } \frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha } \mathcal{I}_{1}(V \chi _{2^{j+2}B}) (z) \,dz. $$

By Hölder’s inequality with \(\frac{1}{{p}'_{\alpha }}+\frac{ \alpha }{s}+\frac{1}{q_{1}}=1\), similar to the estimate of (2), we get

$$\begin{aligned} &\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha } \mathcal{I}_{1}(V \chi _{2^{j+2}B}) (z)\,dz \\ &\quad \lesssim \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl( \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b _{B} \bigr\vert \bigr)^{{p}'_{\alpha }} \,dz \biggr)^{1/{p}'_{\alpha }} \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B}V(z)^{s} \,dz \biggr)^{\alpha /s} \\ &\qquad {} \times \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert \mathcal{I}_{1}(V \chi _{2^{j+2}B}) (z) \bigr\vert ^{q_{1}} \,dz \biggr)^{1/{q_{1}}} \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\alpha +n(1/s-1/q_{1})}[b]_{\theta } \inf_{y\in B}M_{t}f(y) \biggl(1+\frac{2^{j}r}{\rho (x_{0})} \biggr)^{ \theta '+l_{0}\alpha } \\ &\qquad {}\times \biggl(\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B}V(z)^{s}\,dz \biggr)^{1/s} \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\beta }[b]_{\theta }\inf _{y\in B}M_{t}f(y) \biggl(1+\frac{2^{j}r}{\rho (x_{0})} \biggr)^{\theta '+l_{0}(\alpha +1)} \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\beta }[b]_{\theta }\inf _{u\in B}M_{t}f(u). \end{aligned}$$

Then

$$ J_{3}\lesssim [b]_{\theta }\inf_{u\in B}M_{t}f(u). $$

Finally, for \(J_{4}\) we have

$$\begin{aligned} J_{4}&\lesssim \biggl(\frac{\rho (x_{0})}{r} \biggr)^{N}\sum _{j_{0}-1} ^{\infty }2^{-j(\delta +N)} \bigl(2^{j}r\bigr)^{2\beta } \\ &\quad {} \times \frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{ \alpha }\mathcal{I}_{1}(V \chi _{2^{j+2}B}) (z) \,dz. \end{aligned}$$

Notice that

$$\begin{aligned} &\frac{1}{ \vert 2^{j}B \vert } \int _{2^{j}B} \bigl\vert f(z) \bigr\vert \bigl\vert b(z)-b_{B} \bigr\vert V(z)^{\alpha } \mathcal{I}_{1}(V \chi _{2^{j+2}B}) (z)\,dz \\ &\quad \lesssim j\bigl(2^{j}r\bigr)^{-2\beta }[b]_{\theta }\inf _{y\in B}M_{t}f(y) \biggl(1+\frac{2^{j}r}{\rho (x_{0})} \biggr)^{\theta '+l_{0}(\alpha +1)} \\ &\quad \lesssim j2^{j(\theta '+l_{0}(\alpha +1))} \biggl(\frac{\rho (x _{0})}{r} \biggr)^{-\theta '-l_{0}(\alpha +1)} \bigl(2^{j}r\bigr)^{-2\beta }[b]_{ \theta }\inf _{u\in B}M_{t}f(u). \end{aligned}$$

We choose N large enough such that \(N>\theta '+l_{0}(\alpha +1)\), and then

$$ J_{4}\lesssim [b]_{\theta }\inf_{u\in B}M_{t}f(u), $$

which finishes the proof of Lemma 7. □

Now we are in a position to give the proof of Theorem 2.

Proof of Theorem 2

We will prove part (i), and (ii) follows by duality. We start with a function \(f\in L^{p}(\mathbb{R} ^{n})\) with \(p'_{\alpha }< p<\infty \), and by Lemma 6 we have \([b,T_{\beta }^{*}]f\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})\).

By Proposition 3 and Lemma 6 with \(p'_{\alpha }< t< p< \infty \), we have

$$\begin{aligned} \bigl\Vert \bigl[b,T_{\beta }^{*}\bigr]f \bigr\Vert ^{p}_{L^{p}(\mathbb{R}^{n})}&\lesssim \int _{\mathbb{R}^{n}} \bigl\vert M_{\rho ,\delta }\bigl[b,T_{\beta }^{*} \bigr]f \bigr\vert ^{p}\,dx \\ &\lesssim \int _{\mathbb{R}^{n}} \bigl\vert M^{\sharp }_{\rho ,\gamma } \bigl[b,T _{\beta }^{*}\bigr]f \bigr\vert ^{p}\,dx+ \sum_{k} \vert Q_{k} \vert \biggl( \frac{1}{ \vert Q_{k} \vert } \int _{2Q_{k}}\bigl\vert \bigl[b,T_{\beta }^{*}\bigr]f\bigr\vert \biggr)^{p} \\ &\lesssim \int _{\mathbb{R}^{n}} \bigl\vert M^{\sharp }_{\rho ,\gamma } \bigl[b,T _{\beta }^{*}\bigr]f \bigr\vert ^{p}\,dx+[b]^{p}_{\theta } \sum_{k} \int _{2Q_{k}} \bigl\vert M_{t}(f) \bigr\vert ^{p}\,dx. \end{aligned}$$

By Proposition 2 and the boundedness of \(M_{t}\) on \(L^{p}(\mathbb{R}^{n})\), the second term is controlled by \([b]^{p} _{\theta }\|f\|^{p}_{L^{p}(\mathbb{R}^{n})}\). Then, we only need to consider the first term.

Our goal is to find a point-wise estimate of \(M_{\rho ,\gamma }[b,T _{\beta }^{*}]f\). Let \(x\in \mathbb{R}^{n}\) and \(B=B(x_{0},r)\) with \(r<\gamma \rho (x_{0})\) such that \(x\in B\). Write \(f=f_{1}+f_{2}\) with \(f_{1}=f\chi _{2B}\), then

$$ \bigl[b,T_{\beta }^{*}\bigr]f =(b-b_{B})T_{\beta }^{*}f-T_{\beta }^{*} \bigl(f_{1}(b-b _{B})\bigr)-T_{\beta }^{*} \bigl(f_{2}(b-b_{B})\bigr). $$

Then, we need to control the mean oscillation on B of each term that we call \(\mathcal{O}_{1}\), \(\mathcal{O}_{2}\) and \(\mathcal{O}_{3}\).

Let \(t>p'_{\alpha }\), then, by Hölder’s inequality and Lemma 3, we get

$$\begin{aligned} \mathcal{O}_{1}&\lesssim \frac{1}{ \vert B \vert } \int _{B} \bigl\vert (b-b_{B})T_{\beta } ^{*}f \bigr\vert \\ &\lesssim \biggl(\frac{1}{ \vert B \vert } \int _{B} \vert b-b_{B} \vert ^{t'} \biggr)^{1/t'} \biggl(\frac{1}{ \vert B \vert } \int _{B} \bigl\vert T_{\beta }^{*}f \bigr\vert ^{t} \biggr)^{1/t} \\ &\lesssim [b]_{\theta }M_{t}T_{\beta }^{*}f(x_{0}), \end{aligned}$$

since \(r<\gamma \rho (x_{0})\).

To estimate \(\mathcal{O}_{2}\), let \(p'_{\alpha }<\bar{t}<t\) and \(\nu =\frac{\bar{t}t}{t-\bar{t}}\). Then

$$\begin{aligned} \mathcal{O}_{2}&\lesssim \frac{1}{ \vert B \vert } \int _{B} \bigl\vert T_{\beta }^{*} \bigl((b-b _{B})f_{1} \bigr) \bigr\vert \\ &\lesssim \biggl(\frac{1}{ \vert B \vert } \int _{B} \bigl\vert T_{\beta }^{*} \bigl((b-b_{B})f _{1} \bigr) \bigr\vert ^{\bar{t}} \biggr)^{1/\bar{t}} \\ &\lesssim \biggl(\frac{1}{ \vert B \vert } \int _{B} \bigl\vert (b-b_{B})f_{1} \bigr\vert ^{\bar{t}} \biggr) ^{1/\bar{t}} \\ &\lesssim \biggl(\frac{1}{ \vert B \vert } \int _{B} \vert b-b_{B} \vert ^{\nu } \biggr)^{1/ \nu } \biggl(\frac{1}{ \vert B \vert } \int _{2B} \vert f \vert ^{t} \biggr)^{1/t} \\ &\lesssim [b]_{\theta }M_{t}f(x_{0}). \end{aligned}$$

For \(\mathcal{O}_{3}\), note that \(\inf_{y\in B}M_{t}f(y)\leq M_{t}f(x _{0})\), and so by Lemma 7 we get

$$\begin{aligned} \mathcal{O}_{3}&\lesssim \frac{1}{ \vert B \vert ^{2}} \int _{B} \int _{B} \bigl\vert T_{ \beta }^{*} \bigl((b-b_{B})f_{2} \bigr) (x)-T_{\beta }^{*} \bigl((b-b_{B})f _{2} \bigr) (y) \bigr\vert \,dx\,dy \\ &\lesssim [b]_{\theta }M_{t}f(x_{0}). \end{aligned}$$

Thus, we have showed that

$$ \bigl\vert M^{\sharp }_{\rho ,\gamma }\bigl[b,T_{\beta }^{*} \bigr]f \bigr\vert \lesssim [b]_{\theta } \bigl( M_{t}T_{\beta }^{*}f(x)+M_{t}f(x) \bigr). $$

Since \(t< p\), we obtain the desired result. □

Proof of Theorem 3

Let \(f\in H_{\mathcal{L}}^{1}( \mathbb{R}^{n})\). By Proposition 1, we can write \(f= \sum_{j=-\infty }^{\infty }\lambda _{j}a_{j}\), where each \(a_{j}\) is a \((1,q)_{\rho }\)-atom with \(1< q< {p}_{\alpha }\), \(\frac{1}{{p}_{\alpha }}=\frac{ \alpha +1}{q_{0}}-\frac{1}{n}\) and \(\sum_{j=-\infty }^{\infty }| \lambda _{j}|\leq 2\|f\|_{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}\). Suppose \(\operatorname{supp} a_{j}\subset B_{j}=B(x_{j},r_{j})\) with \(r_{j}<\rho (x _{j})\). Write

$$\begin{aligned}{} [b,T_{\beta }]f(x)&= \sum_{j=-\infty }^{\infty } \lambda _{j}[b,T_{ \beta }] a_{j}(x)\chi _{8B_{j}}(x) \\ &\quad {} +\sum_{j:r_{j}\geq {\rho (x_{j})}/{4}}\lambda _{j} \bigl(b(x)-b_{B _{j}} \bigr)T_{\beta }a_{j}(x)\chi _{(8B_{j})^{c}}(x) \\ &\quad {} +\sum_{j:r_{j}< {\rho (x_{j})}/{4}}\lambda _{j} \bigl(b(x)-b_{B_{j}} \bigr)T_{\beta }a_{j}(x)\chi _{(8B_{j})^{c}}(x) \\ &\quad {} -\sum_{j=-\infty }^{\infty }\lambda _{j} T_{\beta }\bigl((b-b_{B_{j}})a _{j}\bigr) (x)\chi _{(8B_{j})^{c}}(x) \\ &= \sum_{i=1}^{4}\sum _{j=-\infty }^{\infty }\lambda _{j}A_{ij}(x). \end{aligned}$$

Note that

$$ \biggl( \int _{B_{j}} \bigl\vert a_{j}(x) \bigr\vert ^{q}\,dx \biggr)^{1/q}\lesssim \vert B_{j} \vert ^{ \frac{1}{q}-1}. $$

By Hölder’s inequality, for \(1< q< {p}_{\alpha } \), and using Theorem 2 we get

$$\begin{aligned} \Vert A_{1,j} \Vert _{L^{1}(\mathbb{R}^{n})} &\lesssim \biggl( \int _{8B _{j}} \bigl\vert [b,T_{\beta }] a_{j}(x) \bigr\vert ^{q}\,dx \biggr)^{\frac{1}{q}}r _{j}^{\frac{n}{q'}} \\ &\lesssim [b]_{\theta }r_{j}^{\frac{n}{q'}} \biggl( \int _{B_{j}} \bigl\vert a _{j}(x) \bigr\vert ^{q}\,dx \biggr)^{1/{q}} \\ &\lesssim [b]_{\theta } \vert B_{j} \vert ^{\frac{1}{q'}+\frac{1}{q}-1} \lesssim [b]_{\theta }. \end{aligned}$$

Thus

$$\begin{aligned} \Biggl\Vert \sum_{j=-\infty }^{\infty }\lambda _{j}A_{1j} \Biggr\Vert _{L^{1}( \mathbb{R}^{n})}&\lesssim \sum _{j=-\infty }^{\infty } \vert \lambda _{j} \vert \Vert A_{1j} \Vert _{L^{1}(\mathbb{R}^{n})} \\ &\lesssim [b]_{\theta }\sum_{j=-\infty }^{\infty } \vert \lambda _{j} \vert \lesssim [b]_{\theta } \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}. \end{aligned}$$

And so

$$ \Biggl\vert \Biggl\{ x\in \mathbb{ R}^{n}: \Biggl\vert \sum _{j=-\infty }^{\infty }\lambda _{j}A_{1j} \Biggr\vert >\frac{\lambda }{4} \Biggr\} \Biggr\vert \lesssim \frac{[b]_{ \theta }}{\lambda } \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}. $$

Since \(z\in B_{j}\), \(x\in 2^{k}B_{j}\setminus 2^{k-1}B_{j}\), we have \(|x-z|\sim |x-x_{j}|\sim 2^{k}r_{j}\), and by Lemma 1 we get

$$ \frac{1}{ (1+\frac{ \vert x-z \vert }{\rho (x)} )^{N}}\lesssim \frac{1}{ (1+\frac{2^{k}r_{j}}{\rho (x_{j})} )^{\frac{N}{k_{0}+1}}}. $$

By Hölder’s inequality, Lemmas 2 and 3, we get

$$\begin{aligned} &\frac{1}{ \vert 2^{k}B_{j} \vert } \int _{2^{k}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{\alpha }\,dx \\ &\quad \lesssim \biggl(\frac{1}{ \vert 2^{k}B_{j} \vert } \int _{2^{k}B_{j}} \bigl\vert b(x)-b _{B_{j}} \bigr\vert ^{(\frac{s}{\alpha })'}\,dx \biggr)^{1/(\frac{s}{\alpha })'} \biggl(\frac{1}{ \vert 2^{k}B_{j} \vert } \int _{2^{k}B_{j}}V(x)^{s}\,dx \biggr)^{ \alpha /s} \\ &\quad \lesssim k[b]_{\theta } \biggl(1+\frac{2^{k}r_{j}}{\rho (x_{j})} \biggr) ^{\theta '} \biggl(\frac{1}{ \vert 2^{k}B_{j} \vert } \int _{2^{k}B_{j}}V(x)\,dx \biggr) ^{\alpha } \\ &\quad \lesssim k[b]_{\theta }\bigl(2^{k}r_{j} \bigr)^{-2\alpha } \biggl(1+\frac{2^{k}r _{j}}{\rho (x_{j})} \biggr)^{\theta '+l_{0}\alpha }. \end{aligned}$$
(5)

Note that \(\frac{1}{{p}'_{\alpha }}+\frac{\alpha }{s}+ \frac{1}{q_{1}}=1\), \(\frac{1}{q_{1}}=\frac{1}{s}-\frac{1}{n} \), so by Hölder’s and Hardy–Littlewood–Sobolev’s inequalities and using the fact that \(V\in \mathit{RH}_{s}\), we obtain

$$\begin{aligned} &\frac{1}{ \vert 2^{k} B_{j} \vert } \int _{2^{k}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{\alpha }\bigl(\mathcal{I}_{1}(V\chi _{2^{k}B}) (x) \bigr)\,dx \\ &\quad \lesssim \biggl(\frac{1}{ \vert 2^{k} B_{j} \vert } \int _{2^{k}B_{j}} \bigl\vert b(x)-b _{B_{j}} \bigr\vert ^{{p}'_{\alpha }}\,dx \biggr)^{1/{p}'_{\alpha }} \biggl(\frac{1}{ \vert 2^{k} B_{j} \vert } \int _{2^{k}B_{j}}V(x)^{s}\,dx \biggr)^{ \alpha /s} \\ &\qquad {} \times \biggl(\frac{1}{ \vert 2^{k} B_{j} \vert } \int _{2^{k}B_{j}}\bigl( \mathcal{I}_{1}(V\chi _{2^{k}B_{j}}) (x)\bigr)^{q_{1}}\,dx \biggr)^{1/q_{1}} \\ &\quad \lesssim [b]_{\theta }k \bigl\vert 2^{k}B_{j} \bigr\vert ^{1/s-1/q_{1}} \biggl(1+\frac{2^{k}r _{j}}{\rho (x_{j})} \biggr)^{\theta '} \biggl(\frac{1}{ \vert 2^{k} B_{j} \vert } \int _{2^{k}B_{j}}V(x)^{s}\,dx \biggr)^{(\alpha +1)/s} \\ &\quad \lesssim [b]_{\theta }k\bigl(2^{k}r_{j} \bigr)^{-2\alpha -1} \biggl(1+\frac{2^{k}r _{j}}{\rho (x_{j})} \biggr)^{\theta '+(\alpha +1)l_{0}}. \end{aligned}$$
(6)

Recall \(\int _{B_{j}}|a_{j}(y)|\,dy\lesssim 1\), \(\beta -\alpha = \frac{1}{2} \) and \(r_{j}/\rho (x_{j})\geq 1/4\). Then, taking N large enough such that \(\frac{N}{k_{0}+1}>\theta '+l_{0}(\alpha +1)\), we get

$$\begin{aligned} & \bigl\Vert A_{2,j}(x) \bigr\Vert _{L^{1}(\mathbb{R}^{n})} \\ &\quad \lesssim \sum_{k\geq 4}\frac{1}{ (1+ \frac{2^{k}r_{j}}{\rho (x)} )^{N}} \frac{1}{(2^{k}r_{j})^{n-2 \beta +1}} \int _{2^{k}B_{j}\setminus 2^{k-1}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{ \alpha }\,dx \int _{B_{j}} \bigl\vert a_{j}(z) \bigr\vert \,dz \\ &\qquad {} +\sum_{k\geq 4}\frac{1}{ (1+\frac{2^{k}r_{j}}{\rho (x)} ) ^{N}}\frac{1}{(2^{k}r_{j})^{n-2\beta }} \\ &\qquad {} \times \int _{2^{k}B_{j}\setminus 2^{k-1}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{ \alpha }\bigl(\mathcal{I}_{1}(V\chi _{2^{k}B}) (x) \bigr)\,dx \int _{B_{j}} \bigl\vert a_{j}(z) \bigr\vert \,dz \\ &\quad \lesssim [b]_{\theta }\sum_{k\geq 4} \frac{k(2^{k}r_{j})^{2\beta -1}}{ (1+\frac{2^{k}r_{j}}{\rho (x_{j})} )^{\frac{N}{k_{0}+1}}} \bigl(2^{k}r_{j}\bigr)^{-2\alpha } \biggl(1+\frac{2^{k}r_{j}}{\rho (x_{j})} \biggr) ^{\theta '+l_{0}\alpha } \\ &\qquad {} +[b]_{\theta }\sum_{k\geq 4}\frac{(2^{k}r_{j})^{2\beta }}{ (1+\frac{2^{k}r _{j}}{\rho (x_{j})} )^{\frac{N}{k_{0}+1}}} \bigl(2^{k}r_{j}\bigr)^{-2 \alpha -1} \biggl(1+ \frac{2^{k}r_{j}}{\rho (x_{j})} \biggr)^{\theta '+( \alpha +1)l_{0}} \\ &\quad \lesssim [b]_{\theta }\sum_{k\geq 3} \frac{k}{(2^{k})^{\frac{N}{k _{0}+1}-\theta '-l_{0}\alpha }} +[b]_{\theta }\sum_{k\geq 3} \frac{k}{(2^{k})^{\frac{N}{k _{0}+1}-\theta '-l_{0}(\alpha +1)}} \\ &\quad \lesssim [b]_{\theta }. \end{aligned}$$

Thus

$$ \Biggl\Vert \sum_{j=-\infty }^{\infty }\lambda _{j}A_{2j} \Biggr\Vert _{L^{1}( \mathbb{R}^{n})}\lesssim [b]_{\theta } \Vert f \Vert _{H_{\mathcal{L}}^{1}( \mathbb{R}^{n})}. $$

Therefore

$$ \Biggl\vert \Biggl\{ x\in \mathbb{ R}^{n}: \Biggl\vert \sum _{j=-\infty }^{\infty }\lambda _{j}A_{2j} \Biggr\vert >\frac{\lambda }{4} \Biggr\} \Biggr\vert \lesssim \frac{[b]_{ \theta }}{\lambda } \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}. $$

When \(x\in 2^{k}B_{j}\setminus 2^{k-1}B_{j}\), and \(z\in B_{j}\), by Lemmas 5 and 1, we have

$$\begin{aligned} \bigl\vert K(x,z) -K(x,x_{j}) \bigr\vert &\lesssim \frac{1}{ (1+\frac{2^{k}r_{j}}{ \rho (x_{j})} )^{N/(k_{0}+1)}}\frac{r_{j}^{\delta }}{(2^{k}r _{j})^{n+\delta -2\beta +1}} \\ &\quad {} +\frac{1}{ (1+\frac{2^{k}r_{j}}{\rho (x_{j})} )^{N/(k _{0}+1)}}\frac{r_{j}^{\delta }}{(2^{k}r_{j})^{n+\delta -2\beta }} \mathcal{I}_{1}(V\chi _{2^{k}B_{j}}) (z), \end{aligned}$$

where \(\delta =2-n/s>0\). Thus, by the vanishing condition of \(a_{j}\), together with (5) and (6), we have

$$\begin{aligned} & \bigl\Vert A_{3,j}(x) \bigr\Vert _{L^{1}(\mathbb{R}^{n})} \\ &\quad \lesssim \sum_{k\geq 4} \int _{2^{k}B_{j}\setminus 2^{k-1}B_{j}} \bigl\vert b(x)-b _{B_{j}} \bigr\vert V(x)^{\alpha } \int _{B_{j}} \bigl\vert K_{\alpha }(x,z)-K_{\alpha }(x,x _{j}) \bigr\vert \bigl\vert a_{j}(z) \bigr\vert \,dz\,dx \\ &\quad \lesssim \sum_{k\geq 3}\frac{1}{ (1+\frac{2^{k}r_{j}}{\rho (x _{j})} )^{\frac{N}{k_{0}+1}}} \frac{r_{j}^{\delta }}{(2^{k}r_{j})^{n+ \delta -2\beta +1}} \int _{2^{k+1}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{\alpha }\,dx \int _{B_{j}} \bigl\vert a_{j}(z) \bigr\vert \,dz \\ &\qquad {} +\sum_{k\geq 3}\frac{1}{ (1+\frac{2^{k}r_{j}}{\rho (x_{j})} ) ^{\frac{N}{k_{0}+1}}}\frac{r_{j}^{\delta }}{(2^{k}r_{j})^{(n+\delta -2 \beta )}} \\ & \qquad {} \times \int _{2^{k+1}B_{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert V(x)^{\alpha }\mathcal{I} _{1}(V\chi _{2^{k}B_{j}}) (x)\,dx \int _{B_{j}} \bigl\vert a_{j}(z) \bigr\vert \,dz \\ &\quad \lesssim [b]_{\theta }\sum_{k\geq 3} \frac{1}{ (1+\frac{2^{k}r _{j}}{\rho (x_{j})} )^{\frac{N}{k_{0}+1}-\theta '-l_{0}\alpha }}\frac{k}{2^{k \delta }} +[b]_{\theta }\sum _{k\geq 3}\frac{1}{ (1+\frac{2^{k}r _{j}}{\rho (x_{j})} )^{\frac{N}{k_{0}+1}-\theta '-l_{0}(\alpha +1)}}\frac{k}{2^{k\delta }} \lesssim [b]_{\theta }. \end{aligned}$$

So that

$$ \Biggl\vert \Biggl\{ x\in \mathbb{ R}^{n}: \Biggl\vert \sum _{j=-\infty }^{\infty }\lambda _{j}A_{3j} \Biggr\vert >\frac{\lambda }{4} \Biggr\} \Biggr\vert \lesssim \frac{[b]_{ \theta }}{\lambda } \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}. $$

Now let us deal with the last part. Since \(r_{j}\leq \rho (x_{j})\), we get

$$\begin{aligned} \bigl\Vert (b-b_{B_{j}})a_{j} \bigr\Vert _{L^{1}(\mathbb{R}^{n})} &\leq \biggl( \int _{B _{j}} \bigl\vert b(x)-b_{B_{j}} \bigr\vert ^{q'}\,dx \biggr)^{1/q'} \biggl( \int _{B_{j}} \bigl\vert a_{j}(x) \bigr\vert ^{q}\,dx \biggr) ^{1/q} \\ &\lesssim [b]_{\theta } \biggl(1+\frac{r_{j}}{\rho (x_{j})} \biggr) ^{\theta '}\lesssim [b]_{\theta }. \end{aligned}$$

Note that

$$\begin{aligned} \bigl\vert A_{4j}(x) \bigr\vert &\leq \sum _{j=-\infty }^{\infty } \vert \lambda _{j} \vert T_{\beta }\bigl( \bigl\vert (b-b _{B_{j}})a_{j} \bigr\vert \bigr) (x)\chi _{(8B_{j})^{c}}(x) \\ &\leq T_{\beta } \Biggl(\sum_{j=-\infty }^{\infty } \bigl\vert \lambda _{j}(b-b _{B_{j}})a_{j} \bigr\vert \Biggr) (x). \end{aligned}$$

By Theorem 1, we know \(T_{\beta }\) is bounded from \(L^{1}(\mathbb{R}^{n})\) into weak \(L^{1}(\mathbb{R}^{n})\). Then

$$\begin{aligned} & \Biggl\vert \Biggl\{ x\in \mathbb{R}^{n}: \Biggl\vert \sum _{j=-\infty }^{\infty }\lambda _{j} A_{4j} \Biggr\vert >\frac{\lambda }{4} \Biggr\} \Biggr\vert \\ &\quad \leq \Biggl\vert \Biggl\{ x\in \mathbb{R}^{n}: \Biggl\vert T_{\beta } \Biggl( \sum_{j=-\infty }^{\infty } \bigl\vert \lambda _{j}(b-b_{B_{j}})a_{j} \bigr\vert \Biggr) (x) \Biggr\vert >\frac{\lambda }{4} \Biggr\} \Biggr\vert \\ &\quad \lesssim \frac{1}{\lambda } \Biggl\Vert \sum_{j=-\infty }^{\infty } \bigl\vert \lambda _{j}(b-b_{B_{j}})a_{j} \bigr\vert \Biggr\Vert _{L^{1}(\mathbb{R}^{n})} \\ &\quad \lesssim \frac{1}{\lambda }\sum_{j=-\infty }^{\infty } \vert \lambda _{j}| \bigl\Vert (b-b_{B_{j}})a_{j} \bigr\Vert _{L^{1}(\mathbb{R}^{n})} \\ &\quad \lesssim \frac{[b]_{\theta }}{\lambda } \Biggl(\sum_{j=-\infty } ^{\infty } \vert \lambda _{j} \vert \Biggr)\lesssim \frac{[b]_{\theta }}{\lambda } \Vert f \Vert _{H_{\mathcal{L}}^{1}(\mathbb{R}^{n})}. \end{aligned}$$

Thus,

$$\begin{aligned} & \Biggl\vert \Biggl\{ x\in \mathbb{R}^{n}: \Biggl\vert \sum _{i=1}^{4}\sum _{j=- \infty }^{\infty }\lambda _{j} A_{ij} \Biggr\vert >\lambda \Biggr\} \Biggr\vert \\ &\quad \lesssim \sum_{i=1}^{4} \Biggl\vert \Biggl\{ x\in \mathbb{R}^{n}: \Biggl\vert \sum _{j=-\infty }^{\infty }\lambda _{j} A_{ij} \Biggr\vert > \frac{\lambda }{4} \Biggr\} \Biggr\vert \\ &\quad \lesssim \frac{[b]_{\theta }}{\lambda } \Vert f \Vert _{H_{\mathcal{L}}^{1}( \mathbb{R}^{n})}. \end{aligned}$$

 □

4 Conclusion

In this paper, we established the \(L^{p}\)-boundedness of commutator operators \([b,T_{\beta }]\) and \([b,T^{*}_{\beta }]\), where \(T_{ \beta }=V^{\alpha }\nabla \mathcal{L}^{-\beta }\), \(\frac{1}{2}< \beta \leq 1\), \(\beta -\alpha =\frac{1}{2} \), and \(b\in \operatorname{BMO}_{\theta }(\rho )\), which is larger than the space \(\operatorname{BMO}(\mathbb{R}^{n})\). At the endpoint, we show that the operator \([b,T_{\beta }]\) is bounded from Hardy space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) continuously into weak \(L^{1}(\mathbb{R}^{n})\). These results enrich the theory of Schrödinger operator.