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Estimates for the commutators of operator \(V^{\alpha }\nabla (-\Delta +V)^{-\beta }\)

Abstract

Let a function b belong to the space \(\operatorname{BMO}_{\theta }(\rho )\), which is larger than the space \(\operatorname{BMO}(\mathbb{R}^{n})\), and let a nonnegative potential V belong to the reverse Hölder class \(\mathit{RH}_{s}\) with \(n/2< s< n\), \(n\geq 3\). Define the commutator \([b,T_{\beta }]f=bT_{ \beta }f-T_{\beta }(bf)\), where the operator \(T_{\beta }=V^{\alpha } \nabla \mathcal{L}^{-\beta }\), \(\beta -\alpha =\frac{1}{2}\), \(\frac{1}{2}< \beta \leq 1\), and \(\mathcal{L}=-\Delta +V\) is the Schrödinger operator. We have obtained the \(L^{p}\)-boundedness of the commutator \([b,T_{\beta }]f\) and we have proved that the commutator is bounded from the Hardy space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) into weak \(L^{1}(\mathbb{R}^{n})\).

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 \(B\lesssim A\).

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 \(\rho (z)\). □

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 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,

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

 □

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.

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The authors are very grateful to the anonymous referees and the editor for their insightful comments and suggestions.

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Hu, Y., Wang, Y. Estimates for the commutators of operator \(V^{\alpha }\nabla (-\Delta +V)^{-\beta }\). J Inequal Appl 2019, 126 (2019). https://doi.org/10.1186/s13660-019-2081-z

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MSC

  • 42B30
  • 35J10
  • 42B35

Keywords

  • Schrödinger operator
  • Hardy space
  • Reverse Hölder class
  • BMO
  • Commutator