$L^p(\mathbb {R}^d)$ Boundedness for a Class of Nonstandard Singular Integral Operators

Abstract

In this paper, let $\Omega$ be homogeneous of degree zero which has vanishing moment of order one, A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, we consider a class of nonstandard singular integral operators, $T_{\Omega ,\,A}$, with rough kernel being of the form $\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\big (A(x)-A(y)-\nabla A(y)(x-y)\big )$. This operator is closely related to the Calderón commutator. We prove that, under the Grafakos-Stefanov minimum size condition $GS_{\beta }(S^{d-1})$ with $2<\beta <\infty$ for $\Omega$, $T_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$ for p with $1+1/(\beta -1)< p < \beta$.

1 Introduction

We will work on $\mathbb {R}^d$, $d\ge 2$. For $x\in \mathbb {R}^d$ and $1\le n\le d$, we denote by $x_n$ the n-th variable of x, and $x'=x/\vert x\vert$. Let $\Omega$ be homogeneous of degree zero, integrable on $S^{d-1}$, the unit sphere in $\mathbb {R}^d$, and satisfy the vanishing moment condition that for all $1\le n\le d$,

$$\begin{aligned} & \int _{S^{d-1}}\Omega (x')x_n'dx=0. \end{aligned}$$

(1.1)

Define the d-dimensional Calderón commutator $\mathscr {C}_{\Omega , a}$ by

$$\begin{aligned} & \mathscr {C}_{\Omega , a}f(x)=\mathrm{p. v.}\int _{\mathbb {R}^d}\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\big (a(x)-a(y)\big )f(y)dy, \end{aligned}$$

where a is a function on $\mathbb {R}^d$ such that $\partial _n a\in L^{\infty }(\mathbb {R}^d)$ for all n with $1\le n\le d$. This operator was introduced by Calderón [2] and plays an important role in the theory of singular integrals. For the progress of the study of Calderón commutator, we refer the references [1, 2, 10, 13, 17, 25,26,27,28, 14, Chapter 8] and the related references therein.

Now let A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, that is, $\partial _nA\in \textrm{BMO}(\mathbb {R}^d)$ for all n with $1\le n\le d$. Let $\Omega$ be homogeneous of degree zero, integrable on $S^{d-1}$, and satisfy the vanishing moment condition (1.1). Define the operator $T_{\Omega , A}$ by

$$\begin{aligned} T_{\Omega , A}f(x)=\mathrm{p. v.}\int _{\mathbb {R}^d}\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy. \end{aligned}$$

(1.2)

This operator is closely related to the d-dimensional Calderón commutator. For the case of $\nabla A\in L^{\infty }(\mathbb {R}^d)$, the $L^p(\mathbb {R}^d)$ boundedness and the endpoint estimates of $T_{\Omega ,\,A}$ can be deduced from the $L^p(\mathbb {R}^d)$ boundedness of Calderón commutator. On the other hand, for the case of $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, $T_{\Omega ,\,A}$ is not a Calderón–Zygmund operator even if $\Omega \in \textrm{Lip}(S^{d-1})$. Cohen [6] first considered the mapping properties of $T_{\Omega ,\,A}$, and proved that if $\Omega \in \textrm{Lip}_\alpha (S^{d-1})$ ($\alpha \in (0,\,1]$), then for $p\in (1,\,\infty )$, $T_{\Omega ,\,A}$ is a bounded operator on $L^p(\mathbb {R}^d)$ with bound $C\Vert \nabla A\Vert _{\textrm{BMO}(\mathbb {R}^d)}$; see also [8] for the $L^p(\mathbb {R}^d)$ boundedness of an operator related to $T_{\Omega ,\,A}$. Hofmann [18] improved the result of Cohen, and showed that $\Omega \in L^{\infty }(S^{d-1})$ is a sufficient condition such that $T_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$. Fairly recently, Hu, Tao, Wang and Xue [22] considered the $L^p(\mathbb {R}^d)$ boundedness of $T_{\Omega , A}$ when $\Omega$ satisfies certain minimum size condition, and established the following estimates.

Theorem 1.1

Let $\Omega$ be homogeneous of degree zero, satisfy the vanishing condition (1.1), A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Suppose that $\Omega \in L(\log L)^{2}( {S}^{d-1})$. Then $T_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$ for all $p\in (1,\,\infty )$.

There is, however, another typical minimum size condition for functions on $S^{d-1}$. Let $\Omega \in L^1(S^{d-1})$ and $\beta \in [1,\,\infty )$, we say that $\Omega \in GS_{\beta }(S^{d-1})$ if

$$\begin{aligned} \sup _{\zeta \in S^{d-1}}\int _{S^{d-1}}\vert \Omega (\theta )\vert \log ^{\beta }\big (\frac{1}{\vert \zeta \cdot \theta \vert }\big )d\theta <\infty . \end{aligned}$$

This size condition was introduced by Grafakos and Stefanov [15], to study the $L^p(\mathbb {R}^d)$ boundedness for the homogeneous singular integral operator defined by

$$\begin{aligned} T_{\Omega }f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\frac{\Omega (x-y)}{\vert x-y\vert ^d}f(y)dy \end{aligned}$$

(1.3)

Grafakos and Stefanov [15] proved that if $\Omega$ is homogeneous of degree zero and has mean value zero on $S^{d-1}$ and $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta >1$, then the operator $T_{\Omega }$ is bounded on $L^p(\mathbb {R}^d)$ for $1+1/\beta<p<1+\beta$. Fan, Guo and Pan [12] improved the result of Grafakos and Stefanov, and proved that $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta >1$ is a sufficient condition such that $T_{\Omega }$ is bounded on $L^p(\mathbb {R}^d)$ for $2\beta /(2\beta -1)<p<2\beta$.

Let $P_{ry'}(x')$ be the Poission kernel on $S^{d-1}$, that is

$$\begin{aligned} P_{ry'}(x')=\frac{1-r^2}{\vert ry'-x'\vert ^d}, \end{aligned}$$

where $0\le r< 1$ and $x',\,y'\in S^{d-1}$. For a function $\Omega \in L^1(S^{d-1})$, we define the radial maximal function

$$\begin{aligned} P^+\Omega (x') = \sup _{0\le r< 1}\left| \int _{S^{d-1}} \Omega (y')P_{rx'}(y')dy' \right| . \end{aligned}$$

The Hardy space $H^1(S^{d-1})$, is a subspace of $L^1(S^{d-1})$ which contains all $L^1(S^{d-1})$ functions $\Omega$ with the finite norms $\Vert \Omega \Vert _{H^1(S^{d-1})} = \Vert P^+\Omega \Vert _{L^1(S^{d-1})}$, see also [9]. As is well known, for $\beta \in [1,\,\infty )$,

$$\begin{aligned} H^1(S^{d-1})\subset L(\log L)^{\beta }(S^{d-1})\subset GS_{\beta }(S^{d-1}). \end{aligned}$$

Moreover, as Grafakos and Stefanov [15] showed,

$$\begin{aligned} \big (\cap _{\beta >1}GS_{\beta }(S^{d-1})\big )\backslash H^1(S^{d-1})\not =\emptyset . \end{aligned}$$

Thus, it is natural to ask if $T_{\Omega ,\,A}$ enjoys a $L^p(\mathbb {R}^d)$ estimate similar to the operator $T_{\Omega }$ defined as (1.3) when $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (1,\,\infty )$. Hu [20] considered this question and proved the following result.

Theorem 1.2

Let $\Omega$ be homogeneous of degree zero which satisfies the vanishing moment condition (1.1), A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Suppose that $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta >3$, then $T_{\Omega ,\,A}$ is bounded on $L^2(\mathbb {R}^d)$.

In this paper, we will improve and extend Theorem 1.2. Our main result can be stated as follows.

Theorem 1.3

Let $\Omega$ be homogeneous of degree zero, satisfy the vanishing moment condition (1.1), A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Suppose that $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta >2$. Then for p with $1+1/(\beta -1)< p < \beta$, $T_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$.

To prove Theorem 1.3, we will first prove that $T_{\Omega , A}$ is bounded on $L^2(\mathbb {R}^d)$ when $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (2,\,\infty )$. To prove the $L^p(\mathbb {R}^d)$ boundedness of $T_{\Omega , A}$, we will show that, there exists a sequence of operators $\{R_{l,A}\}_{l\in \mathbb {N}}$ such that

• (i) for $p\in (1,\,2)$, $R_{l, A}$ is bounded on $L^p(\mathbb {R}^d)$ with bound $Cl^2$;

• (ii) for any $\varepsilon \in (0,\,1)$ and $l\in \mathbb {N}$,

  $$\begin{aligned} \Vert R_{l,A}-T_{\Omega , A}\Vert _{L^2(\mathbb {R}^d)\rightarrow L^2(\mathbb {R}^d)}\lesssim l^{-\varepsilon \beta +2}. \end{aligned}$$

This, via interpolation, leads to the desired $L^p(\mathbb {R}^d)$ boundedness of $T_{\Omega , A}$. We remark that in this paper, we are very much motivated by the work of Chen, Hu and Tao [4], in which the authors established a suitable approximation for the Calderón commutator with rough kernel, see also [30] for the approximation of homogeneous singular integrals with rough kernels. However, the operator we consider in this paper is more rough than the Calderón commutator, and the argument in this paper involves much more complicated estimates and refined decompositions than that in [4].

This paper is organized as follows. In Sect. 2, we establish an endpoint estimate for the operators which will be used in the approximation; we also give some facts about the Luxemburgh norms in this section. In Sect. 3, we prove that $T_{\Omega ,\,A}$ with $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (2,\,\infty )$ can be approximated by a sequence of operators with smooth kernels. Sect. 4 is devoted to the proof of Theorem 1.3.

Throughout this paper, we use the symbol $A\lesssim B$ to denote that there exists a positive constant C such that $A\le CB$. Constant with subscript such as $C_1$, does not change in different occurrences. For any set $E\subset \mathbb {R}^d$, $\chi _E$ denotes its characteristic function. For a cube $I\subset \mathbb {R}^d$ and $\lambda \in (0,\,\infty )$, we use $\ell (I)$ to denote the side length of I, and $\lambda I$ to denote the cube with the same center as I and whose side length is $\lambda$ times that of I. For $x\in \mathbb {R}^d$ and $r>0$, $B(x,\,r)$ denotes the ball centered at x and having radius r. For a suitable function f, we denote $\widehat{f}$ the Fourier transform of f. For locally integrable function f and a cube $I\subset \mathbb {R}^d$, $\langle f\rangle _{I}$ denotes the mean value of f on I, that is, $\langle f\rangle _{I}=\vert I\vert ^{-1}\int _If(y)dy.$

2 A Preliminary $L^p(\mathbb {R}^d)$ Estimate

Let K be a locally integrable function on $\mathbb {R}^d\backslash \{0\}$, A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Let $T_A$ be an $L^2(\mathbb {R}^d)$ bounded operator, and satisfy that, for bounded function f with compact support and a. e. $x\in \mathbb {R}^d\backslash \textrm{supp}\,f$,

$$\begin{aligned} T_Af(x)=\int _{\mathbb {R}^d}K(x-y)\frac{A(x)-A(y)-\nabla A(y)(x-y)}{\vert x-y\vert }f(y)dy.\end{aligned}$$

(2.1)

This operator plays a key role in the approximation of $T_{\Omega , A}$. The main purpose of this section is establish the $L^p(\mathbb {R}^d)$ boundedness for the operator $T_A$ whose kernel K satisfies a minimum size conditions and minimum regularity conditions.

2.1 Some Facts About the Luxemburgh Norms

We list some known facts about the Luxemburgh norms. Details are given in [29]. Let $\Psi :\, [0,\,\infty )\rightarrow [0,\,\infty )$ be Young function, namely, $\Psi$ is convex and continuous on $[0,\,\infty )$, $\Psi (0)=0$ and $\lim _{t\rightarrow \infty }\Psi (t)=\infty$. We always assume that $\Psi$ satisfies a doubling condition, that is, $\Psi (2t)\le C\Psi (t)$ for any $t\in (0,\,\infty )$.

Let $\Psi$ be a Young function, and $Q\subset \mathbb {R}^d$ be a cube. Define the Luxemburg norm $\Vert \cdot \Vert _{L^{\Psi }(Q)}$ by

$$\begin{aligned} \Vert f\Vert _{L^{\Psi }(Q)}=\inf \left\{ \lambda >0:\,\frac{1}{ \vert Q\vert }\int _{Q}\Psi \Big (\frac{\vert f(x)\vert }{\lambda }\Big )dx\le 1\right\} . \end{aligned}$$

It is well known that

$$\begin{aligned} \frac{1}{\vert Q\vert }\int _{Q}\Psi \left( \vert f(x)\vert \right) dx\le 1\Leftrightarrow \Vert f\Vert _{L^{\Psi }(Q)}\le 1, \end{aligned}$$

and

$$\begin{aligned} \Vert f\Vert _{L^{\Psi }(Q)}\le \inf \left\{ \mu +\frac{\mu }{\vert Q\vert }\int _{Q}\Psi \left( \frac{\vert f(x)\vert }{\mu }\right) dx:\,\mu >0\right\} \le 2\Vert f\Vert _{L^{\Psi }(Q)};\nonumber \\ \end{aligned}$$

(2.2)

see see [29, p. 54] and [29, p. 69] respectively. For $p\in [1,\,\infty )$ and $\gamma \in \mathbb {R}$, set $\Psi _{p,\,\gamma }(t)=t^p\log ^{\gamma }(\textrm{e}+t)$. We denote $\Vert f\Vert _{L^{\Psi _{p,\,\gamma }}(Q)}$ as $\Vert f\Vert _{L^p(\log L)^{\gamma },\,Q}$.

Let $\Psi$ be a Young function. $\Psi ^*$, the complementary function of $\Psi$, is defined on $[0,\,\infty )$ by

$$\begin{aligned} \Psi ^*(t)=\sup \{st-\Psi (s): \,s\ge 0\}. \end{aligned}$$

The generalization of Hölder inequality

$$\begin{aligned} \frac{1}{\vert Q \vert }\int _Q\vert f(x)h(x)\vert dx\le \Vert f\Vert _{L^{\Psi }(Q)}\Vert h\Vert _{L^{\Psi ^*}(Q)} \end{aligned}$$

(2.3)

holds for $f\in L^{\Psi }(Q)$ and $h\in L^{\Psi ^*}(Q)$. see [29, p. 6].

For a cube $Q\subset \mathbb {R}^d$ and $\gamma >0$, we also define $\Vert f\Vert _{\textrm{exp}L^{\gamma },\,Q}$ by

$$\begin{aligned} \Vert f\Vert _{\textrm{exp}L^{\gamma },\,Q}=\inf \left\{ t>0:\,\frac{1}{\vert Q\vert }\int _{Q}\textrm{exp}\left( \frac{\vert f(y)\vert }{t}\right) ^{\gamma }dy\le 2\right\} . \end{aligned}$$

As it is well known, for $\Psi (t) = t\log (\textrm{e}+t)$, its complementary function $\Psi ^*(t)\approx \textrm{e}^t-1$. Let $b\in \textrm{BMO}(\mathbb {R}^d)$. The John–Nirenberg inequality tells us that for any $Q\subset \mathbb {R}^d$ and $p\in [1,\,\infty )$,

$$\begin{aligned} \Vert \vert b-\langle b\rangle _Q\vert ^p\Vert _{\textrm{exp}L^{1/p},\,Q}\lesssim \Vert b\Vert ^p_{\textrm{BMO}(\mathbb {R}^d)}. \end{aligned}$$

This, together with the inequality (2.3), shows that

$$\begin{aligned} \frac{1}{\vert Q\vert }\int _{Q}\vert b(x)-\langle b\rangle _Q\vert ^p\vert h(x)\vert ^pdx\lesssim \Vert h\Vert _{L^p(\log L)^p,\,Q}^p\Vert b\Vert _{\textrm{BMO}(\mathbb {R}^d)}^p. \end{aligned}$$

(2.4)

2.2 The $L^p(\mathbb {R}^d)$ Estimate for $T_{A}$

We need a preliminary lemma.

Lemma 2.1

Let A be a function on $\mathbb {R}^d$ with derivatives of order one in $L^q(\mathbb {R}^d)$ for some $q\in (d,\,\infty ]$. Then

$$\begin{aligned} \vert A(x)-A(y)\vert \lesssim \vert x-y\vert \left( \frac{1}{\vert I_x^y\vert }\int _{I_x^y}\vert \nabla A(z)\vert ^qdz\right) ^{\frac{1}{q}}, \end{aligned}$$

where $I_x^y$ is the cube which is centered at x and has side length $2\vert x-y\vert .$

Lemma 2.1 is just Lemma 1.4 in [3].

To obtain the $L^p(\mathbb {R}^d)$ boundedness of $T_{A}$, we need the following endpoint estimate.

Theorem 2.2

Let K be a locally integrable function on $\mathbb {R}^d\backslash \{0\}$, A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Let $T_A$ be an $L^2(\mathbb {R}^d)$ bounded operator with bound no more than 1 and satisfy (2.1). Suppose that

• (i) for each n with $1\le n\le d$, there exists an $L^2(\mathbb {R}^d)$ bounded operator $T^n$ with bound no more than 1 and satisfies that for bounded function f with compact support and a. e. $x\in \mathbb {R}^d\backslash \textrm{supp}\,f$,

  $$\begin{aligned} T^nf(x)=\int _{\mathbb {R}^d}K(x-y)\frac{x_n-y_n}{\vert x-y\vert }f(y)dy; \end{aligned}$$

• (ii) for each R with $0<R<\infty$,

  $$\begin{aligned} \int _{R<\vert x\vert <2R}\vert K(x)\vert dx\le 1; \end{aligned}$$

• (iii) for each $R>0$ and $y\in \mathbb {R}^d$ with $\vert y\vert <R/4$,

  $$\begin{aligned} \sum _{l=2}^{\infty }l\int _{2^{l}R<\vert x-y\vert \le 2^{l+1}R}\vert K(x-y)-K(x)\vert dx\le 1. \end{aligned}$$

Then for $\lambda >0$ and bounded function f with compact support,

$$\begin{aligned} \vert \{x\in \mathbb {R}^d:\, \vert T_Af(x)\vert >\lambda \}\vert \lesssim \int _{\mathbb {R}^d}\frac{\vert f(x)\vert }{\lambda }\log \left( \textrm{e}+\frac{\vert f(x)\vert }{\lambda }\right) dx. \end{aligned}$$

Proof

Theorem 2.2 can be proved by mimicking the proof of Theorem 1 in [23]. For the sake of self-contained, we present the main step of the proof here. Without loss of generality, we assume that $\Vert \nabla A\Vert _{\textrm{BMO}(\mathbb {R}^d)}=1$. For given bounded function f with compact support and $\lambda >0$, we apply the Calderoń-Zygmund decomposition to f at level $\lambda$, and obtain the following decomposition of f

$$\begin{aligned} f=g+b=g+\sum _jb_j, \end{aligned}$$

such that

• (a) $\Vert g\Vert _{L^{\infty }(\mathbb {R}^n)}\lesssim \lambda$ and $\Vert g\Vert _{L^1(\mathbb {R}^n)}\lesssim \Vert f\Vert _{L^1(\mathbb {R}^n)}$;

• (b) for each j, $b_j$ is supported on a cube $Q_j$, and cubes $\{Q_j\}$ are pairwise disjoint, $\int _{Q_j}b_j(x)dx=0$ and $\Vert b_j\Vert _{L^1(\mathbb {R}^n)}\lesssim \lambda \vert Q_j\vert$;

• (c) $\sum _{j}\vert Q_j\vert \lesssim \lambda ^{-1}\Vert f\Vert _{L^1(\mathbb {R}^n)}$.

The inequality (2.2) now tells us that

$$\begin{aligned} \sum _j\vert Q_j\vert \Vert b_j\Vert _{L\log L,\,Q_j}\lesssim & \sum _j\vert Q_j\vert \left( \lambda +\frac{\lambda }{\vert Q_\vert }\int _{Q_j}\frac{\vert f(x)\vert }{\lambda }\log \left( \textrm{e}+\frac{\vert f(x)\vert }{\lambda }\right) dx\right) \nonumber \\\lesssim & \int _{\mathbb {R}^d}\frac{\vert f(x)\vert }{\lambda }\log \left( \textrm{e}+\frac{\vert f(x)\vert }{\lambda }\right) dx. \end{aligned}$$

(2.5)

By the $L^2(\mathbb {R}^d)$ boundedness of $T_{A}$, we deduce that

$$\begin{aligned} \vert \{x\in \mathbb {R}^d:\,\vert T_{A}g(x)\vert >\lambda /2\}\vert \lesssim \lambda ^{-2} \Vert T_Ag\Vert _{L^2(\mathbb {R}^d)}^2\lesssim \lambda ^{-1}\Vert f\Vert _{L^1(\mathbb {R}^d)}.\end{aligned}$$

(2.6)

To estimate $T_{A}b$, we set $E=\cup _{j}4dQ_j$, and

$$\begin{aligned} A_j(y)=A(y)-\sum _{n=1}^d\langle \partial _nA\rangle _{Q_j}y_n. \end{aligned}$$

It then follows that for $x,\,y\in \mathbb {R}^d$,

$$\begin{aligned} A(x)-A(y)-\nabla A(y)(x-y)=A_j(x)-A_j(y)-\nabla A_j(y)(x-y). \end{aligned}$$

For $x\in \mathbb {R}^d\backslash E$, write

$$\begin{aligned} T_Ab(x)= & \sum _j\int _{\mathbb {R}^d}K(x-y)\frac{A_j(x)-A_j(y)}{\vert x-y\vert }b_j(y)dy\\ & -\sum _{n=1}^d\int _{\mathbb {R}^d}K(x-y)\frac{x_n-y_n}{\vert x-y\vert }\sum _{j}(\partial _nA(y)-\langle \partial _n A\rangle _{Q_j})b_j(y)dy\\= & \sum _jT_A^1b_j(x)-\sum _{n=1}^dT^n\Big (\sum _j(\partial _nA-\langle \partial _n A\rangle _{Q_j})b_j\Big )(x). \end{aligned}$$

Recall that $T^n$ is bounded on $L^2(\mathbb {R}^d)$. Our assumption (ii) implies that $T^n$ is also bounded from $L^1(\mathbb {R}^d)$ to $L^{1,\,\infty }(\mathbb {R}^d)$. As in [23, p. 764], an argument involving inequality (2.4) with $p=1$ and (2.5) leads to that

$$\begin{aligned} & \left| \left\{ x\in \mathbb {R}^d:\,\sum _{n=1}^d\Big \vert T^n\Big (\sum _j(\partial _nA-\langle \partial _n A\rangle _{Q_j})b_j\Big )(x)\Big \vert>\lambda /4\right\} \right| \nonumber \\ & \quad \le \sum _{n=1}^d\left| \left\{ x\in \mathbb {R}^d:\,\Big \vert T^n\Big (\sum _j(\partial _nA-\langle \partial _n A\rangle _{Q_j})b_j\Big )(x)\Big \vert >\frac{\lambda }{4d}\right\} \right| \nonumber \\ & \quad \lesssim \frac{1}{\lambda }\sum _{n=1}^d\sum _j\Vert (\partial _nA-\langle \partial _n A\rangle _{Q_j})b_j\Vert _{L^1(\mathbb {R}^d)}\lesssim \frac{1}{\lambda }\sum _j\vert Q_j\vert \Vert b_j\Vert _{L\log L,\,Q_j}\nonumber \\ & \quad \lesssim \int _{\mathbb {R}^d}\frac{\vert f(x)\vert }{\lambda }\log \left( \textrm{e}+\frac{\vert f(x)\vert }{\lambda }\right) dx. \end{aligned}$$

(2.7)

We now estimate $\sum _jT_{A}^1b_j$. For each fixed j, we choose $x^j\in 3Q_j\backslash 2Q_j$. Observe that

$$\begin{aligned} & \left| K(x-y)\frac{A_j(x)-A_j(y)}{\vert x-y\vert }-K(x-x^j)\frac{A_j(x)-A_j(x^j)}{\vert x-x^j\vert }\right| \\ & \quad \le \vert K(x-y)-K(x-x^j)\vert \frac{\vert A_j(x)-A_j(y)\vert }{\vert x-y\vert }\\ & \qquad +\vert K(x-x^j)\vert \left| \frac{A_j(x)-A_j(y)}{\vert x-y\vert }-\frac{A_j(x)-A_j(x^j)}{\vert x-x^j\vert }\right| . \end{aligned}$$

For $x\in \mathbb {R}^d\backslash E$, by the vanishing moment of $b_j$, we have that

$$\begin{aligned} \vert T_A^1b_j(x)\vert\le & \int _{\mathbb {R}^d}\vert K(x-y)-K(x-x^j)\vert \frac{\vert A_j(x)-A_j(y)\vert }{\vert x-y\vert }\vert b_j(y)\vert dy\\ & +\vert K(x-x^j)\vert \vert A_j(x)-A_j(x^j)\vert \int _{\mathbb {R}^d}\frac{\vert y-x^j\vert }{\vert x-y\vert ^2}\vert b_j(y)\vert dy\\ & +\vert K(x-x^j)\vert \int _{\mathbb {R}^d}\frac{\vert A_j(y)-A_j(x^j)\vert }{\vert x-y\vert }\vert b_j(y)\vert dy\\:= & \textrm{I}_j(x)+\textrm{II}_j(x)+\textrm{III}_j(x). \end{aligned}$$

For each $y\in Q_j$, we know that

$$\begin{aligned} \vert \langle \nabla A\rangle _{Q_j}-\langle \nabla A\rangle _{I_{x^j}^y}\vert \lesssim \log \left( \textrm{e}+ \frac{\vert x^j-y\vert }{\ell (Q_j)}\right) . \end{aligned}$$

It then follows from Lemma 2.1 that, for $y\in Q_j$,

$$\begin{aligned} \vert A_j(x^j)-A_j(y)\vert\lesssim & \vert x^j-y\vert \left( \frac{1}{\vert I_{x_j}^y\vert }\int _{I_{x^j}^y}\vert \nabla A(z)-\langle \nabla A\rangle _{Q_j}\vert ^qdx\right) ^{1/q}\\\le & \vert x^j-y\vert \left( \frac{1}{\vert I_{x_j}^y\vert }\int _{I_{x^j}^y}\vert \nabla A(z)-\langle \nabla A\rangle _{I_{x^j}^y}\vert ^qdx\right) ^{1/q}\\ & +\vert x^j-y\vert \vert \langle \nabla A\rangle _{Q_j}-\langle \nabla A\rangle _{I_{x^j}^y}\vert \\\lesssim & \vert x^j-y\vert \left( 1+\log \left( \textrm{e}+ \frac{\vert x^j-y\vert }{\ell (Q_j)}\right) \right) \lesssim \ell (Q_j), \end{aligned}$$

since $\vert x^j-y\vert \approx \ell (Q_j)$. Therefore,

$$\begin{aligned} \int _{\mathbb {R}^d\backslash 4dQ_j}\vert \textrm{III}_j(x)\vert dx\lesssim & \ell (Q_j)\sum _{l=2}^{\infty }l\int _{Q_j}\int _{2^{l+1}dQ_j\backslash 2^ldQ_j}\vert K(x-x^j)\vert \frac{\vert b_j(y)\vert }{\vert x-y\vert }dxdy\\\lesssim & \Vert b_j\Vert _{L^1(\mathbb {R}^d)}. \end{aligned}$$

For $l\ge 2$, $x\in 2^{l+1}dQ_j\backslash 2^ldQ_j$ and $y\in Q_j$, another application of Lemma 2.1 leads to that

$$\begin{aligned} \vert A_j(x)-A_j(y)\vert \lesssim l\vert x-y\vert ,\,\,\vert A_j(x)-A_j(x^j)\vert \lesssim l\vert x-x^j\vert . \end{aligned}$$

This, in turn, implies that

$$\begin{aligned} \int _{\mathbb {R}^d\backslash 4dQ_j}\vert \textrm{I}_j(x)\vert dx\lesssim & \sum _{l=2}^{\infty }l\int _{Q_j}\int _{2^{l+1}dQ_j\backslash 2^ldQ_j}\vert K(x-y)-K(x-x^j)\vert dx\vert b_j(y)\vert dy\\\lesssim & \Vert b_j\Vert _{L^1(\mathbb {R}^d)}, \end{aligned}$$

and

$$\begin{aligned} \int _{\mathbb {R}^d\backslash 4dQ_j}\vert \textrm{II}_j(x)\vert dx\lesssim & \sum _{l=2}^{\infty }l\int _{Q_j}\int _{2^{l+1}dQ_j\backslash 2^ldQ_j}\vert K(x-x^j)\vert \frac{\vert y-x^j\vert }{\vert x-y\vert }dx\vert b_j(y)\vert dy\\\lesssim & \Vert b_j\Vert _{L^1(\mathbb {R}^d)}, \end{aligned}$$

Combining the estimates for $\textrm{I}_j$, $\textrm{II}_j$ and $\textrm{III}_j$ leads to that

$$\begin{aligned} \Big \vert \Big \{x\in \mathbb {R}^d\backslash E:\,\vert \sum _jT_A^1b_j(x)\vert >\lambda /4\Big \}\Big \vert\le & 4\lambda ^{-1}\sum _j\int _{\mathbb {R}^d\backslash 4Q_j}\vert T_A^1b_j(x)\vert dx\\\lesssim & \lambda ^{-1}\Vert f\Vert _{L^1(\mathbb {R}^d)}. \end{aligned}$$

This, along with estimates (2.6)–(2.7) and the fact $\vert E\vert \lesssim \Vert f\Vert _{L^1(\mathbb {R}^d)}$, yields our desired conclusion. $\square$

We are now ready to give the $L^p(\mathbb {R}^d)$ boundedness for $T_{A}$.

Theorem 2.3

Let K be a locally integrable function on $\mathbb {R}^d\backslash \{0\}$, A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Let $T_A$ be an $L^2(\mathbb {R}^d)$ bounded operator with bound no more than 1 and satisfy (2.1). Under the hypothesis of Theorem 2.2, $T_A$ is bounded on $L^p(\mathbb {R}^d)$ for all $p\in (1,\,2]$ with bound C.

By a standard interpolation argument (see the proof of Corollary 1.3 in [22]), Theorem 2.3 follows from Theorem 2.2. We omit the details for brevity.

3 An Approximation of $T_{\Omega , A}$

In this section, we will show that $T_{\Omega ,\,A}$ can be approximated by a sequences of operators with “smooth kernels”. We first recall the definition of Calderón–Zygmund kernel.

Definition 3.1

Let $\Gamma$ be a locally integrable function on $\mathbb {R}^d\backslash \{0\}$. We say that $\Gamma$ is a Calderón–Zygmund kernel, if

• (i) for all $x\in \mathbb {R}^d\backslash \{0\}$,

  $$\begin{aligned} \vert \Gamma (x)\vert \lesssim \frac{1}{\vert x\vert ^d}; \end{aligned}$$

• (ii) for $x,\,y\in \mathbb {R}^d$ with $\vert x\vert \ge 4\vert y\vert$,

  $$\begin{aligned} \vert \Gamma (x-y)-\Gamma (x)\vert \lesssim \frac{\vert y\vert }{\vert x-y\vert ^{d+1}}. \end{aligned}$$

Lemma 3.2

Let $\Gamma$ be a function on $\mathbb {R}^d\backslash \{0\}$ which satisfies the following conditions:

• (i) $\Gamma$ is a Calderón–Zygmund kernel;

• (ii) for all $r,\,R$ with $0<r<R<\infty$ and $1\le n\le d$,

  $$\begin{aligned} \int _{r<\vert x\vert <R}\Gamma (x)x_ndx=0. \end{aligned}$$

Let A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, and $T_{\Gamma , A}$ be the operator defined by

$$\begin{aligned} T_{\Gamma , A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\Gamma (x-y)\frac{A(x)-A(y)-\nabla A(y)(x-y)}{\vert x-y\vert }f(y)dy. \end{aligned}$$

Then for all $p\in (1,\,\infty )$, $T_{\Gamma , A}$ is bounded on $L^p(\mathbb {R}^d)$.

Proof

Let $\mathcal {C}_A$ be the operator defined by

$$\begin{aligned} \mathcal {C}_{A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\Gamma (x-y)\frac{A(x)-A(y)}{\vert x-y\vert }f(y)dy. \end{aligned}$$

As it was pointed out in Theorem 1.1 in [24] that, under the hypothesis of Lemma 3.2, the estimate

$$\begin{aligned} \Vert \mathcal {C}_Af\Vert _{L^r(\mathbb {R}^d)}\lesssim \Vert \nabla A\Vert _{L^q(\mathbb {R}^d)}\Vert f\Vert _{L^p(\mathbb {R}^d)},\end{aligned}$$

(3.1)

holds true for $p\in (1,\,\infty )$ and $q\in (1,\,\infty ]$ with $1/r=1/q+1/p$, see also [1] for the case that K is a homogeneous kernel. With this estimate, repeating the proof of Corollary 1.2 in [6], we then can deduce the $L^p(\mathbb {R}^d)$ ($p\in (1,\,\infty )$) boundedness of $T_{A}$. $\square$

Lemma 3.3

Let $\phi \in C^{\infty }_0(\mathbb {R}^d)$ be a radial function such that $\textrm{supp}\, \phi \subset \{1/4\le \vert \xi \vert \le 4\}$ and

$$\begin{aligned} \sum _{l\in \mathbb {Z}}\phi ^3(2^{-l}\xi )=1,\,\,\,\vert \xi \vert >0. \end{aligned}$$

Let $\Phi =\widehat{\phi }$, A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Define the operator $S_{j; A}$ by

$$\begin{aligned} S_{j; A}f(x)=\int _{\mathbb {R}^d}2^{jd}\Phi (2^j(x-y))\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy. \end{aligned}$$

Then

$$\begin{aligned} \Big \Vert \Big (\sum _{j\in \mathbb {Z}}\vert 2^jS_{j;A}f\vert ^2\Big )^{\frac{1}{2}}\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

(3.2)

and

$$\begin{aligned} \Big \Vert \sum _{j\in \mathbb {Z}}2^jS_{j;A}f_j\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim \Big \Vert \Big (\sum _{j}\vert f_j\vert ^2\Big )^{1/2}\Big \Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(3.3)

Proof

We only prove (3.2), since (3.3) can be deduced from (3.2) by a standard duality argument. On the other hand, by the well known randomization argument (see [11, p. 545]), to prove (3.2), it suffices to prove that for all $\{\varepsilon _j\}_{j\in \mathbb {Z}}$ with $\varepsilon _j=\pm 1$,

$$\begin{aligned} \Big \Vert \sum _{j\in \mathbb {Z}}\varepsilon _j2^jS_{j;\,A}f\Big \Vert _{L^2(\mathbb {R}^d)}\le C \Vert f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

(3.4)

and the bound C is independent of $\{\varepsilon _j\}$.

Let $\varepsilon _j=\pm 1$ ($j\in \mathbb {Z})$, and

$$\begin{aligned} K_1(x)=\sum _{j\in \mathbb {Z}}2^{j(d+1)}\varepsilon _j\Phi (2^jx). \end{aligned}$$

(3.5)

By the fact that

$$\begin{aligned} \vert \Phi (2^jx)\vert \lesssim \frac{1}{(1+\vert 2^jx\vert )^{d+2}}, \end{aligned}$$

we know that for each $x\in \mathbb {R}^d\backslash \{0\}$,

$$\begin{aligned} \vert K_1(x)\vert \le \sum _{j:\,2^j\le \vert x\vert ^{-1}}2^{j(d+1)}\vert \Phi (2^jx)\vert +\sum _{j:\,2^j> \vert x\vert ^{-1}}2^{j(d+1)}\vert \Phi (2^jx)\vert \lesssim \vert x\vert ^{-d-1}. \end{aligned}$$

(3.6)

On the other hand, by the smoothness of $\Phi$, it is easy to verify that for $x,\,h\in \mathbb {R}^d$ with $\vert x\vert \ge 4\vert h\vert$,

$$\begin{aligned} \vert K_1(x+h)-K_1(x)\vert \lesssim \frac{\vert h\vert }{\vert x\vert ^{d+2}}. \end{aligned}$$

(3.7)

Since $\Phi$ is also a radial function, it certainly enjoys vanishing moment of order one. Thus, for all $0<r<R<\infty$ and $1\le n\le d$,

$$\begin{aligned} \int _{r<\vert x\vert <R}K_1(x)x_ndx=0. \end{aligned}$$

(3.8)

Estimates (3.6)-(3.8), via Lemma 3.2, leads to our desired conclusion. $\square$

Lemma 3.4

Let $\phi \in C^{\infty }_0(\mathbb {R}^d)$ be a radial function such that $\textrm{supp}\, \phi \subset \{1/4\le \vert \xi \vert \le 4\}$ and

$$\begin{aligned} \sum _{l\in \mathbb {Z}}\phi ^3(2^{-l}\xi )=1,\,\,\,\vert \xi \vert >0. \end{aligned}$$

Let $S_j$ be the operator defined by

$$\begin{aligned} \widehat{S_jf}(\xi )=\phi (2^{-j}\xi )\widehat{f}(\xi ). \end{aligned}$$

Then

• (i) for $b\in \textrm{BMO}(\mathbb {R}^d)$, we have that

  $$\begin{aligned} \Big \Vert \Big (\sum _{j\in \mathbb {Z}}\vert [b,\,S_j]f\vert ^2\Big )^{1/2}\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}; \end{aligned}$$

  where and in what follows, for a locally integrable function b and a linear operator T, $[b,\,T]$ denotes the commutator defined

  $$\begin{aligned} [b,\,T]f(x)=b(x)Tf(x)-T(bf)(x); \end{aligned}$$

• (ii) for function a on $\mathbb {R}^d$ which satisfies that $\nabla a\in L^{\infty }(\mathbb {R}^d)$, it follows that

  $$\begin{aligned} \Big \Vert \Big (\sum _{j\in \mathbb {Z}}\vert 2^j[a,\,S_j]f\vert ^2\Big )^{1/2}\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Proof

Conclusion (i) is just [19, Lemma 1]. To prove conclusion (ii), let $\Phi =\widehat{\phi }$ and $K_1$ be the function defined by (3.5). Estimates (3.6)-(3.8), via (3.1), leads to conclusion (ii). $\square$

Remark 3.5

Conclusion (ii) of Lemma 3.4 was first proved by Chen and Ding, using a different argument, see [5, Lemma 2.3].

Lemma 3.6

Let $\delta \in (0,\,1)$, $l\in \mathbb {Z}$ and $D>0$ be constants, m be a multiplier such that $\textrm{supp}\, m\subset \{\vert \xi \vert \le D^{-1}2^l\}$, and

$$\begin{aligned} \Vert m\Vert _{L^{\infty }(\mathbb {R}^d)}\le D^{-1}\min \{(\delta 2^l)^2,\,\log ^{-\beta }(\textrm{e}+2^l)\}, \end{aligned}$$

and for all multi-indices $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \Vert \partial ^{\gamma }m\Vert _{L^{\infty }(\mathbb {R}^d)}\le D^{\vert \gamma \vert -1}\max \{1,\,2^{-l\vert \gamma \vert }\}. \end{aligned}$$

Let A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, and $T_{m,\,A}$ be the operator defined by

$$\begin{aligned} T_{m,\,A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\Theta (x-y)(A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy, \end{aligned}$$

with $\Theta$ the inverse Fourier transform of m. Then for any $\varepsilon \in (0,\,1)$,

$$\begin{aligned} \Vert T_{m,A}f\Vert _{L^2(\mathbb {R}^d)}\lesssim \min \{(\delta 2^l)^{\varepsilon /2},\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(3.9)

Proof

The argument here is a variant of the proof of Lemma 3.2 in [4], together with some refined estimates of Luxemburg norms. We assume that $\Vert \nabla A\Vert _{\textrm{BMO}(\mathbb {R}^d)}=1$. Set $E=\min \{(\delta 2^l)^2,\,\log ^{-\beta }(\textrm{e}+2^l)\}$. Let $\phi \in C^{\infty }_0(\mathbb {R}^d)$ be a radial function, $\textrm{supp}\,\phi \subset B(0,\,2)$, $\phi (x)=1$ when $\vert x\vert \le 1$. Set $\varphi (x)=\phi (x)-\phi (2x)$. Then $\textrm{supp}\, \varphi \subset \{1/4\le \vert x\vert \le 4\}$ and

$$\begin{aligned} \sum _{j\in \mathbb {Z}}\varphi (2^{-j}x)\equiv 1,\,\,\vert x\vert >0. \end{aligned}$$

(3.10)

Let $\varphi _j(x)=\varphi (2^{-j}x)$ for $j\in \mathbb {Z}$. Set

$$\begin{aligned} W_{j}(x)=\Theta (x)\varphi _j(x),\,\,j\in \mathbb {Z}. \end{aligned}$$

Let $T_{m,j}$ be the convolution operator with kernel $W_j$. Observing that for all multi-indices $\gamma \in \mathbb {Z}_+^d$, $\partial ^{\gamma }\varphi (0)=0$, we thus have that

$$\begin{aligned} \int _{\mathbb {R}^d}\widehat{\varphi }(\xi )\xi ^{\gamma }d\xi =0. \end{aligned}$$

This, in turn, implies that for all $N\in \mathbb {N}$ and $\xi \in \mathbb {R}^d$,

$$\begin{aligned} \vert \widehat{W_j}(\xi )\vert= & \left| \int _{\mathbb {R}^d}\Big (m(\xi -2^{-j}\eta )-\sum _{\vert \gamma \vert \le N}\frac{1}{\gamma !}\partial ^{\gamma }m(\xi )(2^{-j}\eta )^{\gamma }\Big )\widehat{\varphi }(\eta )d\eta \right| \nonumber \\\lesssim & 2^{-j(N+1)}\sum _{\vert \gamma \vert =N+1}\Vert \partial ^{\gamma }m\Vert _{L^{\infty }(\mathbb {R}^d)} \int _{\mathbb {R}^d}\vert \eta \vert ^{N+1}\vert \widehat{\varphi }(\eta )\vert d\eta \nonumber \\\lesssim & 2^{-j(N+1)}D^{N}\max \{1,\,2^{-l(N+1)}\}. \end{aligned}$$

(3.11)

On the other hand, a trivial computation yields for $j\in \mathbb {Z}$,

$$\begin{aligned} \Vert \widehat{W_j}\Vert _{L^{\infty }(\mathbb {R}^d)}\le \Vert m\Vert _{L^{\infty }(\mathbb {R}^d)}\Vert \widehat{\varphi _j}\Vert _{L^1(\mathbb {R}^d)}\lesssim D^{-1}E. \end{aligned}$$

(3.12)

Interpolation inequalities (3.11) and (3.12) gives us that for $\varepsilon \in (0,\,1)$,

$$\begin{aligned} & \Vert \widehat{W_j}\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 2^{-j(N+1)(1-\varepsilon )}D^{N(1-\varepsilon )-\varepsilon }\max \{1,\,2^{-l(N+1)}\}^{1-\varepsilon } E^{\varepsilon }. \end{aligned}$$

(3.13)

We now prove (3.9). Let $T_{m,j;A}$ be the operator defined by

$$\begin{aligned} T_{m,j;A}f(x)=\int _{\mathbb {R}^d}W_j(x-y)\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy. \end{aligned}$$

For $\varepsilon \in (0,\,1)$, let $F_{\varepsilon }=\min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}$. We claim that for all $j\in \mathbb {Z}$ and $N\in \mathbb {N}$ and $\varepsilon \in (0,\,1)$,

$$\begin{aligned} \Vert T_{m,j;A}f\Vert _{L^2(\mathbb {R}^d)}\lesssim & (2^{-j}D)^{N(1-\varepsilon )-\varepsilon }\log (\textrm{e}+2^jD^{-1})F_{\varepsilon }\nonumber \\ & \quad \times \max \{1,\,2^{-l(N+1)}\}^{1-\varepsilon }\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(3.14)

Observe that $\textrm{supp}\,W_j\subset \{x:\,\vert x\vert \le 2^{j+2}\}$. If I is a cube having side length $2^j$, and $f\in L^2(\mathbb {R}^d)$ with $\textrm{supp}\,f\subset I$, then $T_{m,j}f\subset 100dI$. Therefore, to prove (3.14), we may assume that $\textrm{supp}\, f\subset I$ with I a cube having side length $2^j$. Let $x_0$ be a point on the boundary of 200dI and $A_I^*(y)=A(y)-\sum _{n=1}^d \langle \partial _n A\rangle _{100dI}y_n$, and

$$\begin{aligned} A_I(y)=(A_I^*(y)-A_I^*(x_0))\zeta _I(y), \end{aligned}$$

where $\zeta _I\in C^{\infty }_0(\mathbb {R}^d)$, $\textrm{supp}\,\zeta \subset 150dI$ and $\zeta (x)\equiv 1$ when $x\in 100dI$. Observe that $\Vert \nabla \zeta \Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 2^{-j}$. An application of Lemma 2.1 tells us that for all $y\in 100dI$,

$$\begin{aligned} \vert A_I(y)-A_I(x_0)\vert \lesssim 2^j. \end{aligned}$$

This shows that

$$\begin{aligned} \Vert A_I\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 2^j. \end{aligned}$$

Write

$$\begin{aligned} T_{m,j;A}f(x)=A_I(x) T_{m,j}f(x)-T_{m,\,j}(A_If)(x)-\sum _{n=1}^d[h_n,\,T_{m,j}](f\partial _nA_I)(x), \end{aligned}$$

where $h_n(x)=x_n$ (recall that $x_n$ denotes the nth variable of x). It then follows from (3.13) that

$$\begin{aligned} & \Vert A_IT_{m,j}f\Vert _{L^2(\mathbb {R}^d)}+\Vert T_{m,j}(A_If)\Vert _{L^2(\mathbb {R}^d)}\nonumber \\ & \quad \lesssim (2^{-j}D)^{N(1-\varepsilon )-\varepsilon }\max \{1,\,2^{-j(N+1)}\}^{1-\varepsilon }E^{\varepsilon } \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(3.15)

Applying the John–Nirenberg inequality, we know that

$$\begin{aligned} \Vert \vert \partial _nA_I\vert ^2\Vert _{\textrm{exp}L^{1/2},\,I}\lesssim 1. \end{aligned}$$

Recall that $\textrm{supp}\,[h_n,\,T_{m,j}](f\partial _nA_I)\subset 100dI$. It then follows from inequality (2.4) that

$$\begin{aligned} \Vert [h_n,\,T_{m,j}](f\partial _nA_I)\Vert _{L^2(\mathbb {R}^d)}= & \sup _{\Vert g\Vert _{L^2(\mathbb {R}^d)}\le 1} \left| \int _{\mathbb {R}^d}\partial _nA_I(x)f(x)[h_n,\,T_{m,j}]g(x)dx\right| \\\le & \Vert f\Vert _{L^2(\mathbb {R}^d)}\sup _{\begin{array}{c} \Vert g\Vert _{L^2(\mathbb {R}^d)}\le 1\\ {\textrm{supp}\,g\subset 100dI} \end{array}} \Vert \partial _nA_I[h_n,\,T_{m,j}]g\Vert _{L^2(I)}\\\le & \Vert f\Vert _{L^2(\mathbb {R}^d)}\Big (\vert I\vert \sup _{\begin{array}{c} \Vert g\Vert _{L^2(\mathbb {R}^d)}\le 1\\ {\textrm{supp}\,g\subset 100dI} \end{array}} \Vert [h, T_{m,\,j}]g \Vert _{L^2(\log L)^2,\,I}^2\Big )^{1/2}. \end{aligned}$$

Now let $g\in L^2(\mathbb {R}^d)$ with $\Vert g\Vert _{L^2(\mathbb {R}^d)}\le 1$ and $\textrm{supp}\,g\subset 100dI$. Observe that

$$\begin{aligned} \Vert W_j\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim \Vert m\Vert _{L^1(\mathbb {R}^d)}\lesssim 2^{dl}D^{-d-1}E, \end{aligned}$$

which, via Young’s inequality, implies that

$$\begin{aligned} \Vert T_{m,j}g\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim \Vert W_{j}\Vert _{L^{\infty }(\mathbb {R}^d)}\Vert g\Vert _{L^1(\mathbb {R}^d)}\lesssim 2^{dl}D^{-d-1}E\Vert g\Vert _{L^1(\mathbb {R}^d)}, \end{aligned}$$

and so

$$\begin{aligned} \Vert \chi _{I}[h_n,\,T_{m,j}]g\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 2^{dl}2^{j}D^{-d-1}E\Vert g\Vert _{L^1(\mathbb {R}^d)} \lesssim 2^{dl}2^jD^{-d-1}E2^{dj/2}, \end{aligned}$$

since

$$\begin{aligned} \Vert g\Vert _{L^1(\mathbb {R}^d)}\lesssim \vert I\vert ^{1/2}\Vert g\Vert _{L^2(\mathbb {R}^d)}\lesssim 2^{dj/2}. \end{aligned}$$

On the other hand, we deduce from (3.13) that

$$\begin{aligned} \Vert [h_n,\,T_{m,j}]g\Vert _{L^{2}(\mathbb {R}^d)}\lesssim & 2^j \Vert T_{m,j}g\Vert _{L^{2}(\mathbb {R}^d)}\\\lesssim & (2^{-j}D)^{N(1-\varepsilon )-\varepsilon } \max \{1,\,2^{-l(N+1)}\}^{1-\varepsilon }E^{\varepsilon } \Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Set

$$\begin{aligned} \lambda _0 = \big [(2^{-j}D)^{N(1-\varepsilon )-\varepsilon }\log ^2(\textrm{e}+2^{j}D^{-1})F_{\varepsilon }\max \{1,\,2^{-l(N+1)}\}^{1-\varepsilon }\big ]^2 2^{-jd}. \end{aligned}$$

A straightforward computation tells us that

$$\begin{aligned} & \int _{I}\vert [h_n,\,T_{m,j}]g(x)\vert ^2\log ^2\left( \textrm{e}+\frac{\vert [h_n,\,T_{m,j}]g(x)\vert }{\sqrt{\lambda _0}}\right) dx\\ & \quad \lesssim \Big (\log ^2(\textrm{e}+2^{j}D^{-1})+\max \{1,\,l\}\Big )\int _{I}\vert [h_n,\,T_{m,j}]g(x)\vert ^2dx\\ & \quad \lesssim \big [(2^{-j}D)^{N(1-\varepsilon )-\varepsilon }F_{\varepsilon } \max \{1,\,2^{-l(N+1)}\}^{(1-\varepsilon )}\log (2+2^{j}D^{-1})\big ]^2\\ & \quad \lesssim \lambda _02^{jd}, \end{aligned}$$

since $E^{\varepsilon }\max \{1,\,l\}\le F_{\varepsilon }$, and

$$\begin{aligned} \frac{\Vert \chi _{I}[h_n,\,T_{m,j}]g\Vert _{L^{\infty }(\mathbb {R}^d)}}{\sqrt{\lambda _0}}\lesssim & (2^{j}D^{-1})^{d+1} (2^jD^{-1})^{N(1-\varepsilon )-\varepsilon }2^{dl}F_{\varepsilon }^{-1}. \end{aligned}$$

This tells us that

$$\begin{aligned} \Vert [h_n, T_{m,\,j}]g\Vert _{L^2(\log L)^2,\,I}\lesssim \sqrt{\lambda _0}, \end{aligned}$$

and thus

$$\begin{aligned} \Vert [h_n,\,T_{m,j}](f\partial _nA_I)\Vert _{L^2(\mathbb {R}^d)}\lesssim & (2^{-j}D)^{N(1-\varepsilon )-\varepsilon }\log (\textrm{e}+2^{j}D^{-1})\nonumber \\ & \times F_{\varepsilon }\max \{1,2^{-l(N+1)}\}^{1-\varepsilon }\Vert f\Vert _{L^2(\mathbb {R}^d)}.\nonumber \\ \end{aligned}$$

(3.16)

The estimate (3.14) then follows from (3.15) and (3.16).

We now conclude the proof of Lemma 3.6. It suffices to consider the case $\varepsilon \in (4/5,\,1)$. Let $G_{\varepsilon }=\min \{(\delta 2^l)^{\varepsilon /2},\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}$. For each fixed $\varepsilon \in (4/5,\,1)$, we choose $N_1,\,N_2\in \mathbb {N}$ such that

$$\begin{aligned} \frac{\varepsilon }{1-\varepsilon }<N_1<\frac{5\varepsilon /2-1}{1-\varepsilon },\,\,N_2(1-\varepsilon )-\varepsilon <0. \end{aligned}$$

Observe that

$$\begin{aligned} F_{\varepsilon }\max \{1,2^{-l(N_1+1)}\}^{1-\varepsilon }\lesssim G_{\varepsilon },\,F_{\varepsilon }\max \{1,2^{-l(N_2+1)}\}^{1-\varepsilon }\lesssim G_{\varepsilon }. \end{aligned}$$

A straightforward computation shows that if

$$\begin{aligned} \Vert T_{m,A}f\Vert _{L^2(\mathbb {R}^d)}\lesssim & \sum _{j:\,2^{-j}D\le 1}\Vert T_{m,\,j;\,A}f\Vert _{L^2(\mathbb {R}^d)}+\sum _{j:\,2^{-j}D>1}\Vert T_{m,j;\,A}f\Vert _{L^2(\mathbb {R}^d)}\\\lesssim & \sum _{j:\,2^{-j}D\le 1}(2^{-j}D)^{N_1(1-\varepsilon )-\varepsilon }\log (\textrm{e}+2^jD^{-1})G_{\varepsilon }\Vert f\Vert _{L^2(\mathbb {R}^d)}\\ & +\sum _{j:\,2^{-j}D>1}(2^{-j}D)^{N_2(1-\varepsilon )-\varepsilon }\log (\textrm{e}+2^jD^{-1})G_{\varepsilon }\Vert f\Vert _{L^2(\mathbb {R}^d)}\\\lesssim & G_{\varepsilon }\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

This completes the proof of Lemma 3.6. $\square$

Lemma 3.7

Let $\delta \in (0,\,1)$, $l\in \mathbb {Z}$ and $D>0$ be constants, m be a multiplier such that $\textrm{supp}\, m\subset \{\vert \xi \vert \le D^{-1}2^l\}$, and

$$\begin{aligned} \Vert m\Vert _{L^{\infty }(\mathbb {R}^d)}\le D^{-1}\min \{(\delta 2^l)^2,\,\log ^{-\beta }(\textrm{e}+2^l)\}, \end{aligned}$$

and for all multi-indices $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \Vert \partial ^{\gamma }m\Vert _{L^{\infty }(\mathbb {R}^d)}\le D^{\vert \gamma \vert -1}\max \{1,\,2^{-l\vert \gamma \vert }\}. \end{aligned}$$

Let $T_m$ be the multiplier operator defined by

$$\begin{aligned} \widehat{T_mf}(\xi )=m(\xi )\widehat{f}(\xi ). \end{aligned}$$

Then for any $b\in \textrm{BMO}(\mathbb {R}^d)$ and $\varepsilon \in (0,\,1)$,

$$\begin{aligned} \Vert [b,\,T_m]f\Vert _{L^2(\mathbb {R}^d)}\lesssim D^{-1}\min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}\Vert b\Vert _{\textrm{BMO}(\mathbb {R}^d)}\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Proof

Let $\widetilde{m}(\xi )=Dm(2^{l}D^{-1}\xi )$ and $T_{\widetilde{m}}$ be the multiplier operator with multiplier $\widetilde{m}$. We know that $\textrm{supp}\, \widetilde{m}\subset \{\vert \xi \vert \le 1\}$, and

$$\begin{aligned} \Vert \widetilde{m}\Vert _{L^{\infty }(\mathbb {R}^d)}\le \min \{(\delta 2^l)^2,\,\log ^{-\beta }(\textrm{e}+2^l)\}, \end{aligned}$$

and for all multi-indices $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \Vert \partial ^{\gamma }\widetilde{m}\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 1. \end{aligned}$$

Applying Lemma 2 in [19], we then obtain that

$$\begin{aligned} \Vert [b,\,T_{\widetilde{m}}]f\Vert _{L^2(\mathbb {R}^d)}\lesssim \min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}\Vert b\Vert _{\textrm{BMO}(\mathbb {R}^d)}\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

This, via dilation-invariance, implies our desired conclusion and then completes the proof of Lemma 3.7. $\square$

Theorem 3.8

Let $\delta \in (0,\,1)$ be a constant, $\{\mu _j\}_{j\in \mathbb {Z}}$ be a sequence of functions on $\mathbb {R}^d\backslash \{0\}$. Suppose that for some $\beta \in (2,\,\infty )$,

$$\begin{aligned} \Vert \mu _j\Vert _{L^1(\mathbb {R}^d)}\lesssim 2^{-j},\,\vert \widehat{\mu _j}(\xi )\vert \lesssim 2^{-j}\min \{\vert \delta 2^j\xi \vert ^{2},\,\log ^{-\beta }(\textrm{e}+\vert 2^{j}\xi \vert )\}, \end{aligned}$$

and for all multi-indices $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \Vert \partial ^{\gamma }\widehat{\mu _j}\Vert _{L^{\infty }(\mathbb {R}^d)}\lesssim 2^{j(\vert \gamma \vert -1)}. \end{aligned}$$

Let $\mu (x)=\sum _{j\in \mathbb {Z}}\mu _j(x)$ and $T_{\mu , A}$ be the operator defined by

$$\begin{aligned} T_{\mu , A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\mu (x-y)\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy, \end{aligned}$$

where A is a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Then for any $\varepsilon \in (0,\,1)$,

$$\begin{aligned} \Vert T_{\mu , A}f\Vert _{L^2(\mathbb {R}^d)}\lesssim \log ^{-\varepsilon \beta +2}(\textrm{e}+\delta ^{-1})\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Proof

It suffices to consider the case $\varepsilon \in (1/2,\,1)$. Let T be the operator defined by

$$\begin{aligned} Tf(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\mu (x-y)f(y)dy. \end{aligned}$$

It is easy to verify that for $\xi \in \mathbb {R}^d$,

$$\begin{aligned}\vert \widehat{\mu }(\xi )\vert\lesssim & \vert \xi \vert \sum _{j:\, 2^j>\vert \xi \vert ^{-1}}\log ^{-\beta }(\textrm{e}+\vert 2^j\xi \vert )+\vert \xi \vert ^2\sum _{j:\, 2^j\le \vert \xi \vert ^{-1}}2^{j}\lesssim \vert \xi \vert . \end{aligned}$$

This in turn, gives us that

$$\begin{aligned} \int _{\mathbb {R}^d}\vert \widehat{Tf}(\xi )\vert ^2d\xi \lesssim \Vert f\Vert ^2_{\dot{L}_{1}^2(\mathbb {R}^d)}, \end{aligned}$$

and

$$\begin{aligned} \int _{\mathbb {R}^d}\vert \xi \vert ^{-2}\vert \widehat{Tf}(\xi )\vert ^2d\xi \lesssim \Vert f\Vert ^2_{L^2(\mathbb {R}^d)}, \end{aligned}$$

where $\Vert f\Vert _{\dot{L}^2_1(\mathbb {R}^d)}$ is the homogeneous Sobolev norm defined as

$$\begin{aligned} \Vert f\Vert _{\dot{L}^2_1(\mathbb {R}^d)}^2=\int _{\mathbb {R}^d}\vert \xi \vert ^{2}\vert \widehat{f}(\xi )\vert ^2d\xi . \end{aligned}$$

Let $U_j$ be the convolution operator with kernel $\mu _j$, and $\phi \in C^{\infty }_0(\mathbb {R}^d)$ such that $0\le \phi \le 1$, $\textrm{supp}\,\phi \subset \{1/4\le \vert \xi \vert \le 4\}$ and

$$\begin{aligned} \sum _{l\in \mathbb {Z}}\phi ^3(2^{-l}\xi )=1,\,\,\vert \xi \vert >0. \end{aligned}$$

Set $m_j(\xi )=\widehat{\mu _j}(\xi )$, and $m_j^l(\xi )=m_j(\xi )\phi (2^{j-l}\xi )$. Define the operator $U_j^l$ by

$$\begin{aligned} \widehat{U_j^lf}(\xi )=m_j^l(\xi )\widehat{f}(\xi ). \end{aligned}$$

Let $S_l$ be the multiplier operator defined as in Lemma 3.4. We claim that for functions $f,\,g\in C^{\infty }_0(\mathbb {R}^d)$,

$$\begin{aligned} & \int _{\mathbb {R}^d}g(x)T_{\mu , A}f(x)dx=\int _{\mathbb {R}^d}g(x)\sum _l\sum _{j} (S_{l-j}U_j^lS_{l-j})_Af(x)dx, \end{aligned}$$

(3.17)

where and in what follows,

$$\begin{aligned} (S_{l-j}U_j^lS_{l-j})_Af(x)=\int _{\mathbb {R}^d}L(x-y)\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy, \end{aligned}$$

with L the kernel of the convolution operator $S_{l-j}U_j^lS_{l-j}$. We define $U_{j,A}^l$ similarly. To prove this, let $R>0$ be large enough such that $\textrm{supp}\,f\cup \textrm{supp}\,g\subset B(0,\,R)$. Let $\zeta \in C^{\infty }_0(\mathbb {R}^d)$ such that $0\le \zeta \le 1$, $\zeta \equiv 1$ on $B(0,\,R)$ and $\textrm{supp}\, \zeta \subset B(0,\,2R)$. Set

$$\begin{aligned} A_R(y)=\big (A(y)-\sum _{n=1}^d\langle \partial _nA\rangle _{B(0,\,R)}y_n\big )\zeta (y). \end{aligned}$$

Then

$$\begin{aligned} \int _{\mathbb {R}^d}g(x)T_{\mu , A}f(x)dx= & \int _{\mathbb {R}^d}g(x)A_{R}(x)Tf(x)dx-\int _{\mathbb {R}^d}g(x)T(A_Rf)(x)dx\\ & -\sum _{n=1}^d\int _{\mathbb {R}^d}g(x)[h_n,\,T](f\partial _nA_R)(x)dx, \end{aligned}$$

where the function $h_n(x)=x_n$. Note that $h_nf\partial _nA_R\in L^2(\mathbb {R}^d)$. It then follows that

$$\begin{aligned} \int _{\mathbb {R}^d}g(x)[h_n,\,T](f\partial _nA_R)(x)dx=\int _{\mathbb {R}^d} g(x)\sum _{l}\sum _j[h_n,\,S_{l-j}U_j^lS_{l-j}](f\partial _nA_R)(x)dx. \end{aligned}$$

Since $gA_R$, $fA_R\in L^2(\mathbb {R}^d)$, we also have that

$$\begin{aligned} & \int _{\mathbb {R}^d}g(x)A_R(x)Tf(x)dx=\int _{\mathbb {R}^d}g(x)A_R(x)\sum _l\sum _{j} S_{l-j}U_j^lS_{l-j}f(x)dx, \\ & \quad \int _{\mathbb {R}^d}g(x)T(A_Rf)(x)dx=\int _{\mathbb {R}^d}g(x)\sum _l\sum _{j} S_{l-j}U_j^lS_{l-j}(fA_R)(x)dx. \end{aligned}$$

These three equalities lead to (3.17) directly.

Now we estimate $(S_{l-j}U_j^lS_{l-j})_Af$. Obviously, $\textrm{supp}\,m_j^l\subset \{\vert \xi \vert \le 2^{l-j+2}\}$ and

$$\begin{aligned} \vert m_j^l(\xi )\vert \lesssim 2^{-j}\min \{(\delta 2^{l})^2,\,\log ^{-\beta } (\textrm{e}+2^l)\}.\end{aligned}$$

Furthermore, by the fact that

$$\begin{aligned} \vert \partial ^{\gamma }\phi (2^{j-l}\xi )\vert \lesssim 2^{(j-l)\vert \gamma \vert },\,\,\vert \partial ^{\gamma }m_j(\xi )\vert \lesssim 2^{j(\vert \gamma \vert -1)}, \end{aligned}$$

it then follows that for all $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \vert \partial ^{\gamma }m_j^l(\xi )\vert \lesssim 2^{j(\vert \gamma \vert -1)}\max \{1,\,2^{-\vert \gamma \vert l}\}. \end{aligned}$$

This, via Lemma 3.6, tells us that

$$\begin{aligned} \Vert U_{j,\,A}^lf\Vert _{L^2(\mathbb {R}^d)}\lesssim \min \{(\delta 2^l)^{\frac{\varepsilon }{2}},\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(3.18)

Also, we have that

$$\begin{aligned} \Vert U_{j}^lf\Vert _{L^2(\mathbb {R}^d)}\lesssim 2^{-j}\min \{(\delta 2^{l})^2,\,\log ^{-\beta } (\textrm{e}+2^l)\}\Vert f\Vert _{L^2(\mathbb {R}^d)}.\end{aligned}$$

(3.19)

For fixed $j,\,l\in \mathbb {Z}$, write

$$\begin{aligned}(S_{l-j}U_j^lS_{l-j})_Af(x)= & S_{l-j,A}(U_j^lS_{l-j}f)(x)+S_{j}(U_{j}^lS_{l-j})_Af(x)\\ & +\sum _{n=1}^d[h_n,\,S_{l-j}]([\partial _nA,\,U_{j}^lS_{l-j}]f)(x)\\:= & \textrm{I}_j^lf(x)+\textrm{II}_j^lf(x)+\sum _{n=1}^d\textrm{III}_j^{l,n}f(x). \end{aligned}$$

We now estimate terms $\textrm{I}_j^l$, $\textrm{II}_j^l$ and $\textrm{III}_j^{l,n}$. Inequality (3.3) in Lemma 3.3, along with (3.19) leads to that

$$\begin{aligned} \Big \Vert \sum _jS_{l-j,A}(U_j^lS_{l-j}f)\Big \Vert _{L^2(\mathbb {R}^d)}^2\lesssim & \sum _{j}2^{2(j-l)} \big \Vert U_{j}^lS_{l-j}f\big \Vert _{L^2(\mathbb {R}^d)}^2\\\lesssim & 2^{-2l}\min \{(\delta 2^{l})^2,\,\log ^{-\beta } (\textrm{e}+2^l)\}^2\Vert f\Vert ^2_{L^2(\mathbb {R}^d)}. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _l\Big \Vert \sum _j\textrm{I}_j^lf\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim & \Big (\sum _{l>\log \frac{1}{\sqrt{\delta }}}l^{-\beta }+\delta ^2\sum _{l\le \log \frac{1}{\sqrt{\delta }}}2^l\Big ) \Vert f\Vert _{L^2(\mathbb {R}^d)}\\\lesssim & \log ^{-\beta +1} (\textrm{e}+\delta ^{-1})\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

For each fixed $j, l\in \mathbb {Z}$ and n with $1\le n\le d$, it follows from Lemma 3.7 and (3.19) that

$$\begin{aligned} \big \Vert [\partial _nA,\,U_{j}^lS_{l-j}]f\big \Vert _{L^2(\mathbb {R}^d)}\le & \big \Vert [\partial _nA,\,U_{j}^l]S_{l-j}f\big \Vert _{L^2(\mathbb {R}^d)} +\big \Vert U_{j}^l[\partial _nA,\,S_{l-j}]f\big \Vert _{L^2(\mathbb {R}^d)}\\\lesssim & 2^{-j}\min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}\Vert S_{l-j}f\Vert _{L^2(\mathbb {R}^d)}^2\\ & +2^{-j}\min \{(\delta 2^l)^{2},\log ^{-\beta }(\textrm{e}+2^l)\}\Vert [\partial _nA,S_{l-j}]f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

which, along with Lemma 3.4, implies that

$$\begin{aligned} & \Big \Vert \sum _{j}\vert [h_n,\,S_{l-j}]([\partial _nA,\,U_{j}^lS_{l-j}]f)\vert \Big \Vert _{L^2(\mathbb {R}^d)}^2\lesssim \sum _{j}2^{2(j-l)}\big \Vert [\partial _nA,\,U_{j}^lS_{l-j}]f\big \Vert ^2_{L^2(\mathbb {R}^d)}\\ \\ & \quad \lesssim 2^{-2l}\min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}^2\Vert f\Vert ^2_{L^2(\mathbb {R}^d)}. \end{aligned}$$

Thus,

$$\begin{aligned} \sum _l\Big \Vert \sum _j\textrm{III}_{j}^{l,n}f\Big \Vert _{L^2(\mathbb {R}^d)}\lesssim & \Big (\sum _{l>\log \frac{1}{\sqrt{\delta }}}l^{-\varepsilon \beta +1}+\delta ^2\sum _{l\le \log \frac{1}{\sqrt{\delta }}}2^{(4\varepsilon -2)l}\Big ) \Vert f\Vert _{L^2(\mathbb {R}^d)}\\\lesssim & \log ^{-\varepsilon \beta +2} (\textrm{e}+\delta ^{-1})\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

As for term $\textrm{II}_j^l$, write

$$\begin{aligned} (U_j^lS_{l-j})_Af= & U_{j,A}^lS_{l-j}f+U_{j}^lS_{l-j,\,A}f+\sum _{n=1}^d[h_n,\,U^l_j]([\partial _nA,\,S_{l-j}]f). \end{aligned}$$

It follows from Littlewood–Paley theory and (3.18) that

$$\begin{aligned} \Big \Vert \sum _jS_{l-j}U_{j,\,A}^lS_{l-j}f\Big \Vert _{L^2(\mathbb {R}^d)}^2\lesssim & \sum _j\Vert U_{j,\,A}^lS_{l-j}f\Vert _{L^2(\mathbb {R}^d)}^2 \\\lesssim & \min \{(\delta 2^l)^{\varepsilon /2},\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}^2\Vert f\Vert _{L^2(\mathbb {R}^d)}^2.\end{aligned}$$

Again by Lemma 3.3 and (3.19), we deduce that

$$\begin{aligned} \Big \Vert \sum _jS_{l-j}U_{j}^lS_{l-j,A}f\Big \Vert _{L^2(\mathbb {R}^d)}^2\lesssim & \sum _j\Vert U_{j}^lS_{l-j,A}f\Vert _{L^2(\mathbb {R}^d)}^2 \\\lesssim & 2^{-2l}\min \{(\delta 2^l)^{2},\,\log ^{-\beta }(\textrm{e}+2^l)\}^2\Vert f\Vert _{L^2(\mathbb {R}^d)}^2.\end{aligned}$$

Similar to the term $\textrm{III}_{j}^{l,n}$, we have that for each $1\le n\le d$,

$$\begin{aligned} & \Big \Vert \sum _jS_{l-j}[h_n,\,U^l_j]([\partial _nA,\,S_{l-j}]f)\Big \Vert _{L^2(\mathbb {R}^d)}^2\\ & \quad \lesssim \sum _j\Vert [h_n,\,U^l_j]([\partial _nA,\,S_{l-j}]f)\Vert _{L^2(\mathbb {R}^d)}^2 \\ & \quad \lesssim 2^{-2l}\min \{(\delta 2^l)^{2\varepsilon },\,\log ^{-\varepsilon \beta +1}(\textrm{e}+2^l)\}^2\Vert f\Vert ^2_{L^2(\mathbb {R}^d)}.\end{aligned}$$

Therefore,

$$\begin{aligned} \sum _l\Big \Vert \sum _j\textrm{II}_{j}^lf\Big \Vert _{L^2(\mathbb {R}^d)} \lesssim \log ^{-\varepsilon \beta +2} (\textrm{e}+\delta ^{-1})\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

The estimates for $\textrm{I}_j^l$, $\textrm{II}_j^l$ and $\textrm{III}_{j}^{l,n}$ above, via (3.17), leads to our desired conclusion. $\square$

We are now ready to establish our main result in this section.

Theorem 3.9

Let $\Omega$ be homogeneous of degree zero, satisfy the vanishing moment condition (1.1) and $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (2,\,\infty )$. Let A be a function on $\mathbb {R}^d$ such that $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$. Then there exists a sequence of operators $\{R_{l,A}\}_{l\in \mathbb {N}}$ such that

• (i) $R_{l,\,A}$ is defined as

  $$\begin{aligned} R_{l,A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\widetilde{K}^l(x-y)\frac{A(x)-A(y)-\nabla A(y)(x-y)}{\vert x-y\vert }f(y)dy, \end{aligned}$$

  the function $\widetilde{K}^l$ satisfies the size condition that for $0<R<\infty$,

  $$\begin{aligned} \int _{R<\vert x\vert <2R}\vert \widetilde{K}^l(x)\vert dx\lesssim 1, \end{aligned}$$

  and the regularity that for all $R>0$ and $y\in \mathbb {R}^d$,

  $$\begin{aligned} \sum _{m=2}^{\infty }m\int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\vert \widetilde{K}^l(x-y)-\widetilde{K}^l(x)\vert dx\lesssim l^2; \end{aligned}$$

• (ii) for each fixed n with $1\le n\le d$, the operator $W_l^n$ defined by

  $$\begin{aligned} W_l^nf(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\widetilde{K}^l(x-y)\frac{x_n-y_n}{\vert x-y\vert }f(y)dy \end{aligned}$$

  is bounded on $L^2(\mathbb {R}^d)$ with bound independent of l;

• (iii) for each fixed $\varepsilon \in (0,\,1)$,

  $$\begin{aligned} \Vert R_{l,\,A}-T_{\Omega ,A}\Vert _{L^2(\mathbb {R}^d)}\lesssim l^{-\varepsilon \beta +2}. \end{aligned}$$

  (3.20)

Proof

For $j\in \mathbb {Z}$, let $K_j(x)=\frac{\Omega (x)}{\vert x\vert ^{d+1}}\chi _{\{2^{j-1}\le \vert x\vert <2^j\}}(x)$. Let $\psi \in C^{\infty }_0(\mathbb {R}^d)$ be a nonnegative radial function such that

$$\begin{aligned} \textrm{supp}\, \psi \subset \{x:\,\vert x\vert \le 1/4\},\,\,\,\int _{\mathbb {R}^d}\psi (x)dx=1. \end{aligned}$$

For $j\in \mathbb {Z}$, set $\psi _j(x) = 2^{-dj}\psi (2^{-j}x)$. For a positive integer l, define

$$\begin{aligned} H_l(x)=\sum _{j\in \mathbb {Z}}K_j*\psi _{j-l}(x). \end{aligned}$$

Let $R_l$ be the convolution operator with kernel $H_l$. For a function A on $\mathbb {R}^d$ with $\nabla A\in \textrm{BMO}(\mathbb {R}^d)$, denote

$$\begin{aligned} R_{l,A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}H_l(x-y)\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y)dy. \end{aligned}$$

Now we prove (3.20). Write

$$\begin{aligned} \sum _{j\in \mathbb {Z}}K_j(x)-H_l(x)=\sum _{j\in \mathbb {Z}}\big (K_j(x)-K_j*\psi _{j-l}(x)\big )=:\sum _{j\in \mathbb {Z}}\mu _{j,l}(x). \end{aligned}$$

The fact $\psi$ is radial, implies that, for n with $1\le n\le d$,

$$\begin{aligned} \int _{\mathbb {R}^d}\psi (x)x_ndx=0. \end{aligned}$$

From this we know that for all n with $1\le n\le d$, $\partial _n\widehat{\psi }(0)=0.$ By Taylor series expansion and the fact that $\widehat{\psi }(0)=1$, we deduce that

$$\begin{aligned} \vert \widehat{\psi }(2^{j-l}\xi )-1\vert \lesssim \min \{1,\,\vert 2^{j-l}\xi \vert ^{2}\}. \end{aligned}$$

On the other hand, as it was proved in [15], we know that when $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (1,\,\infty )$,

$$\begin{aligned} \vert \widehat{K_j}(\xi )\vert \lesssim 2^{-j}\min \{1,\,\log ^{-\beta }(\textrm{e}+\vert 2^j\xi \vert )\}, \end{aligned}$$

which, in turn leads to following Fourier transform estimate

$$\begin{aligned} & \vert \widehat{\mu _{j,l}}(\xi )\vert =\vert \widehat{K_j}(\xi )\vert \vert \widehat{\psi }(2^{j-l}\xi )-1\vert \lesssim 2^{-j}\min \{\log ^{-\beta }(\textrm{e}+\vert 2^j\xi \vert ),\,\vert 2^{j-l}\xi \vert ^{2}\}. \end{aligned}$$

(3.21)

On the other hand, a trivial computation shows that for all multi-indices $\gamma \in \mathbb {Z}_+^d$,

$$\begin{aligned} \Vert \partial ^{\gamma }\widehat{K_j}\Vert _{L^{\infty }(\mathbb {R}^{d})}\lesssim \Vert \Omega \Vert _{L^1(S^{d-1})}2^{(\vert \gamma \vert -1)j}, \end{aligned}$$

and so for all $\xi \in \mathbb {R}^d$,

$$\begin{aligned} & \vert \partial ^{\gamma }\widehat{\mu _{j,l}}(\xi )\vert \lesssim \sum _{\gamma _1+\gamma _2=\gamma }\vert \partial ^{\gamma _1}\widehat{K_j}(\xi )\vert \vert \partial ^{\gamma _2}\widehat{\psi }(2^{j-l}\xi )\vert \lesssim \Vert \Omega \Vert _{L^1(S^{d-1})}2^{j(\vert \gamma \vert -1)}.\nonumber \\ \end{aligned}$$

(3.22)

Let $\widetilde{K}^l(x-y)=H_l(x-y)\vert x-y\vert$. The Fourier transforms (3.21) and (3.22), via Theorem 3.8 with $\delta =2^{-l}$, lead to (3.20) directly.

We now verify conclusion (i). For each fixed $R>0$,

$$\begin{aligned} \int _{R<\vert x\vert <2R}\vert H_l(x)\vert dx\lesssim \sum _{j:\,2^j\approx R}\Vert K_j\Vert _{L^1(\mathbb {R}^d)}\Vert \psi _{j-l}\Vert _{L^1(\mathbb {R}^d)}\lesssim R^{-1}. \end{aligned}$$

On the other hand, for $R>0$ and $y\in \mathbb {R}^d$ with $\vert y\vert <R/4$,

$$\begin{aligned} & \int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\big \vert H_l(x-y)\vert x-y\vert -H_l(x)\vert x\vert \big \vert dx\\ & \quad \le \int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\vert H_l(x-y)-H_l(x)\vert \vert x-y\vert dx\\ & \qquad +\vert y\vert \int _{2^{m-1}R<\vert x\vert \le 2^{m+2}R}\vert H_l(x)\vert dx\\ & \quad \lesssim 2^mR\int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\vert H_l(x-y)-H_l(x)\vert dx+\frac{\vert y\vert }{2^mR}. \end{aligned}$$

Observe that

$$\begin{aligned} \Vert \psi _{j-l}(\cdot -y)-\psi _{j-l}(\cdot )\Vert _{L^1(\mathbb {R}^d)}\lesssim \min \{1,\,2^{l-j}\vert y\vert \}. \end{aligned}$$

Young’s inequality now tells us that

$$\begin{aligned} & \int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\vert H_l(x-y)-H_l(x)\vert dx\\ & \quad \lesssim \sum _{j:\,2^j\approx 2^{m+1}R}\Vert K_j\Vert _{L^1(\mathbb {R}^d)} \Vert \psi _{j-l}(\cdot -y)-\psi _{j-l}(\cdot )\Vert _{L^1(\mathbb {R}^d)}\\ & \quad \lesssim (2^mR)^{-1}\min \{1,\,2^{l-m}\}. \end{aligned}$$

This, in turn, implies that

$$\begin{aligned} & \sum _{m=2}^{\infty }m\int _{2^{m}R<\vert x-y\vert \le 2^{m+1}R}\vert \widetilde{K}^l(x-y)-\widetilde{K}^l(x)\vert dx\\ & \quad \lesssim \sum _{m=2}^{\infty }m\min \{1,\,2^{l-m}\}+\sum _{m=2}^{\infty }m2^{-m}\lesssim l^2. \end{aligned}$$

Finally, for each fixed n with $1\le n\le d$, let

$$\begin{aligned} Y_l^nf(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\big (\sum _{j\in \mathbb {Z}}K_j(x-y)-H_l(x-y)\big )(x_n-y_n)f(y)dy. \end{aligned}$$

The estimates (3.21) and (3.22), via [4, Theorem 3.4], state that

$$\begin{aligned} \Vert Y^n_lf\Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

For $1\le n\le d$, let $T_{\Omega }^n$ be the operator defined by

$$\begin{aligned} T_{\Omega }^{n}h(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\frac{\Omega (x-y)(x_n-y_n)}{\vert x-y\vert ^{d+1}}h(y)dy. \end{aligned}$$

(3.23)

It is well known that $T_{\Omega }^n$ is bounded on $L^2(\mathbb {R}^d)$. Note that

$$\begin{aligned}\mathrm{p.\,v.}\int _{\mathbb {R}^d}\sum _{j\in \mathbb {Z}}K_j(x-y)(x_n-y_n)f(y)dy=T_{\Omega }^nf(x). \end{aligned}$$

Therefore, $W_l^n$ is bounded on $L^2(\mathbb {R}^d)$ with bound independent of l. This completes the proof of Theorem 3.9. $\square$

4 Proof of Theorem 1.3

Let $\varphi \in C^{\infty }_0(\mathbb {R}^d)$ be a radial function which satisfies (3.10), $\varphi _j(x)=\varphi (2^{-j}x)$. For each fixed $j\in \mathbb {Z}$, set

$$\begin{aligned} T_{\Omega ,\,A;\,j}f(x)=\int _{\mathbb {R}^d}K_{A,\,j}(x,\,y)f(y)dy, \end{aligned}$$

where

$$\begin{aligned} K_{A,\,j}(x,\,y)=\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\big (A(x)-A(y)-\nabla A(y)(x-y)\big )\varphi _j(\vert x-y\vert ). \end{aligned}$$

Let $\omega \in C^{\infty }_0(\mathbb {R}^d)$ be a radial function, have integral zero and $\textrm{supp}\,\omega \subset B(0,\,1)$. Note that $\widehat{\omega }$ is also a radial function on $\mathbb {R}^d$. Let $Q_s$ be the operator defined by $Q_tf(x)=\omega _t*f(x)$, where $\omega _t(x)=t^{-d}\omega (t^{-1}x)$ for $t>0$. We assume that

$$\begin{aligned} \int ^{\infty }_0[\widehat{\omega }(s)]^4\frac{ds}{s}=1. \end{aligned}$$

The Calderón reproducing formula

$$\begin{aligned} \int ^{\infty }_0Q_s^4\frac{ds}{s}=I \end{aligned}$$

(4.1)

then holds true. Moreover, the classical Littlewood–Paley theory tells us that for all $p\in (1,\,\infty )$,

$$\begin{aligned} & \Big \Vert \Big (\int _0^{\infty }\vert Q_sf\vert ^2\frac{ds}{s}\Big )^{\frac{1}{2}}\Big \Vert _{L^p(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^p(\mathbb {R}^d)}. \end{aligned}$$

(4.2)

It is well know that for $b\in \textrm{BMO}(\mathbb {R}^d)$ and $p\in (1,\,\infty )$,

$$\begin{aligned} \Big \Vert \Big (\int _0^{\infty }\vert [b,\,Q_s]f\vert ^2\frac{ds}{s}\Big )^{\frac{1}{2}}\Big \Vert _{L^p(\mathbb {R}^d)}\lesssim \Vert b\Vert _{\textrm{BMO}(\mathbb {R}^d)}\Vert f\Vert _{L^p(\mathbb {R}^d)}. \end{aligned}$$

(4.3)

For a function $\Omega \in L^1(S^{d-1})$, define the operators $W_{\Omega ,j}$ and $U_{\Omega ,n, j}$ by

$$\begin{aligned} W_{\Omega ,j}h(x)= & \int _{\mathbb {R}^d}\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\varphi _j(x-y)h(y)dy, \\ U_{\Omega ,n, j}h(x)= & \int _{\mathbb {R}^d}\frac{\Omega (x-y)(x_n-y_n)}{\vert x-y\vert ^{d+1}}\varphi _j(x-y)h(y)dy,\,\,1\le n\le d. \end{aligned}$$

Lemma 4.1

Let $\Omega$ be homogeneous of degree zero, and $\Omega \in GS_{\beta }(S^{d-1})$ for some $\beta \in (1,\,\infty )$, then for $j\in \mathbb {Z}$ and $s\in (0, 2^j]$,

$$\begin{aligned} \Vert Q_sW_{\Omega ,\,j}f\Vert _{L^2(\mathbb {R}^d)}\lesssim 2^{-j} \log ^{-\beta }(\textrm{e}+2^j/s)\Vert f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

(4.4)

and

$$\begin{aligned} \Vert Q_sU_{\Omega ,\,n,j}f\Vert _{L^2(\mathbb {R}^d)}\lesssim \log ^{-\beta }(\textrm{e}+2^j/s)\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.5)

Furthermore, for $b\in \textrm{BMO}(\mathbb {R}^d)$, $j\in \mathbb {Z}$ and $s\in (0,\,2^j]$,

$$\begin{aligned} & \big \Vert [b, Q_sU_{\Omega , n,\,j}]f\big \Vert _{L^2(\mathbb {R}^d)}\lesssim \log ^{-\beta +1}(\textrm{e}+ 2^j/s)\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.6)

Proof

Inequalities (4.4) and (4.5) were proved in [4]. We now prove (4.6). We may assume that $\Vert b\Vert _{\textrm{BMO}(\mathbb {R}^d)}=1$. By dilation-invariance, it suffices to consider the case $j=0$ and $s\in (0,\,1]$. Let $\mathbb {R}^d=\cup _lI_l$, where $I_l$ are cubes having disjoint interiors, and side length 1. For each fixed l, let $f_l=f\chi _{I_l}.$ Observing that $\textrm{supp}\,Q_sU_{\Omega ,0}f_l\subset 20dI_l$, and ${Q_sU_{\Omega ,n,0}f_l}$ have bounded overlaps, we then have that

$$\begin{aligned} \Vert [b, Q_sU_{\Omega ,n, 0}]f\Vert _{L^2(\mathbb {R}^d)}^2\lesssim \sum _l\Vert [b, Q_sU_{\Omega ,n, 0}]f_l\Vert _{L^2(\mathbb {R}^d)}^2. \end{aligned}$$

Thus, we may assume that supp$f\subset I$, with I a cube having side length 1. An application of the inequality (2.4) gives us that

$$\begin{aligned} \int _{\mathbb {R}^d}\vert b(x)-\langle b\rangle _I\vert ^2\vert Q_sU_{\Omega , n,0}f(x)\vert ^2dx\lesssim \Vert Q_sU_{\Omega , n,0}f\Vert _{L^2(\log L)^2,20dI}^2. \end{aligned}$$

Now let $\lambda _0=\log ^{-\beta +1}(\textrm{e}+1/s)$, h be a function on $\mathbb {R}^d$ such that $\textrm{supp}\,h\subset 20dI$ and $\Vert h\Vert _{L^2(\mathbb {R}^d)}=1$. Observing that $\Vert h\Vert _{L^1(\mathbb {R}^d)}\lesssim 1$, we then get that

$$\begin{aligned} \Vert Q_sU_{\Omega ,n,0}h\Vert _{L^\infty (\mathbb {R}^d)}\lesssim s^{-d}\Vert U_{\Omega ,n,0}h\Vert _{L^1(\mathbb {R}^d)}\lesssim s^{-d}\Vert h\Vert _{L^1(\mathbb {R}^d)}\lesssim s^{-d}, \end{aligned}$$

and for any $x\in 20dI$ and $s\in (0,\,1]$,

$$\begin{aligned} \frac{\vert Q_sU_{\Omega ,n,0}h(x)\vert }{\lambda _0}\lesssim s^{-d}\log ^{\beta -1}(\textrm{e}+1/s)\lesssim s^{-d-1}. \end{aligned}$$

A straightforward computation involving estimate (4.5) leads to that

$$\begin{aligned} & \int _{20dI}\left( \frac{\vert Q_sU_{\Omega ,n,0}h(x)\vert }{\lambda _0}\right) ^2\log ^2\left( \textrm{e}+\frac{\vert Q_sU_{\Omega ,n,0}h(x)\vert }{\lambda _0}\right) dx\\ & \quad \lesssim \frac{1}{\lambda _0^2}\Vert Q_sU_{\Omega ,n,0}h\Vert _{L^2(\mathbb {R}^d)}^2\log ^2(\textrm{e}+1/s)\lesssim 1, \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert Q_sU_{\Omega ,n,0}h\Vert _{L^2(\log L)^2,20dI}\lesssim \lambda _0. \end{aligned}$$

This, via inequality (2.4), yields

$$\begin{aligned} \Vert \vert b-\langle b\rangle _I\vert Q_sU_{\Omega ,n,0}h\Vert _{L^2(\mathbb {R}^d)} \lesssim \lambda _0\Vert h\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.7)

We also have that

$$\begin{aligned} \Vert Q_sU_{\Omega ,n,0}((b-\langle b\rangle _I)f)\Vert _{L^2(\mathbb {R}^d)}\lesssim \lambda _0\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.8)

In fact, a standard computation leads to that

$$\begin{aligned} & \Vert Q_sU_{\Omega , n,0}((b-\langle b\rangle _I)f)\Vert _{L^2(\mathbb {R}^d)}\\ & \quad =\sup _{\Vert g\Vert _{L^{2}(\mathbb {R}^d)}\le 1}\left| \int _{\mathbb {R}^d}Q_sU_{\Omega ,n, 0}((b-\langle b\rangle _I)f)(x)g(x)dx\right| \\ & \quad =\sup _{\Vert g\Vert _{L^{2}(\mathbb {R}^d)}\le 1}\left| \int _{I}Q_sU_{\Omega ,n, 0}(g\chi _{20dI})(x)(b(x)-\langle b\rangle _I)f(x)dx\right| \\ & \quad \le \sup _{\Vert g\Vert _{L^{2}(\mathbb {R}^d)}\le 1}\Vert f\Vert _{L^{2}(\mathbb {R}^d)}\Vert (b-\langle b\rangle _I)Q_sU_{\Omega ,n, 0}(g\chi _{20dI})\Vert _{L^{2}(\mathbb {R}^d)}, \end{aligned}$$

which, along with (4.7), implies (4.8). Combining estimates (4.7) and (4.8) yields (4.6) for the case of $j=0$, and completes the proof of Lemma 4.1. $\square$

Proof of Theorem 1.3

The procedure follows two steps. At first, we prove the $L^2(\mathbb {R}^d)$ boundedness of $T_{A}$, by following the argument in the proof of Theorem 1.3 in [17], together with some refined decomposition and estimates for $T_{\Omega , A}$. Then we prove the $L^p(\mathbb {R}^d)$ boundedness, using the approximation established in Sect. 3. Again, we assume that $\Vert \nabla A\Vert _{\textrm{BMO}(\mathbb {R}^d)}=1$.

We now prove the $L^2(\mathbb {R}^d)$ boundedness of $T_{\Omega ,\,A}$. By the Calderón reproducing formula (4.1), it suffices to prove that for $f,\,g\in C^{\infty }_0(\mathbb {R}^d)$,

$$\begin{aligned} & \left| \int ^{\infty }_0\int _{0}^t\int _{\mathbb {R}^d}Q_s^4{T}_{\Omega ,\,A}Q_t^4f(x)g(x)dx \frac{ds}{s}\frac{dt}{t}\right| \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^{2}(\mathbb {R}^d)}; \end{aligned}$$

(4.9)

and

$$\begin{aligned} & \left| \int ^{\infty }_0\int _{t}^{\infty }\int _{\mathbb {R}^d}Q_s^4{T}_{\Omega ,\,A}Q_t^4f(x)g(x)dx \frac{ds}{s}\frac{dt}{t}\right| \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

(4.10)

We first prove (4.9). Let $\alpha \in (\frac{d+1}{d+2},\,1)$ be a constant. For each fixed $j\in \mathbb {Z}$, let

$$\begin{aligned} F_{j,1}=\{(s,\,t):\,0<t\le 2^j,\,0<s\le t\}, \end{aligned}$$

$$\begin{aligned} F_{j,2}=\{(s,\,t):\,2^j<t<\infty ,\,0<s\le (2^jt^{\alpha -1})^{1/\alpha }\}, \end{aligned}$$

and

$$\begin{aligned} F_{j,3}=\{(s,\,t):\,2^j<t<\infty ,\,(2^jt^{\alpha -1})^{1/\alpha }<s\le t\}. \end{aligned}$$

For $k=1,\,2,\,3$, set

$$\begin{aligned} \textrm{E}_k(f,\,g)=\sum _{j\in \mathbb {Z}}\int \int \chi _{F_{j,k}}\int _{\mathbb {R}^d}Q_s{T}_{\Omega ,\,A;j}Q_t^4f(x)Q_s^3g(x)dx \frac{ds}{s}\frac{dt}{t}, \end{aligned}$$

with

$$\begin{aligned} T_{\Omega ,A;j}=\int _{\mathbb {R}^d}\frac{\Omega (x-y)}{\vert x-y\vert ^{d+1}}\varphi _j(\vert x-y\vert )\big (A(x)-A(y)-\nabla A(y)(x-y)\big )f(y){d}y. \end{aligned}$$

It was proved in [22] that

$$\begin{aligned} \Vert E_3(f,\,g)\Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert \Omega \Vert _{L^1(S^{d-1})}\Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Then the proof of (4.9) is reduced to proving that for $k=1,\,2$,

$$\begin{aligned} \Vert E_k(f,\,g)\Vert _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.11)

The proofs of (4.11) for $k=1$ and $k=2$ are similar, so we only prove (4.11) for the case of $k=2$. For each fixed $j\in \mathbb {Z}$, let $\{I_l\}$ be the sequence of cubes having disjoint interiors and side lengths $2^j$, such that $\mathbb {R}^d=\bigcup _lI_l.$ For fixed l, let $h_{s,l}(x)=Q_sg(x)\chi _{I_l}(x)$, $\zeta _l\in C_0^\infty (\mathbb {R}^d)$ such that $\text {supp} \,\zeta _l\subset 48 dI_l$, $0\le \zeta _l\le 1$ and $\zeta _l(x)\equiv 1$ when $x\in 32 dI_l$. Let $x^l$ be a point on the boundary of $100 dI_l$. Let

$$\begin{aligned} \tilde{A}_{l}(y)=A(y)-\sum _{m=1}^d\langle \partial _m A\rangle _{I_l}y_m,~~A_{l}(y)=(\tilde{A}_{l}(y)-\tilde{A}_{l}(x^l))\zeta _l(y). \end{aligned}$$

It follows from Lemma 2.1 that for all $y\in \mathbb {R}^d$,

$$\begin{aligned} \vert A_l(y)\vert \lesssim 2^j,\,\,\vert \nabla A_l(y)\vert \lesssim 1+\vert \nabla A(y)-\langle \nabla A\rangle _{I_l}\vert . \end{aligned}$$

(4.12)

Note that for $x\in 48 dI_l$ and $y\in \mathbb {R}^d$ with $\vert x-y\vert \le 2^{j+2}$,

$$\begin{aligned} A(x)-A(y)-\nabla A(y)(x-y)= A_{l}(x)-A_{l}(y)-\nabla A_{l}(y)(x-y). \end{aligned}$$

Write

$$\begin{aligned} {T}_{\Omega ,\,A;j}h(x)=A_lW_{\Omega ,j}h(x)-W_{\Omega ,j}(A_lh)(x)-\sum _{n=1}^dU_{\Omega ,n,j}(\partial _nA_lh)(x). \end{aligned}$$

Set

$$\begin{aligned} \textrm{D}_1= & \left| \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d}A_l W_{\Omega ,j}Q_t^4f(x)Q_s^3h_{s,l}(x)dx\frac{ds}{s}\frac{dt}{t}\right| , \\ \textrm{D}_2= & \left| \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d} W_{\Omega ,j}(A_l Q_t^4f)(x)Q_s^3h_{s,l}(x)dx\frac{ds}{s}\frac{dt}{t}\right| , \end{aligned}$$

and for $1\le n\le d$,

$$\begin{aligned} \textrm{D}_{3,n}=\Big \vert \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d} U_{\Omega ,n,j}(\partial _nA_lQ_t^4f)(x)Q_s^3h_{s,l}(x)dx\frac{ds}{s}\frac{dt}{t}\Big \vert . \end{aligned}$$

It then follows that

$$\begin{aligned} \vert E_2(f,\,g)\vert \le \textrm{D}_1+\textrm{D}_2+\sum _{n=1}^d\textrm{D}_{3,n}. \end{aligned}$$

We first consider term $\textrm{D}_1$. To this aim, we split it into two parts as

$$\begin{aligned} \textrm{D}_1= & \left| \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d}Q_t^4f(x) W_{\Omega ,j}(A_lQ_s^3h_{s,l})(x)dx\frac{ds}{s}\frac{dt}{t}\right| \\\le & \left| \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d}Q_t^4f(x) W_{\Omega ,j}([A_l, Q_s^3]h_{s,l})(x)dx\frac{ds}{s}\frac{dt}{t}\right| \\ & +\left| \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d}Q_t^4f(x) W_{\Omega ,j} Q_s^3(A_lh_{s,l})(x)dx\frac{ds}{s}\frac{dt}{t}\right| \\:= & \textrm{D}_{11}+\textrm{D}_{12}.\end{aligned}$$

An application of Hölder’s inequality leads to that

$$\begin{aligned} \textrm{D}_{11}\lesssim & \left( \sum _{j\in \mathbb {Z}}\sum _l \int \int \chi _{F_{j,2}}\big \Vert \chi _{64dI_l}Q_t^4f\big \Vert ^2_{L^2(\mathbb {R}^d)} (2^{-j}s)^{\frac{1}{2}} \frac{ds}{s}\frac{dt}{t}\right) ^{1/2}\\ & \times \left( \sum _{j\in \mathbb {Z}}\sum _{l} \int \int \chi _{F_{j,2}}\big \Vert W_{\Omega ,\,j}([A_l, Q_s^3]h_{s,l}) \big \Vert ^2_{L^{2}(\mathbb {R}^d)} (2^{-j}s)^{-\frac{1}{2}}\frac{ds}{s}\frac{dt}{t}\right) ^{1/2}\\:= & \textrm{I}_1\textrm{I}_2. \end{aligned}$$

A straightforward computation gives that

$$\begin{aligned}\textrm{I}_1\lesssim & \left\| \left( \int ^{\infty }_{0}\vert Q_t^4f\vert ^2\int _{0}^{t}\sum _{j:\, 2^j\ge s^{\alpha }t^{1-\alpha }}(2^{-j}s)^{\frac{1}{2}} \frac{ds}{s} \frac{dt}{t}\right) ^{1/2}\right\| _{L^2(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Lemma 2.1, along with estimate (4.12), tells us that for $x,\,y\in \mathbb {R}^d$ with $\vert x-y\vert \le s\le 2^j$,

$$\begin{aligned} \vert A_l(x)-A_l(y)\vert\lesssim & \vert x-y\vert \left( \frac{1}{\vert I_x^y\vert }\int _{I_x^y}\vert \nabla A(z)-\langle \nabla A\rangle _{I_l}\vert ^qdz\right) ^{1/q}\\\lesssim & \vert x-y\vert \left( 1+\log \left( \frac{2^j}{\vert x-y\vert }\right) \right) \lesssim 2^j(2^{-j}s)^{\frac{1}{2}}, \end{aligned}$$

since $\Phi (t)=t\log (\textrm{e}+t)$ is increasing and $\Phi (t)\le t^{1/2}$ when $t\le 1$. Therefore,

$$\begin{aligned} \vert [A_l, Q_s^3]h(x)\vert \lesssim \int _{\mathbb {R}^d}\widetilde{\omega }_s(x-y)\vert A_l(x)-A_l(y)\vert \vert h(y)\vert dy\lesssim 2^j (2^{-j}s)^{\frac{1}{2}}Mh(x), \end{aligned}$$

where $\widetilde{\omega }_s(x)=s^{-d}\widetilde{\omega }(s^{-1}x)$ and $\widetilde{\omega }(x)=\omega *\omega *\omega (x)$. Let $M_{\Omega }$ be the operator defined by

$$\begin{aligned} M_{\Omega }h(x)=\sup _{r>0}r^{-d}\int _{\vert x-y\vert <r}\vert \Omega (x-y)h(y)\vert dy. \end{aligned}$$

We then have that

$$\begin{aligned}\textrm{I}_2\lesssim & \left( \sum _{j\in \mathbb {Z}}\sum _{l} \int \int \chi _{F_{j,2}}\big \Vert M_\Omega Mh_{s,l}\big \Vert ^2_{L^{2}(\mathbb {R}^d)} (2^{-j}s)^{\frac{1}{2}}\frac{ds}{s}\frac{dt}{t}\right) ^{1/2}\\\lesssim & \left( \sum _{j\in \mathbb {Z}}\sum _{l} \int \int \chi _{F_{j,2}}\Vert h_{s,l}\Vert ^2_{L^{2}(\mathbb {R}^d)}(2^{-j}s)^{\frac{1}{2}}\frac{ds}{s}\frac{dt}{t} \right) ^{1/2}\\\lesssim & \Vert g \Vert _{L^{2}(\mathbb {R}^d)}, \end{aligned}$$

since

$$\begin{aligned} \int _{s}^{\infty }\sum _{j:\, 2^j\ge s^{\alpha }t^{1-\alpha }}(2^{-j}s)^{\frac{1}{2}}\frac{dt}{t}\lesssim 1. \end{aligned}$$

Therefore,

$$\begin{aligned} \textrm{D}_{11}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g \Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

For term $\textrm{D}_{12}$, another application of Hölder’s inequality yields

$$\begin{aligned} & \textrm{D}_{12}\lesssim \left( \sum _{j\in \mathbb {Z}} \int \int \chi _{F_{j,2}}\big \Vert Q_s^2Q_t^3f\big \Vert ^2_{L^2(\mathbb {R}^d)} \log ^{-\sigma _1}(\textrm{e}+2^j/s) \frac{ds}{s}\frac{dt}{t}\right) ^{1/2}\\ & \quad \quad \times \left\| \left( \sum _{j\in \mathbb {Z}} \int \int \chi _{F_{j,2}}\Big \vert Q_sW_{\Omega ,j}Q_t\Big (\sum _lA_lh_{s,l}\Big )\Big \vert ^2 \log ^{\sigma _1}(\textrm{e}+2^j/{s})\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\right\| _{L^{2}(\mathbb {R}^d)}\\ & \quad :=\textrm{I}_3\textrm{I}_4, \end{aligned}$$

where $1<\sigma _1<2\beta -1$ is a constant. Observe that

$$\begin{aligned} \sum _{j: 2^j\ge s}\log ^{-\sigma _1}(\textrm{e}+2^j/s)\lesssim 1. \end{aligned}$$

It then follows that

$$\begin{aligned} \textrm{I}_3\lesssim \Big \Vert \Big ( \int ^{\infty }_{0}\int _0^{\infty }\vert Q_sQ_t^3f\vert ^2 \frac{ds}{s} \frac{dt}{t}\Big )^{1/2}\Big \Vert _{L^2(\mathbb {R}^d)} \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

From (4.3) and (4.4) in Lemma 4.1, we know that

$$\begin{aligned} \textrm{I}_4\lesssim & \left( \sum _{j\in \mathbb {Z}} \int ^{2^j}_{0}\big \Vert W_{\Omega ,\,j}Q_s\Big (\sum _lA_lh_{s,l}\Big )(x)\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{\sigma _1}(\textrm{e}+2^j/s)\frac{ds}{s}\right) ^{1/2}\\\lesssim & \left( \sum _{j\in \mathbb {Z}} \int _0^{2^j}2^{-j}\Big \Vert \sum _lA_lh_{s,l}\Big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{-2\beta +\sigma _1}(\textrm{e}+2^j/s)\frac{ds}{s}\right) ^{1/2}\\\lesssim & \left( \int ^{\infty }_{0}\Vert Q_sg\Vert ^2_{L^{2}(\mathbb {R}^d)}\sum _{j:\, 2^j\ge s} \log ^{-2\beta +\sigma _1}(\textrm{e}+2^j/s)\frac{ds}{s}\right) ^{1/2}\\\lesssim & \Vert g\Vert _{L^{2}(\mathbb {R}^d)}, \end{aligned}$$

where in the third inequality, we have invoked the fact that the supports of functions $\{A_lh_{s,l}\}_l$ have bounded overlaps, and

$$\begin{aligned} \Big \vert \sum _lA_lh_{s,l}\Big \vert ^2\lesssim 2^j\sum _l\vert h_{s,l}\vert ^2=2^j\vert Q_sg\vert ^2. \end{aligned}$$

Therefore,

$$\begin{aligned} \textrm{D}_{12}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g \Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

We turn our attention to term $\textrm{D}_2$. Observe that $Q_sW_{\Omega ,j}=W_{\Omega , j}Q_s$. It then follows that

$$\begin{aligned} \textrm{D}_2\le & \Big \vert \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d} Q_sW_{\Omega ,j}([A_l,Q_s^2] Q_t^4f)(x)h_{s,l}(x)dx\frac{ds}{s}\frac{dt}{t}\Big \vert \\ & +\Big \vert \sum _j\sum _l\int \int \chi _{F_{j,2}}\int _{\mathbb {R}^d} Q_sW_{\Omega ,j}(A_lQ_s^2Q_t^4f)(x)h_{s,l}(x)dx\frac{ds}{s}\frac{dt}{t}\Big \vert \\:= & \textrm{D}_{21}+\textrm{D}_{22}.\end{aligned}$$

Similar to term $\textrm{D}_{11}$ and term $\textrm{D}_{12}$ respectively,, we have that

$$\begin{aligned} \textrm{D}_{21}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g \Vert _{L^{2}(\mathbb {R}^d)},\,\,\textrm{D}_{22}\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g \Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

To consider $\textrm{D}_{3,n}$, we write

$$\begin{aligned} \partial _nA_lQ_t^4f(x)=[\partial _n{A}, Q_t]Q_t^3f(x)+Q_t[\partial _n{A}, Q_t]Q_t^2f(x)+Q_t^2(\partial _n\widetilde{A}_lQ_t^2)f(x), \end{aligned}$$

and

$$\begin{aligned} \textrm{D}_{3,n}\le & \sum _j\int \int \chi _{F_{j,2}}\Big \vert \sum _l\int _{\mathbb {R}^d}Q_s^3h_{s,l}(x)U_{\Omega ,n,j}([\partial _n{A}, Q_t]Q_t^3f)(x)dx\Big \vert \frac{ds}{s}\frac{dt}{t}\\ & +\sum _j\int \int \chi _{F_{j,2}}\Big \vert \sum _l\int _{\mathbb {R}^d}Q_s^3h_{s,l}(x)U_{\Omega ,n,j}Q_t([\partial _n{A}, Q_t]Q_t^2f)(x)dx\Big \vert \frac{ds}{s}\frac{dt}{t}\\ & +\sum _j\int \int \chi _{F_{j,2}}\Big \vert \sum _l\int _{\mathbb {R}^d}Q_s^3h_{s,l}(x)U_{\Omega ,n,j}Q_t^2(\partial _n\widetilde{A}_lQ_t^2)f(x)dx\Big \vert \frac{ds}{s}\frac{dt}{t}\\:= & \sum _{i=1}^3\textrm{D}_{3,n}^i. \end{aligned}$$

Let $2<\sigma _2<2\beta -2$. It follows from Hölder’s inequality that

$$\begin{aligned} \textrm{D}_{3,n}^1= & \sum _j\int \int \chi _{F_{j,2}}\Big \vert \sum _l\int _{\mathbb {R}^d}Q_s^2U_{\Omega ,n,j}([\partial _n{A}, Q_t]Q_t^3f)(x)Q_sh_{s,l}(x)dx\Big \vert \frac{ds}{s}\frac{dt}{t} \\\lesssim & \left( \sum _j\int \int \chi _{F_{j,2}}\big \Vert Q_s^2U_{\Omega ,n,j}([\partial _n{A}, Q_t]Q_t^3f)\big \Vert ^2_{L^{2}(\mathbb {R}^d)}\log ^{\sigma _2}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\\ & \times \left( \sum _j \int \int \chi _{F_{j,2}} \Big \Vert Q_s\big (\sum _lh_{s,l}\big )\Big \Vert ^2_{L^{2}(\mathbb {R}^d)}\log ^{-\sigma _2}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\right) ^{1/2}\\:= & \textrm{J}_1\textrm{J}_2. \end{aligned}$$

We have that

$$\begin{aligned} \textrm{J}_2\le & \left( \int _0^\infty \Vert Q_s^2g\Vert ^2_{L^{2}(\mathbb {R}^d)}\int _s^\infty \sum _{j:2^j\ge s^\alpha t^{1-\alpha }} \log ^{-\sigma _2}(\textrm{e}+2^j/s)\frac{dt}{t}\frac{ds}{s}\right) ^{\frac{1}{2}}\\\lesssim & \Big \Vert \Big (\int _0^\infty \vert Q_s^2g\vert ^2\frac{ds}{s}\Big )^{\frac{1}{2}}\Big \Vert _{L^{2}(\mathbb {R}^d)}\lesssim \Vert g\Vert _{L^{2}(\mathbb {R}^d)}, \end{aligned}$$

where in the second inequality, we have invoked the fact that

$$\begin{aligned} \int _s^\infty \sum _{j:2^j\ge s^\alpha t^{1-\alpha }} \log ^{-\sigma _2}(\textrm{e}+2^j/s)\frac{dt}{t}\lesssim \int _s^\infty \log ^{-\sigma _2+1}(\textrm{e}+t/s)\frac{dt}{t}\lesssim 1. \end{aligned}$$

On the other hand, it follows from (4.5) in Lemma 4.1 that

$$\begin{aligned} \textrm{J}_1\le & \left( \sum _j \int \int \chi _{F_{j,2}}\log ^{-2\beta +\sigma _2}(\textrm{e}+2^j/s)\Vert [\partial _n{A}, Q_t]Q_t^3f\Vert ^2_{L^2(\mathbb {R}^d)}\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}} \\\lesssim & \left( \int _0^\infty \Vert [\partial _n{A}, Q_t]Q_t^3f\Vert ^2_{L^2(\mathbb {R}^d)}\frac{dt}{t}\sum _{j:2^j\le t}\int _0^{(2^jt^{\alpha -1})^{1/\alpha }}\log ^{-2\beta +\sigma _2}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s}\right) ^{\frac{1}{2}}\\\lesssim & \Vert f\Vert _{L^2(\mathbb {R}^d)}, \end{aligned}$$

since $\beta >2$ and

$$\begin{aligned} \sum _{j:2^j\le t}\int _0^{(2^jt^{\alpha -1})^{1/\alpha }}\log ^{-2\beta +\sigma _2}(\textrm{e}+2^j/s)\frac{ds}{s}\le \sum _{j:2^j\le t}\log ^{-2\beta +\sigma _2+1}\big (\textrm{e}+t/2^j\big )\lesssim 1. \end{aligned}$$

Therefore,

$$\begin{aligned} \textrm{D}_{3,n}^1\lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g \Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

Similarly, we have that

$$\begin{aligned} \textrm{D}_{3,n}^2\lesssim \Vert g\Vert _{L^{2}(\mathbb {R}^d)} \Vert f\Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

To estimate $\textrm{D}_{3,n}^3$, write

$$\begin{aligned} & \int _{\mathbb {R}^d}Q_s^3h_{s,l}(x)U_{\Omega ,n,j}Q_t^2(\partial _n\widetilde{A}_lQ_t^2)f(x)dx\\ & \quad =\int _{\mathbb {R}^d}Q_t^2h_{s,l}(x)U_{\Omega ,n,j}Q_s^3(\partial _n\widetilde{A}_lQ_t^2)f(x)dx\\ & \quad =-\int _{\mathbb {R}^d}Q_t^2h_{s,l}(x)U_{\Omega ,n,j}Q_s^2([\partial _n{A}, Q_s]Q_t^2f)(x)dx\\ & \qquad +\int _{\mathbb {R}^d}Q_t^2h_{s,l}(x)U_{\Omega ,n,j}Q_s^2(\partial _n\widetilde{A}_lQ_sQ_t^2f)(x)dx:=\textrm{I}_1^l(s,\,t)+\textrm{I}_2^l(s,\,t). \end{aligned}$$

For the integral corresponding to $\textrm{I}_l^1$, we choose $\sigma _3$ with $1<\sigma _3<2\beta -1$, and deduce from (4.5) in Lemma 4.1 that

$$\begin{aligned} & \sum _j\int \int \chi _{F_{j,2}}\Big \vert \sum _l \textrm{I}_1^l(s,t)\Big \vert \frac{ds}{s}\frac{dt}{t}\\ & \quad \lesssim \left( \sum _j\int \int \chi _{F_{j,2}}\Vert Q_t^2Q_sg\Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{-\sigma _3}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}} \\ & \qquad \times \left( \sum _j\int \int \chi _{F_{j,2}}\Vert U_{\Omega ,n,j}Q_s^2[\partial _n{A}, Q_s]Q_t^2f\Vert ^2_{L^{2}(\mathbb {R}^d)}\log ^{\sigma _3}\big (\textrm{e}+\frac{2^j}{s}\big )\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\\ & \quad \lesssim \Vert g\Vert _{L^2(\mathbb {R}^d)}\left( \int _0^\infty \int _0^\infty \big \Vert [\partial _n{A}, Q_s]Q_t^2f\big \Vert ^2_{L^2(\mathbb {R}^d)}\sum _{j:2^j\ge s}\log ^{-2\beta +\sigma _3}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}} \\ & \quad \lesssim \Vert g\Vert _{L^2(\mathbb {R}^d)}\left( \int _0^\infty \int _0^\infty \big \Vert [\partial _n{A}, Q_s]Q_t^2f\big \Vert ^2_{L^2(\mathbb {R}^d)}\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\lesssim \Vert f\Vert _{L^{2}(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

To estimate the integral corresponding to $\textrm{I}_2^l$, write

$$\begin{aligned} \textrm{I}_2^l= & \int _{\mathbb {R}^d}Q_sQ_t^2f(x)[\partial _nA, U_{\Omega ,n,j}Q_s]Q_t^2Q_sh_{s,l}(x)dx\\ & +\int _{\mathbb {R}^d}Q_sQ_t^2f(x)U_{\Omega ,n,j}Q_s[\partial _nA, Q_t^2]Q_sh_{s,l}(x)dx\\ & +\int _{\mathbb {R}^d}Q_sQ_t^2f(x)U_{\Omega ,n,j}Q_sQ_t^2[\partial _nA, Q_s]h_{s,l}(x)dx\\ & +\int _{\mathbb {R}^d}Q_sQ_t^2f(x)U_{\Omega ,n,j}Q_sQ_t^2Q_s(\partial _m\widetilde{A}_lh_{s,l})(x)dx\\:= & \textrm{V}_{n,l}^1(s,t)+\textrm{V}_{n,l}^2(s,t)+\textrm{V}_{n,l}^3(s,t)+\textrm{V}_{n,l}^4(s,t). \end{aligned}$$

The estimates for the part of $\textrm{V}_{n,l}^2$ and $\textrm{V}_{n,l}^3$ are similar to the estimate for the term corresponding to $\textrm{I}_{1}^l$, and are omitted. As for $\textrm{V}_{n,l}^4$, we choose $1<\sigma _4<2\beta -3$. It then follows from Lemma 4.1 that

$$\begin{aligned} & \left( \sum _j\int \int \chi _{F_{j,2}}\big \Vert U_{\Omega ,n,j}Q_s Q_t^2(\sum _lQ_s(\partial _n\widetilde{A}_lh_{s,l}))\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{\sigma _4}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\\ & \quad \lesssim \left( \sum _j\int \int \chi _{F_{j,2}}\big \Vert Q_t^2(\sum _lQ_s(\partial _n\widetilde{A}_lh_{s,l}))\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{-2\beta +\sigma _4}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s}\frac{dt}{t}\right) ^{\frac{1}{2}}\\ & \quad \lesssim \left( \sum _j\int _0^{2^j}\big \Vert \sum _lQ_s(\partial _n\widetilde{A}_lh_{s,l})\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{-2\beta +\sigma _4}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s} \right) ^{\frac{1}{2}}. \end{aligned}$$

Let $x\in 48dI_l$ and $q\in (1,\,2)$. A straightforward computation involving Hölder’s inequality and the John-Nirenberg inequality gives us that

$$\begin{aligned} \vert Q_s(\partial _n\widetilde{A}_{l}h)(x)\vert \lesssim M_qh(x)+\log (\textrm{e}+2^j/s)Mh(x), \end{aligned}$$

where $I(x,\,s)$ is the cube centered at x and having side length s. This implies that

$$\begin{aligned} & \sum _j\int _0^{2^j}\big \Vert \sum _lQ_s(\partial _n\widetilde{A}_lh_{s,l})\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{-2\beta +\sigma _4}(\textrm{e}+2^j/s)\frac{ds}{s} \\ & \quad \lesssim \int _0^\infty \Vert Q_sg\Vert ^2_{L^2(\mathbb {R}^d)}\sum _{j:2^j\ge s}\log ^{-2\beta +\sigma _4+2}(\textrm{e}+2^j/s)\frac{ds}{s} \lesssim \Vert g\Vert _{L^{2}(\mathbb {R}^d)}, \end{aligned}$$

since $-2\beta +\sigma _4+2<-1$. Therefore,

$$\begin{aligned} & \sum _j\int \int \chi _{F_{j,2}}\sum _l\vert \textrm{V}_{n,l}^4(s,t)\vert \frac{ds}{s}\frac{dt}{t}\\ & \quad \lesssim \Big (\sum _j\int \int \chi _{F_{j,2}}\Vert Q_sQ_t^2f\Vert ^2_{L^{2}(\mathbb {R}^d)}\log ^{-\sigma _4}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\Big )^{\frac{1}{2}} \\ & \qquad \times \Big (\sum _j\int \int \chi _{F_{j,2}}\big \Vert U_{\Omega ,n,j}Q_s Q_t^2(\sum _lQ_s(\partial _n\widetilde{A}_lh_{s,l}))\big \Vert ^2_{L^{2}(\mathbb {R}^d)} \log ^{\sigma _4}(\textrm{e}+\frac{2^j}{s})\frac{ds}{s}\frac{dt}{t}\Big )^{\frac{1}{2}}\\ & \quad \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Now, we consider the part corresponding to $\textrm{V}_{n,l}^1$. Invoking (4.6) in Lemma 4.1, we deduce that

$$\begin{aligned} & \Big \Vert \Big (\sum _j\int \int \chi _{F_{j,2}}\vert [\partial _nA, U_{\Omega ,n,j}Q_s] Q_t^2Q_s^2g\vert ^2\log ^{\sigma _4}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\Big )^{\frac{1}{2}}\Big \Vert _{L^{2}(\mathbb {R}^d)}\\ & \quad \lesssim \Big (\sum _j\int \int \chi _{F_{j,2}}\big \Vert Q_t^2Q_s^2g\big \Vert ^2_{L^2(\mathbb {R}^d)}\log ^{-2\beta +\sigma _4+2}(\textrm{e}+2^j/s)\frac{ds}{s} \frac{dt}{t}\Big )^{\frac{1}{2}}\\ & \quad \lesssim \Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Therefore,

$$\begin{aligned} & \sum _j\int \int \chi _{F_{j,2}}\sum _l\vert \textrm{V}_{n,l}^1(s,t)\vert \frac{ds}{s}\frac{dt}{t}\\ & \quad \lesssim \Big \Vert \Big (\sum _j\int \int \chi _{F_{j,2}}\vert Q_sQ_t^2f\vert ^2\log ^{-\sigma _4}((\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\Big )^{\frac{1}{2}}\Big \Vert _{L^{2}(\mathbb {R}^d)}\\ & \qquad \times \Big \Vert \Big (\sum _j\int \int \chi _{F_{j,2}}\vert [\partial _nA, U_{\Omega ,n,j}Q_s] Q_t^2Q_s^2g\vert ^2\log ^{\sigma _4}(\textrm{e}+2^j/s)\frac{ds}{s}\frac{dt}{t}\Big )^{\frac{1}{2}}\Big \Vert _{L^{2}(\mathbb {R}^d)}\\ & \quad \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Combining the estimates for $\textrm{I}_l^1$, and $\textrm{V}_{n,l}^i$ ($i=1,\,2,\,3,\,4$), yields

$$\begin{aligned} \textrm{D}_{3,n}^3\lesssim \Vert g\Vert _{L^{2}(\mathbb {R}^d)} \Vert f\Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

which, along with the estimates for $\textrm{D}_1$, $\textrm{D}_2$, $\textrm{D}_{3,n}^1$ and $\textrm{D}_{3,n}^2$ leads to (4.11) with $k=2$. This verifies the inequality (4.9).

Now we turn our attention to inequality (4.10). Let $P_s$ be the operator defined by

$$\begin{aligned} P_s=\int _{s}^{\infty }Q_t^4\frac{dt}{t}. \end{aligned}$$

It was proved in [16] that

$$\begin{aligned} \int _0^{\infty }\Vert P_sf\Vert _{L^2(\mathbb {R}^d)}^2\frac{ds}{s}\le \Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Let $\widetilde{T}_{\Omega ,\,A}$ be the adjoint of $T_{\Omega ,\,A}$, that is,

$$\begin{aligned} \widetilde{T}_{\Omega ,\,A}f(x)=\mathrm{p.\,v.}\int _{\mathbb {R}^d}\frac{\widetilde{\Omega }(y-x)}{\vert x-y\vert ^{d+1}}\big (A(x)-A(y)-\nabla A(x)(x-y)\big )f(y)dy, \end{aligned}$$

with $\widetilde{\Omega }(x)=\Omega (-x)$. Obviously,

$$\begin{aligned} \widetilde{T}_{\Omega ,\,A}h(x)=T_{\widetilde{\Omega },\,A}h(x)+\sum _{n=1}^d[\partial _nA,\, T_{\widetilde{\Omega }}^{n}]h(x), \end{aligned}$$

(4.13)

where $T_{\widetilde{\Omega }}^n$ is defined as in (3.23), but with $\Omega$ replaced by $\widetilde{\Omega }$. As the inequality (4.9), we have that

$$\begin{aligned} \Big \vert \int ^{\infty }_0\int _{0}^t\int _{\mathbb {R}^d}Q_s^4{T}_{\widetilde{\Omega },\,A}Q_t^4f(x)g(x)dx \frac{ds}{s}\frac{dt}{t}\Big \vert \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^{2}(\mathbb {R}^d)}; \end{aligned}$$

For each n with $1\le n\le d$, we know from [21, Theorem 2] that $[\partial _n A,\,T_{\Omega }^n]$ is bounded on $L^p(\mathbb {R}^d)$ provided that $1+1/(\beta - 1)< p < \beta$. A straightforward computation yields

$$\begin{aligned} & \Big \vert \int ^{\infty }_0\int _{0}^t\int _{\mathbb {R}^d}Q_s^4[\partial _nA,\,T_{\widetilde{\Omega }}^n]Q_t^4f(x)g(x)dx \frac{ds}{s}\frac{dt}{t}\Big \vert \\ & \quad =\Big \vert \int ^{\infty }_0\int _{s}^{\infty }\int _{\mathbb {R}^d}Q_t^4f(x)[\partial _nA,\,T_{\widetilde{\Omega }}^n]Q_s^4g(x)dx \frac{ds}{s}\frac{dt}{t}\Big \vert \\ & \quad \lesssim \Big (\int ^{\infty }_0\Vert [\partial _nA,\,T_{\widetilde{\Omega }}^n]Q_s^4g \Vert _{L^2(\mathbb {R}^d)}^2\frac{ds}{s}\Big )^{1/2}\Big (\int ^{\infty }_0 \Vert P_sf\Vert _{L^2(\mathbb {R}^d)}^2\frac{ds}{s}\Big )^{1/2}\\ & \quad \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

Therefore,

$$\begin{aligned} \Big \vert \int ^{\infty }_0\int _{0}^t\int _{\mathbb {R}^d}Q_s^4\widetilde{T}_{{\Omega },\,A}Q_t^4f(x)g(x)dx \frac{ds}{s}\frac{dt}{t}\Big \vert \lesssim \Vert f\Vert _{L^2(\mathbb {R}^d)}\Vert g\Vert _{L^{2}(\mathbb {R}^d)}. \end{aligned}$$

This, via dulaity argument, gives (4.10).

With the $L^2(\mathbb {R}^d)$ boundedness of $T_{\Omega ,\,A}$ in hand, we now verify the $L^p(\mathbb {R}^d)$ boundedness of $T_{\Omega ,\,A}$ for the case of $1+1/(\beta -1)< p < 2$. Let $R_{l,A}$ be the operator defined as in Theorem 3.9, and $\varepsilon \in (0,\,1)$ be a constant which will be chosen later. An application of Theorem 3.9 gives us that

$$\begin{aligned} \big \Vert R_{2^l,A}f-R_{2^{l+1},A}f\big \Vert _{L^2(\mathbb {R}^d)}\lesssim 2^{(-\varepsilon \beta +2)l}\Vert f\Vert _{L^2(\mathbb {R}^d)}. \end{aligned}$$

(4.14)

Therefore, the series

$$\begin{aligned} T_{\Omega ,A}=R_{2,\,A}+\sum _{l=1}^{\infty }(R_{2^{l+1},A}-R_{2^{l},\,A}) \end{aligned}$$

(4.15)

converges in $L^2(\mathbb {R}^d)$ operator norm and for $f,\,g\in C^{\infty }_0(\mathbb {R}^d)$,

$$\begin{aligned} \int _{\mathbb {R}^d} (T_{\Omega , A}-R_{2,A})f(x)g(x)dx=\sum _{l=1}^{\infty }\int _{\mathbb {R}^d} (R_{2^{l+1},A}-R_{2^{l}, A})f(x)g(x)dx. \end{aligned}$$

(4.16)

On the other hand, from Theorem 3.9 we know that $R_{l,A}$ is bounded on $L^2(\mathbb {R}^d)$ with bound independent of l. This, via Theorem 2.3, (ii) and (iii) of Theorem 3.9, shows that for $p\in (1,\,2]$, $R_{l,A}$ is bounded on $L^p(\mathbb {R}^d)$ with bound $Cl^2$. Thus, we have that

$$\begin{aligned} \big \Vert R_{2^l,A}f-R_{2^{l+1},A}f\big \Vert _{L^p(\mathbb {R}^d)}\lesssim 2^{2l}\Vert f\Vert _{L^p(\mathbb {R}^d)},\,\,\,p\in (1,\,2]. \end{aligned}$$

(4.17)

Let $1<p<2$ and $\varrho \in (0,\,1)$. Interpolation between the inequalities (4.14) and (4.17) leads to that

$$\begin{aligned} \big \Vert R_{2^l,A}f-R_{2^{l+1},A}f\big \Vert _{L^p(\mathbb {R}^d)}\lesssim 2^{(-2\varepsilon \beta /p'+2+\varrho )l}\Vert f\Vert _{L^p(\mathbb {R}^d)}. \end{aligned}$$

For each p with $1+1/\beta<p<2$, we choose $\varepsilon >0$ close to 1 sufficiently, and $\varrho >0$ close to 0 sufficiently, such that $2\varepsilon \beta /p'>2+\varrho$, and then obtain that

$$\begin{aligned} \sum _{l=1}^{\infty }\big \Vert R_{2^l,\,A}f-R_{2^{l+1},\,A}f\big \Vert _{L^p(\mathbb {R}^d)}\lesssim \Vert f\Vert _{L^p(\mathbb {R}^d)}. \end{aligned}$$

This, along with (4.16), shows that $T_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$.

It remains to consider the $L^p(\mathbb {R}^d)$ boundedness of $T_{\Omega ,A}$ for the case of $2<p<\beta$. Observe that the operator $T_{\widetilde{\Omega },\,A}$ is also bounded on $L^p(\mathbb {R}^d)$ for $1+1/\beta<p<2$. Thus by (4.13), we know that $\widetilde{T}_{\Omega ,\,A}$, the adjoint operator of $T_{\Omega ,\,A}$, is also bounded on $L^p(\mathbb {R}^d)$ for $1+1/\beta<p<2$, and so ${T}_{\Omega ,\,A}$ is bounded on $L^p(\mathbb {R}^d)$ for $2<p<\beta$. This finishes the proof of Theorem 1.3. $\square$