Vector-valued inequalities for the commutators of rough singular kernels

Abstract

Vector-valued inequalities are considered for the commutator of the singular integral with rough kernel. The results obtained in this paper are substantial improvement and extension of some known results.

MSC: 42B20, 42B25.

1 Introduction

The homogeneous singular integral operator $T_{\mathrm{\Omega }}$ is defined by

$$T_{\mathrm{\Omega }}f(x)=\text{p.v. }{\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Omega }(x-y)}{|x-y{|}^n}f(y)dy,$$

when $\mathrm{\Omega }\in L^1(S^{n-1})$ satisfies the following conditions:

• (a) Ω is a homogeneous function of degree zero on ${\mathbb{R}}^n\setminus \{0\}$, i.e.,

  $$\mathrm{\Omega }(tx)=\mathrm{\Omega }(x)\text{for any }t>0\text{ and }x\in {\mathbb{R}}^n\setminus \{0\}.$$

  (1.1)

• (b) Ω has mean zero on $S^{n-1}$, the unit sphere in ${\mathbb{R}}^n$, i.e.,

  $${\int }_{S^{n-1}}\mathrm{\Omega }\left(x^{\prime }\right)d\sigma \left(x^{\prime }\right)=0.$$

  (1.2)

Using a rotation method, Calderón and Zygmund [1] proved that $T_{\mathrm{\Omega }}$ is bounded in $L^p({\mathbb{R}}^n)$ for $1<p<\mathrm{\infty }$ if Ω is odd or $\mathrm{\Omega }\in L{log}^{+}L(S^{n-1})$. In [2], Grafakos and Stefanov gave a nice survey, which contains a thorough discussion of the history of the operator $T_{\mathrm{\Omega }}$.

For a function $b\in L_{\mathrm{loc}}({\mathbb{R}}^n)$, let A be a linear operator on some measurable function space. Then the commutator between A and b is defined by $[b,A]f(x):=b(x)Af(x)-A(bf)(x)$.

In 1976, Coifman et al. [3] obtained a characterization of $L^p$-boundedness of the commutators $[b,R_j]$ generated by the Reisz transforms $R_j$ ($j=1,\dots ,n$) and a BMO function b. As an application of this characterization, a decomposition theorem of the real Hardy space is given in this paper. Moreover, the authors in [3] proved also that if $\mathrm{\Omega }\in Lip(S^{n-1})$, then the commutator $[b,T_{\mathrm{\Omega }}]$ for $T_{\mathrm{\Omega }}$ and a BMO function b is bounded on $L^p$ for $1<p<\mathrm{\infty }$ which is defined by

$$[b,T_{\mathrm{\Omega }}]f(x)=\text{p.v. }{\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Omega }(x-y)}{|x-y{|}^n}(b(x)-b(y))f(y)dy.$$

In the same paper, Coifman et al. [3] outlined a different approach, which is less direct but shows a close relationship between the weighted inequalities of the operator $T_{\mathrm{\Omega }}$ and the weighted inequalities of the commutator $[b,T_{\mathrm{\Omega }}]$. In 1993, Alvarez et al. [4] developed the idea of [3] and established a generalized boundedness criterion for the commutator of linear operators. The result of Alvarez et al. (see [4], Theorem 2.13) can be stated as follows.

Theorem A ([4])

Let $1<p<\mathrm{\infty }$. If a linear operator T is bounded on $L^p(w)$ for all $w\in A_q$, ($1<q<\mathrm{\infty }$), where $A_q$ denotes the weight class of Muckenhoupt, then for $b\in BMO$, ${\parallel [b,T]f\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel f\parallel }_{L^p}$.

Combining Theorem A with the well-known results by Duoandikoetxea [5] on the weighted $L^p$ boundedness of the rough singular integral $T_{\mathrm{\Omega }}$, we know that if $\mathrm{\Omega }\in L^q(S^{n-1})$ for some $q>1$, then $[b,T_{\mathrm{\Omega }}]$ is bounded on $L^p$ for $1<p<\mathrm{\infty }$. However, it is not clear up to now whether the operator $T_{\mathrm{\Omega }}$ with $\mathrm{\Omega }\in L^1\setminus {\bigcup }_{q>1}{L}^q(S^{n-1})$ is bounded on $L^p(w)$ for $1<p<\mathrm{\infty }$ and all $w\in A_r$ ($1<r<\mathrm{\infty }$). Hence, if $\mathrm{\Omega }\in L^1\setminus {\bigcup }_{q>1}{L}^q(S^{n-1})$, the $L^p$ boundedness of $[b,T_{\mathrm{\Omega }}]$ cannot be deduced from Theorem A. In this case, Hu [6] used the refined Fourier estimate, the Littlewood-Paley decomposition, and the properties of Young functions and got the following result.

Theorem B ([6])

Suppose that $\mathrm{\Omega }\in L{({log}^{+}L)}^2(S^{n-1})$ satisfying (1.1) and (1.2). Then, for $b\in BMO({\mathbb{R}}^n)$ and $1<p<\mathrm{\infty }$, the commutator $[b,T_{\mathrm{\Omega }}]$ is bounded on $L^p({\mathbb{R}}^n)$ with bound $C{\parallel b\parallel }_{BMO}$.

Recently, Chen and Ding [7] gave a sufficient condition which contains ${\bigcup }_{q>1}{L}^q(S^{n-1})$ such that the commutator of convolution operators is bounded on $L^p({\mathbb{R}}^n)$ for $1<p<\mathrm{\infty }$. This condition was introduced by Grafakos and Stefanov in [8], and it is defined by

$$\underset{\xi \in S^{n-1}}{sup}{\int }_{S^{n-1}}|\mathrm{\Omega }(y)|{(ln\frac{1}{|\xi \cdot y|})}^{1+\alpha }d\sigma (y)<\mathrm{\infty },$$

(1.3)

where $\alpha >0$ is a fixed constant. Let $F_{\alpha }(S^{n-1})$ denote the space of all integrable functions Ω on $S^{n-1}$ satisfying (1.3). The result in [7] can be stated as follows.

Theorem C Let Ω be a function in $L^1(S^{n-1})$ satisfying (1.1) and (1.2). If $\mathrm{\Omega }\in F_{\alpha }(S^{n-1})$ for some $\alpha >1$, then $[b,T_{\mathrm{\Omega }}]$ extends to a bounded operator from $L^p$ into itself for $\frac{\alpha +1}{\alpha }<p<\alpha +1$.

The condition (1.3) above has been considered by many authors in the context of rough integral operators. One can consult [9–15] among numerous references for its development and applications. The examples in [8] show that there is the following relationship between $F_{\alpha }(S^{n-1})$ and $H^1(S^{n-1})$ (the Hardy space on $S^{n-1}$):

$$\bigcup _{q>1}{L}^q\left(S^{n-1}\right)\subset \bigcap _{\alpha >0}{F}_{\alpha }\left(S^{n-1}\right)⊈H^1\left(S^{n-1}\right)⊈\bigcup _{\alpha >0}{F}_{\alpha }\left(S^{n-1}\right).$$

On the other hand, for all $\tau \ge 0$, $L{({log}^{+}L)}^{1+\tau }(S^{n-1})\subset H^1(S^{n-1})$. So, for all $\tau \ge 0$, ${\bigcap }_{\alpha >0}{F}_{\alpha }(S^{n-1})⊈L{({log}^{+}L)}^{1+\tau }(S^{n-1})$.

The study of vector-valued inequalities for singular integrals with rough kernels has attracted much attention (for example, see [16]). In 2011, Tang and Wu [17] considered the vector-valued inequalities ($L^p({\ell }^q)$, $L^p({\ell }^q)$), ($1<p,q<\mathrm{\infty }$) of the commutator $[b,T_{\mathrm{\Omega }}]$ with the kernel $\mathrm{\Omega }\in L{({log}^{+}L)}^2(S^{n-1})$ satisfying (1.1) and (1.2). In this paper, we consider the vector-valued inequalities for a class of commutators of singular integrals with $\mathrm{\Omega }\in F_{\alpha }(S^{n-1})$ for some $\alpha >0$. Now we state our result as follows.

Theorem 1.1 Let Ω be a function in $L^1(S^{n-1})$ satisfying (1.1) and (1.2) if $\mathrm{\Omega }\in F_{\alpha }(S^{n-1})$ for some $\alpha >1$. Suppose that $1<p,q<\mathrm{\infty }$ satisfy

• (a) $2\le p,q<\mathrm{\infty }$ and $p\cdot q<2(\alpha +1)$; or

• (b) $2<p<\mathrm{\infty }$, $1<q<2$ and $p\cdot q^{\prime }<2(\alpha +1)$; or

• (c) $1<p,q<2$ and $p^{\prime }\cdot q^{\prime }<2(\alpha +1)$; or

• (d) $1<p<2$, $2<q<\mathrm{\infty }$ and $p^{\prime }\cdot q<2(\alpha +1)$.

Then $[b,T_{\mathrm{\Omega }}]$ extends to a bounded operator from $L^p({\ell }^q)$ into itself.

Corollary 1.2 Let Ω be a function in $L^1(S^{n-1})$ satisfying (1.1) and (1.2). If $\mathrm{\Omega }\in {\bigcap }_{\alpha >1}{F}_{\alpha }(S^{n-1})$, then $[b,T_{\mathrm{\Omega }}]$ extends to a bounded operator from $L^p({\ell }^q)$ into itself for $1<p,q<\mathrm{\infty }$.

This paper is organized as follows. First, in Section 2, we give some definitions, which will be used in the proofs of the main results. In Section 3, we give some preliminary lemmas for the proof of Theorem 1.1. Then, in Section 4, we give the proof of Theorem 1.1. Throughout this paper, the letter C stands for a positive constant which is independent of the essential variables and not necessarily the same one in each occurrence. Moreover, the notations `∨’ and `∧’ denote the Fourier transform and the inverse Fourier transform, respectively. As usual, for $p\ge 1$, $p^{\prime }=p/(p-1)$ denotes the dual exponent of p.

We collect the notation to be used throughout this paper:

$${\parallel \{f_j\}\parallel }_{L^p({\ell }^q)}={\parallel {(\sum _{j\in \mathbb{Z}}|f_j{|}^q)}^{1/q}\parallel }_{L^p};{\parallel f\parallel }_{L^p}={\left({\int }_{{\mathbb{R}}^n}\right|f(x){|}^{p}dx)}^{1/p}.$$

2 Definitions

Firstly, we need to recall some definitions which will be used in the proof of Theorem 1.1.

Let $\phi \in \mathcal{S}({\mathbb{R}}^n)$ be a radial function which is supported in the unit ball and satisfies $\phi (\xi )=1$ for $|\xi |\le \frac{1}{2}$. The function $\psi (\xi )=\phi (\frac{1}{2})-\phi (\xi )$ is supported in $\{\frac{1}{2}\le |\xi |\le 2\}$ and satisfies the identity

$$\sum _{j\in \mathbb{Z}}\psi \left(2^{-j}\xi \right)=1\text{for }\xi \ne 0.$$

We denote by ${\mathrm{\Delta }}_j$ and $G_j$ the convolution operators whose symbols are $\psi (2^{-j}\xi )$ and $\phi (2^{-j}\xi )$, respectively.

The paraproduct of Bony [18] between two functions f, g is defined by

$${\pi }_f(g)=\sum _{j\in \mathbb{Z}}({\mathrm{\Delta }}_{j}f)(G_j-3g).$$

At least formally, we have the following Bony decomposition:

$$fg={\pi }_f(g)+{\pi }_g(f)+R(f,g)\text{with }R(f,g)=\sum _{i\in \mathbb{Z}}\sum _{|k-i|\le 2}({\mathrm{\Delta }}_{i}f)({\mathrm{\Delta }}_{k}g).$$

(2.1)

3 Key lemmas

Let us begin with some lemmas, which will be used in the proof of Theorem 1.1. The first one can be found in [17].

Lemma 3.1 If $\varphi \in \mathcal{S}({\mathbb{R}}^n)$ with $supp(\varphi )\subset x:1/2\le |x|\le 2$ and for $l\in \mathbb{Z}$, define the multiplier operator $S_l$ by $S_{l}f(\xi )=\varphi (2^{-l}\xi )f(\xi )$ and $S_l^2$ by $S_l^{2}f=S_l(S_l)$. Then, for $b\in BMO({\mathbb{R}}^n)$, for any positive integer k and $b\in BMO({\mathbb{R}}^n)$, denote by $S_{l;b;k}$ (respectively $S_{l;b;k}^2$) the kth-order commutator of $S_l$ (respectively $S_l^2$ ). Then, for $1<p,q<\mathrm{\infty }$, we have

$$\begin{array}{rl}\text{(i)}& {\parallel {\left(\sum _{j\in \mathbb{Z}}{(\sum _{l\in \mathbb{Z}}|S_{l;b;k}{f}_j{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{j\in \mathbb{Z}}|f_j{|}^q)}^{1/q}\parallel }_{L^p},\\ \text{(ii)}& {\parallel {\left(\sum _{j\in \mathbb{Z}}{\left(\right|\sum _{l\in \mathbb{Z}}{S}_{l;b;k}{f}_{j,l}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {\left(\sum _{j\in \mathbb{Z}}{(\sum _{l\in \mathbb{Z}}|f_{j,l}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p},\\ \text{(iii)}& {\parallel {\left(\sum _{j\in \mathbb{Z}}{\left(\sum _{l\in \mathbb{Z}}\right|S_{l;b;k}^2{f}_j{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{j\in \mathbb{Z}}|f_j{|}^q)}^{1/q}\parallel }_{L^p},\\ \text{(iv)}& {\parallel {\left(\sum _{j\in \mathbb{Z}}{\left(\right|\sum _{l\in \mathbb{Z}}{S}_{l;b;k}^2{f}_{j,l}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {\left(\sum _{j\in \mathbb{Z}}{(\sum _{l\in \mathbb{Z}}|f_{j,l}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p},\end{array}$$

where C is independent of j and l.

Lemma 3.2 ([19])

Let $1<p,q<\mathrm{\infty }$, ${\{{({\sum }_k|g_{k;j}{|}^2)}^{1/2}\}}_j\in L^p({\ell }^q)$, and $\mathrm{\Omega }\in L^1(S^{n-1})$. Denote ${\sigma }_k(x)=|x{|}^{-n}|\mathrm{\Omega }(x^{\prime })|{\chi }_{\{2^k<|x|\le 2^{k+1}\}}(x)$. Then

$${\parallel {\left(\sum _{j\in \mathbb{Z}}{(\sum _{k\in \mathbb{Z}}|{\sigma }_k\ast g_{k,j}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\le C{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{j\in \mathbb{Z}}{(\sum _{k\in \mathbb{Z}}|g_{k,j}{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p},$$

where C is independent of $\{g_{k,j}\}$.

Lemma 3.3 ([7])

For the multiplier $G_k$ ($k\in \mathbb{Z}$) defined in Section  2 and $b\in BMO({\mathbb{R}}^n)$,

$$|G_{k}b(x)-G_{k}b(y)|\le C\frac{|x-y{|}^{\delta }{2}^{k\delta }}{\delta }{\parallel b\parallel }_{BMO}\mathit{\text{for }}0<\delta <1,$$

where C is independent of k and δ.

Lemma 3.4 ([20])

For any $u\in {\mathcal{S}}^{{}^{\prime }}({\mathbb{R}}^n)$ and $v\in {\mathcal{S}}^{{}^{\prime }}({\mathbb{R}}^n)$, the following properties hold:

• (i) ${\mathrm{\Delta }}_j{\mathrm{\Delta }}_{i}u\equiv 0$ if $|j-i|\ge 2$,

• (ii) ${\mathrm{\Delta }}_j(G_{i-3}{\mathrm{\Delta }}_{i}u)\equiv 0$ if $|j-i|\ge 4$.

If we replace ${\mathrm{\Delta }}_j$ with $S_j$, the above inequalities also hold.

4 Proof of Theorem 1.1

Recall that

$$[b,T_{\mathrm{\Omega }}]f(x)={\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Omega }(x-y)}{|x-y{|}^n}(b(x)-b(y))f(y)dy.$$

Let $\varphi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^n)$ be a radial function such that $0\le \varphi \le 1$, $supp\varphi \subset \{1/2\le |\xi |\le 2\}$ and ${\sum }_{l\in \mathbb{Z}}{\varphi }^3(2^{-l}\xi )=1$ for $|\xi |\ne 0$. Define the multiplier $S_l$ by $\widehat{S_{l}f}(\xi )=\varphi (2^{-l}\xi )\hat{f}(\xi )$. Set

$${\sigma }_j(x)=\frac{\mathrm{\Omega }(x^{\prime })}{|x{|}^n}{\chi }_{\{2^j\le |x|<2^{j+1}\}}(x)$$

for $j\in \mathbb{Z}$. Set

$$m_j(\xi )=\widehat{{\sigma }_j}(\xi ),m_j^l(\xi )=m_j(\xi )\varphi \left(2^{j-l}\xi \right).$$

Define the operator $T_j$ and $T_j^l$ by

$$\widehat{T_{j}f}(\xi )=m_j(\xi )\hat{f}(\xi ),\widehat{T_j^{l}f}(\xi )=m_j^l(\xi )\hat{f}(\xi ).$$

Denote by $[b,T_j]$ and $[b,T_j^l]$ the commutator of $T_j$ and $T_j^l$, respectively. Define the operator $V_l$ by

$$V_{l}h(x)=\sum _{j\in \mathbb{Z}}[b,S_{l-j}{T}_j^l{S}_{l-j}^2]h(x).$$

Then we know

$$[b,T_{\mathrm{\Omega }}]h(x)=\sum _{l\in \mathbb{Z}}{V}_{l}h(x).$$

Then by the Minkowski inequality, we have, for $1<p,q<\mathrm{\infty }$,

$${\parallel {\left(\sum _{s\in \mathbb{Z}}\right|[b,T_{\mathrm{\Omega }}]f_s{|}^q)}^{1/q}\parallel }_{L^p}\le \sum _{l\in \mathbb{Z}}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

So, to prove Theorem 1.1, it suffices to prove that

$$\sum _{l\in \mathbb{Z}}{(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}{\parallel }_{L^p}\le C{(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}{\parallel }_{L^p}.$$

(4.1)

It is well known that for some constant $0<\beta <1$ and any fixed constant $0<\upsilon <1$ (see [7] and [15]),

$${\parallel V_{l}f\parallel }_{L^2}\le C{\parallel b\parallel }_{BMO}{2}^{\beta l}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel f\parallel }_{L^2},l\le 1,$$

and

$${\parallel V_{l}f\parallel }_{L^2}\le C{\parallel b\parallel }_{BMO}{log}^{(-\alpha -1)v+1}(2+2^l){\parallel f\parallel }_{L^2},l\ge 2$$

which gives that

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^2}\le C{\parallel b\parallel }_{BMO}{2}^{\beta l}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^2},l\le 1,$$

(4.2)

and

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^2}\le C{\parallel b\parallel }_{BMO}{log}^{(-\alpha -1)\beta +1}(2+2^l){\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^2},l\ge 2.$$

(4.3)

If we can prove that, for any $1<p,q<\mathrm{\infty }$, $0<\delta <1$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le Cmax\{2,\frac{2^{\delta l}}{\delta }\}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p},$$

(4.4)

where C is independent of l and δ, we may finish the proof of Theorem 1.1. The proof of (4.4) will be postponed. Now, we will use (4.2), (4.3), and (4.4) to prove Theorem 1.1. Since

$$\begin{array}{rl}\sum _{l\in \mathbb{Z}}{(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}{\parallel }_{L^p}\le & \sum _{l\le 1}{(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}{\parallel }_{L^p}\\ +\sum _{l\ge 2}{(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}{\parallel }_{L^p}\\ :=& I_1+I_2,\end{array}$$

we will estimate $I_1$ and $I_2$, respectively. We first estimate $I_1$. For $l\le 1$, taking $q=2$ in (4.4), then interpolating between (4.2) and (4.4), there exists a constant $0<{\theta }_1<1$ such that for $1<p<\mathrm{\infty }$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}\le C{2}^{{\theta }_1\beta l}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}.$$

(4.5)

For $l\le 1$ and any fixed $1<p<\mathrm{\infty }$, interpolating between (4.4) and (4.5), there exists a constant $0<{\theta }_2<1$ such that for $1<q<\mathrm{\infty }$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{2}^{{\theta }_1{\theta }_2\beta l}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Therefore we get, for $1<p,q<\mathrm{\infty }$,

$$\begin{array}{rl}\sum _{l\le 1}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}& \le \sum _{l\le 1}{2}^{{\theta }_1{\theta }_2\beta l}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.\end{array}$$

Next, we will estimate $I_2$ for (a), (b), (c), and (d), respectively. For $2\le l<\mathrm{\infty }$, taking $\delta =1/l$ in (4.4), we get, for any $1<p,q<\mathrm{\infty }$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le Cl{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

(4.6)

Taking $q=2$ in (4.6) gives that for any $1<r<\mathrm{\infty }$, we have

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^r}\le Cl{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^r}.$$

(4.7)

We first treat the case (a) : $2\le p,q<\mathrm{\infty }$ and $p\cdot q<2(\alpha +1)$. Now, for any $p\ge 2$, we take r sufficiently large such that $r>p$ in (4.7). Using the Riesz-Thorin interpolation theorem between (4.3) and (4.7), we have that for any $l\ge 2$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-\theta }{log}^{((-\alpha -1)v+1)\theta }(2+2^l){\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p},$$

where $\theta =\frac{2(r-p)}{p(r-2)}$. We can see that if $r\mapsto \mathrm{\infty }$, then θ goes to $2/p$ and ${log}^{((-\alpha -1)v+1)\theta }(2+2^l)$ goes to ${log}^{((-\alpha -1)v+1)2/p}(2+2^l)$. Therefore, we get

$$\begin{array}{rl}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}& \le C{\parallel b\parallel }_{BMO}{l}^{1-2/p}{log}^{((-\alpha -1)v+1)\frac{2}{p}}(2+2^l){\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}.\end{array}$$

(4.8)

On the other hand, fix p, for any $2\le q<\mathrm{\infty }$, (4.6) also means that for any λ sufficiently large such that $\lambda >q$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^{\lambda })}^{1/\lambda }\parallel }_{L^p}\le Cl{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^{\lambda })}^{1/\lambda }\parallel }_{L^p}.$$

(4.9)

Using the Riesz-Thorin interpolation theorem between (4.8) and (4.9), we have that

$$\begin{array}{rl}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}& \le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}{\theta }_1}{l}^{1-{\theta }_1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}{\theta }_1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p},\end{array}$$

where ${\theta }_1=\frac{2(\lambda -q)}{q(\lambda -2)}$. We can see that if $\lambda \mapsto \mathrm{\infty }$, then ${\theta }_1$ goes to $2/q$ and $l^{1-(\alpha +1)v\frac{2}{p}{\theta }_1}$ goes to $l^{1-(\alpha +1)v\frac{2}{p}\frac{2}{q}}$. This gives that for any fixed $0<\upsilon <1$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}\frac{2}{q}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Thus, by the inequality above, we have, for $p\cdot q<2(\alpha +1)$,

$$\begin{array}{rl}\sum _{l\ge 2}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}& \le C\left(\sum _{l\ge 2}{l}^{1-(\alpha +1)v\frac{2}{p}\frac{2}{q}}\right){\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.\end{array}$$

Next, for the case (b) : $2\le p<\mathrm{\infty }$, $1<q<2$, and $p\cdot q^{\prime }<2(\alpha +1)$. For any $p\ge 2$, we have

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}.$$

(4.10)

Similarly, fix p, for $1<q<2$, (4.6) also means that for any λ sufficiently small such that $1<\lambda <q$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^{\lambda })}^{1/\lambda }\parallel }_{L^p}\le Cl{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^{\lambda })}^{1/\lambda }\parallel }_{L^p}.$$

(4.11)

Using the Riesz-Thorin interpolation theorem between (4.10) and (4.11), we have

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}{\theta }_1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p},$$

where ${\theta }_1=\frac{2(\lambda -q)}{q(\lambda -2)}$. We can see that if $\lambda \mapsto 1$, then ${\theta }_1$ goes to $2/q^{\prime }$ and $l^{1-(\alpha +1)v\frac{2}{p}{\theta }_1}$ goes to $l^{1-(\alpha +1)v\frac{2}{p}\frac{2}{q^{\prime }}}$. This gives that for any fixed $0<\upsilon <1$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p}\frac{2}{q^{\prime }}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Thus, for $2\le p<\mathrm{\infty }$, $1<q<2$, and $p\cdot q^{\prime }<2(\alpha +1)$, we have

$$\sum _{l\ge 2}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Now, for the case (c) : $1<p,q<2$ and $p\cdot q^{\prime }<2(\alpha +1)$. For any $1<p<2$, we take r sufficiently small such that $1<r<p$ in (4.7). Using the Riesz-Thorin interpolation theorem between (4.3) and (4.7), we have that for any $l\ge 2$,

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-\theta }{log}^{((-\alpha -1)v+1)\theta }(2+2^l){\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p},$$

(4.12)

where $\theta =\frac{2(r-p)}{p(r-2)}$. We can see that if $r\mapsto 1$, then θ goes to $2/p^{\prime }$ and ${log}^{((-\alpha -1)v+1)\theta }(2+2^l)$ goes to ${log}^{((-\alpha -1)v+1)2/p^{\prime }}(2+2^l)$. Therefore, we get

$$\begin{array}{rl}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^2)}^{1/2}\parallel }_{L^p}& \le C{\parallel b\parallel }_{BMO}{l}^{1-2/p}{log}^{((-\alpha -1)v+1)\frac{2}{p^{\prime }}}(2+2^l){\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p^{\prime }}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^2)}^{1/2}\parallel }_{L^p}.\end{array}$$

(4.13)

Then, using the previous argument, for any fixed $1<p<2$ and $1<q<2$, we get

$${\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{l}^{1-(\alpha +1)v\frac{2}{p^{\prime }}\frac{2}{q^{\prime }}}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

(4.14)

Thus if $p^{\prime }\cdot q^{\prime }<2(\alpha +1)$, then

$$\sum _{l\ge 2}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Finally, for the case $(d):1<p<2$, $2\le q<\mathrm{\infty }$, and $p^{\prime }\cdot q^{\prime }<2(\alpha +1)$, using the previous argument, we get

$$\sum _{l\ge 2}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Therefore, we prove that

$$I_2\le C{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}$$

for four cases.

Now, we turn our attention to proving (4.4). Since $T_j^l{S}_{l-j}=T_j{S}_{l-j}^2$ for any $j,l\in \mathbb{Z}$, we may write

$$[b,S_{l-j}^2{T}_j^l{S}_{l-j}]f=[b,S_{l-j}^2]\left(T_j{S}_{l-j}^{2}f\right)+S_{l-j}^2[b,T_j]\left(S_{l-j}^{2}f\right)+S_{l-j}^2{T}_j\left([b,S_{l-j}^2]f\right).$$

Thus,

$$\begin{array}{r}{\parallel {(\sum _{s\in \mathbb{Z}}|V_l{f}_s{|}^q)}^{1/q}\parallel }_{L^p}\\ \le {\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}[b,S_{l-j}^2]\left(T_j{S}_{l-j}^2{f}_s\right){|}^q)}^{1/q}\parallel }_{L^p}+{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2{T}_j\left([b,S_{l-j}^2]f_s\right){|}^q)}^{1/q}\parallel }_{L^p}\\ +{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2[b,T_j]\left(S_{l-j}^2{f}_s\right){|}^q)}^{1/q}\parallel }_{L^p}\\ :=L_1+L_2+L_3.\end{array}$$

(4.15)

Below we shall estimate $L_i$ for $i=1,2,3$, respectively. As regards $L_1$, by Lemma 3.1 and Lemma 3.2, we have, for $1<p<\mathrm{\infty }$,

$$\begin{array}{rl}{L}_1& \le C{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|T_j{S}_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|S_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.\end{array}$$

Similarly, we get

$$L_2\le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.$$

Hence, by (4.15), to show (4.4) it remains to give the estimate of $L_3$. We will apply Bony paraproduct to do this. By (2.1),

$$fg={\pi }_f(g)+{\pi }_g(f)+R(f,g).$$

We have

$$\begin{array}{r}b(x)\left(T_j{S}_{l-j}^2{f}_s\right)(x)\\ ={\pi }_{(T_j{S}_{l-j}^2{f}_s)}(b)(x)+R(b,T_j{S}_{l-j}^2{f}_s)(x)+{\pi }_b\left(T_j{S}_{l-j}^2{f}_s\right)(x)\end{array}$$

and

$$b{S}_{l-j}^2{f}_s(x)={\pi }_{(S_{l-j}^2{f}_s)}(b)(x)+R(b,S_{l-j}^2{f}_s)(x)+{\pi }_b\left(S_{l-j}^2{f}_s\right)(x).$$

Then we get

$$\begin{array}{r}[b,T_j]S_{l-j}^2{f}_s(x)\\ =b(x)\left(T_j{S}_{l-j}^2{f}_s\right)(x)-T_j\left(b{S}_{l-j}^2{f}_s\right)(x)\\ =[{\pi }_{(T_j{S}_{l-j}^2{f}_s)}(b)(x)-T_j({\pi }_{(S_{l-j}^2{f}_s)}(b))(x)]+[R(b,T_j{S}_{l-j}^2{f}_s)(x)-T_j\left(R(b,S_{l-j}^2{f}_s)\right)(x)]\\ +[{\pi }_b\left(T_j{S}_{l-j}^2{f}_s\right)(x)-T_j\left({\pi }_b\left(S_{l-j}^2{f}_s\right)\right)(x)].\end{array}$$

Thus

$$\begin{array}{rl}{L}_3\le & {\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2[{\pi }_{(T_j{S}_{l-j}^2{f}_s)}(b)-T_j({\pi }_{(S_{l-j}^2{f}_s)}(b))]{|}^q)}^{1/q}\parallel }_{L^p}\\ +{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2[R(b,T_j{S}_{l-j}^2{f}_s)-T_j\left(R(b,S_{l-j}^2{f}_s)\right)]{|}^q)}^{1/q}\parallel }_{L^p}\\ +{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2[{\pi }_b\left(T_j{S}_{l-j}^2{f}_s\right)-T_j\left({\pi }_b\left(S_{l-j}^2{f}_s\right)\right)]{|}^q)}^{1/q}\parallel }_{L^p}\\ :=& M_1+M_2+M_3.\end{array}$$

(4.16)

• (a) The estimate of $M_1$. Recall that ${\pi }_g(f)={\sum }_{j\in \mathbb{Z}}({\mathrm{\Delta }}_{j}f)(G_{j-3}g)$. For $M_1$, by Lemma 3.4(i), we know ${\mathrm{\Delta }}_i{S}_{k}g=0$ for $g\in {\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ when $|i-k|\ge 3$. Then

  $$\begin{array}{r}{\pi }_{(T_j{S}_{l-j}^2{f}_s)}(b)(x)-T_j({\pi }_{(S_{l-j}^2{f}_s)}(b))(x)\\ =\sum _{|i-(l-j)|\le 2}\{{\mathrm{\Delta }}_i\left(T_j{S}_{l-j}^2{f}_s\right)(x)(G_{i-3}b)(x)-T_j[\left({\mathrm{\Delta }}_i{S}_{l-j}^2{f}_s\right)(G_{i-3}b)](x)\}\\ =\sum _{|i-(l-j)|\le 2}[G_{i-3}b,T_j]\left({\mathrm{\Delta }}_i{S}_{l-j}^2{f}_s\right)(x).\end{array}$$

  (4.17)

Then we get

$$M_1\le \sum _{|k|\le 2}{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2([G_{l-j+k-3}b,T_j]\left({\mathrm{\Delta }}_{l-j+k}{S}_{l-j}^2{f}_s\right)){|}^q)}^{1/q}\parallel }_{L^p}.$$

(4.18)

Without loss of generality, we may assume $k=0$. By Lemma 3.1, we get

$$M_1\le C{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|[G_{l-j-3}b,T_j]\left({\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s\right){|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}.$$

(4.19)

Note that

$$\begin{array}{r}|[G_{l-j-3}b,T_j]\left({\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s\right)(x)|\\ =|{\int }_{2^j\le |x-y|<2^{j+1}}\frac{\mathrm{\Omega }(x-y)}{{|x-y|}^n}(G_{l-j-3}b(x)-G_{l-j-3}b(y)){\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s(y)dy|\\ \le C{\int }_{2^j\le |x-y|<2^{j+1}}\frac{|\mathrm{\Omega }(x-y)|}{|x-y{|}^n}|G_{l-j-3}b(x)-G_{l-j-3}b(y)||{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s(y)|dy.\end{array}$$

By Lemma 3.3, we have, for any $0<\delta <1$,

$$\begin{array}{r}|[G_{l-j-3}b,T_j]{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s(x)|\\ \le C{2}^{(l-j-3)\delta }\frac{|x-y{|}^{\delta }}{\delta }{\parallel b\parallel }_{BMO}{\int }_{2^j\le |x-y|<2^{j+1}}\frac{|\mathrm{\Omega }(x-y)|}{|x-y{|}^n}|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s(y)|dy\\ \le C\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{\int }_{2^j\le |x-y|<2^{j+1}}\frac{|\mathrm{\Omega }(x-y)|}{|x-y{|}^n}|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s(y)|dy\\ =C\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{T}_{|\mathrm{\Omega }|,j}\left(\right|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s\left|\right)(x),\end{array}$$

(4.20)

where

$$T_{|\mathrm{\Omega }|,j}{f}_s(x)={\int }_{2^j\le |x-y|<2^{j+1}}\frac{|\mathrm{\Omega }(x-y)|}{|x-y{|}^n}{f}_s(y)dy,$$

and C is independent of δ and l. Then, by (4.19), (4.20) and applying Lemma 3.2 and Lemma 3.1, we have that for $1<p<\mathrm{\infty }$,

$$\begin{array}{rl}{M}_1& \le C\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|T_{|\mathrm{\Omega }|,j,d}\left(\right|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s\left|\right){|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel \mathrm{\Omega }\parallel }_{L^1}\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel \mathrm{\Omega }\parallel }_{L^1}\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|S_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel \mathrm{\Omega }\parallel }_{L^1}\frac{2^{l\delta }}{\delta }{\parallel b\parallel }_{BMO}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p},\end{array}$$

(4.21)

where C is independent of l and δ.

• (b) The estimate of $M_2$. By Lemma 3.4(i), we know for $|k|\le 2$, ${\mathrm{\Delta }}_{i+k}{S}_{l-j}g=0$ for $g\in {\mathcal{S}}^{{}^{\prime }}({\mathbb{R}}^n)$ when $|i-(l-j)|\ge 5$. Thus

  $$\begin{array}{r}R(b,T_j{S}_{l-j}^2{f}_s)-T_j\left(R(b,S_{l-j}^2{f}_s)\right)(x)\\ =\sum _{i\in \mathbb{Z}}\sum _{|k|\le 2}({\mathrm{\Delta }}_{i}b)(x)\left(T_j{\mathrm{\Delta }}_{i+k}{S}_{l-j}^2{f}_s\right)(x)-T_j(\sum _{i\in \mathbb{Z}}\sum _{|k|\le 2}({\mathrm{\Delta }}_{i}b)\left({\mathrm{\Delta }}_{i+k}{S}_{l-j}^2{f}_s\right))(x)\\ =\sum _{k=-2}^2\sum _{|i-(l-j)|\le 4}(({\mathrm{\Delta }}_{i}b)(x)\left(T_j{\mathrm{\Delta }}_{i+k}{S}_{l-j}^2{f}_s\right)(x)-T_j(({\mathrm{\Delta }}_{i}b)\left({\mathrm{\Delta }}_{i+k}{S}_{l-j}^2{f}_s\right))(x))\\ =\sum _{k=-2}^2\sum _{|i-(l-j)|\le 4}[{\mathrm{\Delta }}_{i}b,T_j]\left({\mathrm{\Delta }}_{i+k}{S}_{l-j}^2{f}_s\right)(x).\end{array}$$

Then we get

$$M_2\le \sum _{|k|\le 6}{\parallel {\left(\sum _{s\in \mathbb{Z}}\right|\sum _{j\in \mathbb{Z}}{S}_{l-j}^2[{\mathrm{\Delta }}_{l-j+k}b,T_j]\left({\mathrm{\Delta }}_{l-j+k}{S}_{l-j}^2{f}_s\right){|}^q)}^{1/q}\parallel }_{L^p}.$$

Without loss of generality, we may assume $k=0$. By the equality above and using Lemma 3.1, ${sup}_{i\in \mathbb{Z}}{\parallel {\mathrm{\Delta }}_i(b)\parallel }_{L^{\mathrm{\infty }}}\le C{\parallel b\parallel }_{BMO}$ (see [21]) and Lemma 3.2, we have, for $1<p<\mathrm{\infty }$,

$$\begin{array}{rl}{M}_2& \le C\underset{i\in \mathbb{Z}}{sup}{\parallel {\mathrm{\Delta }}_i(b)\parallel }_{L^{\mathrm{\infty }}}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|T_{|\mathrm{\Omega }|,j}\left(\right|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s\left|\right){|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|{\mathrm{\Delta }}_{l-j}{S}_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/r}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|S_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.\end{array}$$

(4.22)

• (c) The estimate of $M_3$. Finally, we give the estimate of $M_3$. By Lemma 3.4(ii), we know $S_j({\mathrm{\Delta }}_{i}g{G}_{i-3}h)=0$ for $g,h\in {\mathcal{S}}^{{}^{\prime }}({\mathbb{R}}^n)$ if $|j-i|\ge 5$. We get

  $$\begin{array}{r}{S}_{l-j}^2({\pi }_b\left(T_j{S}_{l-j}^2{f}_s\right)-T_j\left({\pi }_b\left(S_{l-j}^2{f}_s\right)\right))\\ =S_{l-j}^2(\sum _{i\in \mathbb{Z}}({\mathrm{\Delta }}_{i}b)\left(G_{i-3}{T}_j{S}_{l-j}^2{f}_s\right)-T_j(\sum _{i\in \mathbb{Z}}({\mathrm{\Delta }}_{i}b)\left(G_{i-3}{S}_{l-j}^2{f}_s\right)))(x)\\ =\sum _{|i-(l-j)|\le 4}\{S_{l-j}^2(({\mathrm{\Delta }}_{i}b)\left(G_{i-3}{T}_j{S}_{l-j}^2{f}_s\right))(x)-S_{l-j}^2{T}_j(({\mathrm{\Delta }}_{i}b)\left(G_{i-3}{S}_{l-j}^2{f}_s\right))(x)\}.\end{array}$$

Thus, by Lemma 3.1, ${sup}_{i\in \mathbb{Z}}{\parallel {\mathrm{\Delta }}_i(b)\parallel }_{L^{\mathrm{\infty }}}\le C{\parallel b\parallel }_{BMO}$, and Lemma 3.2, we get, for $1<p<\mathrm{\infty }$,

$$\begin{array}{rl}{M}_3& \le C\underset{i\in \mathbb{Z}}{sup}{\parallel {\mathrm{\Delta }}_i(b)\parallel }_{L^{\mathrm{\infty }}}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|T_{|\mathrm{\Omega }|,j}\left(\right|G_{l-j}{S}_{l-j}^2{f}_s\left|\right){|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|G_{l-j}{S}_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {\left(\sum _{s\in \mathbb{Z}}{\left(\sum _{j\in \mathbb{Z}}\right|S_{l-j}^2{f}_s{|}^2)}^{q/2}\right)}^{1/q}\parallel }_{L^p}\\ \le C{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}.\end{array}$$

(4.23)

By (4.16), (4.21)-(4.23), we get

$$L_3\le Cmax\{2,\frac{2^{\delta l}}{\delta }\}{\parallel b\parallel }_{BMO}{\parallel \mathrm{\Omega }\parallel }_{L^1}{\parallel {(\sum _{s\in \mathbb{Z}}|f_s{|}^q)}^{1/q}\parallel }_{L^p}\text{for }l\in \mathbb{Z},$$

where C is independent of δ and l. This establishes the proof of (4.4).