---
abstract: 'We prove that the weak Morrey space $WM^{p}_{q}$ is contained in the Morrey space $M^{p}_{q_{1}}$ for $1\leq q_{1}< q\leq p<\infty$. As applications, we show that if the commutator $[b,T]$ is bounded from $L^p$ to $L^{p,\infty}$ for some $p\in (1,\infty)$, then $b\in \mathrm{BMO}$, where $T$ is a Calderón-Zygmund operator. Also, for $1<p\leq q<\infty$, $b\in \mathrm{BMO}$ if and only if $[b,T]$ is bounded from $M^{p}_{q}$ to $WM_{q}^{p}$. For $b$ belonging to Lipschitz class, we obtain similar results.'
address:
- College of Mathematics and System Sciences Xinjiang University Urumqi 830046 Republic of China
- School of Mathematics and Statistics WuHan University WuHan 430072 Republic of China
author:
- 'Dinghuai Wang, Jiang Zhou$^\ast$ and Wenyi Chen'
title: Characterizations of the BMO and Lipschitz spaces via commutators on weak Lebesgue and Morrey spaces 
---

[^1]

Introduction {#sec:int}
============

Let $T$ be a Calderón-Zygmund operator defined by $$Tf(x)=\mathrm{p.v.}\int_{\mathbb{R}^{n}}K(x-y)f(y)dy,$$ where the kernel $K(x)=\frac{\Omega(x)}{|x|^{n}}$ satisfies the following conditions:

1.  $\Omega$ is homogeneous of degree zero on $\mathbb{R}^{n}$, i.e., $\Omega(\lambda x)=\Omega(x)$ for all $\lambda>0$ and $x\in \mathbb{R}^{n}$;

2.  $\Omega\in C^{\infty}(\mathbb{S}^{n-1})$ and $\int_{\mathbb{S}^{n-1}}\Omega(x)dx=0.$

A locally integrable function $b$ belongs to the $\mathrm{BMO}$ space if $b$ satisfies $$\|b\|_{*}:=\sup_{Q}\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|dx<\infty,$$ where $b_{Q}:=\frac{1}{|Q|}\int_{Q}b(x)dx$ and the supremum is taken over all cubes $Q$ in $\mathbb{R}^{n}$. A well known result of Coifman, Rochberg and Weiss [@CRW] states that the commutator $$[b,T](f):=bT(f)-T(bf)$$ is bounded on some $L^p$, $1<p<\infty$, if and only if $b\in \mathrm{BMO}$. An interesting question is raised. Is $b$ in $\mathrm{BMO}$ if $[b,T]$ is of weak type $(p,p)$ for some $p\in (1,\infty)$? We will give an affirmative answer in this paper.

For $1\leq q\leq p <\infty$, we say that a function $f$ belongs to Morrey space $M^{p}_{q}$ if $$\|f\|_{M^{p}_{q}}:=\sup_{x\in \mathbb{R}^{n},r>0}\frac{1}{|Q(x,r)|^{1/q-1/p}}\bigg(\int_{Q(x,r)}|f(y)|^{q}dy\bigg)^{1/q}<\infty;$$ a function $f$ belongs to weak Morrey space $WM_{q}^{p}$ if $$\|f\|_{WM_{q}^{p}}:=\sup_{x\in \mathbb{R}^{n},r>0}\frac{1}{|Q(x,r)|^{1/q-1/p}}\sup_{\lambda>0}\bigg(\lambda^{q}|\{y\in Q(x,r):|f(y)|>\lambda\}|\bigg)^{1/q}<\infty.$$ Morrey spaces describe local regularity more precisely than $L^{p}$ spaces and can be seen as a complement of $L^{p}$. In fact, $L^p=M^{p}_{p}\subset M^{p}_{q}$ and $WM^{p}_{p}=L^{p,\infty}$ for $1\leq q \leq p<\infty$. In 1997, Ding [@D] showed that $b$ is in $\mathrm{BMO}$ if and only if the commutator $[b,T]$ of Calderón-Zygmund operator $T$ is bounded on Morrey spaces. We will demonstrate here that $b\in \mathrm{BMO}$ if and only if $[b,T]$ is weak bounded on Morrey spaces.

Another subject of this paper is to consider the characterizations of Lipschitz functions via commutators. For $0<\alpha<1$, the Lipschitz space $\mathrm{Lip}_{\alpha}$ is the set of functions $f$ such that $$\|f\|_{\mathrm{Lip}_{\alpha}}:=\sup_{x,h\in \mathbb{R}^{n}\atop_{h\neq 0}}\frac{|f(x+h)-f(x)|}{|h|^{\alpha}}<\infty.$$ In 1978, Janson [@J] proved that, for $1<p<q<\infty$ with $1/q= 1/p-\alpha/n$, $b\in \mathrm{Lip}_{\alpha}$ if and only if $[b,T]$ is bounded from $L^p$ to $L^{q}$. In 1995, using Sobolev-Besov embedding, Paluszyński [@PA] obtained that, for $1<p<\infty$ and $0<\alpha<1$, $b\in \mathrm{Lip}_{\alpha}$ if and only if $[b,T]$ is bounded from $L^p$ to the homogeneous Triebel-Lizorkin spaces $\dot{F}^{\alpha,\infty}_{p}$. Paluszyński’s idea was novel for the study about the boundedness of commutators from $L^p$ to $\dot{F}^{\alpha,\infty}_{p}$ and shed new light on the characterization of Lipschitz space via commutators. Recently, Shi and Lu [@SL] showed that, for $1<q\leq p<\infty$, $1<t\leq s<\infty$, $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$ and $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$, $b\in \mathrm{Lip}_{\alpha}$ if and only if $[b,T]$ is bounded from $M^{p}_{q}$ to $M^{s}_{t}$. We are going to show that, for $1\leq p<q<\infty$ with $1/q=1/p-\alpha/n$, $b\in \mathrm{Lip}_{\alpha}$ if and only if $[b,T]$ is bounded from $L^p$ to $L^{q,\infty}$. Also, for $1\leq q\leq p<\infty$, $1<t\leq s<\infty$, $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$ and $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$, $b\in \mathrm{Lip}_{\alpha}$ if and only if $[b,T]$ is bounded from $M_{q}^{p}$ to $WM_{t}^{s}$.

Throughout this paper, the letter $C$ denotes constants which are independent of main variables and may change from one occurrence to another.

Characterization of $\mathrm{BMO}$ space via commutators
========================================================

In this section, we characterize $\mathrm{BMO}$ space via the boundedness of commutator on (weak) Lebesgue spaces or (weak) Morrey spaces. First of all, we compare with Morrey spaces and weak Morrey spaces.

It is clear that $M^{p}_{q_{1}}$ is contained in $WM_{q_{2}}^{p}$ and $\|\cdot\|_{WM^{p}_{q_{2}}}\leq C\|\cdot\|_{M^{p}_{q_{1}}}$ if $1\leq q_{2}\leq q_{1}\leq p<\infty$. However, for $1\leq q_{1}< q_{2}\leq p<\infty$, one has the reverse inequality as follows.

\[main\] If $1\leq q_{1}< q_{2}\leq p<\infty$, then $WM^{p}_{q_{2}}\subset M^{p}_{q_{1}}$ and $\|f\|_{M^{p}_{q_{1}}}\leq \break 2\big(\frac{q_{1}}{q_{2}-q_{1}}\big)^{\frac{1}{q_{2}}}\|f\|_{WM^{p}_{q_{2}}}$.

Let $f\in WM^{p}_{q_{2}}$. Given a cube $Q \subset \mathbb{R}^n$ and $\lambda>0$, $$\frac{1}{|Q|^{1/q_{2}-1/p}}\big(\lambda^{q_{2}}|\{x\in Q:|f(x)|>\lambda\}|\big)^{1/q_{2}}\leq \|f\|_{WM^{p}_{q_{2}}};$$ that is, $$|\{x\in Q:|f(x)|>\lambda\}|\leq \|f\|^{q_{2}}_{WM^{p}_{q_{2}}}|Q|^{1-q_{2}/p}\lambda^{-q_{2}}.$$ Choose $$N=\|f\|_{WM^{p}_{q_{2}}}|Q|^{-1/p}\Big(\frac{q_{1}}{q_{2}-q_{1}}\Big)^{1/q_{2}}.$$ Thus, $$\begin{aligned}
\int_{Q}|f(x)|^{q_{1}}dx&=&q_{1}\int_{0}^{\infty}\lambda^{q_{1}-1}|\{x\in Q:|f(x)|>\lambda\}|d\lambda\\
&\leq&q_{1}\int_{0}^{N}\lambda^{q_{1}-1}|Q|d\lambda+q_{1}\int_{N}^{\infty}\lambda^{q_{1}-1}\|f\|^{q_{2}}_{WM^{p}_{q_{2}}}|Q|^{1-q_{2}/p}\lambda^{-q_{2}}d\lambda\\
&=&|Q|N^{q_{1}}+\frac{q_{1}}{q_{2}-q_{1}}\|f\|^{q_{2}}_{WM^{p}_{q_{2}}}|Q|^{1-q_{2}/p}N^{q_{1}-q_{2}},\end{aligned}$$ which gives $$\bigg(\int_{Q}|f(y)|^{q_{1}}dy\bigg)^{1/q_{1}}\leq 2\|f\|_{WM^{p}_{q_{2}}}|Q|^{1/q_{1}-1/p}\Big(\frac{q_{1}}{q_{2}-q_{1}}\Big)^{1/q}.$$ Then $$\|f\|_{M^{p}_{q_{1}}}\leq 2\Big(\frac{q_{1}}{q_{2}-q_{1}}\Big)^{1/q_{2}}\|f\|_{WM^{p}_{q_{2}}}$$ and the lemma follows.

\[rem\]For $1\leq q_{1}<q_{2}< p<\infty$. Kozono and Yamazaki [@KY Lemma 1.7] proved that $\|f\|_{M_{q_{2}}^{p}}\leq C\|f\|_{L^{p,\infty}}$. The above Theorem \[main\] yields immediately $$\|f\|_{M^{p}_{q_{1}}}\leq C\|f\|_{WM_{q_{2}}^{p}}\leq C\|f\|_{M^{p}_{q_{2}}}\leq C\|f\|_{WM^{p}_{p}}=C\|f\|_{L^{p,\infty}}.$$

Now we return to our first subject.

\[thm2.1\] Let $1< p<\infty$. The following statements are equivalent:

1.  $b\in \mathrm{BMO}$;

2.  $[b,T]$ is a bounded operator from $L^p$ to $L^p$;

3.  $[b,T]$ is a bounded operator from $L^p$ to $L^{p,\infty}$.

\[thm2.2\] Let $1<q\leq p<\infty$. The following statements are equivalent:

1.  $b\in \mathrm{BMO}$;

2.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $M^{p}_{q}$;

3.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $WM^{p}_{q}$.

0.5cm [*Proof of Theorem \[thm2.1\].*]{} The equivalence of $(1)$ and $(2)$ was proved in \[3\]. By the inequality $\|f\|_{L^{p,\infty}}\leq \|f\|_{L^p}$, it is obvious that $(2)$ implies $(3)$.

To show $(3)\Rightarrow (1)$, we use Paluszyński idea given in [@PA]. For $z_{0}\in \mathbb{R}^{n}\backslash \{0\}$, let $\delta=\frac{|z_{0}|}{2\sqrt{n}}$ and $Q(z_{0},\delta)$ denote the open cube centered at $z_{0}$ with side length $2\delta$. Then $K(x)^{-1}$ has an absolutely convergent Fourier series $$\frac{1}{K(x)}=\sum a_{m}e^{i \langle v_{m},x\rangle}$$ with $\sum |a_{m}|<\infty$, where the exact form of the vectors $v_{m}$ is unrelated. Then, we have the expansion $$\frac{1}{K(x)}=\frac{\delta^{-n}}{K(\delta x)}=\delta^{-n}\sum a_{m}e^{i\langle v_{m},\delta x\rangle} ~ for ~ |x-\frac{z_{0}}{\delta}|<\sqrt{n}.$$ Given cubes $Q=Q(x_{0},r)$ and $Q'=Q(x_{0}-rz_{1},r)$, if $x\in Q$ and $y\in Q'$, then $$\Big|\frac{x-y}{r}-\frac{z_{0}}{\delta}\Big|\leq \Big|\frac{x-x_{0}}{r}\Big|+\Big|\frac{y-(x_{0}-\frac{r z_{0}}{\delta})}{r}\Big|< \sqrt{n}.$$ Let $s(x)=\overline{\mathrm{sgn}(\int_{Q'}(b(x)-b(y))dy)}$. Then $$\begin{aligned}
&&\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx\\
&&=\frac{1}{|Q|}\frac{1}{|Q'|}\int_{Q}\bigg|\int_{Q'}(b(x)-b(y))dy\bigg|dx\\
&&=\frac{1}{|Q|^{2}}\int_{Q}\int_{Q'}s(x)(b(x)-b(y))dydx\\
&&=\frac{1}{|Q|^{2}}\int_{Q}\int_{Q'}s(x)(b(x)-b(y))\frac{r^{n}K(x-y)}{K(\frac{x-y}{r})}dydx\\
&&=\frac{1}{|Q|}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(b(x)-b(y))K(x-y)\sum a_{m}e^{i\langle v_{m}, \delta\frac{x-y}{r}\rangle}s(x)\chi_{Q}(x)\chi_{Q'}(y)dydx\\
&&=\frac{1}{|Q|}\sum a_{m}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(b(x)-b(y))K(x-y)e^{i\langle v_{m},\frac{\delta}{r}x\rangle}
s(x)\chi_{Q}(x)e^{-i\langle v_{m},\frac{\delta}{r}y\rangle}\chi_{Q'}(y)dydx.\end{aligned}$$ Setting $g_{m}(y)=e^{-i\langle v_{m},\frac{\delta}{r}y\rangle}\chi_{Q'}(y)$ and $h_{m}(x)=e^{i\langle v_{m},\frac{\delta}{r}x\rangle}s(x)\chi_{Q}(x)$, we have $$\begin{aligned}
\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx&=&\frac{1}{|Q|}\sum a_{m}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}(b(x)-b(y))K(x-y)g_{m}(y)h_{m}(x)dydx\\
&=&\frac{1}{|Q|}\sum a_{m}\int_{\mathbb{R}^{n}}[b,T](g_{m})(x)h_{m}(x)dx.\end{aligned}$$ Choose $q\in (1,p)$. By Remark \[rem\], $$\begin{aligned}
\label{eq1}
\begin{split}
\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|dx&\leq \frac{2}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx\\
&\leq\frac{C}{|Q|^{1/p}}\sum |a_{m}|\|[b,T](g_{m})\|_{M^{p}_{q}}\\
&\leq\frac{C}{|Q|^{1/p}}\sum |a_{m}|\|[b,T](g_{m})\|_{L^{p,\infty}}\\
&\leq\frac{C\|g_{m}\|_{L^p}}{|Q|^{1/p}}\|[b,T]\|_{L^p\rightarrow L^{p,\infty}}\\
&\leq C\|[b,T]\|_{L^p\rightarrow L^{p,\infty}},
\end{split}\end{aligned}$$ which yields $b\in \mathrm{BMO}$ and $\|b\|_{*}\leq C\|[b,T]\|_{L^p\rightarrow L^{p,\infty}}$. Hence, the proof of Theorem \[thm2.1\] is completed.

0.5cm

[*Proof of Theorem \[thm2.2\].*]{} We use a same argument as the proof of Theorem \[thm2.1\] except choosing $q_{1}\in(1,q)$ and replacing by $$\begin{aligned}
\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|dx&\leq& \frac{2}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx\\
&\leq&\frac{C}{|Q|^{1/p}}\sum |a_{m}|\|[b,T](g_{m})\|_{M^{p}_{q_{1}}}\\
&\leq&\frac{C}{|Q|^{1/p}}\sum |a_{m}|\|[b,T](g_{m})\|_{WM^{p}_{q}}\\
&\leq&\frac{C\|g_{m}\|_{M^p_q}}{|Q|^{1/p}}\|[b,T]\|_{M^p_q\rightarrow WM^{p}_{q}}\\
&\leq&C\|[b,T]\|_{M^p_q\rightarrow WM^{p}_{q}}.\end{aligned}$$ Then $b\in \mathrm{BMO}$ and $\|b\|_{*}\leq C\|[b,T]\|_{M^p_q\rightarrow WM^{p}_{q}}$. This completes the proof of Theorem \[thm2.2\].

Characterization of $\mathrm{Lip}_{\alpha}$ via commutators
===========================================================

We give a lemma that can be used to prove a characterization of Lipschitz functions.

\[lem3.1\] For $0<\alpha<1$ and $1<q\leq \infty$, $$\begin{aligned}
\|b\|_{\mathrm{Lip}_{\alpha}}&\approx \sup_{Q}\frac{1}{|Q|^{1+\alpha/n}}\int_{Q}|b(x)-b_{Q}|dx\\
&\approx
\begin{cases}\displaystyle
\sup_{Q}\frac{1}{|Q|^{\alpha/n}}\bigg(\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|^{q}dx\bigg)^{1/q} &\text{if}\ \ 1<q<\infty\\
\displaystyle\mathop\mathrm{ess~sup}_{Q}\frac{|b(x)-b_{Q}|}{|Q|^{\alpha/n}} &\text{if}\ \ q=\infty
\end{cases},\end{aligned}$$ where the supremum is taken over all cubes $Q\subset\mathbb{R}^{n}$ and $\approx$ means equivalence.

The first equivalence can be found in [@DS pages 14 and 38], and the second equivalence can be found in [@JTW].

The first result of this section is

\[thm3.1\] Let $0<\alpha<1$, $1\leq p<\frac{n}{\alpha}$ and $1/q=1/p-\alpha/n$. The following statements are equivalent:

1.  $b\in \mathrm{Lip}_{\alpha}$;

2.  $[b,T]$ is a bounded operator from $L^p$ to $L^{q,\infty}$.

$(1)\Rightarrow (2)$: Let $b\in \mathrm{Lip}_{\alpha}$. Then $$\begin{aligned}
\label{eq2}
\begin{split}
\big|[b,T](f)(x)\big|&=\bigg|\int_{\mathbb{R}^{n}}(b(x)-b(y))f(y)K(x-y)dy\bigg|\\
&\leq\|b\|_{\mathrm{Lip}_{\alpha}}\int_{\mathbb{R}^{n}}\frac{|f(y)|}{|x-y|^{n-\alpha}}dy\\
&\leq\|b\|_{\mathrm{Lip}_{\alpha}}I_{\alpha}(|f|)(x),
\end{split}\end{aligned}$$ which implies $$\|[b,T]f\|_{L^{q,\infty}}\leq \|b\|_{\mathrm{Lip}_{\alpha}}\|I_{\alpha}(|f|)\|_{L^{q,\infty}}\leq C\|b\|_{\mathrm{Lip}_{\alpha}}\|f\|_{L^{p}}.$$

$(2)\Rightarrow (1)$: We follow the method in the proof of Theorem \[thm2.1\] except choosing $q_{1}\in(1,q)$ and replacing by $$\begin{aligned}
\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|dx&\leq& \frac{2}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx\\
&\leq&\frac{C}{|Q|^{1/q}}\sum |a_{m}|\|[b,T](g_{m})\|_{M^{q}_{q_{1}}}\\
&\leq&\frac{C}{|Q|^{1/q}}\sum |a_{m}|\|[b,T](g_{m})\|_{L^{q,\infty}}\\
&\leq&\frac{C\|g_{m}\|_{L^p}}{|Q|^{1/q}}\|[b,T]\|_{L^p\rightarrow L^{q,\infty}}\\
&\leq&C|Q|^{\alpha/n}\|[b,T]\|_{L^p\rightarrow L^{q,\infty}}.\end{aligned}$$ From Lemma \[lem3.1\] we conclude that $b\in \mathrm{Lip}_{\alpha}$ and $\|b\|_{\mathrm{Lip}_{\alpha}}\leq C\|[b,T]\|_{L^p\rightarrow L^{q,\infty}}$. Hence, the proof of Theorem \[thm3.1\] is completed.

Using Theorem \[thm3.1\] and Janson’s result [@J], we immediately have

Let $0<\alpha<1$, $1< p<\frac{n}{\alpha}$ and $1/q=1/p-\alpha/n$. The following statements are equivalent:

1.  $b\in \mathrm{Lip}_{\alpha}$;

2.  $[b,T]$ is a bounded operator from $L^p$ to $L^{q}$;

3.  $[b,T]$ is a bounded operator from $L^p$ to $L^{q,\infty}$.

We give two remarkable results about the boundedness of fractional integral operator $I_{\alpha}$ on Morrey spaces, where $I_{\alpha}$ is defined by $$I_{\alpha}f(x)=\int_{\mathbb{R}^n}\frac{f(y)}{|x-y|^{n-\alpha}}dy,\qquad x\in\mathbb{R}^{n} \ and \ 0<\alpha<n.$$

\[P\][(Peetre [@PE])]{} Let $0<\alpha<n$, $1<q\leq p<\infty$ and $1<t\leq s<\infty$. If $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$ and $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$, then $\|I_{\alpha}f\|_{M^{s}_{t}}\leq C\|f\|_{M^{p}_{q}}$.

\[A\][(Adams [@A])]{} Let $0<\alpha<n$, $1<q\leq p<\infty$ and $1<l\leq s<\infty$. If $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$ and $\frac{l}{s}=\frac{q}{p}$, then $\|I_{\alpha}f\|_{M^{s}_{l}}\leq C\|f\|_{M^{p}_{q}}$.

We will use the above two theorems to show other characterizations of Lipschitz functions.

\[thm3.2\] Let $0<\alpha<1$, $1\leq q\leq p<\infty$, $1<t\leq s<\infty$, $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$ and $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$. The following statements are equivalent:

1.  $b\in \mathrm{Lip}_{\alpha}$;

2.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $WM^{s}_{t}$.

\[thm3.3\] Let $0<\alpha<1$, $1< q\leq p<\infty$, $1<t\leq s<\infty$, $1<l\leq s<\infty$, $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$, $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$ and $\frac{l}{s}=\frac{q}{p}$,. The following statements are equivalent:

1.  $b\in \mathrm{Lip}_{\alpha}$;

2.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $M^{s}_{l}$;

3.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $M^{s}_{t}$;

4.  $[b,T]$ is a bounded operator from $M^{p}_{q}$ to $WM^{s}_{t}$.

[*Proof of Theorem \[thm3.2\].*]{} $(1)\Rightarrow (2)$: For $q>1$, Theorem \[P\] and yield $$\|[b,T]f\|_{WM^{s}_{t}}\leq \|b\|_{\mathrm{Lip}_{\alpha}}\|I_{\alpha}(|f|)\|_{WM_{t}^{s}}\leq \|b\|_{\mathrm{Lip}_{\alpha}}\|I_{\alpha}(|f|)\|_{M_{t}^{s}}\leq C\|b\|_{\mathrm{Lip}_{\alpha}}\|f\|_{M^{p}_{q}}.$$ For $q=1$, it follows from [@T Theorem 1.4] that $$\|I_{\alpha}(f)\|_{WM_{1}^{s}}\leq C\|f\|_{M^{p}_{q}},$$ which gives $$\|[b,T]f\|_{WM^{s}_{1}}\leq \|b\|_{\mathrm{Lip}_{\alpha}}\|I_{\alpha}(|f|)\|_{WM_{1}^{s}}\leq C\|b\|_{Lip_{\alpha}}\|f\|_{M^{p}_{q}}.$$

$(2)\Rightarrow (1)$: A same argument as the proof of Theorem \[thm2.1\] except choosing $q_{1}\in(1,t)$ and replacing by $$\begin{aligned}
\frac{1}{|Q|}\int_{Q}|b(x)-b_{Q}|dx&\leq& \frac{2}{|Q|}\int_{Q}|b(x)-b_{Q'}|dx\\
&\leq&\frac{C}{|Q|^{1/s}}\sum |a_{m}|\|[b,T](g_{m})\|_{M^{s}_{q_{1}}}\\
&\leq&\frac{C}{|Q|^{1/s}}\sum |a_{m}|\|[b,T](g_{m})\|_{WM^{s}_{t}}\\
&\leq&C|Q|^{\alpha/n}\|[b,T]\|_{M^p_q\rightarrow WM^s_t}.\end{aligned}$$ implies $b\in \mathrm{Lip}_{\alpha}$ due to Lemma \[lem3.1\].

0.5cm [*Proof of Theorem \[thm3.3\].*]{} $(1)\Rightarrow (2)$: It follows from and Theorem \[A\] that $[b,T]$ is bounded from $M^{p}_{q}$ to $M^{s}_{l}$.

$(2)\Rightarrow (3)$: Since $\frac{1}{s}=\frac{1}{p}-\frac{\alpha}{n}$, $\frac{1}{t}=\frac{1}{q}-\frac{\alpha}{n}$ and $\frac{l}{s}=\frac{q}{p}$, we have $$\frac{1}{q}-\frac{1}{t}=\frac{1}{p}-\frac{1}{s}=\frac{q}{p}\Big(\frac{1}{q}-\frac{1}{l}\Big)\leq \frac{1}{q}-\frac{1}{l},$$ and hence $t\leq l$. Then $ \|\cdot\|_{M^{s}_{t}}\leq\|\cdot\|_{M^{s}_{l}}$.

It is obvious for $(3)\Rightarrow (4)$.

$(4)\Rightarrow (1)$: By Theorem \[thm3.2\], we obtain $b\in \rm{Lip}_{\alpha}$. Hence, the proof of Theorem \[thm3.3\] is completed. $\hfill$

0.5cm

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[^1]: The research was supported by National Natural Science Foundation of China (Grant No.11261055).\
    Corresponding author, Email: zhoujiangshuxue@126.com.
