# Hardy operators and the commutators on Hardy spaces

## Abstract

In this paper, the boundedness of the classic Hardy operator and its adjoint on Hardy spaces is obtained. We also discuss the boundedness for the commutators generated by the classic Hardy operator and its adjoint with $$BMO$$ and $$CMO(\mathbb{R}^{+})$$ functions on Hardy spaces.

## Introduction

Let P and Q be the classical Hardy operator and its adjoint on $$\mathbb{R}^{+}=(0,+\infty )$$,

\begin{aligned} &Pf(x)=\frac{1}{x} \int _{0}^{x}f(y)\,dy, \qquad Qf(x)= \int _{x}^{\infty } \frac{f(y)}{y}\,dy. \end{aligned}

It is well known that, for $$p>1$$, $$\|Pf\|_{L^{p}(\mathbb{R}^{+})}\leq p'\|f\|_{L^{p}(\mathbb{R}^{+})}$$ and $$\|Qf\|_{L^{p}(\mathbb{R}^{+})}\leq p\|f\|_{L^{p}(\mathbb{R}^{+})}$$, where $$p'=p/(p-1)$$. For the boundedness of P and Q on $$L^{1}(\mathbb{R}^{+})$$, we have that P is bounded from $$L^{1}(\mathbb{R}^{+})$$ to $$L^{1,\infty }(\mathbb{R}^{+})$$ but not bounded on $$L^{1}(\mathbb{R}^{+})$$ and Q is bounded on $$L^{1}(\mathbb{R}^{+})$$.

For earlier development of this kind of inequality and many applications in analysis, see [2, 47, 10].

Let $$b\in L_{loc}(\mathbb{R}^{+})$$, the commutators of Hardy operators P and its adjoint Q are defined by

\begin{aligned}& [b,P]f(x)=b(x)Pf(x)-P(bf) (x), \\& [b,Q]f(x)=b(x)Qf(x)-Q(bf) (x). \end{aligned}

Long and Wang  established Hardy’s integral inequalities for commutators generated by P and Q with $$CMO$$ function. Li, Zhang, and Xue in  obtained some two-weight inequalities for commutators generated by P and Q with $$CMO$$ function. Zhao, Fu, and Lu in  studied the boundedness on Hardy spaces for n-dimensional Hardy operators and the commutators.

In this paper, we discuss the boundedness on the Hardy spaces for the Hardy operator P, its adjoint operator Q, and their commutators with $$BMO$$ and $$CMO$$ functions.

## The boundedness of P and Q on Hardy spaces

Because $$\mathbb{R}^{+}$$ is a space of homogeneous type with distance $$d(x,y)=|x-y|$$ and Lebesgue measure, we can define the Hardy space as Coifman and Weiss in . We begin with the definitions of the atom, the molecule, and the Hardy space on $$\mathbb{R}^{+}$$.

A function $$a\in L^{\infty }(\mathbb{R}^{+})$$ is called $$(1,\infty )$$-atom if it satisfies the following conditions: (1) $$\operatorname{supp}(a)\subset (x_{0},x_{0}+r]$$, where $$x_{0}\geq 0$$, $$r>0$$; (2) $$\| a\|_{L^{\infty }}\leqslant r^{-1}$$; (3) $$\int a(x)\,dx=0$$.

Let $$\varepsilon >0$$, we say that a function M on $$\mathbb{R}^{+}$$ is a ε-molecule centered at $$x_{0}$$ if

$$\biggl( \int _{\mathbb{R}^{+}} \bigl\vert M(x) \bigr\vert ^{2}\,dx \biggr) \biggl( \int _{ \mathbb{R}^{+}} \bigl\vert M(x) \bigr\vert ^{2} \vert x-x_{0} \vert ^{1+\varepsilon }\,dx \biggr)^{1/ \varepsilon }\leq 1$$

and

$$\int _{\mathbb{R}^{+}}M(x)\,dx=0.$$

The atomic Hardy space $$H^{1}(\mathbb{R}^{+})$$ is defined by

$$H^{1}\bigl(\mathbb{R}^{+}\bigr)=\biggl\{ f\in L^{1} \bigl(\mathbb{R}^{+}\bigr):f(x)=\sum_{k} \lambda _{k}a_{k}(x), \text{ and } \sum _{k} \vert \lambda _{k} \vert < \infty \biggr\} ,$$

where each $$a_{k}$$ is a $$(1,\infty )$$-atom. We set the $$H^{1}(\mathbb{R}^{+})$$ norm of f by

$$\Vert f \Vert _{H^{1}}=\inf \Biggl\{ \sum _{k}^{\infty }\vert \lambda _{k}\vert \Biggr\} ,$$

where the infimum is taken over all the decompositions of $$f=\sum_{k} \lambda _{k}a_{k}$$ as above.

### Remark 2.1

P is not bounded on $$H^{1}(\mathbb{R}^{+})$$. In fact, if we take $$f_{0}(x)=\frac{1}{4}\chi _{(0,1]}-\frac{1}{4}\chi _{(1,2]}$$, then we have

$$Pf_{0}(x)= \textstyle\begin{cases} \frac{1}{4} , &0< x \leqslant 1 , \\ \frac{1}{2x}-\frac{1}{4} , &1< x\leqslant 2, \\ 0 , & 2< x . \end{cases}$$

Obviously, $$f_{0}\in H^{1}(\mathbb{R}^{+})$$; however $$\int _{0}^{+\infty }Pf_{0}(x)\,dx\neq 0$$, therefore P is not bounded on $$H^{1}(\mathbb{R}^{+})$$.

### Lemma 2.2

()

If M is a ε-molecule centered at $$x_{0}$$, then $$M\in H^{1}(\mathbb{R}^{+})$$. Moreover, $$\|M\|_{H^{1}}$$ depends only on the constant ε.

### Theorem 2.3

Q is bounded on $$H^{1}(\mathbb{R}^{+})$$.

### Proof

Assume that a is a $$(1,\infty )$$-atom of $$H^{1}(\mathbb{R}^{+})$$ with $$\operatorname{supp}(a)\subset (x_{0},x_{0}+r]$$, $$x_{0}\geq 0$$, $$r>0$$. It is enough to prove that Qa is a 1-molecule.

\begin{aligned} & \int _{\mathbb{R}^{+}} \bigl\vert Qa(x) \bigr\vert ^{2}\,dx \cdot \int _{\mathbb{R}^{+}} \bigl\vert Qa(x) \bigr\vert ^{2} \vert x-x_{0} \vert ^{2}\,dx \\ &\quad \leq \int _{x_{0}}^{x_{0}+r} \frac{1}{r^{2}} \biggl\vert \int _{x}^{x_{0}+r} \frac{1}{y}\,dy \biggr\vert ^{2}\,dx\cdot \int _{x_{0}}^{x_{0}+r} \frac{1}{r^{2}} \biggl\vert \int _{x}^{x_{0}+r} \frac{1}{y}\,dy \biggr\vert ^{2} \vert x-x_{0} \vert ^{2}\,dx \\ &\quad \leq \frac{1}{r^{2}} \int _{x_{0}}^{x_{0}+r} \biggl(\log \frac{x_{0}+r}{x} \biggr)^{2}\,dx\cdot \int _{x_{0}}^{x_{0}+r} \biggl(\log \frac{x_{0}+r}{x} \biggr)^{2}\,dx \\ &\quad \leq 4 \end{aligned}

and

\begin{aligned} \int _{0}^{\infty }Qa(x)\,dx&= \int _{0}^{\infty } \int _{x}^{\infty }\frac{a(y)}{y}\,dy \,dx = \int _{0}^{\infty } \int _{0}^{y}\frac{a(y)}{y}\,dy \\ &= \int _{0}^{\infty }a(y)\,dx \,dy=0. \end{aligned}

Hence is a 1-molecule. Using Lemma 2.2, we have $$\|Qa\|_{H^{1}}\leq C$$, C depends only on the constant ε. This ends the proof. □

### Theorem 2.4

P and Q are bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$.

### Proof

It is sufficient to prove that

$$\Vert Pa \Vert _{L^{1}}\leqslant C, \qquad \Vert Qa \Vert _{L^{1}}\leqslant C,$$

where a is a $$(1,\infty )$$-atom and C is independent of a.

We first prove that $$\int _{0}^{\infty }|Pa(x)|\,dx\leq C$$, where a is a $$(1,\infty )$$-atom of $$H^{1}(\mathbb{R}^{+})$$ and C is independent of a. Suppose that $$\operatorname{supp}a\subset (x_{0},x_{0}+r]$$, $$x_{0}\geq 0$$, $$r>0$$. Using conditions (2) and (3) of a, we have

\begin{aligned} & \int _{0}^{\infty } \bigl\vert Pa(x) \bigr\vert \,dx \\ &\quad = \int _{0}^{x_{0}} \biggl\vert \frac{1}{x} \int _{0}^{x}a(y)\,dy \biggr\vert \,dx+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{0}^{x}a(y)\,dy \biggr\vert \,dx+ \int _{x_{0}+r}^{\infty } \biggl\vert \frac{1}{x} \int _{0}^{x} a(y)\,dy \biggr\vert \,dx \\ &\quad \leq \int _{x_{0}}^{x_{0}+r}\frac{1}{x} \int _{0}^{x} \bigl\vert a(y) \bigr\vert \,dy \,dx \\ &\quad \leq \int _{x_{0}}^{x_{0}+r}\frac{1}{x} \int _{0}^{x}\frac{1}{r}\,dy \,dx=1. \end{aligned}

For the operator Q, by Lemma 2.2, we can get that Q is bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$. Here we give a direct proof. Suppose that a is a $$(1,\infty )$$-atom of $$H^{1}(\mathbb{R}^{+})$$, $$\operatorname{supp}a\subset (x_{0},x_{0}+r]$$ and $$x_{0}>0$$. If $$x_{0}=0$$, the proof is more simple.

\begin{aligned} & \int _{0}^{\infty } \bigl\vert Qa(x) \bigr\vert \,dx \\ &\quad = \int _{0}^{x_{0}} \biggl\vert \int _{x}^{\infty }\frac{a(y)}{y}\,dy \biggr\vert \,dx+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \int _{x}^{\infty }\frac{a(y)}{y}\,dy \biggr\vert \,dx+ \int _{x_{0}+r}^{\infty } \biggl\vert \int _{x}^{\infty }\frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\quad \leq \int _{0}^{x_{0}} \biggl\vert \int _{x_{0}}^{x_{0}+r}\frac{a(y)}{y}\,dy \biggr\vert \,dx+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \int _{x}^{\infty }\frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\quad \leq x_{0} \int _{x_{0}}^{x_{0}+r}\frac{ \vert a(y) \vert }{y}\,dy+ \frac{1}{r} \int _{x_{0}}^{x_{0}+r} \int _{x}^{x_{0}+r}\frac{1}{y}\,dy \,dx \\ &\quad \leq 1+\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \int _{x_{0}}^{y}\frac{1}{y}\,dx \,dy \\ &\quad = 1+\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \biggl(1-\frac{x_{0}}{y} \biggr) \,dy\leq 2. \end{aligned}

The proof is complete. □

## The boundedness for commutators on Hardy spaces

In this section, we give some results on the boundedness for the commutators generated by P and Q with $$BMO$$ and $$CMO$$ functions on the Hardy spaces and other Hardy-type spaces.

Let $$b\in L_{loc}(\mathbb{R}^{+})$$, we say that $$b\in BMO(\mathbb{R}^{+})$$ if, for any interval $$I\subset \mathbb{R}^{+}$$,

$$\sup_{I}\frac{1}{ \vert I \vert } \int _{I} \bigl\vert b(y)-b_{I} \bigr\vert \,dy= \Vert b \Vert _{BMO(\mathbb{R}^{+})} < \infty ,$$

where $$b_{I}=\frac{1}{|I|}\int _{I}b(x)\,dx$$ and $$|I|$$ is the length of I. The definition of $$BMO$$ function and John–Nirenberg inequality for the $$BMO$$ function on spaces of homogeneous type imply that, for any $$p\geq 1$$, the following are true:

\begin{aligned} \Vert b \Vert _{BMO(\mathbb{R}^{+})} \approx &\sup_{I}\inf _{c\in \mathbb{C}} \biggl(\frac{1}{ \vert I \vert } \int _{I} \bigl\vert b(y)-c \bigr\vert ^{p}\,dy \biggr)^{1/p} \\ \approx &\sup_{I} \biggl(\frac{1}{ \vert I \vert } \int _{I} \bigl\vert b(y)-b_{I} \bigr\vert ^{p}\,dy \biggr)^{1/p}. \end{aligned}

### Remark 3.1

Let $$b\in BMO(\mathbb{R}^{+})$$, then $$[b,P]$$ is not bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$. In fact, if we take

$$b(x)= \textstyle\begin{cases} 1 , &0< x\leq 1, \\ -1 , &1< x\leqslant 2, \\ 0 , &2< x, \end{cases}\displaystyle \qquad f(x)= \textstyle\begin{cases} 1 , &0< x\leq 1, \\ -1 , &1< x\leq 2, \\ 0 , &2< x, \end{cases}$$

it is easy to see that $$b\in BMO(\mathbb{R}^{+})$$ for any $$1\leq p<\infty$$ and $$f\in H^{1}(\mathbb{R}^{+})$$, then

$$[b,P]f(x)=\frac{1}{x} \int _{0}^{x}\bigl[b(x)-b(y)\bigr]f(y)\,dy= \textstyle\begin{cases} 0 , &0< x\leq 1, \\ -\frac{2}{x} , &1< x. \end{cases}$$

We have

$$\int _{0}^{+\infty } \bigl\vert [b,P]f(x) \bigr\vert \,dx= \int _{1}^{+\infty }\frac{2}{x}\,dx=+ \infty .$$

So, the commutator $$[b,P]$$ is not bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$.

### Theorem 3.2

Let $$b\in BMO(\mathbb{R}^{+})$$, then $$[b,P]$$ is bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1,\infty }(\mathbb{R}^{+})$$. More precisely,

$$\bigl\Vert [b,P]f \bigr\Vert _{L^{1,\infty }(\mathbb{R}^{+})}\leq C \Vert b \Vert _{BMO(\mathbb{R}^{+})} \Vert f \Vert _{H^{1}(\mathbb{R}^{+})}.$$

### Proof

It is sufficient to prove that, for any $$\lambda >0$$,

$$\lambda \bigl\vert \bigl\{ x\in {\mathbb{R}^{+}}: \bigl\vert [b,P]a(x) \bigr\vert >\lambda \bigr\} \bigr\vert \leq C \Vert b \Vert _{BMO},$$

where a is a $$(1,\infty )$$-atom of $$H^{1}(\mathbb{R}^{+})$$, C is independent of a.

Let $$\operatorname{supp}(a)\subset (x_{0},x_{0}+r]$$ for $$r>0$$. Then

\begin{aligned}{} [b,P]a(x)&=\frac{1}{x} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)a(y)\,dy \\ &=\frac{1}{x} \int _{0}^{x}\bigl(b(x)-b_{(x_{0},x_{0}+r]}\bigr)a(y) \,dy+\frac{1}{x} \int _{0}^{x}\bigl(b_{(x_{0},x_{0}+r]}-b(y)\bigr)a(y) \,dy \\ &=\mathrm{J}_{1}+\mathrm{J}_{2}. \end{aligned}

So we have

$$\bigl\vert \bigl\{ x\in {\mathbb{R}^{+}}: \bigl\vert [b,P]a(x) \bigr\vert >\lambda \bigr\} \bigr\vert \leq \biggl\vert \biggl\{ x\in \mathbb{R}^{+}: \vert \mathrm{J}_{1} \vert \geq \frac{\lambda }{2}\biggr\} \biggr\vert + \biggl\vert \biggl\{ x\in \mathbb{R}^{+}: \vert \mathrm{J}_{2} \vert \geq \frac{\lambda }{2}\biggr\} \biggr\vert .$$

Using Chebyshev’s inequality yields

\begin{aligned} \frac{\lambda }{2} \biggl\vert \biggl\{ x\in \mathbb{R}^{+}: \vert \mathrm{J}_{1} \vert > \frac{\lambda }{2}\biggr\} \biggr\vert &\leq \int _{0}^{+\infty } \vert \mathrm{J}_{1} \vert \,dx \\ &= \int _{0}^{x_{0}+r} \vert \mathrm{J}_{1} \vert \,dx+ \int _{x_{0}+r}^{+\infty } \vert \mathrm{J}_{1} \vert \,dx \\ &=\mathrm{J}_{11}+\mathrm{J}_{12}. \end{aligned}

Since P is bounded on $$L^{p}(\mathbb{R})$$ for all $$1< p<\infty$$, using Hölder’s inequality, we obtain

\begin{aligned} \mathrm{J}_{11}&= \int _{x_{0}}^{x_{0}+r} \bigl\vert \bigl(b(x)-b_{(x_{0},x_{0}+r]} \bigr)Pa(x) \bigr\vert \,dx \\ &\leq \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-b_{(x_{0},x_{0}+r]} \bigr\vert ^{p}\,dx \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert Pa(x) \bigr\vert ^{p'}\,dx \biggr)^{1/p'} \\ &\leq p^{p'}r^{1/p} \biggl(\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-b_{(x_{0},x_{0}+r]} \bigr\vert ^{p}\,dx \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert a(x) \bigr\vert ^{p'}\,dx \biggr)^{1/p'} \\ &\leq C \Vert b \Vert _{BMO}, \end{aligned}

where we exploited condition (2) of the $$(1,\infty )$$-atom a.

If $$x\in (x_{0}+r,\infty )$$, using condition (3) of the $$(1,\infty )$$-atom a, we have

$$Pa(x)=\frac{1}{x} \int _{0}^{x}a(y)\,dy=\frac{1}{x} \int _{0}^{x_{0}+r}a(y)\,dy=0.$$

So, $$\mathrm{J}_{12}=0$$.

We next estimate the term $$|\{x\in \mathbb{R}^{+}: |\mathrm{J}_{2}|\geq \lambda /2\}|$$, which is divided into two parts: $$|\{x\in (0,x_{0}+r]: |\mathrm{J}_{2}|\geq \lambda /2\}|$$ and $$|\{x\in (x_{0}+r,\infty ): |\mathrm{J}_{2}|\geq \lambda /2\}|$$. Moreover,

\begin{aligned} &\frac{\lambda }{2} \biggl\vert \biggl\{ x\in (0,x_{0}+r]: \vert \mathrm{J}_{2} \vert \geq \frac{\lambda }{2}\biggr\} \biggr\vert \\ &\quad \leq \int _{0}^{x_{0}+r} \vert \mathrm{J}_{2} \vert \,dx \\ &\quad = \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{x_{0}}^{x} \bigl(b_{(x_{0},x_{0}+r]}-b(y) \bigr)a(y) \,dy \biggr\vert \,dx \\ &\quad \leq \int _{x_{0}}^{x_{0}+r}\frac{1}{x} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert ^{p}\,dy \biggr)^{\frac{1}{p}} \biggl( \int _{x_{0}}^{x} \bigl\vert a(y) \bigr\vert ^{p'}\,dy \biggr)^{ \frac{1}{p'}}\,dx \\ &\quad \leq r^{1/p-1} \int _{x_{0}}^{x_{0}+r}\frac{1}{(x-x_{0})^{1/p}} \biggl(r^{-1} \int _{x_{0}}^{x_{0}+r} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert ^{p}\,dy \biggr)^{\frac{1}{p}}\,dx \\ &\quad \leq C \Vert b \Vert _{BMO}. \end{aligned}

For the last term, we have

\begin{aligned} & \biggl\vert \biggl\{ x\in (x_{0}+r,\infty ): \vert \mathrm{J}_{2} \vert \geq \frac{\lambda }{2}\biggr\} \biggr\vert \\ &\quad = \biggl\vert \biggl\{ x\in (x_{0}+r,\infty ): \frac{1}{x} \biggl\vert \int _{x_{0}}^{x_{0}+r} \bigl(b_{(x_{0},x_{0}+r]}-b(y) \bigr)a(y)\,dy \biggr\vert \geq \frac{\lambda }{2} \biggr\} \biggr\vert \\ &\quad \leq \biggl\vert \biggl\{ x:x_{0}+r< x\leq \frac{2}{\lambda } \biggl\vert \int _{x_{0}}^{x_{0}+r} \bigl(b_{(x_{0},x_{0}+r]}-b(y) \bigr)a(y)\,dy \biggr\vert \biggr\} \biggr\vert \\ &\quad \leq \frac{2}{\lambda } \biggl\vert \int _{x_{0}}^{x_{0}+r} \bigl(b_{(x_{0},x_{0}+r]}-b(y) \bigr)a(y)\,dy \biggr\vert \\ &\quad \leq \frac{2}{\lambda } \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert ^{p}\,dy \biggr)^{\frac{1}{p}} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert a(y) \bigr\vert ^{p'}\,dy \biggr)^{\frac{1}{p'}} \\ &\quad \leq \frac{2}{\lambda }\cdot \Vert b \Vert _{BMO}. \end{aligned}

Combining all the above estimates, we complete the proof. □

### Theorem 3.3

Let $$b\in BMO(\mathbb{R}^{+})$$, then $$[b,Q]$$ is bounded from $$H^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$. More precisely, there exists a constant $$C>0$$ such that, for any $$f\in H^{1}(\mathbb{R}^{+})$$,

$$\bigl\Vert [b,Q]f \bigr\Vert _{L^{1}}\leq C \Vert b \Vert _{BMO} \Vert f \Vert _{H^{1}}.$$

### Proof

It is sufficient to prove that

$$\bigl\Vert [b,Q]a \bigr\Vert _{L^{1}}\leqslant C \Vert b \Vert _{BMO},$$

where a is a $$(1,\infty )$$-atom and C is independent of a.

Using the condition $$\operatorname{supp}(a)\subset (x_{0},x_{0}+r]$$, $$x_{0}\geq 0$$, $$r>0$$, we have

\begin{aligned} &\int _{0}^{+\infty } \bigl\vert [b,Q]a \bigr\vert \,dx\\ &\quad = \int _{0}^{x_{0}} \biggl\vert \int _{x}^{ \infty } \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\qquad {}+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \int _{x}^{+\infty } \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx+ \int _{x_{0}+r}^{\infty } \biggl\vert \int _{x}^{ \infty } \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\quad = \int _{0}^{x_{0}} \biggl\vert \int _{x}^{\infty } \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx + \int _{x_{0}}^{x_{0}+r} \biggl\vert \int _{x}^{+ \infty } \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\quad =\mathrm{K}_{1}+\mathrm{K}_{2}. \end{aligned}

Suppose that $$x_{0}>0$$. If $$x_{0}=0$$, the proof is similar and simple. For $$\mathrm{K}_{1}$$, we have the following estimate:

\begin{aligned} \mathrm{K}_{1}&\leq \int _{0}^{x_{0}} \biggl\vert \int _{x_{0}}^{x_{0}+r} \bigl(b(x)-c \bigr)\frac{a(y)}{y} \,dy \biggr\vert \,dx+ \int _{0}^{x_{0}} \biggl\vert \int _{x_{0}}^{x_{0}+r} \bigl(c-b(y) \bigr)\frac{a(y)}{y} \,dy \biggr\vert \,dx \\ &\leq \int _{0}^{x_{0}}\bigl|b(x)-c\bigr| \biggl\vert \int _{x_{0}}^{x_{0}+r} \frac{ \vert a(y) \vert }{y}\,dy \biggr\vert \,dx+r^{-1} \int _{x_{0}}^{x_{0}+r} \bigl\vert c-b(y) \bigr\vert \,dy \\ &\leq \frac{1}{x_{0}} \biggl( \int _{0}^{x_{0}} \bigl\vert b(x)-c \bigr\vert \,dx \biggr) \int _{x_{0}}^{x_{0}+r} \bigl\vert a (x) \bigr\vert \,dx+r^{-1} \int _{x_{0}}^{x_{0}+r} \bigl\vert c-b(y) \bigr\vert \,dy \\ &\leq \frac{1}{x_{0}} \int _{0}^{x_{0}} \bigl\vert b(x)-c \bigr\vert \,dx+r^{-1} \int _{x_{0}}^{x_{0}+r} \bigl\vert c-b(y) \bigr\vert \,dy, \end{aligned}

where $$c\in \mathbb{C}$$ is an arbitrary complex number. Hence

\begin{aligned} \mathrm{K}_{1}&\leq \inf_{c\in \mathbb{C}}\frac{1}{x_{0}} \int _{0}^{x_{0}} \bigl\vert b(x)-c \bigr\vert \,dx+ \inf_{c\in \mathbb{C} }r^{-1} \int _{x_{0}}^{x_{0}+r} \bigl\vert c-b(y) \bigr\vert \,dy \\ &\leq 2 \Vert b \Vert _{BMO}. \end{aligned}

For $$\mathrm{K}_{2}$$, using the $$L^{p}$$-boundedness of Q and condition (2) of a, we get

\begin{aligned} \mathrm{K}_{2}&= \int _{x_{0}}^{x_{0}+r} \biggl\vert \int _{x}^{x_{0}+r} \bigl(b(x)-b(y) \bigr) \frac{a(y)}{y}\,dy \biggr\vert \,dx \\ &\leq \int _{x_{0}}^{x_{0}+r} \bigl\vert \bigl(b(x)-b_{(x_{0},x_{0}+r]} \bigr)Qa(x) \bigr\vert \,dx \\ &\quad {} + \int _{x_{0}}^{x_{0}+r} \int _{x}^{x_{0}+r} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert \frac{ \vert a(y) \vert }{y}\,dy \,dx \\ &\leq \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-b_{(x_{0},x_{0}+r]} \bigr\vert ^{p}\,dy \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert Qa(x) \bigr\vert ^{p'}\,dy \biggr)^{1/{p'}} \\ &\quad {}+ \int _{x_{0}}^{x_{0}+r} \int _{x_{0}}^{y} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert \frac{ \vert a(y) \vert }{y}\,dx \,dy \\ &\leq \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-b_{(x_{0},x_{0}+r]} \bigr\vert ^{p}\,dy \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert a(x) \bigr\vert ^{p'}\,dy \biggr)^{1/{p'}} \\ &\quad {}+\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \bigl\vert b_{(x_{0},x_{0}+r]}-b(y) \bigr\vert \,dy \\ &\leq C \Vert b \Vert _{BMO}. \end{aligned}

Combining all the above estimates, we complete the proof. □

Perez in  introduced a kind of Hardy spaces associated with $$BMO$$ functions and proved the boundedness on these Hardy spaces for the commutators of singular integrals with $$BMO$$ functions. Now we define similar Hardy spaces on $$\mathbb{R}^{+}$$. Let $$b\in BMO(\mathbb{R}^{+})$$, the function a is called b-atom if there is an interval $$(x_{0},x_{0}+r]$$, $$x_{0}\geq 0$$, $$r>0$$, satisfying

\begin{aligned}& {\mathrm{(i)}}\quad \operatorname{supp}(a) \subset (x_{0},x_{0}+r], \qquad {\mathrm{(ii)}}\quad \Vert a \Vert _{L^{\infty }}\leqslant {r}^{-1}, \\& {\mathrm{(iii)}}\quad \int a(y)\,dy=0 \quad \text{and} \quad {\mathrm{(iv)}}\quad \int a(y)b(y)\,dy=0. \end{aligned}

The space $$H_{b}^{1}(\mathbb{R}^{+})$$ consists of the subspace of $$L^{1}(\mathbb{R}^{+})$$ of functions f, which can be written as $$f=\sum_{j}\lambda _{j}a_{j}$$, where $$a_{j}$$ are b-atoms and $$\lambda _{j}$$ are complex numbers with $$\sum_{j}|\lambda _{j}|<\infty$$. The $$H_{b}^{1}(\mathbb{R}^{+})$$ norm of f is defined by

$$\Vert f \Vert _{H_{b}^{1}(\mathbb{R}^{+})}=\inf \biggl\{ \sum _{j} \vert \lambda _{j} \vert \biggr\} ,$$

where the infimum has taken over all the decompositions of $$f=\sum_{j}\lambda _{j}a_{j}$$ as above.

### Theorem 3.4

Let $$b\in BMO(\mathbb{R}^{+})$$, then $$[b,P]$$ is bounded from $$H_{b}^{1}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$.

### Proof

We only need to prove that, for any b-atom a, we have

$$\bigl\Vert [b,P]a \bigr\Vert _{L^{1}}\leqslant C \Vert b \Vert _{BMO}.$$

Using conditions (iii), (iv) and $$\operatorname{supp}(a)\subset (x_{0},x_{0}+r]$$, $$x_{0}\geq 0$$, $$r>0$$, we have

\begin{aligned} &\int _{0}^{+\infty } \bigl\vert [b,P]a \bigr\vert \,dx\\ &\quad = \int _{0}^{x_{0}} \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)a(y)\,dy \biggr\vert \,dx \\ &\qquad {}+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)a(y)\,dy \biggr\vert \,dx+ \int _{x_{0}+r}^{\infty } \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)a(y)\,dy \biggr\vert \,dx \\ &\quad = \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)a(y)\,dy \biggr\vert \,dx \\ &\quad \leq \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(b(x)-c\bigr)a(y)\,dy \biggr\vert \,dx+ \int _{x_{0}}^{x_{0}+r} \biggl\vert \frac{1}{x} \int _{0}^{x}\bigl(c-b(y)\bigr)a(y)\,dy \biggr\vert \,dx \\ &\quad =\mathrm{L}_{1}+\mathrm{L}_{2}, \end{aligned}

where $$c\in \mathbb{C}$$ is an arbitrary complex number.

For $$\mathrm{L}_{1}$$, let $$p>1$$, using Hölder’s inequality, the $$L^{p'}$$ boundedness of the Hardy operator P, and conditions (ii) of b-atom a, we have

\begin{aligned} \mathrm{L}_{1} &\leq \int _{x_{0}}^{x_{0}+r} \bigl\vert \bigl(b(x)-c\bigr)Pa(x) \bigr\vert \,dx \\ &\leq \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-c \bigr\vert ^{p}\,dx \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert Pa(x) \bigr\vert ^{p'}\,dx \biggr)^{1/p'} \\ &\leq p' \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-c \bigr\vert ^{p}\,dx \biggr)^{1/p} \biggl( \int _{x_{0}}^{x_{0}+r} \bigl\vert a(x) \bigr\vert ^{p'}\,dx \biggr)^{1/p'} \\ &\leq p' \biggl(\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-c \bigr\vert ^{p}\,dx \biggr)^{1/p}. \end{aligned}

For $$\mathrm{L}_{2}$$, using Hölder’s inequality and conditions (ii) of b-atom a, we obtain

\begin{aligned} \mathrm{L}_{2} &\leq \int _{x_{0}}^{x_{0}+r} \biggl(\frac{1}{x} \int _{0}^{x} \bigl\vert b(y)-c \bigr\vert ^{p}\,dy \biggr)^{1/p} \biggl(\frac{1}{x} \int _{x_{0}}^{x} \bigl\vert a(y) \bigr\vert ^{p'}\,dy \biggr)^{1/p'}\,dx \\ &\leq \frac{1}{r} \int _{x_{0}}^{x_{0}+r} \biggl(\frac{1}{x} \int _{0}^{x} \bigl\vert b(y)-c \bigr\vert ^{p}\,dy \biggr)^{1/p}\,dx. \end{aligned}

Hence

\begin{aligned} \mathrm{L}_{1}+\mathrm{L}_{2}&\leq p'\inf _{c\in \mathbb{C}} \biggl(\frac{1}{r} \int _{x_{0}}^{x_{0}+r} \bigl\vert b(x)-c \bigr\vert ^{p}\,dx \biggr)^{1/p}+\frac{1}{r} \int _{x_{0}}^{x_{0}+r}\inf_{c\in \mathbb{C}} \biggl( \frac{1}{x} \int _{0}^{x} \bigl\vert b(y)-c \bigr\vert ^{p}\,dy \biggr)^{1/p}\,dx \\ &\leq C \Vert b \Vert _{BMO}. \end{aligned}

This ends the proof. □

Let $$1\leq p<\infty$$, we say that $$b\in CMO^{p}(\mathbb{R}^{+})$$ if

$$\Vert b \Vert _{CMO^{p}}=\sup_{r>0} \biggl( \frac{1}{r} \int _{0}^{r} \bigl\vert b(y)-b_{(0,r]} \bigr\vert ^{p}\,dy \biggr)^{1/p}< \infty .$$

By the definition of $$CMO^{p}$$ function, for any $$p\geq 1$$, we have

\begin{aligned} \Vert b \Vert _{CMO^{p}} \approx &\sup_{r>0}\inf _{c\in \mathbb{C}} \biggl( \frac{1}{r} \int _{0}^{r} \bigl\vert b(y)-c \bigr\vert ^{p}\,dy \biggr)^{1/p}. \end{aligned}

It is easy to see $$BMO(\mathbb{R}^{+}) \subsetneqq CMO^{p}(\mathbb{R}^{+})$$, where $$1\leq p<\infty$$. $$CMO^{q}(\mathbb{R}^{+}) \subsetneqq CMO^{p}(\mathbb{R}^{+})$$ for $$1\leq p< q<\infty$$.

Let $$1< p\leq \infty$$, a function a is called central $$(1,p)$$-atom, if it satisfies the following conditions: (1) $$\operatorname{supp}(a)\subset (0,r]$$, where $$r>0$$; (2) $$\| a\|_{L^{p}} \leqslant r^{1/p-1}$$; (3) $$\int a(x)\,dx=0$$.

The central atomic Hardy space $$H_{c}^{1.p}(\mathbb{R}^{+})$$ is defined by

$$H_{c}^{1,p}\bigl(\mathbb{R}^{+}\bigr)=\biggl\{ f\in L^{1}\bigl(\mathbb{R}^{+}\bigr): f(x)=\sum _{k} \lambda _{k}a_{k}(x), \text{ and } \sum_{k} \vert \lambda _{k} \vert < \infty \biggr\} ,$$

where each $$a_{k}$$ is a central $$(1,p)$$-atom. We set the $$H_{c}^{1,p}(\mathbb{R}^{+})$$ norm of f by

$$\Vert f \Vert _{H_{c}^{1,p}}=\inf \Biggl\{ \sum _{k}^{\infty }\vert \lambda _{k}\vert \Biggr\} ,$$

where the infimum is taken over all the decompositions of $$f=\sum_{k} \lambda _{k}a_{k}$$ as above.

Similar to the proof in García-Cuerva , we can obtain that the dual space of $$H_{c}^{1.p}(\mathbb{R}^{+})$$ is $$CMO^{p'}(\mathbb{R}^{+})$$ for $$1< p\leq \infty$$.

Taking the same example in Remark 3.1, we can show that, for $$1< p<\infty$$ and $$b\in CMO^{p}(\mathbb{R}^{+})$$, the commutator $$[b,P]$$ is not bounded from $$H_{c}^{1,p'}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$. Similar to the proof of Theorem 3.2 and Theorem 3.3, we can prove the following theorem and omit the details here.

### Theorem 3.5

Let $$1< p<\infty$$ and $$b\in CMO^{p}(\mathbb{R}^{+})$$, then $$[b,P]$$ is bounded from $$H_{c}^{1,p'}(\mathbb{R}^{+})$$ to $$L^{1,\infty }(\mathbb{R}^{+})$$ and $$[b,Q]$$ bounded from $$H_{c}^{1,p'}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$.

Let $$p\geq 1$$ and $$b\in CMO^{p}(\mathbb{R}^{+})$$, the function a is called central $$(1,p,b)$$-atom if there exists $$r>0$$, satisfying $${\mathrm{(i).}}$$ $$\operatorname{supp}(a) \subset (0,r]$$, $${\mathrm{(ii).}}$$ $$\|a\|_{L^{p}} \leq {r}^{1/p-1}$$, $${\mathrm{(iii).}}$$ $$\int a(y)\,dy=0$$, $${\mathrm{(iv).}}$$ $$\int a(y)b(y)\,dy=0$$.

The space $$H_{b,c}^{1,p}(\mathbb{R}^{+})$$ consists of the subspace of $$L^{1}(\mathbb{R}^{+})$$ of functions f, which can be written as $$f=\sum_{j}\lambda _{j}a_{j}$$, where $$a_{j}$$ are central $$(1,p,b)$$-atoms and $$\lambda _{j}$$ are complex numbers with $$\sum_{j}|\lambda _{j}|<\infty$$. The $$H_{b,c}^{1,p}(\mathbb{R}^{+})$$ norm of f is defined by

$$\Vert f \Vert _{H_{b,c}^{1,p}}=\inf \biggl\{ \sum _{j} \vert \lambda _{j} \vert \biggr\} ,$$

where the infimum has taken over all the decompositions of $$f=\sum_{j}\lambda _{j}a_{j}$$ as above.

It is easy to see $$H_{b,c}^{1,p}(\mathbb{R}^{+})\subsetneqq H_{c}^{1,p}(\mathbb{R}^{+})$$. Similar to the proof of Theorem 3.4, we can obtain the following and omit the details here.

### Theorem 3.6

Let $$1< p<\infty$$ and $$b\in CMO^{p}(\mathbb{R}^{+})$$, then $$[b,P]$$ is bounded from $$H_{b,c}^{1,p'}(\mathbb{R}^{+})$$ to $$L^{1}(\mathbb{R}^{+})$$.

Not applicable.

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## Acknowledgements

The authors express deep thanks to the referee for valuable comments and suggestions.

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This work was supported by the doctor’s fund of Hebei Normal University (No. L2018B32).

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Correspondence to Wenming Li.

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