Weighted inequalities for fractional Hardy operators and commutators

Abstract

In this paper, we introduce a fractional maximal operators $N_{\alpha }$ on $(0,\infty )$ associated to the fractional Hardy operator $P_{\alpha }$ and its dual $Q_{\alpha }, 0\leq \alpha <1$, and obtain some characterizations for the one-weight and two-weight inequalities for $N_{\alpha }$. We also give some $A_{p}$ type sufficient conditions for the two-weight inequalities for the fractional Hardy operators, the dual operators and the commutators of the fractional Hardy operators with CMO functions.

1 Introduction

Let $P_{\alpha }$ and $Q_{\alpha }$ be the fractional Hardy operator and its adjoint on $(0,\infty )$,

$$\begin{aligned} P_{\alpha }f(x)=\frac{1}{x^{1-\alpha }} \int _{0}^{x}f(y)\,dy,\qquad Q_{\alpha }f(x)= \int _{x}^{\infty }\frac{f(y)}{y^{1-\alpha }}\,dy, \end{aligned}$$

where $0\leq \alpha <1$.

When $\alpha =0$, we denote $P_{0}$ as P and $Q_{0}$ as Q. P and Q are the Hardy operator and its adjoint. Hardy [6, 7] established the Hardy integral inequalities

$$ \int _{0}^{\infty } \bigl\vert Pf(y) \bigr\vert ^{p}\,dy\leq p^{\prime p} \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}\,dy,\quad p>1, $$

and

$$ \int _{0}^{\infty } \bigl\vert Qf(y) \bigr\vert ^{p}\,dy\leq p^{p} \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}\,dy,\quad p>1, $$

where $p'=p/(p-1)$.

The two inequalities above go by the name of Hardy’s integral inequalities. For earlier development of this kind of inequality and many applications in analysis, see [2, 8, 13].

The fractional Calderón operator $S_{\alpha }$ is defined as $S_{\alpha }=P_{\alpha }+Q_{\alpha }$. When $\alpha =0$, $S_{0}$ is denoted S, and S is the Calderón operator, which plays a significant role in the theory of real interpolation; see [1]. Next, we introduce the fractional maximal operators $N_{\alpha }$ related to the fractional Calderón operator. Given a measurable function f on $(0,\infty )$, the fractional maximal operator $N_{\alpha }$ is defined as

$$\begin{aligned} N_{\alpha }f(x)=\sup_{t>x}\frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy,\quad x>0. \end{aligned}$$

Notice that $N_{\alpha }f$ is a decreasing function, and that $|P_{\alpha }f|\leq N_{\alpha }f \leq S_{\alpha }(|f|)$, for any f. Indeed, for any f and $t>x$ we have

$$\begin{aligned} &\bigl\vert P_{\alpha }f(x) \bigr\vert \leq \sup_{t>x} \frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy \leq N_{\alpha }f(x), \\ &\frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy \leq \frac{1}{x^{1-\alpha }} \int _{0}^{x} \bigl\vert f(y) \bigr\vert \,dy+ \int _{x}^{t}\frac{ \vert f(y) \vert }{y^{1-\alpha }}\,dy \leq S_{\alpha }f(x). \end{aligned}$$

Notice that $N_{\alpha }f\leq S_{\alpha }f$ for nonnegative f.

For $\alpha =0$, $N_{0}$ is denoted as N. Duoandikoetxea, Martin-Reyes and Ombrosi in [3] introduced the maximal operator N related to the Calderón operator and studied the weighted inequalities for N. Li, Zhang and Xue [9] obtained some two-weight inequalities for N.

For $1< p\leq q<\infty$, we say a weight w satisfies the $A_{p,q,0}$ condition, denoted as $w\in A_{p,q,0}$, if

$$ [w]_{p,q,0}=\sup_{t>0} \biggl(\frac{1}{t} \int _{0}^{t}w(y)^{q}\,dy \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}w(y)^{-p'}\,dy \biggr)^{1/p'}< \infty. $$

For $p=1< q<\infty$, we write $A_{1,q,0}$ for the class of nonnegative functions w such that

$$ [w]_{1,q,0}=\sup_{t>0} \biggl(\frac{1}{t} \int _{0}^{t}w(y)^{q}\,dy \biggr)^{1/q} \biggl(\mathrm{ess}\sup_{x\in (0,t)} \frac{1}{w(y)} \biggr)< \infty. $$

For $1< p\leq q<\infty$, we say a pair of weights $(u,v)$ satisfies the two-weight $A_{p,q}$ condition, denoted $(u,v)\in A_{p,q}$, if

$$\begin{aligned}{} [u,v]_{p,q}=\sup_{t>0} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{q}\,dy \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-p'}\,dy \biggr)^{1/p'}< \infty. \end{aligned}$$

For $p=1$ we write $(u,v)\in A_{1,q}$, if

$$\begin{aligned}{} [u,v]_{1,q}=\sup_{t>0} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{q}\,dy \biggr)^{1/q} \biggl(\mathrm{ess}\sup_{x\in (0,t)} \frac{1}{v(y)} \biggr)< \infty. \end{aligned}$$

Let $1\leq p<\infty$, we say b is a one-side dyadic $\mathrm{CMO}^{p}$ function, if

$$\begin{aligned} \sup_{j\in \mathbb{Z}} \biggl(\frac{1}{2^{j}} \int _{0}^{2^{j}} \bigl\vert b(y)-b_{(0,2^{j}]} \bigr\vert ^{p}\,dy \biggr)^{1/p}= \Vert b \Vert _{\mathrm{CMO}^{p}}< \infty, \end{aligned}$$

where $b_{(0,2^{j}]}=\frac{1}{2^{j}}\int _{0}^{2^{j}}f(x)\,dx$, we then say that $b\in \mathrm{CMO}^{p}$.

It is easy to see $\mathrm{BMO}(0,\infty )$ ⫋ $\mathrm{CMO}^{p}$, where $1\leq p<\infty$. $\mathrm{CMO}^{q}$ ⫋ $\mathrm{CMO}^{p}$ for $1\leq p< q< \infty$.

Let b be a locally integrable function on $(0,\infty )$, we define the commutators of the fractional Calderón operator $S_{\alpha }$ with b as $S_{\alpha }^{b}=P_{\alpha }^{b}+Q_{\alpha }^{b}$, where

$$\begin{aligned} P_{\alpha }^{b}f(x)=\frac{1}{x^{1-\alpha }} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)f(y)\,dy,\qquad Q_{\alpha }^{b}f(x)= \int _{x}^{\infty }\frac{(b(x)-b(y))f(y)}{y ^{1-\alpha }}\,dy. \end{aligned}$$

For $\alpha =0$, $P_{\alpha }^{b}$ and $Q_{\alpha }^{b}$ be denoted as $P^{b}$ and $Q^{b}$, respectively. Long and Wang [10] established Hardy’s integral inequalities for commutators generated by P and Q with one-sided dyadic CMO functions, and for $0<\alpha <1$ proved that the two commutators of $P_{\alpha }^{b}$ and $Q_{\alpha }^{b}$ are bounded from $L^{p}(\mathbb{R_{+}})$ to $L^{q}(\mathbb{R_{+}})$ with the function b in one-side dyadic $\mathrm{CMO}^{\max {(q,p')}}$, where $1< p< q<\infty$, $\frac{1}{q}=\frac{1}{p}-\alpha$. Fu [4], Zheng and Fu [14] showed some boundedness properties for $P_{\alpha }^{b}$ and $Q_{\alpha }^{b}$, respectively. Li, Zhang and Xue [9] obtained some $A_{p}$ type sufficient conditions such that the two-weight inequalities are true for P, Q, $P^{b}$ and $Q^{b}$. The commutator estimates for Hardy operator are actual in view of applications in PDE; see Mamedov and Brahimov [11].

In this paper, we discuss the one-weight and two-weight inequalities of operators $N_{\alpha }$, $P_{\alpha }$, $Q_{\alpha }$, $P_{\alpha } ^{b}$, $Q_{\alpha }^{b}$, and get the following results.

Theorem 1.1

For $1\leq p<\frac{1}{\alpha }$, $0\le \alpha <1$, $\frac{1}{q}= \frac{1}{p}-\alpha$, $N_{\alpha }$ is bounded from $L^{p}(w^{p})$ to $L^{q,\infty }(w^{q})$ if and only if $w\in A_{p,q,0}$. More precisely,

$$ \sup_{\lambda >0}\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w(y)^{q}\,dy \biggr)^{1/q}\leq [w]_{p,q,0} \Vert f \Vert _{L^{p}(w^{p})}. $$

(1)

For $1< p<\frac{1}{\alpha }$, $0\leq \alpha <1$, $\frac{1}{q}= \frac{1}{p}-\alpha$, $N_{\alpha }$ is bounded from $L^{p}(w^{p})$ to $L^{q}(w^{q})$ if and only if $w\in A_{p,q,0}$. Moreover,

$$ \Vert N_{\alpha }f \Vert _{L^{q}(w^{q})}\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \Vert f \Vert _{L^{p}(w^{p})}. $$

(2)

Theorem 1.2

For $1\leq p<\frac{1}{\alpha }$, $0\leq \alpha <1$, $\frac{1}{q}= \frac{1}{p}-\alpha$, $N_{\alpha }$ is bounded from $L^{p}(v^{p})$ to $L^{q,\infty }(u^{q})$ if and only if $(u,v)\in A_{p,q}$. More precisely,

$$\begin{aligned} \sup_{\lambda >0}\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} u(y)^{q}\,dy \biggr)^{1/q}\leq [u,v]_{p,q} \Vert f \Vert _{L^{p}(v^{p})}. \end{aligned}$$

Theorem 1.3

For $1< p<\infty$, $0\leq \alpha <1$, $0< q<\infty$, $N_{\alpha }$ is bounded from $L^{p}(v^{p})$ to $L^{q}(u^{q})$, if and only if, for any $t>0$, $(u,v)$ satisfies

$$ \biggl( \int _{0}^{t}\bigl[N_{\alpha } \bigl(v^{-p'}\chi _{(0,t)}\bigr) (y)\bigr]^{q}u(y)^{q} \,dy \biggr)^{1/q}\leq C \biggl( \int _{0}^{t}v(y)^{-p'}\,dy \biggr)^{1/p}< \infty. $$

(3)

But for $1< p<\frac{1}{\alpha }$, $0\leq \alpha <1$, $\frac{1}{q}= \frac{1}{p}-\alpha$, $N_{\alpha }$ is not bounded from $L^{p}(v^{p})$ to $L^{q}(u^{q})$ if $(u,v)\in A_{p,q}$, the proof is similar to the case for the Hardy–Littlewood maximal function on $\mathbb{R}^{n}$; see [5]. Notice that $|P_{\alpha }f|\leq N_{\alpha }f \leq S_{ \alpha }(|f|)$, by Theorem 1.1, we see that $(u,v)\in A_{p,q}$, is necessary but not sufficient for $S_{\alpha }$ to be bounded from $L^{p}(v^{p})$ to $L^{q}(u^{q})$.

Theorem 1.4

Let $1< p< q<\infty$, $0\leq \alpha <1$.

• (a) Let $(u,v)$ be a pair of weights for which there exists $r>1$ such that, for every $t>0$,

  $$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{q}\,dy \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-rp'}\,dy \biggr)^{1/rp'}\leq C< \infty. $$

  (4)

  Then

  $$ \biggl( \int _{0}^{\infty } \bigl\vert P_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$

  (5)

• (b) Let $(u,v)$ be a pair of weights for which there exists $r>1$ such that, for every $t>0$,

  $$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{rq}\,dy \biggr)^{1/rq} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-p'}\,dy \biggr)^{1/p'}\leq C< \infty. $$

  (6)

  Then

  $$ \biggl( \int _{0}^{\infty } \bigl\vert Q_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$

  (7)

Theorem 1.5

Let $1< p< q<\infty$, $0\leq \alpha <1$, $b\in CMO^{r'\max {\{q,p'\}}}$, and $(u,v)$ be a pair of weights for which there exists $r>1$ such that, for every $t>0$,

$$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{rq}\,dy \biggr)^{1/rq} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-rp'}\,dy \biggr)^{1/rp'}\leq C< \infty. $$

(8)

Then

$$ \biggl( \int _{0}^{\infty } \bigl\vert P_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p} $$

(9)

and

$$ \biggl( \int _{0}^{\infty } \bigl\vert Q_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$

(10)

2 The proofs of Theorem 1.1, Theorem 1.2 and Theorem 1.3

In order to prove the theorems, we need the fractional maximal operator $N_{\alpha }^{g}$ associated to a fixed positive measurable function g. For $0\le \alpha <1$, we defined $N_{\alpha }^{g}$ as

$$ N_{\alpha }^{g}f(x)=\sup_{t>x} \frac{\int _{0}^{t} \vert f(y) \vert g(y)\,dy}{( \int _{0}^{t}g(y)\,dy)^{1-\alpha }}. $$

Mamedov and Zeren [12] obtained the two-weight inequalities for this maximal operator in the Lebesgue spaces with variable exponent. When $\alpha =0$, we denote $N_{\alpha }^{g}$ as $N^{g}$. Duoandikoetxea, Martin-Reyes and Ombrosi in [3] obtained the following lemma for $N^{g}$.

Lemma 2.1

Let $0\le \alpha <1$ and g be a nonnegative measurable function such that $0< g(0,t)=\int _{0}^{t}g(x)\,dx<\infty$, for all $t>0$.

• (i) $N_{\alpha }^{g}$ is of weak type $(1,\frac{1}{1-\alpha })$ with respect to the measure $g(t)\,dt$. Actually,

  $$ \biggl( \int _{\{x:N_{\alpha }^{g}f(x)>\lambda \}} g(y)\,dy \biggr)^{1- \alpha }\leq \frac{1}{\lambda } \int _{\{x:N_{\alpha }^{g}f(x)>\lambda \}} \bigl\vert f(y) \bigr\vert g(y) \,dy, $$

  (11)

  for all $\lambda >0$ and all measurable functions f.

• (ii) $N_{\alpha }^{g}$ is of strong type $(p,q)$, $1< p<\frac{1}{\alpha }$, $\frac{1}{q}=\frac{1}{p}-\alpha$, with respect to the measure $g(t)\,dt$. More precisely,

  $$\begin{aligned} \biggl( \int _{0}^{\infty } \bigl\vert N_{\alpha }^{g}f(y) \bigr\vert ^{q}g(y)\,dy \biggr)^{1/q} \leq C(p,\alpha ) \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}g(y)\,dy \biggr)^{1/p}, \end{aligned}$$

  in which the constant $C(p,\alpha )$ independent of f and g.

Proof

By standard interpolation arguments, it suffices to prove (i) since by the Hölder inequality, we have

$$\begin{aligned} \frac{\int _{0}^{t} \vert f(x) \vert g(x)\,dx}{(\int _{0}^{t}g(x)\,dx)^{1-\alpha }} & \leq \frac{(\int _{0}^{t} \vert f(x) \vert ^{1/\alpha }g(x)\,dx)^{\alpha }(\int _{0} ^{t}g(x)\,dx)^{1-\alpha }}{(\int _{0}^{t}g(x)\,dx)^{1-\alpha }} \\ &= \biggl( \int _{0}^{t} \bigl\vert f(x) \bigr\vert ^{1/\alpha }g(x)\,dx \biggr)^{\alpha } \\ &\leq \Vert f \Vert _{L^{1/\alpha }(g)}. \end{aligned}$$

Thus we obtain $\|N_{\alpha }^{g}f\|_{\infty }\leq \|f\|_{L^{1/\alpha }(g)}$.

Observe that $N_{\alpha }^{g}f$ is decreasing and continuous. Therefore, if $\{t:N_{\alpha }^{g}(f)(t)>\lambda \}$ is not empty, then it is either a bounded interval $(0,d)$ or all of $(0,\infty )$. In the first case

$$ \lambda \biggl( \int _{0}^{d}g(x)\,dx \biggr)^{1-\alpha }= \int _{0}^{d} \bigl\vert f(x) \bigr\vert g(x) \,dx, $$

(12)

whereas in the second case we have

$$\begin{aligned} \lambda \biggl( \int _{0}^{\infty }g(x)\,dx \biggr)^{1-\alpha }\leq \int _{0} ^{\infty } \bigl\vert f(x) \bigr\vert g(x) \,dx. \end{aligned}$$

Thus we obtain (11). Notice that if $g(0,\infty )=\int _{0}^{\infty }g(x)\,dx=+\infty$ and f is integrable with respect to g, only the first case is possible and the equality holds. □

Proof of Theorem 1.1

Let us prove first the necessity of $A_{p,q,0}$ for the weak-type inequality.

(i) For $1< p<\frac{1}{\alpha }$, let $E_{k}=\{x:w(x)>1/k \}$ and $w_{k}=w\chi _{E_{k}}$. Take $f=w_{k}^{-p'}\chi _{(0,t)}$. Then

$$ N_{\alpha }f(x)\geq \frac{1}{t^{1-\alpha }} \int _{0}^{t}w_{k}^{-p'}, $$

(13)

for $0< x< t$. Thus, $(0,t)\subset \{x:N_{\alpha }f(x)>\lambda \}$ taking as λ the right-hand side of (13). By $N_{\alpha }$ is bounded from $L^{p}(w^{p})$ to $L^{q,\infty }(w^{q})$, we have

$$\begin{aligned} \biggl( \int _{0}^{t}w^{q} \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}w_{k} ^{-p'} \biggr) &=\frac{1}{t^{\alpha }} \biggl( \int _{0}^{t}w^{q} \biggr)^{1/q} \cdot \lambda \\ &\leq \frac{1}{t^{\alpha }} \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w^{q} \biggr)^{1/q}\cdot \lambda \\ &\leq \frac{C}{t^{\alpha }} \biggl( \int _{0}^{t}w_{k}^{-p'p}w^{p} \biggr)^{1/p} \\ &=\frac{C}{t^{\alpha }} \biggl( \int _{0}^{t}w_{k}^{-p'} \biggr)^{1/p}. \end{aligned}$$

Thus we obtain

$$\begin{aligned} \biggl(\frac{1}{t} \int _{0}^{t}w^{q} \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0} ^{t}w_{k}^{-p'} \biggr)^{1/p'}\leq C. \end{aligned}$$

Letting k tend to infinity, $w\in A_{p,q,0}$ follows.

(ii) For $p=1$, let $(0,s)$ be any interval, for any interval $(t_{1},t_{2})\subset (0,s)$. Taking $f=\chi _{(t_{1},t_{2})}$. Then

$$ N_{\alpha }f(x)\geq \frac{1}{s^{1-\alpha }} \int _{0}^{s} \vert \chi _{(t_{1},t_{2})} \vert \,dy=\frac{t_{2}-t_{1}}{s^{1-\alpha }}, $$

(14)

for $0< x< s$. Thus, $(0,s)\subset \{x:N_{\alpha }f(x)>\lambda \}$ taking as λ the right-hand side of (14). By $N_{\alpha }$ is bounded from $L^{p}(w^{p})$ to $L^{q,\infty }(w^{q})$, we have

$$\begin{aligned} \lambda \biggl( \int _{0}^{s}w^{q} \biggr)^{1/q} \leq \lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w^{q} \biggr)^{1/q} \leq C \int _{t_{1}}^{t_{2}}w. \end{aligned}$$

Thus we have

$$\begin{aligned} \biggl(\frac{1}{s} \int _{0}^{s}w^{q} \biggr)^{1/q} \leq \frac{C}{t_{2}-t_{1}} \int _{t_{1}}^{t_{2}}w. \end{aligned}$$

By the Lebesgue differentiation theorem, for any $y\in (0,s)$, we have

$$\begin{aligned} \biggl(\frac{1}{s} \int _{0}^{s}w^{q} \biggr)^{1/q} \leq Cw(y). \end{aligned}$$

Thus, $w\in A_{1,q,0}$ follows.

For the sufficient, arguing as in the proof of the Lemma 2.1, we have (12) with $g\equiv 1$, that is,

$$\begin{aligned} d^{1-\alpha }\lambda = \int _{0}^{d} \bigl\vert f(y) \bigr\vert \,dy. \end{aligned}$$

Then

$$\begin{aligned} &\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w(y)^{q}\,dy \biggr)^{1/q}\\ &\quad=\lambda \biggl( \int _{0}^{d}w(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq \frac{1}{d^{1-\alpha }} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{0}^{d}w(y)^{-p'}\,dy \biggr)^{1/p'} \biggl( \int _{0}^{d}w(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq [w]_{p,q,0} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{1/p}, \end{aligned}$$

and (1) follows.

Now we prove (2). If $w\in A_{p,q,0}, 0< x< t$, $\sigma =w^{-p'}$ and $\delta =w^{q}$, we have

$$\begin{aligned} \biggl(\frac{1}{t^{1-\alpha }} \int _{0}^{t}f(y)\,dy \biggr)^{\frac{1}{(1- \alpha )p'}} &\leq [w]_{p,q,0} \biggl(\frac{\int _{0}^{t}f(y)\,dy}{(\int _{0}^{t}\sigma (y)\,dy)^{(1-\alpha )}} \biggr)^{\frac{1}{(1-\alpha )p'}} \frac{( \int _{0}^{t}\,dy)^{1/q}}{(\int _{0}^{t}\delta (y)\,dy)^{1/q}} \\ &\leq [w]_{p,q,0}\frac{(\int _{0}^{t} \vert N_{\alpha }^{\sigma }(f\sigma ^{-1})(y) \vert ^{\frac{q}{(1-\alpha )p'}}\,dy)^{1/q}}{(\int _{0}^{t}\delta (y)\,dy)^{1/q}} \\ &\leq [w]_{p,q,0} \bigl\vert N^{\delta }\bigl(\delta ^{-1} \bigl\vert N_{\alpha }^{\sigma }\bigl(f \sigma ^{-1}\bigr) \bigr\vert ^{\frac{q}{(1-\alpha )p'}}\bigr) (x) \bigr\vert ^{1/q}. \end{aligned}$$

Therefore,

$$ \bigl\vert N_{\alpha }f(x) \bigr\vert ^{q}\leq [w]_{p,q,0}^{(1-\alpha )p'q} \bigl\vert N^{\delta }\bigl(\delta ^{-1} \bigl\vert N_{\alpha }^{\sigma }\bigl(f\sigma ^{-1}\bigr) \bigr\vert ^{\frac{q}{(1- \alpha )p'}}\bigr) (x) \bigr\vert ^{(1-\alpha )p'}. $$

Consequently, using Lemma 2.1,

$$\begin{aligned} \int _{0}^{\infty } \bigl\vert N_{\alpha }f(y) \bigr\vert ^{q}w(y)^{q}\,dy &\leq C[w]_{p,q,0} ^{(1-\alpha )p'q} \int _{0}^{\infty } \bigl\vert N_{\alpha }^{\sigma } \bigl(f\sigma ^{-1}\bigr) (y) \bigr\vert ^{q} \sigma (y)\,dy \\ &\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \sigma ^{-1}(y) \bigr\vert ^{p}\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

This ends the proof of the theorem. □

Proof of Theorem 1.2

For $1\leq p<\frac{1}{\alpha }$, the proof for the necessity of $A_{p,q}$ for the weak-type inequality is standard and similar to the proof of Theorem 1.1, we omitted here. For the sufficient, we observe that $N_{\alpha }f$ is decreasing and continuous. Therefore, if $\{x:N_{\alpha }f(x)>\lambda \}$ is not empty, then it is a bounded interval $(0,d)$, thus

$$\begin{aligned} d^{1-\alpha }\lambda = \int _{0}^{d} \bigl\vert f(y) \bigr\vert \,dy. \end{aligned}$$

Then

$$\begin{aligned} &\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} u(y)^{q}\,dy \biggr)^{1/q}\\ &\quad=\lambda \biggl( \int _{0}^{d}u(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq \frac{1}{d^{1-\alpha }} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{0}^{d}v(y)^{-p'}\,dy \biggr)^{1/p'} \biggl( \int _{0}^{d}u(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq [u,v]_{p,q} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p}. \end{aligned}$$

This ends the proof. □

Proof of Theorem 1.3

Denote $\sigma =v^{-p'}$. The necessity of (3) follows by a standard argument if we substitute $f=\sigma \chi _{(0,t)}$ into $\|N_{\alpha }f\|_{L^{q}(u ^{q})}\leq C\|f\|_{L^{p}(v^{p})}$.

To show that (3) is sufficient, fix a bounded nonnegative function f with compact support. Since $N_{\alpha }f$ is decreasing and continuous, for each $k\in \mathbb{Z}$, if $\{x\in (0,\infty ):N _{\alpha }f(x)>2^{k}\}$ is not empty, then there exists $d_{k}$ such that $\{x\in (0,\infty ):N_{\alpha }f(x)>2^{k}\}=(0,d_{k})$. Thus $0< d_{k+1}\leq d_{k}$, $\varOmega _{k}=\{x\in (0,\infty ):2^{k}< N_{\alpha }f(x)\leq 2^{k+1}\}=[d_{k+1},d_{k})$ and

$$\begin{aligned} 2^{k}\,d_{k}^{1-\alpha }= \int _{0}^{d_{k}}f(y)\,dy. \end{aligned}$$

Fix a large integer $K>0$, which will go to infinity later, and let $\varLambda _{K}=\{k\in \mathbb{Z}:|k|\leq K\}$. We have

$$\begin{aligned} \mathcal{I}_{K} &= \int _{\bigcup _{k=-K}^{K}\varOmega _{K}} \bigl(N_{\alpha }f(y) \bigr)^{q}u(y)^{q}\,dy \leq \sum_{k=-K}^{K}2^{(k+1)q} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \\ &=2^{q}\sum_{k=-K}^{K} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d _{k}^{1-\alpha }} \int _{0}^{d_{k}}f(y)\,dy \biggr)^{q} \\ &=2^{q}\sum_{k=-K}^{K} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d _{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q} \biggl(\frac{\int _{0}^{d_{k}}(f\sigma ^{-1})(y)\sigma (y)\,dy}{\int _{0}^{d_{k}}\sigma (y)\,dy} \biggr)^{q} \\ &=2^{q} \int _{\mathbb{Z}} T_{K}\bigl(f\sigma ^{-1}\bigr)^{q}\,d\nu, \end{aligned}$$

where ν is the measure on $\mathbb{Z}$ given by

$$\begin{aligned} \nu (k)= \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d_{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q}, \end{aligned}$$

and, for every measurable function h, the operator $T_{K}$ is defined by

$$\begin{aligned} T_{K}h(k)=\frac{\int _{0}^{d_{k}}h(y)\sigma (y)\,dy}{\int _{0}^{d_{k}} \sigma (y)\,dy}\chi _{\varLambda _{K}}(k). \end{aligned}$$

If we prove that $T_{K}$ is uniformly bounded from $L^{p}((0,\infty ), \sigma )$ to $L^{q}(\mathbb{Z},\nu )$ independently of K, we shall obtain

$$\begin{aligned} \mathcal{I}_{K}&\leq C \int _{\mathbb{Z}} T_{K}\bigl(f\sigma ^{-1}\bigr)^{q}\,d \nu \\ &\leq C \biggl( \int _{0}^{\infty }\bigl[\bigl(f\sigma ^{-1} \bigr) (y)\bigr]^{p}\sigma (y)\,dy \biggr)^{q/p}=C \biggl( \int _{0}^{\infty }f(y)^{p}v(y)^{p} \,dy \biggr)^{q/p}. \end{aligned}$$

The uniformity in K of this estimate and the monotone convergence theorem will lead to the desired inequality.

Now we prove that $T_{K}$ is a bounded operator from $L^{p}((0,\infty ),\sigma )$ to $L^{q}(\mathbb{Z},\nu )$. It is clear that $T_{K}:L ^{\infty }((0,\infty ),\sigma )\to L^{\infty }(\mathbb{Z},\nu )$ with constant less than or equal to 1. The Marcinkiewicz interpolation theorem says that it is enough to prove the uniform boundedness of the operators $T_{K}$ from $L^{1}((0,\infty ),\sigma )$ to $L^{q/p,\infty }(\mathbb{Z},\nu )$. For this purpose, fix $h\ge 0$, a bounded function with compact support, and put $F_{\lambda }=\{k\in \mathbb{Z}:T_{K}h(k)> \lambda \}=\{|k|\leq K:T_{K}h(k)>\lambda \}$, and $k_{0}=\min \{k:k \in F_{\lambda }\}$. Using (3), we have

$$\begin{aligned} \nu (F_{\lambda }) &=\sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}} \biggl(\frac{1}{d_{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q}u(x)^{q} \,dx \\ &\leq \sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}}\bigl(N_{\alpha }( \sigma \chi _{(0,d_{k})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq \sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}}\bigl(N_{\alpha }( \sigma \chi _{(0,d_{k_{0}})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq \int _{0}^{d_{k_{0}}}\bigl(N_{\alpha }(\sigma \chi _{(0,d_{k_{0}})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq C \biggl( \int _{0}^{d_{k_{0}}}\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C \biggl(\frac{1}{\lambda } \int _{0}^{d_{k_{0}}}h(y)\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C \biggl(\frac{1}{\lambda } \int _{0}^{\infty }h(y)\sigma (y)\,dy \biggr)^{q/p}, \end{aligned}$$

where the constant C does not depend on K. This ends the proof. □

3 The proofs of Theorem 1.4 and Theorem 1.5

Proof of Theorem 1.4

We first prove (5). By the Hölder inequality and condition (4), we have

$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert P_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \frac{1}{x ^{1-\alpha }} \int _{0}^{x}f(y)\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1- \alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}f(y)\,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}}u(x)^{q}\,dx \\ &\qquad{}\times \Biggl(\sum_{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-p'}\,dy \biggr)^{1/p'} \Biggr)^{q} \\ &\quad\leq \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}}u(x)^{q}\,dx \\ &\qquad{}\times \Biggl(\sum_{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}}^{2^{k+1}}1\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ &\quad\leq C\sum_{j=-\infty }^{\infty } \Biggl(\sum _{k=-\infty }^{j}2^{ \frac{(k-j)}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ &\quad\leq C\sum_{j=-\infty }^{\infty } \Biggl(\sum _{k=-\infty }^{j}2^{ \frac{(k-j)}{2r'}} \Biggr)^{q/p'} \Biggl(\sum_{k=-\infty }^{j}2^{ \frac{(k-j)p}{2r'p'}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ &\quad\leq C \Biggl(\sum_{j=-\infty }^{\infty }\sum _{k=-\infty }^{j}2^{ \frac{(k-j)p}{2r'p'}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ &\quad\leq C \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

Next we will prove the following inequalities (7). By the Hölder inequality and condition (6), we have

$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert Q_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \int _{x}^{ \infty }\frac{f(y)}{y^{1-\alpha }}\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j} ^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \biggl( \int _{2^{j}}^{2^{j+1}}1\,dx \biggr)^{1/r'} \\ &\qquad{}\times \Biggl(\sum_{k=j}^{\infty } \frac{1}{2^{k(1-\alpha )}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}} ^{2^{k+1}}v(y)^{-p'}\,dy \biggr)^{1/p'} \Biggr)^{q} \\ &\quad\leq C\sum_{j=-\infty }^{\infty } \Biggl(\sum _{k=j}^{\infty }2^{ \frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ &\quad\leq C\sum_{j=-\infty }^{\infty } \Biggl(\sum _{k=j}^{\infty }2^{ \frac{(j-k)p'}{2r'q}} \Biggr)^{q/p'} \Biggl(\sum_{k=j}^{\infty }2^{ \frac{(j-k)p}{2r'q}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ &\quad\leq C \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

 □

Lemma 3.1

([10])

Let $b\in \mathrm{CMO}^{1}$, $j,k\in \mathbb{Z}$, then

$$\begin{aligned} \bigl\vert b(t)-b_{(0,2^{j+1}]} \bigr\vert \leq \bigl\vert b(t)-b_{(0,2^{k+1}]} \bigr\vert +2 \vert j-k \vert \Vert b \Vert _{\mathrm{CMO} ^{1}}. \end{aligned}$$

Proof of Theorem 1.5

We first prove (9). We have

$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert P_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \frac{1}{x ^{1-\alpha }} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)f(y)\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1- \alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b(y) \bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq 2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1-\alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b _{(0,2^{j+1}]}\bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\qquad{}+2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1- \alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert b(y)-b_{(0,2^{j+1}]})f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad=2^{q/q'}(\mathrm {I}+\mathrm{II}). \end{aligned}$$

For I, by the Hölder inequality and condition (8), we have

$$\begin{aligned} \mathrm{I} ={}&\sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}} \bigl\vert b(x)-b_{(0,2^{j+1}]} \bigr\vert ^{q}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty } ^{j} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert \,dy \Biggr)^{q} \\ \leq{}& \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \biggl( \int _{2^{j}}^{2^{j+1}} \bigl\vert b(x)-b_{(0,2^{j+1}]} \bigr\vert ^{r'q}\,dx \biggr)^{1/r'} \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \\ &{}\times \Biggl(\sum_{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}}^{2^{k+1}}\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty }\frac{2^{j+1}}{2^{jq(1- \alpha )}} \biggl( \frac{1}{2^{j+1}} \int _{0}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \\ &{}\times \Biggl(\sum_{k=-\infty }^{j}2^{\frac{j+1}{rp'}+\frac{k+1}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \frac{1}{2^{j+1}} \int _{0}^{2^{j+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)}{2r'}} \Biggr)^{q/p'} \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)p}{2r'p'}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

For II, by Lemma 3.1, we have

$$\begin{aligned} {\mathrm{II}} \leq {}&\sum_{j=-\infty }^{\infty } \frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=-\infty }^{j} \int _{2^{k}} ^{2^{k+1}}\bigl\vert \bigl(b(y)-b_{(0,2^{k+1}]} \bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &{}+\sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}} \Biggl\vert \sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}2(j-k) \Vert b \Vert _{\mathrm{CMO}^{1}}\bigl\vert f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ ={}&\mathrm{II_{1}+II_{2}}. \end{aligned}$$

For $\mathrm{II_{1}}$, by the Hölder inequality and condition (8), we have

$$\begin{aligned} \mathrm{II_{1}}\leq{}& \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert b(y)-b _{(0,2^{k+1}]} \bigr\vert ^{r'p'}\,dy \biggr)^{1/r'p'} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty }\frac{2^{j+1}}{2^{jq(1- \alpha )}} \Biggl(\sum _{k=-\infty }^{j}2^{\frac{(k+1)}{r'p'}+ \frac{(j+1)}{rp'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

For $\mathrm{II_{2}}$, we have

$$\begin{aligned} {\mathrm{II_{2}}} ={}&C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q} \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j}(j-k) \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert \,dy \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q}\sum _{j=-\infty }^{\infty }\frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j}(j-k) \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}} ^{2^{k+1}}\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}(j-k)2^{\frac{(k-j)}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

Now we prove (10). We have

$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert Q_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \int _{x}^{ \infty }\frac{(b(x)-b(y))f(y)}{y^{1-\alpha }}\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j} ^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b(y) \bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq 2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b _{(0,2^{j+1}]}\bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\qquad{}+2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(y)-b _{(0,2^{j+1}]}\bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\quad =2^{q/q'}( \mathrm{J}+\mathrm{JJ}). \end{aligned}$$

For J, by the Hölder inequality and condition (8), we have

$$\begin{aligned} {\mathrm{J}} ={}&\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \bigl\vert b(x)-b _{(0,2^{j+1}]} \bigr\vert ^{q}u(x)^{q}\,dx \Biggl(\sum _{k=j}^{\infty }\frac{1}{2^{k(1- \alpha )}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert \,dy \Biggr)^{q} \\ \leq {}&\sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}}^{2^{j+1}} \bigl\vert b(x)-b _{(0,2^{j+1}]} \bigr\vert ^{r'q}\,dx \biggr)^{1/r'} \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \Biggl(\sum_{k=j}^{\infty } \frac{1}{2^{k(1-\alpha )}} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}} ^{2^{k+1}}\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty }2^{ \frac{j+1}{r'}} \biggl( \int _{0}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \\ &{}\times \Biggl(\sum_{k=j}^{\infty }2^{-k(1-\alpha )+\frac{k+1}{rp'}+ \frac{k+1}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \frac{1}{2^{k+1}} \int _{0}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty }2^{ \frac{j+1}{r'}} \Biggl(\sum _{k=j}^{\infty }2^{\frac{-(k+1)}{r'q}}\cdot \bigl(2^{k+1} \bigr)^{( \frac{1}{q}+\alpha -\frac{1}{p})} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \\ &{}\times \biggl(\frac{1}{2^{k+1}} \int _{0}^{2^{k+1}}u(x)^{rq}\,dx \biggr)^{1/rq} \biggl(\frac{1}{2^{k+1}} \int _{0}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j} ^{\infty }2^{\frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j} ^{\infty }2^{\frac{(j-k)p'}{2r'q}} \Biggr)^{q/p'} \Biggl(\sum _{k=j}^{\infty }2^{\frac{(j-k)p}{2r'q}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

For JJ, by Lemma 3.1, we have

$$\begin{aligned} {\mathrm{JJ}} \leq{} &\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(y)-b _{(0,2^{k+1}]}\bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &{}+\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j} ^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}2(k-j) \Vert b \Vert _{\mathrm{CMO}^{1}}\bigl\vert f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ ={}&\mathrm{JJ}_{1}+\mathrm{JJ}_{2}. \end{aligned}$$

For $\mathrm{JJ}_{1}$, we have

$$\begin{aligned} {\mathrm{JJ}_{1}} \leq {}&\sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}} ^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r}{2}^{j/r'} \Biggl(\sum_{k=j}^{\infty } \frac{1}{2^{k(1- \alpha )}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert b(y)-b_{(0,2^{k+1}]} \bigr\vert ^{r'p'}\,dy \biggr)^{1/r'p'} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j}^{\infty }2^{\frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

For $\mathrm{JJ}_{2}$, we have

$$\begin{aligned} {\mathrm{JJ}_{2}} \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q} \sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \biggl( \int _{2^{j}} ^{2^{j+1}}1\,dx \biggr)^{1/r'} \Biggl( \sum_{k=j}^{\infty }\frac{(k-j)}{2^{k(1- \alpha )}} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}} ^{2^{k+1}}1\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j}^{\infty }(k-j)2^{\frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$

This ends the proof. □