1 Introduction
Let \(P_{\alpha }\) and \(Q_{\alpha }\) be the fractional Hardy operator and its adjoint on \((0,\infty )\),
where \(0\leq \alpha <1\).
$$\begin{aligned} P_{\alpha }f(x)=\frac{1}{x^{1-\alpha }} \int _{0}^{x}f(y)\,dy,\qquad Q_{\alpha }f(x)= \int _{x}^{\infty }\frac{f(y)}{y^{1-\alpha }}\,dy, \end{aligned}$$
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
and
where \(p'=p/(p-1)\).
$$ \int _{0}^{\infty } \bigl\vert Pf(y) \bigr\vert ^{p}\,dy\leq p^{\prime p} \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}\,dy,\quad p>1, $$
$$ \int _{0}^{\infty } \bigl\vert Qf(y) \bigr\vert ^{p}\,dy\leq p^{p} \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}\,dy,\quad p>1, $$
Anzeige
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
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
Notice that \(N_{\alpha }f\leq S_{\alpha }f\) for nonnegative f.
$$\begin{aligned} N_{\alpha }f(x)=\sup_{t>x}\frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy,\quad x>0. \end{aligned}$$
$$\begin{aligned} &\bigl\vert P_{\alpha }f(x) \bigr\vert \leq \sup_{t>x} \frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy \leq N_{\alpha }f(x), \\ &\frac{1}{t^{1-\alpha }} \int _{0}^{t} \bigl\vert f(y) \bigr\vert \,dy \leq \frac{1}{x^{1-\alpha }} \int _{0}^{x} \bigl\vert f(y) \bigr\vert \,dy+ \int _{x}^{t}\frac{ \vert f(y) \vert }{y^{1-\alpha }}\,dy \leq S_{\alpha }f(x). \end{aligned}$$
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
$$ [w]_{p,q,0}=\sup_{t>0} \biggl(\frac{1}{t} \int _{0}^{t}w(y)^{q}\,dy \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}w(y)^{-p'}\,dy \biggr)^{1/p'}< \infty. $$
Anzeige
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
where \(b_{(0,2^{j}]}=\frac{1}{2^{j}}\int _{0}^{2^{j}}f(x)\,dx\), we then say that \(b\in \mathrm{CMO}^{p}\).
$$\begin{aligned} \sup_{j\in \mathbb{Z}} \biggl(\frac{1}{2^{j}} \int _{0}^{2^{j}} \bigl\vert b(y)-b_{(0,2^{j}]} \bigr\vert ^{p}\,dy \biggr)^{1/p}= \Vert b \Vert _{\mathrm{CMO}^{p}}< \infty, \end{aligned}$$
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,
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,
$$ \sup_{\lambda >0}\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w(y)^{q}\,dy \biggr)^{1/q}\leq [w]_{p,q,0} \Vert f \Vert _{L^{p}(w^{p})}. $$
(1)
$$ \Vert N_{\alpha }f \Vert _{L^{q}(w^{q})}\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \Vert f \Vert _{L^{p}(w^{p})}. $$
(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
\(t>0\),
Then
$$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{q}\,dy \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-rp'}\,dy \biggr)^{1/rp'}\leq C< \infty. $$
(4)
$$ \biggl( \int _{0}^{\infty } \bigl\vert P_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$
(5)
(b)
Let
\((u,v)\)
be a pair of weights for which there exists
\(r>1\)
such that, for every
\(t>0\),
Then
$$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{rq}\,dy \biggr)^{1/rq} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-p'}\,dy \biggr)^{1/p'}\leq C< \infty. $$
(6)
$$ \biggl( \int _{0}^{\infty } \bigl\vert Q_{\alpha }f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$
(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
\(t>0\),
Then
and
$$ t^{(1/q+\alpha -1/p)} \biggl(\frac{1}{t} \int _{0}^{t}u(y)^{rq}\,dy \biggr)^{1/rq} \biggl(\frac{1}{t} \int _{0}^{t}v(y)^{-rp'}\,dy \biggr)^{1/rp'}\leq C< \infty. $$
(8)
$$ \biggl( \int _{0}^{\infty } \bigl\vert P_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p} $$
(9)
$$ \biggl( \int _{0}^{\infty } \bigl\vert Q_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \biggr)^{1/q} \leq C \biggl( \int _{0}^{\infty } \bigl\vert f(x) \bigr\vert ^{p}v(x)^{p}\,dx \biggr)^{1/p}. $$
(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
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 \(N^{g}\).
$$ N_{\alpha }^{g}f(x)=\sup_{t>x} \frac{\int _{0}^{t} \vert f(y) \vert g(y)\,dy}{( \int _{0}^{t}g(y)\,dy)^{1-\alpha }}. $$
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,
for all
\(\lambda >0\)
and all measurable functions
f.
$$ \biggl( \int _{\{x:N_{\alpha }^{g}f(x)>\lambda \}} g(y)\,dy \biggr)^{1- \alpha }\leq \frac{1}{\lambda } \int _{\{x:N_{\alpha }^{g}f(x)>\lambda \}} \bigl\vert f(y) \bigr\vert g(y) \,dy, $$
(11)
(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,
in which the constant
\(C(p,\alpha )\)
independent of
f
and
g.
$$\begin{aligned} \biggl( \int _{0}^{\infty } \bigl\vert N_{\alpha }^{g}f(y) \bigr\vert ^{q}g(y)\,dy \biggr)^{1/q} \leq C(p,\alpha ) \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}g(y)\,dy \biggr)^{1/p}, \end{aligned}$$
Proof
By standard interpolation arguments, it suffices to prove (i) since by the Hölder inequality, we have
Thus we obtain \(\|N_{\alpha }^{g}f\|_{\infty }\leq \|f\|_{L^{1/\alpha }(g)}\).
$$\begin{aligned} \frac{\int _{0}^{t} \vert f(x) \vert g(x)\,dx}{(\int _{0}^{t}g(x)\,dx)^{1-\alpha }} & \leq \frac{(\int _{0}^{t} \vert f(x) \vert ^{1/\alpha }g(x)\,dx)^{\alpha }(\int _{0} ^{t}g(x)\,dx)^{1-\alpha }}{(\int _{0}^{t}g(x)\,dx)^{1-\alpha }} \\ &= \biggl( \int _{0}^{t} \bigl\vert f(x) \bigr\vert ^{1/\alpha }g(x)\,dx \biggr)^{\alpha } \\ &\leq \Vert f \Vert _{L^{1/\alpha }(g)}. \end{aligned}$$
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
whereas in the second case we have
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. □
$$ \lambda \biggl( \int _{0}^{d}g(x)\,dx \biggr)^{1-\alpha }= \int _{0}^{d} \bigl\vert f(x) \bigr\vert g(x) \,dx, $$
(12)
$$\begin{aligned} \lambda \biggl( \int _{0}^{\infty }g(x)\,dx \biggr)^{1-\alpha }\leq \int _{0} ^{\infty } \bigl\vert f(x) \bigr\vert g(x) \,dx. \end{aligned}$$
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
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
Thus we obtain
Letting k tend to infinity, \(w\in A_{p,q,0}\) follows.
$$ N_{\alpha }f(x)\geq \frac{1}{t^{1-\alpha }} \int _{0}^{t}w_{k}^{-p'}, $$
(13)
$$\begin{aligned} \biggl( \int _{0}^{t}w^{q} \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0}^{t}w_{k} ^{-p'} \biggr) &=\frac{1}{t^{\alpha }} \biggl( \int _{0}^{t}w^{q} \biggr)^{1/q} \cdot \lambda \\ &\leq \frac{1}{t^{\alpha }} \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w^{q} \biggr)^{1/q}\cdot \lambda \\ &\leq \frac{C}{t^{\alpha }} \biggl( \int _{0}^{t}w_{k}^{-p'p}w^{p} \biggr)^{1/p} \\ &=\frac{C}{t^{\alpha }} \biggl( \int _{0}^{t}w_{k}^{-p'} \biggr)^{1/p}. \end{aligned}$$
$$\begin{aligned} \biggl(\frac{1}{t} \int _{0}^{t}w^{q} \biggr)^{1/q} \biggl(\frac{1}{t} \int _{0} ^{t}w_{k}^{-p'} \biggr)^{1/p'}\leq C. \end{aligned}$$
(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
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
Thus we have
By the Lebesgue differentiation theorem, for any \(y\in (0,s)\), we have
Thus, \(w\in A_{1,q,0}\) follows.
$$ N_{\alpha }f(x)\geq \frac{1}{s^{1-\alpha }} \int _{0}^{s} \vert \chi _{(t_{1},t_{2})} \vert \,dy=\frac{t_{2}-t_{1}}{s^{1-\alpha }}, $$
(14)
$$\begin{aligned} \lambda \biggl( \int _{0}^{s}w^{q} \biggr)^{1/q} \leq \lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w^{q} \biggr)^{1/q} \leq C \int _{t_{1}}^{t_{2}}w. \end{aligned}$$
$$\begin{aligned} \biggl(\frac{1}{s} \int _{0}^{s}w^{q} \biggr)^{1/q} \leq \frac{C}{t_{2}-t_{1}} \int _{t_{1}}^{t_{2}}w. \end{aligned}$$
$$\begin{aligned} \biggl(\frac{1}{s} \int _{0}^{s}w^{q} \biggr)^{1/q} \leq Cw(y). \end{aligned}$$
For the sufficient, arguing as in the proof of the Lemma 2.1, we have (12) with \(g\equiv 1\), that is,
Then
and (1) follows.
$$\begin{aligned} d^{1-\alpha }\lambda = \int _{0}^{d} \bigl\vert f(y) \bigr\vert \,dy. \end{aligned}$$
$$\begin{aligned} &\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} w(y)^{q}\,dy \biggr)^{1/q}\\ &\quad=\lambda \biggl( \int _{0}^{d}w(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq \frac{1}{d^{1-\alpha }} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{0}^{d}w(y)^{-p'}\,dy \biggr)^{1/p'} \biggl( \int _{0}^{d}w(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq [w]_{p,q,0} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{1/p}, \end{aligned}$$
Now we prove (2). If \(w\in A_{p,q,0}, 0< x< t\), \(\sigma =w^{-p'}\) and \(\delta =w^{q}\), we have
Therefore,
Consequently, using Lemma 2.1,
This ends the proof of the theorem. □
$$\begin{aligned} \biggl(\frac{1}{t^{1-\alpha }} \int _{0}^{t}f(y)\,dy \biggr)^{\frac{1}{(1- \alpha )p'}} &\leq [w]_{p,q,0} \biggl(\frac{\int _{0}^{t}f(y)\,dy}{(\int _{0}^{t}\sigma (y)\,dy)^{(1-\alpha )}} \biggr)^{\frac{1}{(1-\alpha )p'}} \frac{( \int _{0}^{t}\,dy)^{1/q}}{(\int _{0}^{t}\delta (y)\,dy)^{1/q}} \\ &\leq [w]_{p,q,0}\frac{(\int _{0}^{t} \vert N_{\alpha }^{\sigma }(f\sigma ^{-1})(y) \vert ^{\frac{q}{(1-\alpha )p'}}\,dy)^{1/q}}{(\int _{0}^{t}\delta (y)\,dy)^{1/q}} \\ &\leq [w]_{p,q,0} \bigl\vert N^{\delta }\bigl(\delta ^{-1} \bigl\vert N_{\alpha }^{\sigma }\bigl(f \sigma ^{-1}\bigr) \bigr\vert ^{\frac{q}{(1-\alpha )p'}}\bigr) (x) \bigr\vert ^{1/q}. \end{aligned}$$
$$ \bigl\vert N_{\alpha }f(x) \bigr\vert ^{q}\leq [w]_{p,q,0}^{(1-\alpha )p'q} \bigl\vert N^{\delta }\bigl(\delta ^{-1} \bigl\vert N_{\alpha }^{\sigma }\bigl(f\sigma ^{-1}\bigr) \bigr\vert ^{\frac{q}{(1- \alpha )p'}}\bigr) (x) \bigr\vert ^{(1-\alpha )p'}. $$
$$\begin{aligned} \int _{0}^{\infty } \bigl\vert N_{\alpha }f(y) \bigr\vert ^{q}w(y)^{q}\,dy &\leq C[w]_{p,q,0} ^{(1-\alpha )p'q} \int _{0}^{\infty } \bigl\vert N_{\alpha }^{\sigma } \bigl(f\sigma ^{-1}\bigr) (y) \bigr\vert ^{q} \sigma (y)\,dy \\ &\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \sigma ^{-1}(y) \bigr\vert ^{p}\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C[w]_{p,q,0}^{(1-\alpha )p'q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}w(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
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
Then
This ends the proof. □
$$\begin{aligned} d^{1-\alpha }\lambda = \int _{0}^{d} \bigl\vert f(y) \bigr\vert \,dy. \end{aligned}$$
$$\begin{aligned} &\lambda \biggl( \int _{\{x:N_{\alpha }f(x)>\lambda \}} u(y)^{q}\,dy \biggr)^{1/q}\\ &\quad=\lambda \biggl( \int _{0}^{d}u(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq \frac{1}{d^{1-\alpha }} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{0}^{d}v(y)^{-p'}\,dy \biggr)^{1/p'} \biggl( \int _{0}^{d}u(y)^{q}\,dy \biggr)^{1/q} \\ &\quad\leq [u,v]_{p,q} \biggl( \int _{0}^{d} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p}. \end{aligned}$$
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
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
where ν is the measure on \(\mathbb{Z}\) given by
and, for every measurable function h, the operator \(T_{K}\) is defined by
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
The uniformity in K of this estimate and the monotone convergence theorem will lead to the desired inequality.
$$\begin{aligned} 2^{k}\,d_{k}^{1-\alpha }= \int _{0}^{d_{k}}f(y)\,dy. \end{aligned}$$
$$\begin{aligned} \mathcal{I}_{K} &= \int _{\bigcup _{k=-K}^{K}\varOmega _{K}} \bigl(N_{\alpha }f(y) \bigr)^{q}u(y)^{q}\,dy \leq \sum_{k=-K}^{K}2^{(k+1)q} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \\ &=2^{q}\sum_{k=-K}^{K} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d _{k}^{1-\alpha }} \int _{0}^{d_{k}}f(y)\,dy \biggr)^{q} \\ &=2^{q}\sum_{k=-K}^{K} \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d _{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q} \biggl(\frac{\int _{0}^{d_{k}}(f\sigma ^{-1})(y)\sigma (y)\,dy}{\int _{0}^{d_{k}}\sigma (y)\,dy} \biggr)^{q} \\ &=2^{q} \int _{\mathbb{Z}} T_{K}\bigl(f\sigma ^{-1}\bigr)^{q}\,d\nu, \end{aligned}$$
$$\begin{aligned} \nu (k)= \int _{d_{k+1}}^{d_{k}}u(y)^{q}\,dy \biggl( \frac{1}{d_{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q}, \end{aligned}$$
$$\begin{aligned} T_{K}h(k)=\frac{\int _{0}^{d_{k}}h(y)\sigma (y)\,dy}{\int _{0}^{d_{k}} \sigma (y)\,dy}\chi _{\varLambda _{K}}(k). \end{aligned}$$
$$\begin{aligned} \mathcal{I}_{K}&\leq C \int _{\mathbb{Z}} T_{K}\bigl(f\sigma ^{-1}\bigr)^{q}\,d \nu \\ &\leq C \biggl( \int _{0}^{\infty }\bigl[\bigl(f\sigma ^{-1} \bigr) (y)\bigr]^{p}\sigma (y)\,dy \biggr)^{q/p}=C \biggl( \int _{0}^{\infty }f(y)^{p}v(y)^{p} \,dy \biggr)^{q/p}. \end{aligned}$$
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
where the constant C does not depend on K. This ends the proof. □
$$\begin{aligned} \nu (F_{\lambda }) &=\sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}} \biggl(\frac{1}{d_{k}^{1-\alpha }} \int _{0}^{d_{k}}\sigma (y)\,dy \biggr)^{q}u(x)^{q} \,dx \\ &\leq \sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}}\bigl(N_{\alpha }( \sigma \chi _{(0,d_{k})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq \sum_{k\in F_{\lambda }} \int _{d_{k+1}}^{d_{k}}\bigl(N_{\alpha }( \sigma \chi _{(0,d_{k_{0}})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq \int _{0}^{d_{k_{0}}}\bigl(N_{\alpha }(\sigma \chi _{(0,d_{k_{0}})})\bigr) (x)^{q}u(x)^{q}\,dx \\ &\leq C \biggl( \int _{0}^{d_{k_{0}}}\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C \biggl(\frac{1}{\lambda } \int _{0}^{d_{k_{0}}}h(y)\sigma (y)\,dy \biggr)^{q/p} \\ &\leq C \biggl(\frac{1}{\lambda } \int _{0}^{\infty }h(y)\sigma (y)\,dy \biggr)^{q/p}, \end{aligned}$$
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
For I, by the Hölder inequality and condition (8), we have
For II, by Lemma 3.1, we have
For \(\mathrm{II_{1}}\), by the Hölder inequality and condition (8), we have
For \(\mathrm{II_{2}}\), we have
$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert P_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \frac{1}{x ^{1-\alpha }} \int _{0}^{x}\bigl(b(x)-b(y)\bigr)f(y)\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1- \alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b(y) \bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq 2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1-\alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b _{(0,2^{j+1}]}\bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\qquad{}+2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \frac{1}{2^{j(1- \alpha )}}\sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}\bigl\vert b(y)-b_{(0,2^{j+1}]})f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad=2^{q/q'}(\mathrm {I}+\mathrm{II}). \end{aligned}$$
$$\begin{aligned} \mathrm{I} ={}&\sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}} \bigl\vert b(x)-b_{(0,2^{j+1}]} \bigr\vert ^{q}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty } ^{j} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert \,dy \Biggr)^{q} \\ \leq{}& \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \biggl( \int _{2^{j}}^{2^{j+1}} \bigl\vert b(x)-b_{(0,2^{j+1}]} \bigr\vert ^{r'q}\,dx \biggr)^{1/r'} \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \\ &{}\times \Biggl(\sum_{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}}^{2^{k+1}}\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty }\frac{2^{j+1}}{2^{jq(1- \alpha )}} \biggl( \frac{1}{2^{j+1}} \int _{0}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \\ &{}\times \Biggl(\sum_{k=-\infty }^{j}2^{\frac{j+1}{rp'}+\frac{k+1}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \frac{1}{2^{j+1}} \int _{0}^{2^{j+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)}{2r'}} \Biggr)^{q/p'} \Biggl( \sum_{k=-\infty }^{j}2^{\frac{(k-j)p}{2r'p'}} \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \Biggr)^{q/p} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'q}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
$$\begin{aligned} {\mathrm{II}} \leq {}&\sum_{j=-\infty }^{\infty } \frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=-\infty }^{j} \int _{2^{k}} ^{2^{k+1}}\bigl\vert \bigl(b(y)-b_{(0,2^{k+1}]} \bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &{}+\sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}} \Biggl\vert \sum _{k=-\infty }^{j} \int _{2^{k}}^{2^{k+1}}2(j-k) \Vert b \Vert _{\mathrm{CMO}^{1}}\bigl\vert f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ ={}&\mathrm{II_{1}+II_{2}}. \end{aligned}$$
$$\begin{aligned} \mathrm{II_{1}}\leq{}& \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1-\alpha )}} \int _{2^{j}} ^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert b(y)-b _{(0,2^{k+1}]} \bigr\vert ^{r'p'}\,dy \biggr)^{1/r'p'} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty }\frac{2^{j+1}}{2^{jq(1- \alpha )}} \Biggl(\sum _{k=-\infty }^{j}2^{\frac{(k+1)}{r'p'}+ \frac{(j+1)}{rp'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{}& C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
$$\begin{aligned} {\mathrm{II_{2}}} ={}&C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q} \sum_{j=-\infty }^{\infty }\frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j}(j-k) \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert \,dy \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q}\sum _{j=-\infty }^{\infty }\frac{1}{2^{jq(1- \alpha )}} \int _{2^{j}}^{2^{j+1}}u(x)^{q}\,dx \Biggl(\sum _{k=-\infty }^{j}(j-k) \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}} ^{2^{k+1}}\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl( \sum_{k=-\infty }^{j}(j-k)2^{\frac{(k-j)}{r'p'}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
Now we prove (10). We have
For J, by the Hölder inequality and condition (8), we have
For JJ, by Lemma 3.1, we have
For \(\mathrm{JJ}_{1}\), we have
For \(\mathrm{JJ}_{2}\), we have
This ends the proof. □
$$\begin{aligned} & \int _{0}^{\infty } \bigl\vert Q_{\alpha }^{b} f(x) \bigr\vert ^{q}u(x)^{q}\,dx \\ &\quad=\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \biggl\vert \int _{x}^{ \infty }\frac{(b(x)-b(y))f(y)}{y^{1-\alpha }}\,dy \biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq \sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j} ^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b(y) \bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q}\,dx \\ &\quad\leq 2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(x)-b _{(0,2^{j+1}]}\bigr)f(y) \bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\qquad{}+2^{q/q'}\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(y)-b _{(0,2^{j+1}]}\bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &\quad =2^{q/q'}( \mathrm{J}+\mathrm{JJ}). \end{aligned}$$
$$\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}$$
$$\begin{aligned} {\mathrm{JJ}} \leq{} &\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j}^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}\bigl\vert \bigl(b(y)-b _{(0,2^{k+1}]}\bigr)f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ &{}+\sum_{j=-\infty }^{\infty } \int _{2^{j}}^{2^{j+1}} \Biggl\vert \sum _{k=j} ^{\infty }\frac{1}{2^{k(1-\alpha )}} \int _{2^{k}}^{2^{k+1}}2(k-j) \Vert b \Vert _{\mathrm{CMO}^{1}}\bigl\vert f(y)\bigr\vert \,dy \Biggr\vert ^{q}u(x)^{q} \,dx \\ ={}&\mathrm{JJ}_{1}+\mathrm{JJ}_{2}. \end{aligned}$$
$$\begin{aligned} {\mathrm{JJ}_{1}} \leq {}&\sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}} ^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r}{2}^{j/r'} \Biggl(\sum_{k=j}^{\infty } \frac{1}{2^{k(1- \alpha )}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert b(y)-b_{(0,2^{k+1}]} \bigr\vert ^{r'p'}\,dy \biggr)^{1/r'p'} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j}^{\infty }2^{\frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
$$\begin{aligned} {\mathrm{JJ}_{2}} \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{1}}^{q} \sum_{j=-\infty }^{\infty } \biggl( \int _{2^{j}}^{2^{j+1}}u(x)^{rq}\,dx \biggr)^{1/r} \biggl( \int _{2^{j}} ^{2^{j+1}}1\,dx \biggr)^{1/r'} \Biggl( \sum_{k=j}^{\infty }\frac{(k-j)}{2^{k(1- \alpha )}} \\ &{}\times \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \biggl( \int _{2^{k}}^{2^{k+1}}v(y)^{-rp'}\,dy \biggr)^{1/rp'} \biggl( \int _{2^{k}} ^{2^{k+1}}1\,dy \biggr)^{1/r'p'} \Biggr)^{q} \\ \leq {}&C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q}\sum _{j=-\infty }^{\infty } \Biggl(\sum_{k=j}^{\infty }(k-j)2^{\frac{(j-k)}{r'q}} \biggl( \int _{2^{k}}^{2^{k+1}} \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{1/p} \Biggr)^{q} \\ \leq{} &C \Vert b \Vert _{\mathrm{CMO}^{r'p'}}^{q} \biggl( \int _{0}^{\infty } \bigl\vert f(y) \bigr\vert ^{p}v(y)^{p}\,dy \biggr)^{q/p}. \end{aligned}$$
Competing interests
The authors declare that they have no competing interests.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.