# Two-weight norm inequalities for fractional integral operators with $$A_{\lambda,\infty}$$ weights

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Research

## Abstract

In this paper, we introduce a new class of weights, the $$A_{\lambda, \infty}$$ weights, which contains the classical $$A_{\infty}$$ weights. We prove a mixed $$A_{p,q}$$$$A_{\lambda,\infty}$$ type estimate for fractional integral operators.

## Keywords

Fractional integral operators Two-weight norm inequalities Fractional maximal operators $$A_{\lambda,\infty}$$ weights

42B20

## 1 Introduction and the main results

Fractional integral operators and the associated maximal functions are very useful tools in harmonic analysis and PDE, especially in the study of differentiability or smoothness properties of functions. Recall that, for $$0<\lambda <n$$, the fractional integral operator $$I_{\lambda }$$ of a locally integrable function f defined on $$\mathbb{R}^{n}$$ is given by
$$I_{\lambda }f(x):= \int _{\mathbb{R}^{n}}\frac{f(y)}{ \vert x-y \vert ^{\lambda }}\,dy.$$
And the fractional maximal function $$M_{\lambda }$$ is defined by
$$M_{\lambda }f(x) := \sup_{Q\ni x}\frac{1}{ \vert Q \vert ^{\lambda /n}} \int _{Q} \bigl\vert f(y) \bigr\vert \,dy,$$
where the supremum is taken over all cubes Q in $$\mathbb{R}^{n}$$ with sides parallel to the axes. We refer to [1, 2, 3, 4, 5] for more results on fractional integral operators.
For $$1< p,q<\infty$$, we call a locally integrable positive function $$w(x)$$ defined on $$\mathbb{R}^{n}$$ a weight belongs to $$A_{p,q}( \mathbb{R}^{n})$$ if
$$[w]_{A_{p,q}}:=\sup_{Q} \biggl( \frac{1}{ \vert Q \vert } \int _{Q} w(x)^{q}\,dx \biggr) \biggl( \frac{1}{ \vert Q \vert } \int _{Q} w(x)^{-p'}\,dx \biggr)^{q/p'}< \infty .$$
In , Muckenhoupt and Wheeden showed that, for $$1< p< n/(n-\lambda )$$ and $$1/q+1/p'=\lambda /n$$, the fractional integral operator $$I_{\lambda }$$ is bounded from $$L^{p}(w^{p})$$ to $$L^{q}(w^{q})$$ if and only if w belongs to $$A_{p,q}$$. They also proved that the fractional maximal function $$M_{\lambda }$$ is bounded from $$L^{p}(w^{p})$$ to $$L^{q}(w^{q})$$ under the same conditions on the weights. Lacey, Moen, Pérez and Torres  proved the sharp weighted bound for fractional integral operators. Specifically,
$$\Vert I_{\lambda } \Vert _{L^{p}(w^{p})\rightarrow L^{q}(w^{q})}\leq C_{n,p}[w]_{A _{p,q}}^{\frac{\lambda }{n}\max \{1,\frac{p'}{q}\}}.$$
And the sharp weighted bound for the fractional maximal function was proved by Pradolini and Salinas , i.e.,
$$\Vert M_{\lambda } \Vert _{L^{p}(w^{p})\rightarrow L^{q}(w^{q})}\leq C_{n,p}[w]_{A _{p,q}}^{{\frac{\lambda }{n}}\cdot \frac{p'}{q}}.$$
(1.1)
Hytönen and Lacey  introduced a different approach to improving the sharp $$A_{p}$$ estimates for Calderón–Zygmund operators using a mixed $$A_{p}$$$$A_{\infty }$$ condition. Cruz-Uribe and Moen  studied the corresponding problem for fractional integral operators. Recall that w is said to be a weight in $$A_{\infty }$$ if
$$[w]_{A_{\infty }}:=\sup_{Q}\frac{1}{w(Q)} \int _{Q}M(1_{Q}w) (x)\,dx< \infty ,$$
where M is the Hardy–Littlewood maximal function and $$w(Q):=\int _{Q}w(x)\,dx$$. There are several equivalent definitions of the $$A_{\infty }$$ weights. For example $$w\in A_{\infty }'$$ if
$$[w]_{A_{\infty }'}:=\sup_{Q}\exp \biggl( \frac{1}{ \vert Q \vert } \int _{Q}-\log \bigl(w(x) \bigr)\,dx \biggr) \biggl( \frac{1}{ \vert Q \vert } \int _{Q}w(x)\,dx \biggr)< \infty .$$
In , Fujii proved that $$w\in A_{\infty }$$ if and only if $$w\in A_{\infty }'$$. It is well known that $$w\in A_{\infty }$$ if and only if $$w\in A_{p}$$ for some $$p>1$$, here $$A_{p}$$ denotes the class of Muckenhoupt weights for which
$$[w]_{A_{p}}:=\sup_{Q} \biggl( \frac{1}{ \vert Q \vert } \int _{Q}w(x)\,dx \biggr) \biggl( \frac{1}{ \vert Q \vert } \int _{Q}w(x)^{1-p'}\,dx \biggr)^{p-1}< \infty .$$
Sbordone and Wik  showed that
$$[w]_{A_{\infty }'}=\lim_{p\rightarrow \infty }[w]_{A_{p}}.$$
Hytönen and Pérez  showed that $$[w]_{A_{\infty }}\lesssim [w]_{A_{\infty }'}$$, and in fact $$[w]_{A_{\infty }}$$ can be substantially smaller.

In this paper, we introduce a new class of weights, called the $$A_{\lambda ,\infty }$$ weights, which is defined with the fractional maximal function.

### Definition 1.1

Given $$0<\lambda <n$$, $$A_{\lambda ,\infty }$$ consists of all locally integrable functions $$w(x)$$ on $$\mathbb{R}^{n}$$ for which
$$[w]_{A_{\lambda ,\infty }}:= \sup_{Q}\frac{1}{w(Q)} \bigl\Vert M_{\lambda } (w 1_{Q} )\cdot 1_{Q} \bigr\Vert _{n/\lambda }< \infty .$$

We show that $$A_{\infty }$$ is a subset of $$A_{\lambda ,\infty }$$. Specifically, we have the following results.

### Theorem 1.2

For any$$0<\lambda <n$$and$$w\in A_{\infty }$$, we have$$w\in A_{ \lambda ,\infty }$$and
$$[w]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda }}\leq C_{n,\lambda }[w]_{A _{\infty }}.$$

With $$A_{\lambda ,\infty }$$ weights, we give a mixed two-weight estimate of fractional integral operators.

### Theorem 1.3

Letλ, pandqbe constants such that$$0<\lambda <n$$and$$1/q+1/p'=\lambda /n$$. For any$$w\in A_{p,q}$$, set$$\mu = w^{q}$$and$$\sigma =w^{-p'}$$. Then
$$\bigl\Vert I_{\lambda }(\cdot \sigma ) \bigr\Vert _{L^{p}(\sigma )\rightarrow L^{q}( \mu )} \lesssim [w]_{A_{p,q}}^{\frac{1}{q}} \bigl([\mu ]_{A_{\lambda , \infty }}^{\frac{n}{\lambda p'}}+[ \sigma ]_{A_{\lambda ,\infty }}^{\frac{n}{ \lambda q}} \bigr).$$
The above theorem suggests us to generalize the $$A_{p,q}$$ condition for a pair of weights. Given $$1< p,q<\infty$$, we say that a pair of weights $$(\mu ,\sigma )$$ is in the class $$A_{p,q}$$ if
$$[\mu ,\sigma ]_{A_{p,q}} := \sup_{Q} \biggl( \frac{1}{ \vert Q \vert } \int _{Q}\mu (x)\,dx \biggr) \biggl(\frac{1}{ \vert Q \vert } \int _{Q}\sigma (x)\,dx \biggr)^{q/p'}< \infty .$$
With this notation, we can generalize Theorem 1.3 as follows.

### Theorem 1.4

Letλ, pandqbe constants such that$$0<\lambda <n$$and$$1/q+1/p'=\lambda /n$$. For any pair of weights$$(\mu ,\sigma )\in A _{p,q}$$withμ, $$\sigma \in A_{\lambda ,\infty }$$, we have
$$\bigl\Vert I_{\lambda }(\cdot \sigma ) \bigr\Vert _{L^{p}(\sigma )\rightarrow L^{q}( \mu )} \lesssim [\mu ,\sigma ]_{A_{p,q}}^{\frac{1}{q}} \bigl([\mu ]_{A _{\lambda ,\infty }}^{\frac{n}{\lambda p'}}+[\sigma ]_{A_{\lambda , \infty }}^{\frac{n}{\lambda q}} \bigr).$$

The paper is organized as follows. In Sect. 2, we collect some preliminary results. And in Sect. 3, we give proofs for the main results.

## 2 Preliminaries

In this section, we introduce some preliminary results.

Let $$\mathscr {D}$$ be a set consisting of cubes in $$\mathbb{R}^{n}$$. Recall that $$\mathscr {D}$$ is said to be a general dyadic grid if it satisfies the following three conditions:
1. 1.

for any $$Q\in \mathscr {D}$$, its side length $$l(Q)$$ is of the form $$2^{k}$$ for some $$k\in \mathbb{Z}$$;

2. 2.

$$Q_{1}\cap Q_{2}\in \{Q_{1},Q_{2},\emptyset \}$$ for any $$Q_{1},Q _{2} \in \mathscr {D}$$;

3. 3.

the cubes of a fixed side length $$2^{k}$$ form a partition of $$\mathbb{R}^{n}$$.

Given a general dyadic grid $$\mathscr {D}$$, we call a subset $$\mathcal{S} \subset \mathscr {D}$$ a sparse family in $$\mathscr {D}$$ if it satisfies
\begin{aligned} \biggl\vert \bigcup_{Q'\in \mathcal{S}, Q'\varsubsetneq Q}Q' \biggr\vert \leq \frac{1}{2} \vert Q \vert , \quad \forall Q\in \mathcal{S}. \end{aligned}
For any $$Q\in \mathcal{S}$$, denote
\begin{aligned} E(Q):=Q\Big\backslash \biggl(\bigcup_{Q'\in \mathcal{S}, Q'\varsubsetneq Q}Q' \biggr). \end{aligned}
We see from the definition of the sparse family that $$|E(Q)|\geq \frac{1}{2}|Q|$$ for any $$Q\in \mathcal{S}$$.
Below we will make extensive use of the dyadic grids
$$\mathscr {D}^{\alpha }:=\bigl\{ 2^{-k}\bigl([0,1)^{n}+m+(-1)^{k} \alpha \bigr):k\in \mathbb{Z},m \in Z^{n}\bigr\} ,\quad \alpha \in \biggl\{ 0, \frac{1}{3},\frac{2}{3} \biggr\} ^{n}.$$
Hytönen, Lacey and Pérez  proved the following result.

### Lemma 2.1

(Three-lattice lemma)

For any cube$$Q\subset \mathbb{R}^{n}$$, there exists a shifted dyadic cube
$$R\in \mathscr {D}^{\alpha }=\bigl\{ 2^{-k}\bigl([0,1 )^{n}+m+(-1)^{k}\alpha \bigr):k\in \mathbb{Z},m\in \mathbb{Z}^{n}\bigr\}$$
for some$$\alpha \in \{0,\frac{1}{3},\frac{2}{3}\}^{n}$$, such that$$Q\subseteq R$$and$$\ell (R)\leq 6\ell (Q)$$.

Let Q be any cube in $$\mathbb{R}^{n}$$. A dyadic child of Q is any of the $$2^{n}$$ cubes obtained by partitioning Q by n “median hyperplanes” (i.e., the hyperplanes parallel to the faces of Q and dividing each edge into two equal parts).

Passing from Q to its children, then to the children of the children, etc., we obtain a standard dyadic lattice $$\mathcal{D}(Q)$$ of subcubes of Q.

We refer to Lerner and Nazarov  for more properties of the dyadic lattice.

### 2.3 Dyadic fractional integral operators

Given $$0<\lambda <n$$ and a general dyadic grid $$\mathscr {D}$$ in $$\mathbb{R} ^{n}$$, we define the dyadic fractional integral operator $$I^{\mathscr {D}}_{ \lambda }$$ by
$$I^{\mathscr {D}}_{\lambda } f(x):=\sum_{Q\in \mathscr {D}} \frac{1}{ \vert Q \vert ^{\lambda /n}} \int _{Q}f(y)\,dy \cdot 1_{Q}(x).$$
For a sparse family $$\mathcal{S}\subseteq \mathscr {D}$$, the sparse dyadic fractional integral operators $$I^{\mathcal{S}}_{\lambda }$$ are defined similarly. Cruz-Uribe and Moen  proved the following two propositions.

### Proposition 2.2

Given$$0<\lambda <n$$and a nonnegative functionf, then$$I_{\lambda }f$$is pointwise equivalent to a linear combination of dyadic fractional integral operators, i.e.,
$$I_{\lambda }f(x)\simeq \sum_{\alpha \in \{0,1/3,2/3\}^{n}} I_{\lambda }^{\mathscr {D}^{\alpha }}f(x).$$

### Proposition 2.3

Given a bounded, nonnegative functionfwith compact support and a dyadic grid$$\mathscr {D}$$, there exists a sparse family$$\mathcal {S}$$such that, for allλwith$$0<\lambda <n$$, we have
$$I_{\lambda }^{\mathscr {D}}f(x)\lesssim I_{\lambda }^{\mathcal {S}}f(x).$$

### 2.4 Testing condition

Let λ, p and q be constants such that $$0<\lambda <n$$ and $$1< p\leq q<\infty$$. Lacey, Sawyer and Uriarte-Tuero  reduced the proof of the boundedness for $$I_{\lambda }^{\mathcal {S}}(\cdot \sigma )$$ from $$L^{p}(\sigma )$$ to $$L^{q}(\mu )$$ to the boundedness of the testing condition
$$\varTheta ^{\mathscr{D}}_{\mu ,\sigma } :=\sup_{R\in \mathscr{D}} \frac{ \Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \Vert _{L^{q}( \mu )}}{\sigma (R)^{1/p}}, \qquad \varTheta ^{\mathscr{D}}_{\sigma ,\mu } :=\sup _{R\in \mathscr{D}}\frac{ \Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\mu 1_{R}) \Vert _{L^{p'}( \sigma )}}{\mu (R)^{1/q'}},$$
where the operator $$I^{\mathcal{S}(R)}_{\lambda }$$ is defined by
$$I^{\mathcal{S}(R)}_{\lambda }f(x):= \sum_{Q\in \mathcal{S},Q\subseteq R} \frac{1}{ \vert Q \vert ^{\lambda /n}} \int _{Q}f\,dy\cdot 1_{Q}(x).$$

### Proposition 2.4

([15, Theorem 1.11])

Suppose thatλ, pandqare constants such that$$0<\lambda <n$$and$$1< p\leq q<\infty$$. Let$$\mathscr {D}$$be a dyadic grid and let$$\mathcal {S}$$be a sparse subset of$$\mathscr {D}$$. For any pair of weights$$(\mu ,\sigma )$$, we have
$$\bigl\Vert I^{\mathcal{S}}_{\lambda }(\cdot \sigma ) \bigr\Vert _{L^{p}(\sigma )\rightarrow L^{q}(\mu )}\simeq \varTheta ^{\mathscr{D}}_{\mu ,\sigma }+ \varTheta ^{\mathscr{D}}_{\sigma ,\mu }.$$

## 3 Proofs of the main results

First, we show that a weight in $$A_{p,q}$$ is associated with a weight in some $$A_{\lambda ,\infty }$$.

### Theorem 3.1

Suppose thatλ, pandqare constants such that$$0<\lambda <n$$and$$1/q+1/p'=\lambda /n$$. Let$$w\in A_{p,q}$$and set$$\mu = w^{q}$$. Then$$\mu \in A_{\lambda ,\infty }$$and
$$[\mu ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda }} \leq C_{n,p} [w]_{A _{p,q}}.$$

### Proof

Since $$w\in A_{p,q}$$, we have $$w^{-1}\in A_{q',p'}$$ and
$$\bigl[w^{-1}\bigr]_{A_{q',p'}}=\sup_{Q} \biggl(\frac{1}{ \vert Q \vert } \int _{Q} w^{-p'}(x)\,dx \biggr) \biggl( \frac{1}{ \vert Q \vert } \int _{Q} w^{q}(x)\,dx \biggr)^{p'/q} = [w]_{A_{p,q}} ^{\frac{p'}{q}}.$$
Using (1.1), this gives us
\begin{aligned} \Vert M_{\lambda } \Vert _{L^{q'}(w^{-q'})\rightarrow L^{p'}(w^{-p'})}\leq{} & C _{n,p} \bigl[w^{-1}\bigr]_{A_{q',p'}}^{{\frac{\lambda }{n}}\cdot \frac{q}{p'}} = C_{n,p} [w]_{A_{p,q}}^{\frac{\lambda }{n}}. \end{aligned}
Fix some cube $$Q\in \mathbb{R}^{n}$$. Since $$\frac{1}{p'}+\frac{1}{q}=\frac{ \lambda }{n}$$, we see from Hölder’s inequality that
\begin{aligned} \bigl\Vert M_{\lambda }(\mu 1_{Q})\cdot 1_{Q} \bigr\Vert _{n/\lambda }= {}& \bigl\Vert M_{\lambda }( \mu 1_{Q})\cdot w^{-1}\cdot w 1_{Q} \bigr\Vert _{n/\lambda } \\ \leq {}& \bigl\Vert M_{\lambda }(\mu 1_{Q})\cdot w^{-1} 1_{Q} \bigr\Vert _{p'}\cdot \Vert w1_{Q} \Vert _{q} \\ = {}& \bigl\Vert M_{\lambda }(\mu 1_{Q})\cdot 1_{Q} \bigr\Vert _{L^{p'}(w^{-p'})}\cdot \mu (Q)^{1/q} \\ \leq {}& C_{n,p} [w]_{A_{p,q}}^{\frac{\lambda }{n}} \Vert \mu 1_{Q} \Vert _{L ^{q'}(w^{-q'})}\cdot \mu (Q)^{1/q} \\ ={} &C_{n,p} [w]_{A_{p,q}}^{\frac{\lambda }{n}}\mu (Q). \end{aligned}
Hence
$$[\mu ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda }} \lesssim [w]_{A _{p,q}}.$$
This completes the proof. □

Next we show that $$A_{\infty }$$ is contained in $$A_{\lambda ,\infty }$$ for any $$0<\lambda <n$$.

### Proof of Theorem 1.2

Fix some cube $$Q\subseteq \mathbb{R}^{n}$$. Let
$$Q^{\beta }:=Q+\ell (Q)\beta ,\quad \beta \in \biggl\{ -\frac{2}{3},- \frac{1}{3},0,\frac{1}{3},\frac{2}{3}\biggr\} ^{n}.$$
By translations and dilations, we see from Lemma 2.1 that for any cube $$K\subseteq Q$$ there exists some $$R\in \mathcal{D}(Q ^{\beta })$$ for some $$\beta \in \{-\frac{2}{3},-\frac{1}{3},0, \frac{1}{3},\frac{2}{3}\}^{n}$$ such that $$R\in \mathcal{D}(Q^{\beta })$$ and $$\ell (R)\leq 6\ell (K)$$. Hence
$$M_{\lambda }(w1_{Q}) (x)\cdot 1_{Q}(x)\leq C_{n,\lambda } \max_{\beta \in \{-\frac{2}{3},-\frac{1}{3},0,\frac{1}{3},\frac{2}{3} \}^{n}} M_{\lambda ,\mathcal{D}(Q^{\beta })}(w1_{Q}) (x)\cdot 1_{Q}(x),$$
where
$$M_{\lambda ,\mathcal{D}(Q^{\beta })}(w1_{Q}) (x):= \sup_{\substack{K\ni x \\ K\in \mathcal{D}(Q^{\beta })}} \frac{1}{ \vert K \vert ^{\lambda /n}} \int _{K}w(x)1_{Q}(x)\,dx.$$
So it sufficient to estimate
$$\frac{1}{w(Q)^{n/\lambda }} \int _{Q} \bigl\vert M_{\lambda ,\mathcal{D}(Q^{ \beta })}(w1_{Q}) (x) \bigr\vert ^{n/\lambda }\,dx, \quad \beta \in \biggl\{ - \frac{2}{3},- \frac{1}{3},0,\frac{1}{3},\frac{2}{3}\biggr\} ^{n}.$$
Among each $$\mathcal{D}(Q^{\beta })$$, a subset of principal cubes $$\mathscr {P}^{\beta }=\bigcup_{m=0}^{\infty }\mathscr {P}_{m}^{\beta }$$ is constructed as follows: $$\mathscr{P}_{0}^{\beta }=\{Q^{\beta }\}$$, and then inductively $$\mathscr{P}_{m+1}^{\beta }$$ consists of all maximal $$P'\in \mathcal{D}(Q^{\beta })$$ such that
$$\frac{w(P'\cap Q)}{ \vert P' \vert ^{\lambda /n}}> 2^{\lambda /n}\cdot 3^{\lambda } \frac{w(P\cap Q)}{ \vert P \vert ^{\lambda /n}}$$
for some $$P\in \mathscr{P}_{m}^{\beta }$$ with $$P\supset P'$$.
Since $$\lambda /n<1$$, we see from the definition of $$\mathscr {P}^{\beta }$$ that, for any $$P\in \mathscr{P}_{m}^{\beta }$$,
\begin{aligned} \biggl(\sum_{P'\in \mathscr{P}_{m+1}^{\beta },P'\subset P} \frac{ \vert P' \vert }{ \vert P \vert } \biggr)^{\lambda /n}\leq{} & \sum_{P'\in \mathscr{P}_{m+1}^{\beta },P'\subset P} \frac{ \vert P' \vert ^{\lambda /n}}{ \vert P \vert ^{\lambda /n}} \\ \leq{} & \frac{1}{2^{\lambda /n}\cdot 3^{\lambda }} \sum_{P'\in \mathscr{P}_{m+1}^{\beta },P'\subset P} \frac{w(P'\cap Q)}{w(P \cap Q)} \\ \leq {}& \frac{1}{2^{\lambda /n}\cdot 3^{\lambda }}. \end{aligned}
That is,
$$\sum_{P'\in \mathscr{P}_{m+1}^{\beta },P'\subset P} \bigl\vert P' \bigr\vert \leq \frac{1}{2 \cdot 3^{n}} \vert P \vert .$$
(3.1)
For any $$P\in \mathscr{P}^{\beta }$$, we denote
$$E(P):= P\Big\backslash \bigcup_{P'\in \mathscr{P}^{\beta },P'\subsetneq P}P'.$$
By (3.1), we have
$$\bigl\vert E(P) \bigr\vert \geq \biggl(1-\frac{1}{2\cdot 3^{n}} \biggr) \vert P \vert .$$
By the definition of $$Q^{\beta }$$ and $$\mathcal{D}(Q^{\beta })$$, for any $$R\in \mathcal{D}(Q^{\beta })$$ with $$R\cap Q\neq \emptyset$$, we have $$|R\cap Q|\geq \frac{1}{3^{n}}|R|$$. For any $$P\in \mathscr {P}^{\beta }$$, it is easy to see that $$P\mathrel{\cap} Q\neq \emptyset$$. Combining with $$|E(P)| \geq (1-\frac{1}{2\cdot 3^{n}})|P|$$, we obtain
$$\bigl\vert E(P)\cap Q \bigr\vert \geq \frac{1}{2\cdot 3^{n}} \vert P \vert .$$
Hence
\begin{aligned} &\frac{1}{w(Q)^{n/\lambda }} \int _{Q} \bigl\vert M_{\lambda ,\mathcal{D}(Q ^{\beta })}(w1_{Q}) (x) \bigr\vert ^{n/\lambda }\,dx \\ &\quad \leq 2\cdot 3^{n} \frac{1}{w(Q)^{n/\lambda }} \sum _{P\in \mathscr{P}^{\beta }}w(P\cap Q)^{n/\lambda } \\ &\quad \leq 2\cdot 3^{n} \frac{1}{w(Q)^{n/\lambda }} \sum _{P\in \mathscr{P}^{\beta }}w(P\cap Q)\cdot \omega (Q)^{\frac{n}{ \lambda }-1} \\ &\quad \leq 2^{2}\cdot 3^{2n} \frac{1}{w(Q)}\sum _{P\in \mathscr{P}^{ \beta }}\frac{ \vert E(P)\cap Q \vert }{ \vert P \vert }\cdot w(P\cap Q) \\ &\quad \leq 2^{2}\cdot 3^{2n} \frac{1}{w(Q)} \int _{Q}M(w1_{Q}) (x)\,dx \\ &\quad \leq 2^{2}\cdot 3^{2n} [w]_{A_{\infty }}. \end{aligned}
Now we get the conclusion as desired. □

Given a locally integrable function f, a Borel measure ν and a cube Q, we denote $$\langle f\rangle _{Q}:=\frac{1}{|Q|}\int _{Q}f(x)\,dx$$ and $$\langle f\rangle _{Q}^{\nu }:=\frac{1}{\nu (Q)}\int _{Q}f(x)\,d\nu (x)$$. The following result is used in the proof of Theorem 1.4.

### Proposition 3.2

([16, Proposition 2.2])

Let$$1< s<\infty$$, νbe a positive Borel measure and
$$\phi =\sum_{Q\in \mathscr{D}}\alpha _{Q}1_{Q},\qquad \phi _{Q} = \sum_{Q'\subset Q}\alpha _{Q'}1_{Q'}.$$
Then we have
$$\Vert \phi \Vert _{L^{s}(\nu )}\simeq \biggl(\sum _{Q\in \mathscr{D}}\alpha _{Q}\bigl( \langle \phi _{Q} \rangle _{Q}^{\nu }\bigr)^{s-1} \nu (Q) \biggr)^{1/s}.$$

To prove Theorem 1.4, we also need the following lemma.

### Lemma 3.3

([17, Lemma 5.2])

For all$$\gamma \in [0,1)$$, we have$$\sum_{L:L\subseteq P}\langle w \rangle _{L}^{\gamma }|L|\lesssim \langle w \rangle _{P}^{\gamma }|P|$$.

We are now ready to give a proof of Theorem 1.4.

### Proof of Theorem 1.4

By Propositions 2.2, 2.3 and 2.4, it suffices to show that
\begin{aligned}& \varTheta ^{\mathscr{D}}_{\mu ,\sigma } :=\sup_{R\in \mathscr{D}} \frac{ \Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \Vert _{L^{q}( \mu )}}{\sigma (R)^{1/p}} \lesssim [\mu ,\sigma ]_{A_{p,q}}^{ \frac{1}{q}} \cdot [\sigma ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda q}}, \\& \varTheta ^{\mathscr{D}}_{\sigma ,\mu } :=\sup_{R\in \mathscr{D}} \frac{ \Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\mu 1_{R}) \Vert _{L^{p'}( \sigma )}}{\mu (R)^{1/q'}} \lesssim [\mu ,\sigma ]_{A_{p,q}}^{ \frac{1}{q}} \cdot [\mu ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda p'}}. \end{aligned}

There are two cases.

Case 1:$$q\geq 2$$. By Proposition 3.2, we have
\begin{aligned} & \bigl\Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \bigr\Vert _{L ^{q}(\mu )}^{q} \\ &\quad \simeq \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{\sigma (Q)}{ \vert Q \vert ^{\frac{\lambda }{n}}} \biggl( \frac{1}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \frac{1}{ \vert Q' \vert ^{\lambda /n}}\sigma \bigl(Q'\bigr)\mu \bigl(Q' \bigr) \biggr)^{q-1}\cdot \mu (Q) \\ &\quad = \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl( \frac{1}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}^{\frac{q'}{q}}\langle \sigma \rangle _{Q'}^{\frac{q'}{p'}} \langle \mu \rangle _{Q'}^{1-\frac{q'}{q}} \bigl\vert Q' \bigr\vert ^{\frac{1}{q'}} \langle \sigma \rangle _{Q'}^{1-\frac{q'}{p'}} \bigl\vert Q' \bigr\vert ^{\frac{1}{p}} \biggr)^{q-1} \cdot \mu (Q) \\ &\quad \leq [\mu ,\sigma ]_{A_{p,q}}^{\frac{q'}{q}(q-1)} \sum _{\substack{Q\in \mathcal{S}\\Q\subset R}} \frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl(\frac{1}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}^{1-\frac{q'}{q}} \bigl\vert Q' \bigr\vert ^{\frac{1}{q'}} \langle \sigma \rangle _{Q'} ^{1-\frac{q'}{p'}} \bigl\vert Q' \bigr\vert ^{\frac{1}{p}} \biggr)^{q-1}\cdot \mu (Q) \\ &\quad = [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl(\frac{1}{\mu (Q)} \sum _{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}^{1-\frac{q'}{q}} \bigl\vert Q' \bigr\vert ^{\frac{1}{q'}} \cdot \langle \sigma \rangle _{Q'}^{1-\frac{q'}{p'}} \bigl\vert Q' \bigr\vert ^{\frac{1}{p}} \biggr)^{q-1}\cdot \mu (Q) \\ &\quad \leq [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl(\frac{1}{\mu (Q)} \biggl( \sum _{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}^{p'(1-\frac{q'}{q})} \bigl\vert Q' \bigr\vert ^{\frac{p'}{q'}} \biggr)^{\frac{1}{p'}} \\ &\qquad {} \times \biggl(\sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \sigma \rangle _{Q'}^{p(1-\frac{q'}{p'})} \bigl\vert Q' \bigr\vert \biggr)^{\frac{1}{p}} \biggr)^{q-1} \cdot \mu (Q). \end{aligned}
Since $$\frac{p'}{q'}>1$$, this give us
$$\biggl(\sum_{\substack{Q'\in \mathcal{S} \\ Q'\subset Q}} \langle \mu \rangle _{Q'}^{p'(1-\frac{q'}{q})} \bigl\vert Q' \bigr\vert ^{ \frac{p'}{q'}} \biggr)^{\frac{1}{p'}} \leq \biggl( \sum _{\substack{Q'\in \mathcal{S} \\ Q'\subset Q}} \langle \mu \rangle _{Q'}^{q'(1-\frac{q'}{q})} \bigl\vert Q' \bigr\vert \biggr)^{ \frac{1}{q'}}.$$
Note that $$q\geq 2$$. We have
$$0\leq q'\biggl(1-\frac{q'}{q}\biggr)< 1,\qquad 0\leq p\biggl(1- \frac{q'}{p'}\biggr)< 1.$$
It follows from Lemma 3.3 that
$$\sum_{\substack{Q'\in \mathcal{S} \\ Q'\subset Q}} \langle \mu \rangle _{Q'}^{q'(1-\frac{q'}{q})} \bigl\vert Q' \bigr\vert \lesssim \langle \mu \rangle _{Q}^{q'(1-\frac{q'}{q})} \vert Q \vert$$
and
$$\sum_{\substack{Q'\in \mathcal{S} \\ Q'\subset Q}} \langle \sigma \rangle _{Q'}^{p(1-\frac{q'}{p'})} \bigl\vert Q' \bigr\vert \lesssim \langle \sigma \rangle _{Q}^{p(1-\frac{q'}{p'})} \vert Q \vert .$$
Hence
\begin{aligned} & \bigl\Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \bigr\Vert _{L ^{q}(\mu )}^{q} \\ & \quad \lesssim [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl(\frac{1}{\mu (Q)} \biggl( \sum _{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}^{p'(1-\frac{q'}{q})} \bigl\vert Q' \bigr\vert ^{\frac{p'}{q'}} \biggr)^{\frac{1}{p'}} \\ &\qquad {} \times \biggl(\sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \sigma \rangle _{Q'}^{p(1-\frac{q'}{p'})} \bigl\vert Q' \bigr\vert \biggr)^{\frac{1}{p}} \biggr)^{q-1} \cdot \mu (Q) \\ &\quad \lesssim [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \frac{\sigma (Q)}{ \vert Q \vert ^{\frac{ \lambda }{n}}} \biggl(\frac{1}{\mu (Q)} \bigl( \langle \mu \rangle _{Q} ^{q'(1-\frac{q'}{q})} \vert Q \vert \bigr)^{\frac{1}{q'}} \bigl(\langle \sigma \rangle _{Q}^{p(1-\frac{q'}{p'})} \vert Q \vert \bigr)^{\frac{1}{p}} \biggr)^{q-1} \cdot \mu (Q) \\ &\quad = [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \sigma (Q)^{\frac{q}{p}}. \end{aligned}
Notice that
$$\frac{q}{p}\cdot \frac{\lambda }{n}=\frac{q}{p}\cdot \biggl(\frac{1}{q}+ \frac{1}{p'} \biggr) =1+ \frac{q-p}{p'p}>1.$$
This gives us $$\frac{q}{p}>\frac{n}{\lambda }$$.
Since $$\mathcal{S}$$ is a sparse family and $$|E(Q)|>\frac{1}{2}|Q|$$, we have
\begin{aligned} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\sigma (Q)^{\frac{q}{p}} \leq {}& \sigma (R)^{\frac{q}{p}-\frac{n}{\lambda }} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\sigma (Q)^{\frac{n}{ \lambda }} \\ \leq {}& 2\cdot \sigma (R)^{\frac{q}{p}-\frac{n}{\lambda }} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \bigl\vert E(Q) \bigr\vert \cdot \frac{ \sigma (Q)^{\frac{n}{\lambda }}}{ \vert Q \vert } \\ \leq {}& 2\cdot \sigma (R)^{\frac{q}{p}-\frac{n}{\lambda }} \int _{R} \bigl(M_{\lambda }(\sigma 1_{R}) \bigr)^{\frac{n}{\lambda }} \\ \leq {}& 2[\sigma ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda }}\sigma (R)^{\frac{q}{p}}. \end{aligned}
So
$$\bigl\Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \bigr\Vert _{L^{q}( \mu )}^{q} \lesssim [ \mu ,\sigma ]_{A_{p,q}}\cdot [\sigma ]_{A_{ \lambda ,\infty }}^{\frac{n}{\lambda }} \sigma (R)^{\frac{q}{p}}.$$
Taking the supreme over all cubes $$R\in \mathscr {D}$$, we obtain
$$\varTheta ^{\mathscr{D}}_{\mu ,\sigma }\lesssim [\mu ,\sigma ]_{A_{p,q}} ^{\frac{1}{q}}\cdot [\sigma ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda q}}.$$
Case 2:$$1< q<2$$. In this case, we have $$0\leq 1-\frac{q}{p'}<1$$. Using Proposition 3.2 and Lemma 3.3, we get
\begin{aligned} & \bigl\Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \bigr\Vert _{L ^{q}(\mu )}^{q} \\ &\quad \simeq \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{1}{ \vert Q \vert ^{\frac{ \lambda }{n}}}\sigma (Q) \biggl(\frac{1}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \frac{1}{ \vert Q' \vert ^{\lambda /n}}\sigma \bigl(Q'\bigr)\mu \bigl(Q' \bigr) \biggr)^{q-1}\cdot \mu (Q) \\ &\quad = \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{1}{ \vert Q \vert ^{\frac{ \lambda }{n}}}\sigma (Q) \biggl(\frac{1}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \mu \rangle _{Q'}\langle \sigma \rangle _{Q'}^{\frac{q}{p'}} \cdot \langle \sigma \rangle _{Q'}^{1-\frac{q}{p'}} \bigl\vert Q' \bigr\vert ^{2-\frac{\lambda }{n}} \biggr)^{q-1} \cdot \mu (Q) \\ &\quad \leq [\mu ,\sigma ]_{A_{p,q}}^{q-1} \sum _{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{1}{ \vert Q \vert ^{\frac{ \lambda }{n}}}\sigma (Q) \biggl( \frac{ \vert Q \vert ^{1-\frac{\lambda }{n}}}{\mu (Q)} \sum_{\substack{Q'\in \mathcal{S}\\Q'\subset Q}} \langle \sigma \rangle _{Q'}^{1-\frac{q}{p'}} \bigl\vert Q' \bigr\vert \biggr)^{q-1}\cdot \mu (Q) \\ &\quad \lesssim [\mu ,\sigma ]_{A_{p,q}}^{q-1} \sum _{\substack{Q\in \mathcal{S}\\Q\subset R}}\frac{1}{ \vert Q \vert ^{\frac{ \lambda }{n}}}\sigma (Q) \biggl( \frac{ \vert Q \vert ^{1-\frac{\lambda }{n}}}{\mu (Q)}\langle \sigma \rangle _{Q} ^{1-\frac{q}{p'}} \vert Q \vert \biggr)^{q-1}\cdot \mu (Q) \\ &\quad = [\mu ,\sigma ]_{A_{p,q}}^{q-1} \sum _{\substack{Q\in \mathcal{S}\\Q\subset R}} \langle \mu \rangle _{Q}^{2-q} \langle \sigma \rangle _{Q}^{\frac{q}{p'}(2-q)}\cdot \sigma (Q)^{\frac{q}{p}} \\ &\quad \lesssim [\mu ,\sigma ]_{A_{p,q}} \sum_{\substack{Q\in \mathcal{S}\\Q\subset R}} \sigma (Q)^{\frac{q}{p}}. \end{aligned}
We see from the arguments in Case 1 that
$$\sum_{\substack{Q\in \mathcal{S} \\ Q\subset R}}\sigma (Q)^{\frac{q}{p}}\leq 2[ \sigma ]_{A_{\lambda , \infty }}^{\frac{n}{\lambda }}\sigma (R)^{\frac{q}{p}}.$$
Hence
$$\bigl\Vert 1_{R}I^{\mathcal{S}(R)}_{\lambda }(\sigma 1_{R}) \bigr\Vert _{L^{q}( \mu )}^{q} \lesssim [ \mu ,\sigma ]_{A_{p,q}}\cdot [\sigma ]_{A_{ \lambda ,\infty }}^{\frac{n}{\lambda }} \sigma (R)^{\frac{q}{p}}.$$
Taking the supreme over all cubes $$R\in \mathscr {D}$$, we obtain
$$\varTheta ^{\mathscr{D}}_{\mu ,\sigma }\lesssim [\mu ,\sigma ]_{A_{p,q}} ^{\frac{1}{q}}\cdot [\sigma ]_{A_{\lambda ,\infty }}^{\frac{n}{\lambda q}}.$$
The estimates of $$\varTheta ^{\mathscr{D}}_{\sigma ,\mu }$$ can be proved with the symmetry. This completes the proof. □

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