ZHANG Xingsong¹, WEI Mingquan², YAN Dunyan¹
1. School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China;
2. School of Mathematics and Statistics, Xinyang Normal University, Xinyang 464000, Henan, China

Received 23 December 2019; Revised 2 March 2020
Foundation items: Supported by the National Nature Science Foundation of China (11871452), Project of Henan Provincial Department of Education (18A110028), and the Nanhu Scholar Program for Young Scholars of Xinyang Normal University
Corresponding author: WEI Mingquan, E-mail: weimingquan11@mails.ucas.ac.cn

Abstract: In this paper, we calculate the norm of the Hausdorff operator $\mathscr{H}_Φ$ defined on the Morrey space with power weights L^{p, λ}($\mathbb{R}^n$, |x|^αdx) and the homogeneous central Morrey space with power weights $\dot B$^{p, λ}($\mathbb{R}^n$, |x|^αdx), respectively. We also extend our results to the product Hausdorff operator $\mathscr{H}$_Φ^m.
Keywords: Hausdorff operatorproduct Hausdorff operatorMorrey space
带幂权Morrey空间上Hausdorff算子的最佳界
张兴松¹, 魏明权², 燕敦验¹
1. 中国科学院大学数学科学学院, 北京 100049;
2. 信阳师范学院数学与统计学院, 河南信阳 464000
摘要: 分别得到定义在带幂权Morrey空间L^{p, λ}（$\mathbb{R}^n$，|x|^αdx）和带幂权齐次中心Morrey空间$\dot B$^{p, λ}（$\mathbb{R}^n$，|x|^αdx）上的Hausdorff算子$\mathscr{H}_Φ$的范数.并把这些结果推广到乘积Hausdorff算子$\mathscr{H}$_Φ^m.
关键词: Hausdorff算子乘积Hausdorff算子Morrey空间
Hausdorff operator^[1] was first introduced in 1917. As is well known, the Hausdorff operator includes many famous operators such as Hardy operator, adjoint Hardy operator, Cesaro operator and Hardy-Littlewood-Polya operator (see the examples below). Especially, Hardy operator as a kind of very important average operator is widely studied by many mathematicians. Researchers have built a relatively complete and mature theory about Hardy operator. Naturally, an in-depth study on Hausdorff operator is of great significance. In the recent years, Hausdorff operator and its variations have been widely studied by many researchers. For example, Chen et al.^[2-3] considered the boundedness properties of Hausdorff operator on Euclidean spaces, such as the Lebesgue spaces L^p, the Hardy spaces H^p and the Herz type spaces. For the sake of convenience, one can refer to Refs. [4-8] for more details of the recent progress on Hausdorff operators. In 2015, Gao et al.^[9] studied the boundness properties of the (fractional) Hausdorff operators on the Lebesgue spaces L^p with powers. It is the starting point of our research.
1 Preliminaries and main resultsWe first recall the classical one-dimensional Hausdorff operator. For a function ? defined on ${\mathbb{R}}^ + $=(0, ∞), the one-dimensional Hausdorff operator is defined by

$h_{\phi}(f)(x)=\int_{0}^{\infty} \frac{\phi(y)}{y} f\left(\frac{x}{y}\right) \mathrm{d} y, \quad x \in \mathbb{R}.$

It is worth mentioning that if we choose different ?, then we will get different operators. Here we present several important examples that have been extensively studied.

$\begin{aligned}&h_{\phi}(f)(x)= \\&\left\{\begin{array}{l}\frac{1}{x} \int_{0}^{x} f(y) \mathrm{d} y, x \neq 0, \text { if } \phi(y)=\frac{\chi_{(1, \infty)}(y)}{y} ; \\\int_{x}^{\infty} \frac{f(y)}{y} \mathrm{~d} y, \text { if } \phi(y)=\chi_{(0,1)}(y) ; \\\int_{0}^{1} \frac{(1-y)^{\delta-1}}{y} f\left(y^{-1} x\right) \mathrm{d} y, \text { if } \phi(y)=\frac{\chi_{(0,1)}(y)}{(1-y)^{-\delta}}; \\\int_{0}^{\infty} \frac{f(y)}{\max (y, x)} \mathrm{d} y, \text { if } \phi(y)=\chi_{(0,1)}(y)+\frac{\chi_{(1, \infty)}(y)}{y}.\end{array}\right.\end{aligned}$

Here, with the different choice of ?, h_? represents the famous Hardy operator, adjoint Hardy operator, Cesaro operator and Hardy-Littlewood-Polya operator, respectively.
High-dimensional Hausdorff operators have several versions (see Refs. [2, 4, 6]). In 2003, Anderson^[10] studied the n-dimensonal Hausdorff operator

$H_{\Phi} f(x)=\int_{\mathbb{R}^{n}} \frac{\Phi(x /|y|)}{|y|^{n}} f(y) \mathrm{d} y,$

where Φ is a function defined on ${\mathbb{R}}^n$. In 2015, Gao et al.^[9] studied the boundedness of H_Φ on Lebesgue spaces with power weights and gave the sufficient conditions for the boundedness of the (fractional) Hausdorff operators H_Φ on the Lebesgue spaces with power weights. For some special cases, these conditions are sufficient and necessary.
In this paper we will consider the following n-dimensional Hausdorff operator. Give a nonnegative function Φ defined on ${\mathbb{R}}^n$. The n-dimensional Hausdorff operator is defined by

$\mathscr{H}_{\Phi} f(x)=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}} f\left(\frac{x}{|y|}\right) \mathrm{d} y.$

Our main purpose of this paper is to study the sharp bound of the Hausdorff operators ${\mathscr{H}}_Φ$ on the Morrey spaces with power weights.
Before stating our main results, we give some ordinary notations as follows. For a∈ ${\mathbb{R}}^n$ and R>0, Q(a, R) denotes the cube centered at a with side length R and B(a, R) denotes the ball centered at a with radius R. For a nonnegative function ω, ω(A) denotes $\int_A {\omega (x){\rm{d}}x} $, where x∈ ${\mathbb{R}}^n$, A is a measurable subset of ${\mathbb{R}}^n$ and dx is the Lebesgue measure of ${\mathbb{R}}^n$.
Definition 1.1??Let 1≤p < ∞, -1/p≤λ < 0, ω=ω(x)=|x|^α and α≥0. The Morrey space with power weights L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx) is defined by

$\begin{aligned}&L^{p, \lambda}\left(\mathbb{R}^{n},|x|{ }^{\alpha} \mathrm{d} x\right) \\&:=\left\{f \in L_{\mathrm{loc}}^{p}\left(\mathbb{R}^{n},|x|{ }^{\alpha} \mathrm{d} x\right):\right. \\&\left.\quad\|f\|_{L^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)}<+\infty\right\},\end{aligned}$

where

$\begin{aligned}&\|f\|_{L^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)}= \\&\sup \limits_{a \in \mathbb{R}^{n}, R>0}\left(\frac{1}{\omega(Q(a, R))^{1+p \lambda}} \int_{Q(a, R)}|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} .\end{aligned}$

Definition 1.2??Let 1≤p < ∞, -1/p≤λ < 0, ω=ω(x)=|x|^α and α≥0. The homogeneous central Morrey space with power weights ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx) is defined by

$\begin{aligned}&\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|{ }^{\alpha} \mathrm{d} x\right) \\&:=\left\{f \in L_{\mathrm{loc}}^{p}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right):\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)}<+\infty\right\},\end{aligned}$

where

$\begin{aligned}&\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d}x\right)}= \\&\sup \limits_{R>0}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} .\end{aligned}$

Now we formulate our two main results about ${\mathscr{H}}_Φ$.
Theorem 1.1??Let 1≤p < ∞, -1/p≤λ < 0 and α≥0. Then ${\mathscr{H}}_Φ$ is bounded from ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx) to ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx) iff

$C:=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(\alpha+n) \lambda}} \mathrm{d} y<\infty.$

Furthermore, $\left\| {{\mathscr{H}}_Φ} \right\|_{{{\dot B}^{p, \lambda }}({\mathbb{R}}^n , |x{|^\alpha }{\rm{d}}x) \to {{\dot B}^{p, \lambda }}({\mathbb{R}}^n, |x{|^\alpha }{\rm{d}}x)} = C$.
However, when we consider the boundedness of ${\mathscr{H}}_Φ$ on L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx), we can only give a sufficient condition. Because of the appearance of power weights, the usual rotation method is not enough for us to get the necessary condition. Hence we state our partial result as follows.
Theorem 1.2??Let 1≤p < ∞, -1/p≤λ < 0 and α≥0. Then ${\mathscr{H}}_Φ$ is bounded from L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx) to L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx), provided that

$C:=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(\alpha+n) \lambda}} \mathrm{d} y<\infty.$

Proposition 1.1??Let 1≤p < ∞ and -1/p≤λ < 0. Then ${\mathscr{H}}_Φ$ is bounded from L^{p, λ}( ${\mathbb{R}}^n$, dx) to L^{p, λ}( ${\mathbb{R}}^n$, dx) iff

$C:=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+n\lambda}} \mathrm{d} y<\infty.$

Furthermore, $\left\| {{\mathscr{H}}_Φ} \right\|_{{{L}^{p, \lambda }}({\mathbb{R}}^n , {\rm{d}}x) \to {{L}^{p, \lambda }}({\mathbb{R}}^n, {\rm{d}}x)} = C$.
Our results can be extended to product spaces. Let m, n_i∈${\mathbb{N}}^ + $, 1≤i≤m. Φ(y₁, …, y_m) is a nonnegative function defined on ${\mathbb{R}}^{n_1}$×…× ${\mathbb{R}}^{n_m}$. The product Hausdorff operator is defined by

$\begin{aligned}\mathscr{H}_{\Phi}^{m}(f)\left(x_{1}, \cdots, x_{m}\right)= \int_{\mathbb{R}^{n_1}} \ldots \int_{\mathbb{R}^{n_m}} \frac{\Phi\left(y_{1}, \cdots, y_{m}\right)}{\left|y_{1}\right|^{n_{1}} \cdots\left|y_{m}\right|^{n_{m}}} \times \\& f\left(\frac{x_{1}}{\left|y_{1}\right|}, \cdots, \frac{x_{m}}{\left|y_{m}\right|}\right) \mathrm{d} y_{1} \cdots \mathrm{d} y_{m},\end{aligned}$

where x=(x₁, …, x_m)∈ ${\mathbb{R}}^{n_1}$×…× ${\mathbb{R}}^{n_m}$ and y=(y₁, …, y_m)∈ ${\mathbb{R}}^{n_1}$×…× ${\mathbb{R}}^{n_m}$.
For the sake of convenience, we use the following notations.
Let n =(n₁, …, n_m), ${\mathbb{R}}^\mathit{\boldsymbol{n}}$∶=( ${\mathbb{R}}^{n_1}$, …, ${\mathbb{R}}^{n_m}$). α =(α₁, …, α_m), |x|^α=|x|₁^α₁…|x_m|^α_m.
Correspondingly, we define the product Morrey spaces with power weights.
Definition 1.3??Let 1≤p < ∞, m∈ ${\mathbb{N}}^{+}$, -1/p≤λ_i < 0, λ=(λ₁, …, λ_m), ω_i(x_i)=|x_i|^α_i, α_i≥0 and 1≤i≤m. The product Morrey space with power weights L^{p, λ}( ${\mathbb{R}}^\mathit{\boldsymbol{n}}$, |x|^αdx) is defined by

$\begin{aligned}&L^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right) \\&:=\left\{f \in L_{\mathrm{loc}}^{p}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right):\|f\|_{L^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)}\right. \\& \ \ \ \ \ \ <+\boldsymbol{\infty}\},\end{aligned}$

where

$\begin{aligned}&{\|f\|_{L^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x| ^{\boldsymbol{\alpha}}{\mathrm{d}x}\right)}=\sup \limits_{{a_{i} \in \mathbb{R}{}^{n_{i}}{R_{i}}>0}\atop{1 \leqslant i \leqslant m}}}{\left(\frac{1}{\prod\nolimits_{i=1}^{m} \omega_{i}\left(Q\left(a_{i}, R_{i}\right)\right)^{1+p \lambda_{i}}} \times \right.}\\&\left.\int_{Q\left(a_{1}, R_{1}\right)} \cdots \int_{Q\left(a_{m}, R_{m}\right)}|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}}.\end{aligned}$

Definition 1.4??Let 1≤p < ∞, m∈ ${\mathbb{N}}^{+}$, -1/p≤λ_i < 0, λ=(λ₁, …, λ_m), ω_i(x_i)=|x_i|^α_i α_i≥0 and 1≤i≤m. The product homogeneous central Morrey space with power weights ${\dot B^{p, \mathit{\boldsymbol{\lambda }} }}$( ${\mathbb{R}}^{\mathit{\boldsymbol{n}}}$, |x|^αdx) is defined by

$\begin{aligned}&\dot{B}{}^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}}, \ \ |x|{ }^{\boldsymbol{\alpha}} \mathrm{d} x\right) \\&:=\left\{f \in L_{\mathrm{loc}}^{p}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right):\|f\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}{\left(R^{\boldsymbol{n}},|x|\right.}^{\boldsymbol{\alpha}}{\mathrm{d}x)}}<+\boldsymbol{\infty}\right\},\end{aligned}$

where

$\begin{aligned}&\|f\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} {\mathrm{d} x)}}=\sup \limits_{R_{i}>0,1 \leqslant i \leqslant m}\left(\frac{1}{\prod\nolimits_{i=1}^{m} \omega_{i}\left(B\left(0, R_{i}\right)\right)^{1+p \lambda_{i}}}\right) \times \\&\ \ \left.\int_{B\left(0, R_{1}\right)} \cdots \int_{B\left(0, R_{m}\right)}|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}}.\end{aligned}$

Now we formulate our two main results about ${\mathscr{H}}_\Phi ^m$.
Theorem 1.3??Let 1≤p < ∞, -1/p≤λ_i < 0, λ=(λ₁, …, λ_m).α_i≥0 and 1≤i≤m. Then ${\mathscr{H}}_\Phi ^m$ is bounded from ${\dot B^{p, \mathit{\boldsymbol{\lambda }} }}$( ${\mathbb{R}}^{\mathit{\boldsymbol{n}}}$, |x|^αdx)to ${\dot B^{p, \mathit{\boldsymbol{\lambda }}}}$( ${\mathbb{R}}^{\mathit{\boldsymbol{n}}}$, |x|^αdx)iff

$\begin{aligned}&C_{m}:=\int_{\mathbb{R}^{n_{1}}} \cdots \int_{\mathbb{R}^{n_{m}}} \times \\&\frac{\Phi\left(y_{1}, \cdots, y_{m}\right)}{\left|y_{1}\right|^{n_{1}+\left(\alpha_{1}+n_{1}\right) \lambda_{1}} \cdots\left|y_{m}\right|^{n_{m}+\left(\alpha_{m}+n_{m}\right) \lambda_{m}}} \mathrm{~d} y<\boldsymbol{\infty}.\end{aligned}$

Furthermore, $\left\| {{\mathscr{H}}_Φ} \right\|_{{{\dot B}^{p, \mathit{\boldsymbol{\lambda }} }}({\mathbb{R}}^{\mathit{\boldsymbol{n}}} , |x{|^{\mathit{\boldsymbol{\alpha}} } }{\rm{d}}x) \to {{\dot B}^{p, \mathit{\boldsymbol{\lambda }} }}({\mathbb{R}}^{\mathit{\boldsymbol{n}}}, |x{|^{\mathit{\boldsymbol{\alpha}} } }{\rm{d}}x)} = C$.
Theorem 1.4 Let 1≤p < ∞, -1/p≤λ_i < 0, λ=(λ₁, …, λ_m), α_i≥0 and 1≤i≤m. Then ${\mathscr{H}}_\Phi ^m$ is bounded from L^{p, λ}($\mathbb{R}^{\mathit{\boldsymbol{n}}}$, |x|^αdx) to L^{p, λ}($\mathbb{R}^{\mathit{\boldsymbol{n}}}$, |x|^αdx)if

$\begin{aligned}C_{m}&:=\int_{\mathbb{R}^{n_{1}}} \cdots \int_{\mathbb{R}^{n_{m}}} \times \\&\frac{\Phi\left(y_{1}, \cdots, y_{m}\right)}{\left|y_{1}\right|^{n_{1}+\left(\alpha_{1}+n_{1}\right) \lambda_{1}} \cdots\left|y_{m}\right|^{n_{m}+\left(\alpha_{m}+n_{m}\right) \lambda_{m}}} \mathrm{~d} y<\boldsymbol{\infty}.\end{aligned}$

2 Proof of main resultsProof of Theorem 1.1??We first prove the sufficiency. Since $C:\int_{{\mathbb{R}}^n} {\frac{{\Phi (y)}}{{|y{|^{n + (\alpha + n)\lambda }}}}} {\rm{d}}y < \infty $, it follows from Minkowski inequality that

$\begin{aligned}&\left\|\mathscr{H}_{\Phi}(f)\right\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)} \\&=\sup \limits_{R>0}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}\left|\mathscr{H}_{\Phi}(f)(x)\right|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \\&=\sup \limits_{R>0}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}\left|\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}} f\left(\frac{x}{|y|}\right) \mathrm{d} y\right|^{p} \times\right. \\&\left.\quad|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}}\\&\leqslant \sup \limits_{R>0} \int_{\mathbb{R}^{n}}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}\left|\frac{\Phi(y)}{|y|^{n}} f\left(\frac{x}{|y|}\right)\right|^{p} \times\right.\\&\quad\left.|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \mathrm{~d} y \\&= \sup \limits_{R>0} \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}\left|f\left(\frac{x}{|y|}\right)\right|^{p} \times\right.\\&\quad\left.|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \mathrm{~d} y \quad(x \mapsto|y| x)\\&= \sup \limits_{R>0} \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(n+\alpha) \lambda}}\left(\frac{1}{\omega(B(0, R /|y|))^{1+p \lambda}} \int_{B(0, R|y|)} \times\right.\\&\quad\left.|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right){ }^{\frac{1}{p}} \mathrm{~d} y \\&\leqslant \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(n+\alpha) \lambda}}\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)} \\&= C\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha}\mathrm{d} x\right)}.\end{aligned}$

(1)

Hence $\mathscr{H}_{\Phi}$ is bounded from ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx) to ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx).
Next we prove the necessity. Since $\mathscr{H}_{\Phi}$ is bounded from ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx) to ${\dot B^{p, \lambda }}$( ${\mathbb{R}}^n$, |x|^αdx), we have

$\left\|\mathscr{H}_{\Phi}\right\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right) \rightarrow \dot{B}{}^{p, \lambda}(\mathbb{R}^{n},|x| ^{\alpha}{\mathrm{d} x})}<\infty.$

From (1), we can get that

$\left\|\mathscr{H}_{\Phi}\right\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right) \rightarrow \dot{B}{}^{p, \lambda}(\mathbb{R}^{n},|x| ^{\alpha}{\mathrm{d} x})} \le C.$

(2)

If we take g(x)=|x|^(α+n)λ, it is easy to check that g belongs to ${\dot B^{p, \lambda }}$($\mathbb{R}^n$, |x|^αdx) and

$\mathscr{H}_{\Phi}(g)(x)={Cg}(x).$

(3)

In fact, we have that

$\mathscr{H}_{\Phi}(g)(x)=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}} \frac{|x|^{(\alpha+n) \lambda}}{|y|^{(\alpha+n) \lambda}} \mathrm{d} y={Cg}(x).$

It follows that

$\begin{aligned}&\|g\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)}\\&=\sup \limits_{R>0}\left(\frac{1}{\omega(B(0, R))^{1+p \lambda}} \int_{B(0, R)}|g(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \\&=\sup \limits_{R>0}\left(\frac{1}{\left(\frac{\omega_{n-1}}{\alpha+n} R^{\alpha+n}\right)^{1+p \lambda}} \int_{B(0, R)}|x|^{(\alpha+n) \lambda p+\alpha} \mathrm{d} x\right) \\&=\left(\left(\frac{\alpha+n}{\omega_{n-1}}\right)^{1+p \lambda} \frac{\omega_{n-1}}{(\alpha+n)(1+p \lambda)}\right)^{\frac{1}{p}}<\infty.\end{aligned}$

It implies from (3) that

$\left\|\mathscr{H}_{\Phi}\right\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d}x\right) \rightarrow \dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)} \geqslant C .$

(4)

Combining (2) with (4) yields that

$\left\|\mathscr{H}_{\Phi}\right\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d}x\right) \rightarrow \dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)}=C<\infty.$

Proof of Theorem 1.2??Since

$C:=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(\alpha+n) \lambda}} \mathrm{d} y<\infty,$

it follows from Minkowski inequality that

$\begin{aligned}&\left\|\mathscr{H}_{\Phi}(f)\right\|_{L^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)} \\&=\sup \limits_{a \in \mathbb{R}^{n}, R>0}\left(\frac{1}{\omega(Q(a, R))^{1+p \lambda}} \int_{Q(a, R)}\left|\mathscr{H}_{\Phi}(f)(x)\right|^{p} \times\right. \\&\left.\quad|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \\&=\sup \limits_{a \in \mathbb{R}^{n}, R>0}\left(\frac{1}{\omega(Q(a, R))^{1+p \lambda}} \int_{Q(a, R)} \left| \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}} \times\right.\right.\\&\left.\left. \quad f\left(\frac{x}{|y|}\right) \mathrm{d} y\right|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \\&\leqslant \left.\quad \sup \limits_{a \in \mathbb{R}^{n}, R>0} \int_{\mathbb{R}^{n}}\left(\frac{1}{\omega(Q(a, R))^{1+p \lambda}} \int_{Q(a, R)} \right| \frac{\Phi(y)}{|y|^{n}} \times\right. \\&\left.\left.\quad f\left(\frac{x}{|y|}\right) \mathrm{d} y\right|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \mathrm{~d} y \\&= \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}}\left(\frac{1}{\omega(Q(a, R))^{1+p \lambda}} \int_{Q(a, R)}\left|f\left(\frac{x}{|y|}\right)\right|^{p} \times\right. \\&\left.\quad|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \mathrm{~d} y(x \mapsto|y| x)\\&=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}} \left( \frac{\omega\left(Q\left(\frac{a}{|y|}, \frac{R}{|y|}\right)\right)^{1+p \lambda}}{\omega(Q(a, R))^{1+p \lambda}} \frac{1}{\omega\left(Q\left(\frac{a}{|y|}, \frac{R}{|y|}\right)\right)^{1+p \lambda}} \times\right. \\&\left.\quad \int_{Q\left(\frac{a}{|y|}, \frac{R}{|y|}\right)}|f(x)|^{p}|x|^{\alpha}|y|^{\alpha+n} \mathrm{~d} x\right)^{\frac{1}{p}} \mathrm{~d} y \\&=\int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n}}\left( |y|^{-(\alpha+n) p \lambda} \frac{1}{\omega\left(Q\left(\frac{a}{|y|}, \frac{R}{|y|}\right)\right)^{1+p \lambda}} \times\right.\\&\left.\quad \int_{Q\left(\frac{a}{|y|}, \frac{R}{|y|}\right)}|f(x)|^{p}|x|^{\alpha} \mathrm{d} x\right)^{\frac{1}{p}} \mathrm{~d} y \\&\leqslant\ \int_{\mathbb{R}^{n}} \frac{\Phi(y)}{|y|^{n+(n+\alpha) \lambda}}\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d} x\right)} \\&=C\|f\|_{\dot{B}{}^{p, \lambda}\left(\mathbb{R}^{n},|x|^{\alpha} \mathrm{d}x\right)}.\end{aligned}$

Hence $\mathscr{H}_{\Phi}$ is bounded from L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx) to L^{p, λ}( ${\mathbb{R}}^n$, |x|^αdx).
Proof of Proposition 1.1??By Theorem 1.2 we merely need to prove the necessity. More precisely, we need to find a function g∈L^{p, λ}( ${\mathbb{R}}^n$, dx) such that

$\frac{\left\|\mathscr{H}_{\Phi}(g)\right\|_{L^{p, \lambda}\left(\mathbb{R}^{n}, \mathrm{~d} x\right)}}{\|g\|_{L^{p, \lambda}\left(\mathbb{R}^{n}, \mathrm{~d} x\right)}}=C.$

Taking g(x)=|x|^nλ, by Ref.[11], we can easily check g∈L^{p, λ}( ${\mathbb{R}}^n$, dx). For the sake of completeness, we give the specific proof. We consider two cases:
Case 1: if |a|>2R, then |x|>R. For $ - \frac{1}{p} < \lambda < 0$, we have

$\begin{aligned}&\quad \frac{1}{|Q(a, R)|{ }^{1+p \lambda}} \int_{Q(a, R)}|x|{ }^{n \lambda p} \mathrm{~d} x \\&\leqslant \frac{1}{|Q(a, R)|^{1+p \lambda}} \int_{Q(a, R)} R^{n \lambda p} \mathrm{~d} x=1.\end{aligned}$

Case 2: if |a| < R, then Q(a, R)?Q(0, 3R), we have

$\begin{aligned}&\frac{1}{|Q(a, R)|^{1+p \lambda}} \int_{Q(a, R)}|x|{ }^{n \lambda p} \mathrm{~d} x \\&\leqslant \frac{1}{|Q(a, R)|^{1+p \lambda}} \int_{Q(0,3 R)} R^{n \lambda p} \mathrm{~d} x \\&=\frac{|Q(0,3 R)|^{1+p \lambda}}{|Q(a, R)|^{1+p \lambda}} \frac{1}{|Q(0,3 R)|^{1+p \lambda}} \int_{Q(0,3 R)}|x|^{n \lambda p} \mathrm{~d} x \\&=3^{n(1+p \lambda)} \frac{1}{|Q(0,3 R)|^{1+p \lambda}} \int_{Q(0,3 R)}|x|{ }^{n \lambda p} \mathrm{~d} x \\&\leqslant 3^{n(1+p \lambda)}(\sqrt{n})^{n \lambda p} \\&<+\infty .\end{aligned}$

Hence, g∈L^{p, λ}( ${\mathbb{R}}^n$, dx). Since $\mathscr{H}_{\Phi}$(g)(x)=Cg(x), the proof is finished.
Proof of Theorem 1.3??We just prove the theorem for m=2. For m>3, the method is similar. We first prove the sufficiency.
Since C₂ < ∞, by the definitions of $\mathscr{H}_\Phi ^m$ and ${\dot B^{p, \mathit{\boldsymbol{\lambda }} }}$($\mathbb{R}^{\mathit{\boldsymbol{n}}}$, |x|^αdx), it follows from Minkowski inequality that

$\begin{aligned}&\left\|\mathscr{H}_{\Phi}^{2}(f)\right\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)} \\&=\sup \limits_{R_{1}, R_{2}>0}\left(\frac{1}{\prod\nolimits_{i=1}^{2} \omega_{i}\left(B\left(0, R_{i}\right)\right)^{1+p \lambda_{i}}} \int_{B\left(0, R_{1}\right)} \int_{B\left(0, R_{2}\right)} \times\right. \\&\left.\quad\left|\mathscr{H}_{\Phi}^{2}(f)\left(x_{1}, x_{2}\right)\right|^{p}|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)^{\frac{1}{p}} \\&=\sup \limits_{R_{1}, R_{2}>0}\left(\frac{1}{\prod\nolimits_{i=1}^{2} \omega_{i}\left(B\left(0, R_{i}\right)\right)^{1+p \lambda_{i}}} \int_{B\left(0, R_{1}\right)} \int_{B\left(0, R_{2}\right)} \left| \int_{\mathbb{R}}{ }_{n_{1}} \times\right. \right.\\&\left.\left.\quad \int_{\mathbb{R}^{n_{2}}} \frac{\Phi\left(y_{1}, y_{2}\right)}{\left|y_{1}\right|^{n_{1}}\left|y_{2}\right|^{n_{2}}} f\left(\frac{x_{1}}{\left|y_{1}\right|}, \frac{x_{2}}{\left|y_{2}\right|}\right) \mathrm{d} y_{1} \mathrm{~d} y_{2}\right|^{p}|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)^{\frac{1}{p}}\\&\leqslant \sup \limits_{R_{1}, R_{2}>0} \int_{\mathbb{R}^{n_{1}}} \int_{\mathbb{R}^{n_{2}}} \frac{\Phi\left(y_{1}, y_{2}\right)}{\left|y_{1}\right|^{n_{1}}\left|y_{2}\right|^{n_{2}}}\left(\frac{1}{\prod\nolimits_{i=1}^{2} \omega_{i}\left(B\left(0, R_{i}\right)\right)^{1+p \lambda_{i}}} \times\right. \\&\quad \int_{\mathbb{R}^{n_{1}}} \int_{\mathbb{R}^{n_{2}}}\left|f\left(\frac{x_{1}}{\left|y_{1}\right|}, \frac{x_{2}}{\left|y_{2}\right|}\right)\right|^{p}\left|x_{1}\right|^{\alpha_{1}}\left|x_{2}\right|^{\alpha_{2}} \times \\&\left.\quad \mathrm{d} x_{1} \mathrm{~d} x_{2}\right)^{\frac{1}{p}} \mathrm{~d} y_{1} \mathrm{~d} y_{2}\\&=\sup \limits_{R_{1}, R_{2}>0} \int_{\mathbb{R}^{n_{1}}} \int_{\mathbb{R}^{n_{2}}} \frac{\Phi\left(y_{1}, y_{2}\right)}{\left|y_{1}\right|^{n_{1}+\left(\alpha_{1}+n_{1}\right) \lambda_{1}}\left|y_{2}\right|^{n_{2}+\left(\alpha_{2}+n_{2}\right) \lambda_{2}}} \times\\&\quad\left(\frac{1}{\prod\nolimits_{i=1}^{2} \omega_{i}\left(B\left(0, \frac{R}{\left|y_{i}\right|}\right)\right)^{1+p \lambda_{i}}} \int_{B\left(0, \frac{R_{1}}{\mid y_{1}\mid}\right)} \int_{B\left(0, \frac{R_{2}}{\left|y_{2}\right|}\right)} \times\right.\\&\quad\left.\left|f\left(x_{1}, x_{2}\right)\right|^{p}\left|x_{1}\right|^{\alpha_{1}}\left|x_{2}\right|^{\alpha_{2}} \mathrm{~d} x_{1} \mathrm{~d} x_{2}\right)^{\frac{1}{p}} \mathrm{~d} y_{1} \mathrm{~d} y_{2} \text {. }\end{aligned}$

Then we get

$\begin{aligned}&\left\|\mathscr{H}_{\Phi}^{2}(f)\right\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right) }\\&\leqslant \sup \limits_{R_{1}, R_{2}>0} \int_{\mathbb{R}^{n_{1}}} \int_{\mathbb{R}^{n_{2}}} \frac{\Phi\left(y_{1}, y_{2}\right)}{\left|y_{1}\right|^{n_{1}+\left(\alpha_{1}+n_{1}\right) \lambda_{1}}\left|y_{2}\right|^{n_{2}+\left(\alpha_{2}+n_{2}\right) \lambda_{2}}} \times \\&\quad\|f\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)} \\&=C_{2}\|f\|_{\dot{B}{}^{p, \boldsymbol{\lambda}}\left(\mathbb{R}^{\boldsymbol{n}},|x|^{\boldsymbol{\alpha}} \mathrm{d} x\right)}.\end{aligned}$

Hence $\mathscr{H}_\Phi ^2$ is bounded from ${\dot B^{p, \mathit{\boldsymbol{\lambda }} }}$($\mathbb{R}^{\mathit{\boldsymbol{n}}}$, |x|^αdx)to ${\dot B^{p, \mathit{\boldsymbol{\lambda }} }}$($\mathbb{R}^{\mathit{\boldsymbol{n}}}$, |x|^αdx).
Next we prove the necessity. Take

$g\left(x_{1}, x_{2}\right)=\left|x_{1}\right|{ }^{\left(\alpha_{1}+n_{1}\right) \lambda_{1}}\left|x_{2}\right|^{\left(\alpha_{2}+n_{2}\right) \lambda_{2}}.$

By the similar consideration and computation of the necessity of Theorem 1.1, we can obtain our conclusion immediately.
Proof of Theorem 1.4 The proof of Theorem 1.4 is similar to the proof of Theorem 1.2, so we omit the details.

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