Optimal constants for a class of Hausdorff operators on Lebesgue spaces

Abstract

As the extension of classical Hardy operator and Cesàro operator, Hausdorff operator plays an important role in the harmonic analysis, so it is significant to discuss the boundedness of this kind of operator on various function spaces. The article explores the boundedness of a kind of Hausdorff operators on Lebesgue spaces and calculates the optimal constants for the operators to be bounded on such spaces. In addition, the paper also obtains the necessary and sufficient for a kind of multilinear Hausdorff operators to be bounded on Lebesgue spaces and their optimal constants.

Keywords

Hausdorff operators / multilinear operators / Lebesgue spaces / optimal constants

1 Introduction and Main Results

The study of Hausdorff operators originated from the summation of series, see reference [11]. Given a function \mathrm{\Phi } defined on the interval (0,+\mathrm{\infty }), the one-dimensional Hausdorff operator is defined as

H_{\mathrm{\Phi }}(f)(x)={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Phi }(t)}{t}f\left(\frac{x}{t}\right)\text{d}t={\int }_0^{\mathrm{\infty }}\frac{\mathrm{\Phi }\left(\frac{x}{t}\right)}{t}f(t)\text{d}t,x\in \mathbb{R}.

Hausdorff operators include Hardy operators, Cesàro operators, Hardy-Littlewood-Polya operators and other well-known operators. For example, when \mathrm{\Phi }(t)=\frac{{\chi }_{(1,\mathrm{\infty })}(t)}{t},

H_{\mathrm{\Phi }}(f)(x)=H(f)(x)=\frac{1}{x}{\int }_0^{x}f(t)\text{d}t,x\ne 0.

This is the famous Hardy operator. In 1920, Hardy [9] obtained the classic Hardy inequality:

and proved \frac{p}{p-1} is the optimal constant. If \mathrm{\Phi }(t)={\chi }_{(0,1)}(t), then we have

H_{\mathrm{\Phi }}(f)(x)=H^{\ast }(f)(x)={\int }_x^{\mathrm{\infty }}\frac{f(t)}{t}\text{d}t.

This is the adjoint operator of Hardy operator. When \mathrm{\Phi }(t)={\chi }_{(0,1)}(t)(1-t{)}^{\delta } and \mathrm{\Phi }(t)={\chi }_{(0,1)}(t)+\frac{{\chi }_{(1,+\mathrm{\infty })}(t)}{t}, H_{\mathrm{\Phi }} is the well-known Cesàro operator C_{\delta } and Hardy-Littlewood-Polya operator P, respectively. They are defined as

C_{\delta }f(x)={\int }_0^1{t}^{-1}(1-t{)}^{\delta -1}f\left(t^{-1}x\right)\text{d}t,

P(f)(x)=Hf(x)+H^{\ast }f(x)={\int }_0^{\mathrm{\infty }}\frac{f(t)}{max\{t,x\}}\text{d}t.

There are several extended situations for the high-dimensional Hausdorff operators, which can be found in the literature [2‒4]. Assume that \mathrm{\Phi }(t) is a radial function on {\mathbb{R}}^{+}, then n-dimensional Hausdorff operator is defined as

{\mathcal{H}}_{\mathrm{\Phi },A}f(x)={\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Phi }(y)}{|y{|}^n}f(A(y)x)\text{d}y,

where A(y) is an n × n matrix and for almost everywhere y\in \mathrm{s}\mathrm{u}\mathrm{p}\mathrm{p}\mathrm{\Phi }, there is detA(y)\ne 0. When A(y)=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\frac{1}{|y|},\frac{1}{|y|},\dots ,\frac{1}{|y|}), we denote {\mathcal{H}}_{\mathrm{\Phi },A}={\mathcal{H}}_{\mathrm{\Phi }} , that is

{\mathcal{H}}_{\mathrm{\Phi }}f(x)={\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Phi }(y)}{|y{|}^n}f\left(\frac{x}{|y|}\right)\text{d}y.

Hausdorff operator not only includes Hardy operator, Cesàro operator, Hardy-Littlewood-Polya operator and other famous operators, but also has important applications in the fields of convergence and divergence of series, Fourier analysis, geometric analysis and so on (see [1, 4, 10, 11]). Therefore, in recent years, the study of Hausdorff operator has attracted the attention of many researchers around the world [3, 6-8, 12-16].

Assume that \mathrm{\Phi } is a locally integrable function on {\mathbb{R}}^n, for any vector \overrightarrow{\alpha }=\left({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\right), where {\alpha }_i\in \mathbb{R}, a class of Hausdorff operator {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }} is defined as

{\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}f(x)={\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Phi }(y)}{|y{|}^n}f\left(\frac{x_1}{|y{|}^{{\alpha }_1}},\frac{x_2}{|y{|}^{{\alpha }_2}},\dots ,\frac{x_n}{|y{|}^{{\alpha }_n}}\right)\text{d}y,x=\left(x_1,x_2,\dots ,x_n\right).

Let A(y)=\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}\left(\frac{1}{|y{|}^{{\alpha }_1}},\frac{1}{|y{|}^{{\alpha }_2}},\dots ,\frac{1}{|y{|}^{{\alpha }_n}}\right). Then {\mathcal{H}}_{\mathrm{\Phi },A}={\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}. When \overrightarrow{\alpha }=(1,1,\dots ,1), {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}={\mathcal{H}}_{\mathrm{\Phi }}.

It is easy to see that from Minkowski’s inequality and variable substitution, if

then for any 1⩽p⩽\mathrm{\infty },{\mathcal{H}}_{\mathrm{\Phi }} is bounded on L^p space. By simple calculation, when \mathrm{\Phi } is a non-negative function, the above condition is also necessary conditions for {\mathcal{H}}_{\mathrm{\Phi }} to be bounded on the L^p space. A natural question is what is a sufficient and necessary condition for {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }} to be bounded on L^p space? This paper will answer this question and further calculate the bounded norm for the operator

{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}.

On the other hand, in recent years, the study of boundedness of multilinear operators on various function spaces has become one of the important research directions in the field of harmonic analysis. In 2012, Chen et al. [5] defined the multilinear Hausdorff operators S_{\mathrm{\Phi }} and obtained their boundedness on the L^pspace. Assume that \mathrm{\Phi }\left(s_1,s_2,\dots ,s_m\right) is a locally integrable function on {\mathbb{R}}^{+}\times {\mathbb{R}}^{+}\times \cdots \times {\mathbb{R}}^{+}, then S_{\mathrm{\Phi }} is defined by

S_{\mathrm{\Phi }}\left(f_1,f_2,\dots ,f_m\right)(x)={\int }_{{\mathbb{R}}^{nm}}\frac{\mathrm{\Phi }\left(\frac{x}{\left|u_1\right|},\frac{x}{\left|u_2\right|},\dots ,\frac{x}{\left|u_m\right|}\right)}{{\left|u_1\right|}^n{\left|u_2\right|}^n\cdots {\left|u_m\right|}^n}\prod _{j=1}^m{f}_j\left(u_j\right)\text{d}{u}_1\text{d}{u}_2\cdots \text{d}{u}_m.

In 2014, Fan and Zhao [7] further studied the boundedness for two classes of fractional multilinear Hausdorff operators on Lebesgue spaces. Assume that vector \overrightarrow{u}=\left(u_1,u_2,\dots ,u_m\right),u_i\in {\mathbb{R}}^{n_i}, denote \text{d}\overrightarrow{u}=\text{d}{u}_1\text{d}{u}_2\cdots \text{d}{u}_m, vector \overrightarrow{\beta }=\left({\beta }_1,{\beta }_2,\dots ,{\beta }_m\right), {\beta }_i\in \mathbb{R} and \beta \in \mathbb{R}, two class of fractional multilinear Hausdorff operators are defined as

R_{\mathrm{\Phi },\overrightarrow{\beta }}\left(f_1,f_2,\dots ,f_m\right)(x)={\int }_{{\mathbb{R}}^{n_1}}{\int }_{{\mathbb{R}}^{n_2}}\cdots {\int }_{{\mathbb{R}}^{n_m}}\frac{\mathrm{\Phi }\left(\frac{x}{\left|u_1\right|},\frac{x}{\left|u_2\right|},\dots ,\frac{x}{\left|u_m\right|}\right)}{{\prod }_{i=1}^m{\left|u_i\right|}^{n_i-{\beta }_i}}\prod _{i=1}^m{f}_i\left(u_i\right)\text{d}\overrightarrow{u};

S_{\mathrm{\Phi },\beta }\left(f_1,f_2,\dots ,f_m\right)(x)={\int }_{{\mathbb{R}}^{n_1}}{\int }_{{\mathbb{R}}^{n_2}}\cdots {\int }_{{\mathbb{R}}^{n_m}}\frac{\mathrm{\Phi }\left(\frac{x}{|\overrightarrow{u}|}\right)}{|\overrightarrow{u}{|}^{{\sum }_{i=1}^m{n}_i-\beta }}\prod _{i=1}^m{f}_i\left(u_i\right)\text{d}\overrightarrow{u},

where |\overrightarrow{u}|=\sqrt{{\left|u_1\right|}^2+{\left|u_2\right|}^2+\cdots +{\left|u_m\right|}^2}.

Inspired by [7], this paper will consider another kind of multilinear Hausdorff operator. Assume that \mathrm{\Phi } is a locally integrable function on {\mathbb{R}}^{nm}, for any x\in {\mathbb{R}}^n,\overrightarrow{\beta }=\left({\beta }_1,{\beta }_2,\dots ,{\beta }_m\right) with {\beta }_i\in \mathbb{R}, multilinear Hausdorff operator is defined as

\begin{array}{rl}& {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }}\left(f_1,f_2,\dots ,f_m\right)(x)\\ & ={\int }_{{\mathbb{R}}^{nm}}\frac{\mathrm{\Phi }\left(y_1,y_2,\dots ,y_m\right)}{{\prod }_{i=1}^m{\left|y_i\right|}^n}\prod _{i=1}^m{f}_i\left(\frac{x}{{\left|y_i\right|}^{{\beta }_i}}\right)\text{d}{y}_1\text{d}{y}_2\cdots \text{d}{y}_m,y_i\in {\mathbb{R}}^n.\end{array}

When \overrightarrow{\beta }=(1,1,\dots ,1), denote {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }}={\mathcal{S}}_{\mathrm{\Phi },m}, that is

{\mathcal{S}}_{\mathrm{\Phi },m}\left(f_1,f_2,\dots ,f_m\right)(x)={\int }_{{\mathbb{R}}^{nm}}\frac{\mathrm{\Phi }\left(y_1,y_2,\dots ,y_m\right)}{{\prod }_{i=1}^m{\left|y_i\right|}^n}\prod _{i=1}^m{f}_i\left(\frac{x}{\left|y_i\right|}\right)\text{d}{y}_1\text{d}{y}_2\cdots \text{d}{y}_m.

This paper will discuss the necessary and sufficient condition for the boundedness of the {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }} on Lebesgue space and its optimal constant. It is worth mentioning that, different from literature [5] and [14], the conclusions of this paper do not need to be restricted to radial functions for the function \mathrm{\Phi } in the operator {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }}. The main theorem in this paper will be given below:

Theorem 1.1　 Let 1⩽p⩽\mathrm{\infty }, if \mathrm{\Phi }(y) is a non-negative function on {\mathbb{R}}^n, denote

\mathcal{A}:={\int }_{{\mathbb{R}}^n}\mathrm{\Phi }(y)|y{|}^{-n+\sum _{i=1}^n\frac{{\alpha }_i}{p}}\text{d}y.

Then {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }} is bounded operator on the L^p({\mathbb{R}}^n) space if and only if

Further, we have

{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p({\mathbb{R}}^n)\to L^p({\mathbb{R}}^n)}=\mathcal{A}.

Theorem 1.2　 Let 1⩽p,p_1,p_2,\dots ,p_m⩽\mathrm{\infty },\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}\cdots +\frac{1}{p_m}. Suppose that \mathrm{\Phi }(y_1,y_2,\dots ,y_m) is a non-negative function on {\mathbb{R}}^{nm}, denote

\mathcal{B}:={\int }_{{\mathbb{R}}^{nm}}\mathrm{\Phi }(y_1,y_2,\dots ,y_m)\prod _{i=1}^m|y_i{|}^{-n+\frac{n{\beta }_i}{p_i}}\text{d}{y}_1\text{d}{y}_2\cdots \text{d}{y}_m.

Then {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }} is bounded operator from the space L^{p_1}({\mathbb{R}}^n)\times L^{p_2}({\mathbb{R}}^n)\times \cdots \times L^{p_m}({\mathbb{R}}^n) to L^p({\mathbb{R}}^n) if and only if

Furthermore, we have

\Vert {\mathcal{S}}_{\mathrm{\Phi },m,\overrightarrow{\beta }}{\Vert }_{L^{p_1}({\mathbb{R}}^n)\times L^{p_2}({\mathbb{R}}^n)\times \cdots \times L^{p_m}({\mathbb{R}}^n)\to L^p({\mathbb{R}}^n)}=\mathcal{B}.

2 Proof of Theorems

Proof of Theorem 1.1　Firstly, we will prove the sufficiency. By Minkowski’s inequality and variable substitution, we can get

\begin{array}{rl}\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}f{\Vert }_{L^p({\mathbb{R}}^n)}& \le {\displaystyle {\int }_{{\mathbb{R}}^n}{\left({\displaystyle {\int }_{{\mathbb{R}}^n}{\left|\frac{\mathrm{\Phi }(y)}{|y{|}^n}f\left(\frac{x_1}{|y{|}^{{\alpha }_1}},\frac{x_2}{|y{|}^{{\alpha }_2}},\dots ,\frac{x_n}{|y{|}^{{\alpha }_n}}\right)\right|}^p}\text{d}x\right)}^{\frac{1}{p}}}\text{d}y\\ & =\mathcal{A}\Vert f{\Vert }_{L^p({\mathbb{R}}^n)}.\end{array}

So that, when , {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }} is bounded on L^p({\mathbb{R}}^n) space, and

(2.1)\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}{\Vert }_{L^p({\mathbb{R}}^n)\to L^p({\mathbb{R}}^n)}\le \mathcal{A}.

Secondly, we will prove the necessity. Let f_k(x_1,x_2,\dots ,x_n)=x_1^{-\frac{1+\frac{1}{k}}{p}}{\chi }_{\{x_1>1\}}(x_1)\cdots x_n^{-\frac{1+\frac{1}{k}}{p}}{\chi }_{\{x_n>1\}}(x_n), where k>1. By standard calculation, we have

(2.2)\Vert f_k{\Vert }_{L^p({\mathbb{R}}^n)}=k^{\frac{n}{p}}\text{，}

and

{\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}{f}_k(x)={\left(x_1^{-(1+\frac{1}{k})}\cdots x_n^{-(1+\frac{1}{k})}\right)}^{\frac{1}{p}}{\int }_{{\mathbb{R}}^n}\frac{\mathrm{\Phi }(y)}{|y{|}^n}|y{|}^{\frac{{\sum }_{i=1}^n{\alpha }_i(1+\frac{1}{k})}{p}}\prod _{i=1}^n{\chi }_{\left\{\frac{x_i}{|y{|}^{{\alpha }_i}}>1\right\}}\left(\frac{x_i}{|y{|}^{{\alpha }_i}}\right)\text{d}y.

If {\alpha }_i=0, for all i=1,2,\dots ,n, the theorem holds obviously. Now, we prove the theorem in three situations.

Case 1　If {\alpha }_i>0, let’s assume . Then for any i, there is . So, when x_i>k(k>1), we get

\begin{array}{rl}{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}{f}_k\Vert }_{L^p\left({\mathbb{R}}^n\right)}& ⩾({\int }_{x_1⩾k}\dots {{\int }_{x_n⩾k}\left|{\int }_{\left|y\right|⩽min\left\{{x_1}^{\frac{1}{{\alpha }_1}},\dots ,{x_n}^{\frac{1}{{\alpha }_n}}\right\}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\right|}^p\\ & \cdot x_1^{-\left(1+\frac{1}{k}\right)}\cdots x_n^{-\left(1+\frac{1}{k}\right)}\text{d}x{)}^{\frac{1}{p}}\\ & ⩾({\int }_{x_1⩾k}\cdots {{\int }_{x_n⩾k}\left|{\int }_{\left|y\right|⩽k^{\frac{1}{{\alpha }_n}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\right|}^p{x}_1^{-\left(1+\frac{1}{k}\right)}\\ & \cdots x_n^{-\left(1+\frac{1}{k}\right)}\text{d}x{)}^{{}^{\frac{1}{p}}}.\end{array}

Therefore, combining (2.2), there is

\begin{array}{rl}{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}{f}_k\Vert }_{L^p\left({\mathbb{R}}^n\right)}& ⩾{\left({\int }_{x_1⩾k}\cdots {\int }_{x_n⩾k}{x}_1^{-\left(1+\frac{1}{k}\right)}\cdots x_n^{-\left(1+\frac{1}{k}\right)}\text{d}x\right)}^{\frac{1}{p}}\\ & {\int }_{\left|y\right|⩽k^{\frac{1}{{\alpha }_n}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\\ & =k^{-\frac{1}{k}\cdot \frac{n}{p}}{k}^{\frac{n}{p}}{\int }_{\left|y\right|⩽k^{\frac{1}{{\alpha }_n}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\\ & =k^{-\frac{1}{k}\cdot \frac{n}{p}}{\Vert f_k\Vert }_{L_p({\mathbb{R}}^n)}{\int }_{\left|y\right|⩽k^{\frac{1}{{\alpha }_n}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y.\end{array}

So,

{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}⩾k^{-\frac{1}{k}\cdot \frac{n}{p}}{\int }_{\left|y\right|⩽k^{\frac{1}{{\alpha }_n}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y.

Let k\to +\mathrm{\infty }, (using k^{\frac{1}{k}}\to 1 and k^{\frac{1}{{\alpha }_n}}\to +\mathrm{\infty }). We have

(2.3){\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}⩾{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }\left(y\right){\left|y\right|}^{-n+\sum _{i=1}^n\frac{{\alpha }_i}{p}}\text{d}y.

Case 2　If , then \left|y\right|>{x_i}^{\frac{1}{{\alpha }_i}}. So, when x_i>k(k>1), we obtain

\begin{array}{rl}{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}{f}_k\Vert }_{L^p\left({\mathbb{R}}^n\right)}& ⩾{\left({\int }_{x_1⩾k}\cdots {\int }_{x_n⩾k}{x}_1^{-\left(1+\frac{1}{k}\right)}\cdots x_n^{-\left(1+\frac{1}{k}\right)}\text{d}x\right)}^{\frac{1}{p}}\\ & \cdot {\int }_{\left|y\right|⩾max\left\{{x_1}^{\frac{1}{{\alpha }_1}},\dots ,{x_n}^{\frac{1}{{\alpha }_n}}\right\}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\\ & ⩾k^{-\frac{1}{k}\cdot \frac{n}{p}}{k}^{\frac{n}{p}}{\int }_{\left|y\right|>k^{\frac{1}{{\alpha }_1}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y\\ & =k^{-\frac{1}{k}\cdot \frac{n}{p}}{\Vert f_k\Vert }_{L_p({\mathbb{R}}^n)}{\int }_{\left|y\right|>k^{\frac{1}{{\alpha }_1}}}\frac{\mathrm{\Phi }\left(y\right)}{{\left|y\right|}^n}{\left|y\right|}^{\frac{{\sum }_{i=1}^n{\alpha }_i\left(1+\frac{1}{k}\right)}{p}}\text{d}y.\end{array}

Let k\to +\mathrm{\infty } (using k^{\frac{1}{k}}\to 1 and k^{\frac{1}{{\alpha }_n}}\to 0), we have

(2.4){\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}⩾{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }\left(y\right){\left|y\right|}^{-n+{\sum }_{i=1}^n\frac{{\alpha }_i}{p}}\text{d}y.

Case 3　If , combining the above two situations, there is

So,

Let k\to +\mathrm{\infty }. Note: k^{\frac{1}{k}}\to 1, k^{\frac{1}{{\alpha }_i}}\to +\mathrm{\infty } and k^{\frac{1}{{\alpha }_{i+1}}}\to 0, we get

(2.5){\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}⩾{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }\left(y\right){\left|y\right|}^{-n+{\sum }_{i=1}^n\frac{{\alpha }_i}{p}}\text{d}y.

Combining (2.3), (2.4), (2.5), for any n-dimensional vector \overrightarrow{\alpha }, we have

(2.6){\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}⩾\mathcal{A}.

Because {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }} is bounded on L^p\left({\mathbb{R}}^{\text{n}}\right) space, therefore . Combined with (2.1) and (2.6), it is further obtained:

{\Vert {\mathcal{H}}_{\mathrm{\Phi },\overrightarrow{\alpha }}\Vert }_{L^p\left({\mathbb{R}}^n\right)\to L^p\left({\mathbb{R}}^n\right)}=\mathcal{A}.

Theorem 1 has been proved.

Proof of Theorem 1.2　Firstly, we will prove the sufficiency. Without losing generality, let’s assume m=2. By using the Minkowski inequality, Hölde inequality, we obtain

\begin{array}{rl}& {\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^p\left({\mathbb{R}}^n\right)}={\left({{\int }_{{\mathbb{R}}^n}\left|{\int }_{{\mathbb{R}}^{2n}}\frac{\mathrm{\Phi }\left(y_1,y_2\right)}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{f}_1\left(\frac{x}{{\left|y_1\right|}^{{\beta }_1}}\right)f_2\left(\frac{x}{{\left|y_2\right|}^{{\beta }_2}}\right)\text{d}{y}_1\text{d}{y}_2\right|}^p\text{d}x\right)}^{\frac{1}{p}}\\ & ⩽{\int }_{{\mathbb{R}}^{2n}}\frac{\left|\mathrm{\Phi }\left(y_1,y_2\right)\right|}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{\left({{\int }_{{\mathbb{R}}^n}\left|f_1\left(\frac{x}{{\left|y_1\right|}^{{\beta }_1}}\right)f_2\left(\frac{x}{{\left|y_2\right|}^{{\beta }_2}}\right)\right|}^p\text{d}x\right)}^{\frac{1}{p}}\text{d}{y}_1\text{d}{y}_2\\ & ⩽{\int }_{{\mathbb{R}}^{2n}}\frac{\left|\mathrm{\Phi }\left(y_1,y_2\right)\right|}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{\left({{\int }_{{\mathbb{R}}^n}\left|f_1\left(\frac{x}{{\left|y_1\right|}^{{\beta }_1}}\right)\right|}^{p_1}\text{d}x\right)}^{\frac{1}{p_1}}{\left({\int }_{{\mathbb{R}}^n}{\left|f_2\left(\frac{x}{{\left|y_2\right|}^{{\beta }_2}}\right)\right|}^{p_2}\text{d}x\right)}^{\frac{1}{p_2}}\text{d}{y}_1\text{d}{y}_2\\ & ={\int }_{{\mathbb{R}}^{2n}}\left|\mathrm{\Phi }\left(y_1,y_2\right)\right|{\left|y_1\right|}^{-n+\frac{n{\beta }_1}{p_1}}{\left|y_2\right|}^{-n+\frac{n{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2{\Vert f_1\Vert }_{L^{p_1}\left({\mathbb{R}}^n\right)}{\Vert f_2\Vert }_{L^{p_2}\left({\mathbb{R}}^n\right)}.\end{array}

Note that \mathrm{\Phi } is non-negative and , so {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }} is bounded in L^{p_1}\left({\mathbb{R}}^n\right)\times L^{p_2}\left({\mathbb{R}}^n\right) to L^p\left({\mathbb{R}}^n\right) and satisfies:

(2.7){\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\Vert }_{L^{p_1}\times L^{p_2}\to L^p}⩽\mathcal{B}.

Next, we will prove the necessity. We take f_1(x)=|x{|}^{-\frac{n+\frac{1}{k}}{p_1}}{\chi }_{\{|x|>1\}}(x),f_2(x)=|x{|}^{-\frac{n+\frac{1}{k}}{p_2}}{\chi }_{\{|x|>1\}}(x). Calculated by polar coordinate transformation:

{\Vert f_1\Vert }_{L^{p_1}\left({\mathbb{R}}^{\text{n}}\right)}={\omega }_n^{\frac{1}{p_1}}{k}^{\frac{1}{p_1}},{\Vert f_2\Vert }_{L^{p_2}\left({\mathbb{R}}^{\text{n}}\right)}={\omega }_n^{\frac{1}{p_2}}{k}^{\frac{1}{p_2}},

where {\omega }_n denotes the area of the unit sphere. Since \frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}, we have

(2.8){\Vert f_1\Vert }_{L^{p_1}\left({\mathbb{R}}^{\text{n}}\right)}{\Vert f_2\Vert }_{L^{p_2}\left({\mathbb{R}}^{\text{n}}\right)}={\omega }_n^{\frac{1}{p}}{k}^{\frac{1}{p}},

and

If {\beta }_1={\beta }_2=0, then the theorem clearly holds. Therefore, we prove the theorem in three cases.

Case 4　If {\beta }_1,{\beta }_2>0, then . So, when \left|x\right|>k,we have

Thus, combining (2.8), we have

Let k\to +\mathrm{\infty } (it is easy to know that k^{\frac{1}{k}}\to 1, k^{\frac{1}{{\beta }_1}},k^{\frac{1}{{\beta }_2}}\to \mathrm{\infty }). We have

{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^p}⩾{\Vert f_1\Vert }_{L^{p_1}}{\Vert f_2\Vert }_{L^{p_2}}{\int }_{{\mathbb{R}}^{\text{n}}}{\int }_{{\mathbb{R}}^{\text{n}}}\mathrm{\Phi }\left(y_1,y_2\right){\left|y_1\right|}^{-n+\frac{n{\beta }_1}{p_1}}{\left|y_2\right|}^{-n+\frac{n{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2.

So,

{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^{p_1}\times L^{p_2}\to L^p}⩾{\int }_{{\mathbb{R}}^{\text{n}}}{\int }_{{\mathbb{R}}^{\text{n}}}\mathrm{\Phi }\left(y_1,y_2\right){\left|y_1\right|}^{-n+\frac{n{\beta }_1}{p_1}}{\left|y_2\right|}^{-n+\frac{n{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2.

Case 5　If , then \left|y_1\right|>{\left|x\right|}^{\frac{1}{{\beta }_1}}, \left|y_2\right|>{\left|x\right|}^{\frac{1}{{\beta }_2}}. So, when \left|x\right|>k, we have

\begin{array}{rl}& {\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^p}^p⩾{\int }_{|x|>k}|x{|}^{-\left(n+\frac{1}{k}\right)}\\ & {\left|{\int }_{\left|y_2\right|>k^{\frac{1}{{\beta }_2}}}{\int }_{\left|y_1\right|>k^{\frac{1}{{\beta }_1}}}\frac{\mathrm{\Phi }\left(y_1,y_2\right)}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{\left|y_1\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_1}{p_1}}{\left|y_2\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2\right|}^p\text{d}x\\ & =k^{-1-\frac{1}{k}}{\omega }_n{\left|{\int }_{\left|y_2\right|>k^{\frac{1}{{\beta }_2}}}{\int }_{\left|y_1\right|>k^{\frac{1}{{\beta }_1}}}\frac{\mathrm{\Phi }\left(y_1,y_2\right)}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{\left|y_1\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_1}{p_1}}{\left|y_2\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2\right|}^p.\end{array}

So,

\begin{array}{rl}{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^p}& ⩾{\left(k^{-\frac{1}{k}}\right)}^{\frac{1}{p}}{\Vert f_1\Vert }_{L^{p_1}}{\Vert f_2\Vert }_{L^{p_2}}{\int }_{\left|y_2\right|>k^{\frac{1}{{\beta }_2}}}{\int }_{\left|y_1\right|>k^{\frac{1}{{\beta }_1}}}\\ & \frac{\mathrm{\Phi }\left(y_1,y_2\right)}{{\left|y_1\right|}^n{\left|y_2\right|}^n}{\left|y_1\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_1}{p_1}}{\left|y_2\right|}^{\left(n+\frac{1}{k}\right)\frac{{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2.\end{array}

Let k\to +\mathrm{\infty } (it is easy to know that k^{\frac{1}{k}}\to 1 and k^{\frac{1}{{\alpha }_1}},k^{\frac{1}{{\alpha }_2}}\to 0). We have

{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^p}⩾{\Vert f_1\Vert }_{L^{p_1}}{\Vert f_2\Vert }_{L^{p_2}}{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }\left(y_1,y_2\right){\left|y_1\right|}^{-n+\frac{n{\beta }_1}{p_1}}{\left|y_2\right|}^{-n+\frac{n{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2.

So,

{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\left(f_1,f_2\right)\Vert }_{L^{p_1}\times L^{p_2}\to L^p}⩾{\int }_{{\mathbb{R}}^n}{\int }_{{\mathbb{R}}^n}\mathrm{\Phi }\left(y_1,y_2\right){\left|y_1\right|}^{-n+\frac{n{\beta }_1}{p_1}}{\left|y_2\right|}^{-n+\frac{n{\beta }_2}{p_2}}\text{d}{y}_1\text{d}{y}_2.

Case 6　If there are some {\beta }_i>0 and some , without losing generality, we assume that . Combining the above two cases, similar to the proof of Case 3 in Theorem 1.1, we will not repeat it here.

Combining the above three cases, we have

(2.9){\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\Vert }_{L^{p_1}\times L^{p_2}\to L^p}⩾\mathcal{B}.

Further combining (2.7), there are:

{\Vert {\mathcal{S}}_{\mathrm{\Phi },2,\overrightarrow{\beta }}\Vert }_{L^{p_1}\times L^{p_2}\to L^p}=\mathcal{B}.

Theorem 1.2 is proved.

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