In 1984,Carton-Lebrun and Fosset^[1] defined the weighted Hardy-Littlewood average operator U_ψ

${{U}_{\psi }}f\left( x \right)=\int\limits_{0}^{1}{{}}f\left( xt \right)\psi \left( t \right)dt,$

(1)

where ψ:[0,1]→[0,∞) is a function. Evidently the operator U_ψ deeply depends on the nonnegative function ψ. For example,when n=1 and ψ(x)=1 for x∈［0,1],the operator H_ψ is just reduced to the classical Hardy operator

$Hf\left( x \right)=\frac{1}{x}\int\limits_{0}^{x}{{}}f\left( t \right)dt,$

(2)

for x≠0. Consequently,U_ψ is the more extensive Hardy operator and sometimes is called the generalized Hardy operator.

The classical Hardy operator H is bounded on L^p($\mathbb{R}$). That is,for 1<p≤∞,

${{\left\| HF \right\|}_{{{L}^{P}}\left( \mathbb{R} \right)}}\le \frac{p}{p-1}{{\left\| f \right\|}_{{{L}^{P}}\left( \mathbb{R} \right)}}$

holds,where the constant $\frac{p}{p-1}$ is best possible.

In Ref.^[2],Xiao considered the generalized Hardy operator U_ψ and obtained the following theorem.

Theorem A  Suppose that ψ:[0,1]→[0,∞) is a nonnegative function and p∈［1,∞]. Then the operator U_ψ is bounded on L^p($\mathbb{R}$ⁿ) if and only if

$\int\limits_{0}^{1}{{}}{{t}^{-\frac{1}{p}}}\psi \left( t \right)dt<\infty .$

(3)

Moreover,if the inequality (3) holds,then the operator norm of U_ψ on L^p($\mathbb{R}$ⁿ) is given by

$\|{{U}_{\psi }}{{\|}_{{{L}^{p}}({{\mathbb{R}}^{n}})\to {{L}^{p}}({{\mathbb{R}}^{n}})}}=\int\limits_{0}^{1}{{}}{{t}^{-\frac{1}{p}}}\psi \left( t \right)dt;$

(4)

and U_ψ:BMO($\mathbb{R}$ⁿ)→BMO($\mathbb{R}$ⁿ) exists as a bounded operator if and only if

$\int\limits_{0}^{1}{{}}\psi \left( t \right)dt<\infty .$

(5)

Moreover,if the inequality (5) holds,then the operator norm of U_ψ on BMO($\mathbb{R}$ⁿ) is given by

$\|{{U}_{\psi }}{{\|}_{BMO({{\mathbb{R}}^{n}})\to BMO({{\mathbb{R}}^{n}})}}=\int\limits_{0}^{1}{{}}\psi \left( t \right)dt.$

(6)

Recently,Chen et al.^[3] studied the operator U_ψ defined as

$\begin{align} & {{U}_{\psi }}f\left( x \right)=\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}} \\ & f({{x}_{1}}{{t}_{1}},\ldots ,{{x}_{n}}{{t}_{n}})\psi ({{t}_{1}},\ldots ,{{t}_{n}})d{{t}_{1}}\ldots d{{t}_{n}} \\ \end{align}$

(7)

where f is defined on $\mathbb{R}$ⁿ and ψ:[0,1]ⁿ→[0,∞). U_ψ is an operator defined on the one dimensional product space. In Ref.［3],the following theorem was obtained.

Theorem B  Suppose that ψ:[0,1]ⁿ→[0,∞) is a nonnegative function and p∈［1,∞]. Then,the operator U_ψ is bounded on L^p($\mathbb{R}$ⁿ) if and only if

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}{{\left( {{t}_{1}}\cdots {{t}_{n}} \right)}^{-\frac{1}{p}}}\psi \left( {{t}_{1}},\cdots ,{{t}_{n}} \right)dt<\infty .$

(8)

Moreover,if the inequality (8) holds,then the operator norm of U_ψ on L^p($\mathbb{R}$ⁿ) is given by

${{\left\| {{U}_{\psi }} \right\|}_{{{L}^{p}}({{\mathbb{R}}^{n}})\to {{L}^{p}}({{\mathbb{R}}^{n}})}}=\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}{{\left( {{t}_{1}}\cdots {{t}_{n}} \right)}^{-\frac{1}{p}}}\psi \left( {{t}_{1}},\cdots ,{{t}_{n}} \right)dt.$

(9)

Campanato space ε^α,p($\mathbb{R}$ⁿ) was first introduced by Campanato in Ref.^[4]. The definition of Campanato space is as follows.

Definition A  Let -∞<α<∞ and 0<p<∞. A locally integrable function f is said to belong to Campanato space ε^α,p($\mathbb{R}$ⁿ) if there exists some constant C>0 such that for any cube Q⊂$\mathbb{R}$ⁿ with all the sides parallel to the axes,

$\frac{1}{{{\left| Q \right|}^{\frac{\alpha }{n}}}}{{\left( \frac{1}{\left| Q \right|}{{\int }_{Q}}{{\left| f\left( x \right)-{{f}_{Q}} \right|}^{p}}dx \right)}^{1/q}}\le C.$

(10)

The minimal constant C is defined to be the ε^α,p($\mathbb{R}$ⁿ) norm of f and denoted by ‖f‖_{ε^α,p($\mathbb{R}$ⁿ)}.

In Ref.^[5],Zhao et al. studied the boundedness for U_ψ on the space ε^α,p($\mathbb{R}$ⁿ). Theorem C was obtained.

Theorem C  Suppose that ψ:[0,1]→[0,∞) is a nonnegative function and p∈［1,∞),-np≤α<1. Then U_ψ is a bounded operator on ε^α,p($\mathbb{R}$ⁿ) if and only if

$\int\limits_{0}^{1}{{}}{{t}^{\alpha }}\psi \left( t \right)dt<\infty .$

(11)

Moreover,

${{\left\| {{U}_{\psi }} \right\|}_{{{\varepsilon }^{\alpha ,p}}({{\mathbb{R}}^{n}})\to {{\varepsilon }^{\alpha ,p}}({{\mathbb{R}}^{n}})}}=\int\limits_{0}^{1}{{}}{{t}^{\alpha }}\psi \left( t \right)dt.$

(12)

Motivated by the previous studies^{[2-3, 5]},we devoted ourselves to investigating the boundedness of the operators U_ψ at the endpoint. A simple computation implies that U_ψ is not bounded on BMO($\mathbb{R}$ⁿ) and thus is not bounded on ε^α,p($\mathbb{R}$ⁿ),since ε^α,p($\mathbb{R}$ⁿ) equals to BMO($\mathbb{R}$ⁿ) as α=0. It is necessary for us to find some new spaces to replace BMO($\mathbb{R}$ⁿ) and ε^α,p($\mathbb{R}$ⁿ). In this work,we mainly consider this question and introduce two new spaces RMO($\mathbb{R}$ⁿ) and R^α,p($\mathbb{R}$ⁿ). We shall give their definitions as follows.

1 Some definitions

Before we put forward our main results,some useful definitions will be given. First we introduce the spaces RMO($\mathbb{R}$ⁿ) and R^α,p($\mathbb{R}$ⁿ) corresponding to BMO($\mathbb{R}$ⁿ) and ε^α,p($\mathbb{R}$ⁿ),respectively.

Definition 1.1  Let f∈L_loc($\mathbb{R}$ⁿ). We say that f∈RMO($\mathbb{R}$ⁿ) if and only if

${{\left\| f \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}:=\underset{R\in {{\mathbb{R}}^{n}}}{\mathop{\sup }}\,\frac{1}{\left| R \right|}\int_{R}{{}}\left| f\left( x \right)-{{f}_{R}} \right|dx<\infty ,$

(13)

where the supremum is taken over all rectangles with all the sides parallel to the axes and f_R denotes the average of f over R,i.e.,f_R=$\frac{1}{\left| R \right|}$∫_Rf(x)dx.

Definition 1.2  Suppose that f∈L_loc($\mathbb{R}$ⁿ),-∞<α<∞ and 0<p<∞. We say that f∈R^α,p($\mathbb{R}$ⁿ) if and only if

$\frac{1}{{{\left| R \right|}^{\frac{\alpha }{n}}}}{{\left( \frac{1}{\left| R \right|}{{\int }_{R}}{{\left| f\left( x \right)-{{f}_{R}} \right|}^{p}}dx \right)}^{1/q}}\le C.$

(14)

holds for all rectangle R∈$\mathbb{R}$ⁿ. The minimal constant C is defined to be the R^α,p($\mathbb{R}$ⁿ) norm of f and denoted by ‖f‖_{R^α,p($\mathbb{R}$ⁿ)}.

Definition 1.3  Suppose that the measurable function f is defined on $\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n and φ:[0,1]ⁿ→[0,∞). The operator

${{U}_{\psi }}f\left( x \right)=\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}$

(15)

is called the generalized Hardy operator on the high dimensional product space,where xⁱ∈$\mathbb{R}$^m_i with i=1,2,…,n.

Obviously,‖·‖_RMO forms a norm if we define RMO as the quotient space of all equivalent classes of functions whose difference is a constant. By Definition 1.1 and 1.2,it is not difficult for us to deduce that the space BMO($\mathbb{R}$ⁿ) strictly contains RMO($\mathbb{R}$ⁿ) and RMO($\mathbb{R}$ⁿ)⊃L^∞($\mathbb{R}$ⁿ).

Moreover,we conclude that the space R^α,p($\mathbb{R}$ⁿ) is strictly contained in ε^α,p($\mathbb{R}$ⁿ) and equals to the space RMO($\mathbb{R}$ⁿ) when α=0 as well.

2 Main results and their proofs

First,we study the boundedness of the operators U_ψ defined on the space RMO as in (13) and the space R^α,p($\mathbb{R}$ⁿ) as in (14),and so does the operator U_φ.

Theorem 2.1  Let ψ:［0,1]ⁿ→[0,+∞) be a function. Then U_ψ:RMO($\mathbb{R}$ⁿ)→RMO($\mathbb{R}$ⁿ) exists as a bounded operator if and only if

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt<\infty .$

(16)

Moreover,when (16) holds,the operator norm of U_ψ on RMO($\mathbb{R}$ⁿ) is given by

$\|{{U}_{\psi }}{{\|}_{RMO({{\mathbb{R}}^{n}})\to RMO({{\mathbb{R}}^{n}})}}=\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt.$

(17)

Proof  In what follows,for each t=(t₁,…,t_n)∈$\mathbb{R}$ⁿ with t_i>0,i=1,…,n and rectangle R=［a₁,b₁]×[a₂,b₂]×…×[a_n,b_n]∈$\mathbb{R}$ⁿ,define tR=[t₁a₁,t₁b₁]×[t₂a₂,t₂b₂]×…×[t_na_n,t_nb_n]. Assume that (16) holds. If f∈RMO($\mathbb{R}$ⁿ),then,for any rectangle R,it follows from the Fubini’s Theorem that

$\begin{align} & {{\left( {{U}_{\psi }}f \right)}_{R}}=\frac{1}{\left| R \right|}{{\int }_{R}}({{U}_{\psi }}f)\left( y \right)dy \\ & =\frac{1}{\left| R \right|}{{\int }_{R}}\int_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}f\left( {{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}} \right)\psi \left( t \right)dtdy \\ & =\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\frac{1}{\left| R \right|}{{\int }_{R}}f\left( {{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}} \right)dy\psi \left( t \right)dtdy \\ & =\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\frac{1}{\left| tR \right|}{{\int }_{tR}}f\left( {{z}_{1}},\ldots ,{{z}_{n}} \right)dy\psi \left( t \right)dt \\ & =\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}{{f}_{tR}}\psi \left( t \right)dt, \\ \end{align}$

(18)

where in the last equality we use the variable substitution z_i=t_iy_i with i=1,2,…,n. We conclude from the Minkowski’s integral inequality that

$\begin{align} & \frac{1}{\left| R \right|}{{\int }_{R}}\left| \left( {{U}_{\psi }}f \right)\left( y \right)-{{\left( {{U}_{\psi }}f \right)}_{R}} \right|dy \\ & =\frac{1}{\left| R \right|}{{\int }_{R}}\left| \left( {{U}_{\psi }}f \right)\left( y \right)-\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}{{f}_{tR}}\psi \left( t \right)dt \right|dy \\ & =\frac{1}{\left| R \right|}{{\int }_{R}}\left| \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}} \right. \\ & \left. \left( f\left( {{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}} \right)-{{f}_{tR}} \right)\psi \left( t \right)dt \right|dy\le \\ & \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\left( \frac{1}{\left| R \right|}{{\int }_{R}}\left| \left( {{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}} \right)-{{f}_{tR}} \right|dy \right)\psi \left( t \right)dt= \\ & \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\left( \frac{1}{\left| tR \right|}{{\int }_{tR}}\left| f\left( {{z}_{1}},\ldots ,{{z}_{n}} \right)-{{f}_{tR}} \right|dz \right)\psi \left( t \right)dt \\ & \le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt{{\left\| f \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}. \\ \end{align}$

(19)

The inequality (19) shows that

${{\left\| {{U}_{\psi }}f \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}\le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt{{\left\| f \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}.$

Thus we have

${{\left\| {{U}_{\psi }} \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}\le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt$

(20)

Conversely,if U_ψ is bounded on RMO($\mathbb{R}$ⁿ),then we can choose

${{f}_{0}}\left( x \right)=\left\{ \begin{matrix} \begin{matrix} 1, & x\in \mathbb{R}_{l}^{n}, \\ \end{matrix} \\ \begin{matrix} -1, & x\in \mathbb{R}_{r}^{n}, \\ \end{matrix} \\ \end{matrix} \right.$

(21)

where $\mathbb{R}$_lⁿ and $\mathbb{R}$_rⁿ denote the left and right halves of $\mathbb{R}$ⁿ respectively. In fact,$\mathbb{R}$_lⁿ and $\mathbb{R}$_rⁿ are separated by the hyperplane x₁=0,where x₁ is the first coordinate of x∈$\mathbb{R}$ⁿ. At this point,a simple computation leads to

$\left( {{U}_{\psi }}{{f}_{0}} \right)\left( x \right)=\left\{ \begin{matrix} \begin{matrix} \int_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt, & x\in \mathbb{R}_{l}^{n}, \\ \end{matrix} \\ \begin{matrix} -\int_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt, & x\in \mathbb{R}_{r}^{n}. \\ \end{matrix} \\ \end{matrix} \right.$

(22)

That is to say,

$\left( {{U}_{\psi }}{{f}_{0}} \right)\left( x \right)={{f}_{0}}\left( x \right)\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt.$

(23)

By the definition off₀,we clearly have f₀∈RMO($\mathbb{R}$ⁿ) with ‖f‖_{RMO($\mathbb{R}$ⁿ)}≠0,and so does U_ψf₀ by (23). Obviously,(23) implies that

${{\left\| {{U}_{\psi }}{{f}_{0}} \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}\le {{\left\| {{f}_{0}} \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt.$

(24)

Consequently,it follows from (24) that

${{\left\| {{U}_{\psi }} \right\|}_{RMO\left( {{\mathbb{R}}^{n}} \right)}}\ge \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt$

(25)

Combining (20) with (25) yields the conclusion in (17).

Using the almost same method,we obtain the following results.

Corollary 2.1  Let φ:[0,1]ⁿ→[0,+∞) be a function. Then U_φ:RMO($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n)→RMO($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n) exists as a bounded operator if and only if

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt<\infty .$

(26)

Moreover,when (25) holds,the operator norm of U_φ on RMO($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n) is given by

${{\left\| {{U}_{\psi }} \right\|}_{RMO\left( {{\mathbb{R}}^{{{m}_{_{1}}}}}\times \cdots \times {{\mathbb{R}}^{{{m}_{n}}}} \right)}}\to RMO\left( {{\mathbb{R}}^{{{m}_{_{1}}}}}\times \cdots \times {{\mathbb{R}}^{{{m}_{n}}}} \right)=\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\psi \left( t \right)dt.$

(27)

Next we will consider the boundedness of the operators U_ψ on R^α,p($\mathbb{R}$ⁿ).

Theorem 2.2  Suppose that -$\frac{n}{p}$<α<∞ and 1≤p<∞. Let ψ:［0,1]ⁿ→[0,+∞) be a function. If

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt<\infty ,$

(28)

then U_ψ is bounded from R^α,p($\mathbb{R}$ⁿ) to R^α,p($\mathbb{R}$ⁿ) and the following inequality

${{\left\| {{U}_{\psi }} \right\|}_{{{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)\to {{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)}}\le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt$

(29)

holds.

Proof  Assume

$\int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt<\infty $

holds. Noting the equality (18),it follows from the Minkowski’s integral inequality that

$\begin{align} & \frac{1}{{{\left| R \right|}^{\frac{\alpha }{n}}}}{{\left( \frac{1}{\left| R \right|}{{\int }_{R}}{{\left| \left( {{U}_{\psi }}f \right)\left( y \right)-{{({{U}_{\psi }}f)}_{R}} \right|}^{p}}dy \right)}^{\frac{1}{p}}} \\ & =\frac{1}{{{\left| R \right|}^{\frac{\alpha }{n}}}}\left( \frac{1}{\left| R \right|} \right.{{\int }_{R}} \\ & {{\left. {{\left| ({{U}_{\psi }}f)\left( y \right)-\int_{0}^{1}{{}}\cdots \int_{0}^{1}{{}}{{f}_{tR}}\psi \left( t \right)d{{t}^{1p}} \right|}^{p}}dy \right)}^{\frac{1}{p}}} \\ & =\frac{1}{{{\left| R \right|}^{\frac{\alpha }{n}}}}\left( \frac{1}{\left| R \right|} \right.{{\int }_{R}}\left| \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}} \right. \\ & {{\left. \left( f\left( {{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}} \right)-{{f}_{tR}} \right)\psi \left( t \right)dt \right|}^{p}}dy{{)}^{\frac{1}{p}}} \\ & \le \frac{1}{{{\left| R \right|}^{\frac{\alpha }{n}}}}\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}} \\ & {{\left( \frac{1}{\left| R \right|}{{\int }_{R}}{{\left| f({{t}_{1}}{{y}_{1}},\ldots ,{{t}_{n}}{{y}_{n}})-{{f}_{tR}} \right|}^{p}}dy \right)}^{\frac{1}{p}}}\psi \left( t \right)dt \\ & =\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\frac{1}{{{\left| tR \right|}^{\frac{\alpha }{n}}}}\left( \frac{1}{\left| tR \right|}{{\int }_{tR}} \right. \\ & {{\left| f({{z}_{1}},\ldots ,{{z}_{n}})-{{f}_{tR}} \right|}^{p}}dy{{)}^{\frac{1}{p}}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt \\ & \le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt{{\left\| f \right\|}_{{{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)}}. \\ \end{align}$

(30)

The inequality (30) implies that

${{\left\| {{U}_{\psi }} \right\|}_{{{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)}}\le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt{{\left\| f \right\|}_{{{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)}}.$

Naturally we have

${{\left\| {{U}_{\psi }} \right\|}_{{{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)\to {{R}^{\alpha ,p}}\left( {{\mathbb{R}}^{n}} \right)}}\le \int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\alpha /n}\psi \left( t \right)dt.$

Now we formulate the similar conclusion on the high dimensional product space.

Corollary 2.2  Suppose that -$\frac{n}{p}$<α<∞ and 1≤p<∞. Let φ:［0,1]ⁿ→[0,+∞) be a function. if

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\frac{\alpha {{m}_{i}}}{n}}\psi \left( t \right)dt<\infty ,$

(31)

then U_φ is bounded on R^α,p($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n),and the operator norm of U_φ is no more than

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{\frac{\alpha {{m}_{i}}}{n}}\psi \left( t \right)dt.$

Using the same method as in the proof of Theorem B,the obtain the following corollary.

Corollary 2.3  Suppose that φ:［0,1]ⁿ→[0,∞) is a nonnegative function and p∈［1,∞]. Then the operator U_φ is bounded on L^p($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n) if and only if

$\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{-\frac{{{m}_{i}}}{p}}\psi \left( {{t}_{1}},\ldots ,{{t}_{n}} \right)dt<\infty .$

(32)

Moreover,if the inequality (32) holds,then the operator norm of U_φ on L^p($\mathbb{R}$^m₁×…×$\mathbb{R}$^m_n) is given by

$\begin{align} & {{\left\| {{U}_{\psi }} \right\|}_{Lp\left( {{\mathbb{R}}^{{{m}_{1}}}}\times \cdots \times {{\mathbb{R}}^{{{m}_{n}}}} \right)\to Lp\left( {{\mathbb{R}}^{{{m}_{1}}}}\times \cdots \times {{\mathbb{R}}^{{{m}_{n}}}} \right)}} \\ & =\int\limits_{0}^{1}{{}}\cdots \int\limits_{0}^{1}{{}}\prod\limits_{i=1}^{n}{{}}t_{i}^{-\frac{{{m}_{i}}}{p}}\psi \left( {{t}_{1}},\ldots ,{{t}_{n}} \right)dt. \\ \end{align}$

(33)