魏明权
, 燕敦验
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中国科学院大学数学科学学院, 北京 100049
摘要: 研究乘积空间上的一类算子
Uψf(
x)=
$\int\limits_0^1 {} \cdots \int\limits_0^1 {} $f(
x1t1,…,
xntn)ψ(
t1,…,
tn)
dt1…
dtn在RMO(
$\mathbb{R}$n)上有界的充分必要条件,这个条件完全依赖于定义在[0,1]×…×[0,1]上的非负函数
ψ。还给出了
Uψ的算子范数。此外,还把这个结果推广到高维乘积空间。
关键词: 广义哈代算子乘积空间RMO(
$\mathbb{R}$n)
In 1984,Carton-Lebrun and Fosset
[1] defined the weighted Hardy-Littlewood average operator
Uψ${{U}_{\psi }}f\left( x \right)=\int\limits_{0}^{1}{{}}f\left( xt \right)\psi \left( t \right)dt,$ | (1) |
where
ψ:[0,1]→[0,∞) is a function. Evidently the operator
Uψ 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,
Uψ 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
Uψ and obtained the following theorem.
Theorem A? Suppose that
ψ:[0,1]→[0,∞) is a nonnegative function and p∈[1,∞]. Then the operator
Uψ 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
Uψ 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
Uψ: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
Uψ 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
Uψ 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,∞).
Uψ 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
Uψ is bounded on
Lp(
$\mathbb{R}$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
Uψ 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
εα,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
εα,p(
$\mathbb{R}$n) norm of
f and denoted by ‖
f‖
εα,p($\mathbb{R}$n).
In Ref.
[5],Zhao et al. studied the boundedness for
Uψ 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
Uψ is a bounded operator on
εα,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
Uψ at the endpoint. A simple computation implies that
Uψ is not bounded on BMO(
$\mathbb{R}$n) and thus is not bounded on
εα,p(
$\mathbb{R}$n),since
εα,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
εα,p(
$\mathbb{R}$n). In this work,we mainly consider this question and introduce two new spaces RMO(
$\mathbb{R}$n) and
Rα,p(
$\mathbb{R}$n). We shall give their definitions as follows.
1 Some definitionsBefore we put forward our main results,some useful definitions will be given. First we introduce the spaces RMO(
$\mathbb{R}$n) and
Rα,p(
$\mathbb{R}$n) corresponding to BMO(
$\mathbb{R}$n) and
εα,p(
$\mathbb{R}$n),respectively.
Definition 1.1? Let
f∈
Lloc($\mathbb{R}$
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
fR denotes the average of f over R,i.e.,
fR=
$\frac{1}{\left| R \right|}$∫
Rf(x)
dx.
Definition 1.2 ?Suppose that
f∈
Lloc(
$\mathbb{R}$n),-∞<
α<∞ and 0<
p<∞. We say that f∈
Rα,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
Rα,p(
$\mathbb{R}$n) norm of
f and denoted by ‖
f‖
Rα,p($\mathbb{R}$n).
Definition 1.3? Suppose that the measurable function f is defined on
$\mathbb{R}$m1×…×
$\mathbb{R}$mn 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∈
$\mathbb{R}$mi 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
Rα,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 proofsFirst,we study the boundedness of the operators
Uψ 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
Uφ.
Theorem 2.1? Let
ψ:[0,1]
n→[0,+∞) be a function. Then
Uψ: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
Uψ 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,…,
tn)∈
$\mathbb{R}$n with
ti>0,
i=1,…,
n and rectangle
R=[
a1,
b1]×[
a2,
b2]×…×[
an,
bn]∈
$\mathbb{R}$n,define
tR=[
t1a1,
t1b1]×[
t2a2,
t2b2]×…×[
tnan,
tnbn].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
zi=
tiyi 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
Uψ 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 of
f0,we clearly have
f0∈RMO(
$\mathbb{R}$n) with ‖f‖
RMO($\mathbb{R}$n)≠0,and so does
Uψf0 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
Uφ:RMO(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn)→RMO(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn) 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
Uφ on RMO(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn) 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
Uψ 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
Uψ is bounded from
Rα,p(
$\mathbb{R}$n) to
Rα,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
Uφ is bounded on
Rα,p(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn),and the operator norm of
Uφ 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
Uφ is bounded on L
p(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn) 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
Uφ on L
p(
$\mathbb{R}$m1×…×
$\mathbb{R}$mn) 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) |
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