1 Introduction
In recent years, the study of function spaces associated with Hermite operators has inspired great interest. Dziubański [7] introduced the Hardy space \(H_{L}^{p}(\mathbb {R}^{d})\), \(0< p\leq1\), by using the heat maximal function and established its atomic characterization. Dziubański et al. [8] and Yang et al. [20] introduced and studied some BMO spaces and Morrey–Campanato spaces associated with operators. Deng et al. [5] introduced the space \(\mathit{VMO}_{L}(\mathbb {R}^{d})\) and proved that \((\mathit{VMO}_{L}(\mathbb {R}^{d}))^{*}=H_{L}^{1}(\mathbb {R}^{d})\). Moreover, recently, Jiang et al. in [14] defined the predual spaces of Banach completions of Orlicz–Hardy spaces associated with operators. Bui et al. [3] considered the Besov and Triebel–Lizorkin spaces associated with Hermite operators.
One of the main purposes of studying the function spaces is to give the equivalent characterizations of them, for example, square functions characterizations for Hardy spaces [10], Carleson measure characterizations for BMO spaces [8] or Morry–Campanato spaces [6]. The aim of this paper is to give characterizations of the dual spaces and predual spaces of the Hardy spaces \(H_{L}^{p}(\mathbb {R}^{d})\) by a new version of Carleson measure. Now, let us review some known facts about the function spaces for L.
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Let L be the basic Schrödinger operator in \(\mathbb {R}^{d}\), \(d\geq 1\), the harmonic oscillator \(L=-\Delta+|x|^{2}\). Let \(\{T_{t}^{L}\}_{t>0}\) be a semigroup of linear operators generated by −L and \(K_{t}^{L}(x,y)\) be their kernels. The Feynman–Kac formula implies that
Dziubański [7] defined Hardy space \(H_{L}^{p}(\mathbb{R}^{d})\), \(0< p\leq1\) as
where
The norm of Hardy space \(H_{L}^{p}(\mathbb{R}^{d})\) is defined by \(\|f\|_{H_{L}^{p}}=\|Mf\|_{L^{p}}\).
$$ 0\leq K_{t}^{L}(x,y)\leq\widetilde{T_{t}}(x,y)=(4 \pi t)^{-\frac{d}{2}}\exp \biggl(-\frac{|x-y|^{2}}{4t} \biggr). $$
(1)
$$H_{L}^{p}\bigl(\mathbb{R}^{d}\bigr)= \bigl\{ f\in \mathcal{S}'\bigl(\mathbb{R}^{d}\bigr): Mf\in L^{p}\bigl(\mathbb{R}^{d}\bigr) \bigr\} , $$
$$Mf(x)=\sup_{t>0} \bigl\vert T_{t}^{L}f(x) \bigr\vert . $$
Remark 1
For simplicity, we just consider the case of \(\frac{d}{d+1}< p\leq1\) in this paper. But all of our results hold for \(0< p\leq1\).
Let \(\rho(x)=\frac{1}{1+|x|}\) be the auxiliary function defined in [17]. This auxiliary function plays an important role in the estimates of the operators and in the description of the spaces associated with L. Then, for \(\frac{d}{d+1}< p\leq1\) and \(1\leq q\leq\infty\), a function a is an \(H_{L}^{p, q}\)-atom for the Hardy space \(H_{L}^{p}(\mathbb {R}^{d})\) associated with a ball \(B(x_{0},r)\) if
The atomic quasi-norm in \(H_{L}^{p}(\mathbb {R}^{d})\) is defined by
where the infimum is taken over all decompositions \(f=\sum c_{j}a_{j}\) and \(a_{j}\) are \(H_{L}^{p, q}\)-atoms.
$$\begin{aligned} (1)&\quad \operatorname{supp} a\subset B(x_{0},r), \\ (2)&\quad \|a\|_{L^{q}}\leq \bigl\vert B(x_{0},r) \bigr\vert ^{\frac{1}{q}-\frac{1}{p}}, \\ (3)&\quad \mbox{if }r< \rho(x_{0}), \mbox{then } \int a(x)\,dx=0. \end{aligned}$$
$$\|f\|_{L\text{-atom},q}=\inf \Bigl\{ \Bigl(\sum|c_{j}|^{p} \Bigr)^{1/p} \Bigr\} , $$
The atomic decomposition for \(H_{L}^{p}(\mathbb{R}^{d})\) is as follows (see [7]).
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Proposition 1
Let
\(\frac{d}{d+1}< p\leq1\), we have that the norms
\(\|f\|_{H_{L}^{p}}\)
and
\(\|f\|_{L\textit{-}\mathrm{atom},q}\)
are equivalent, that is, there exists a constant
\(C>0\)
such that
where
\(1\leq q\leq\infty\).
$$C^{-1}\|f\|_{H_{L}^{p}}\leq\|f\|_{L\textit{-}\mathrm{atom},q}\leq C \|f \|_{H_{L}^{p}}, $$
Definition 1
Let \(0\leq\alpha<1\), a locally integrable function g on \(\mathbb {R}^{d}\) belongs to \(\Lambda_{\alpha}^{L}\) if and only if \(\|g\|_{\Lambda_{\alpha}^{L}}<\infty\), where
and
$$ \|g\|_{\Lambda_{\alpha}^{L}}= \sup_{B\subset\mathbb {R}^{d}} \biggl\{ |B |^{-\frac{\alpha}{d}} \biggl( \int_{B} \bigl\vert g-g(B, x_{0}) \bigr\vert ^{2}\frac{dx}{|B|} \biggr)^{1/2} \biggr\} $$
$$ g(B,x_{0})= \textstyle\begin{cases} \frac{1}{ | B(x_{0},r) |} \int_{B(x_{0},r)} g(y) \,dy, &\mbox{if } r< \rho(x_{0}), \\ 0, &\mbox{if } r \geq\rho(x_{0}). \end{cases} $$
The duality of \(H_{L}^{p}(\mathbb {R}^{d})\) and \(\Lambda_{d(1/p-1)}^{L}\) can be found in [12] or [20].
In order to give the Carleson measure characterization of \(\Lambda _{d(1/p-1)}^{L}\), we need some notations of the tent spaces (cf. [4]).
Let \(0< p<\infty\) and \(1\leq q\leq\infty\). Then the tent space \(T^{p}_{q}\) is defined as the space of functions f on \(\mathbb {R}^{d+1}_{+}\) so that
and
where \(\Gamma(x)\) is the standard cone whose vertex is \(x\in\mathbb {R}^{d}\), i.e.,
Assume that \(B(x_{0},r)\) is a ball in \(\mathbb {R}^{d}\), its tent B̂ is defined by \(\widehat{B}=\{(x,t):|x-x_{0}|\leq r-t\}\). A function \(a(x,t)\) that is supported in a tent B̂, B is a ball in \(\mathbb {R}^{d}\), is said to be an atom in the tent space \(T^{p}_{2}\) if it satisfies
The atomic decomposition of \(T^{p}_{2}\) is stated as follows.
$$\biggl( \int_{\Gamma(x)} \bigl\vert f(y,t) \bigr\vert ^{q} \frac{dy\,dt}{t^{d+1}} \biggr)^{1/q}\in L^{p}\bigl(\mathbb {R}^{d}\bigr), \quad \text{when } 1\leq q< \infty, $$
$$\sup_{(y,t)\in\Gamma(x)} \bigl\vert f(y,t) \bigr\vert \in L^{p}\bigl(\mathbb {R}^{d}\bigr),\quad \text{when } q=\infty, $$
$$\Gamma(x)=\bigl\{ (y,t):|y-x|< t\bigr\} . $$
$$\biggl( \int_{\widehat{B}} \bigl|a(x,t) \bigr|^{2}\frac{dx\,dt}{t} \biggr)^{1/2}\leq |B|^{1/2-1/p}. $$
Proposition 2
When
\(0< p\leq1\), then every
\(f\in T^{p}_{2}\)
can be written as
\(f=\sum\lambda_{k}a_{k}\), where
\(a_{k}\)
are atoms and
\(\sum|\lambda_{k}|^{p}\leq C\|f\|_{T^{p}_{2}}^{p}\).
Let
where
Assume \(0< p\leq1\), we say a function \(f\in T_{2}^{p,\infty}\) belongs to the space \(T_{2, 0}^{p,\infty}\) if f satisfies \(\eta_{1}(f)=\eta_{1}2(f)=\eta_{3}(f)=0\), where
$$T_{2}^{p,\infty}= \bigl\{ f(x,t): \text{measurable on } \mathbb {R}_{+}^{d+1} \text{ and } \|f\|_{T_{2}^{p,\infty}}< \infty \bigr\} , $$
$$\|f\|_{T_{2}^{p,\infty}}=\sup_{B\subset\mathbb{R}^{d}}\frac {1}{|B|^{1/p-1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f(x,t) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}. $$
$$\begin{aligned}& \eta_{1}(f)=\lim_{r\rightarrow0}\sup_{B\subset\mathbb{R}^{d} r_{B}< r} \frac{1}{|B|^{1/p-1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f(x,t) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}; \\& \eta_{2}(f)=\lim_{r\rightarrow\infty}\sup_{B\subset\mathbb{R}^{d} r_{B}\geq r} \frac{1}{|B|^{1/p-1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f(x,t) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}; \\& \eta_{3}(f)=\lim_{r\rightarrow\infty}\sup_{B\subset B(0,r)^{c}} \frac{1}{|B|^{1/p-1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f(x,t) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}. \end{aligned}$$
Let \(\widetilde{T_{2}^{p}}=\{F(x,t)=\sum_{i}\lambda_{i}a_{i}(x,t): a_{i}(x,t) \text{are } T_{2}^{p} \text{ atoms and } \sum_{i}|\lambda_{i}|^{p}<\infty\}\) and \(\|F\|_{\widetilde{T_{2}^{p}}}=\inf\{ (\sum_{i}|\lambda_{i}|^{p} )^{1/p}: F(x,t)=\sum_{i}\lambda_{i}a_{i}(x,t)\}\). Then \(\widetilde{T_{2}^{p}}\) is a Banach space. In fact, it is the completeness of \(T_{2}^{p}\). Especially, \(\widetilde{T_{2}^{1}}=T_{2}^{1}\).
In [19], the author proved the following result.
Proposition 3
Let
\(0< p\leq1\). Then
$$\bigl(T_{2, 0}^{p,\infty} \bigr)^{*}=\widetilde{T_{2}^{p}}, \qquad \bigl(T_{2}^{p} \bigr)^{*}=T_{2}^{p,\infty}. $$
Let \(\{P_{t}^{L}\}_{t>0}\) be the semigroup of linear operators generated by \(-\sqrt{L}\) and \(D_{t}^{L}f(x)=t|\nabla P_{t}^{L}f |(x)\), where \(\nabla=(\partial_{t},\partial_{x_{1}},\ldots,\partial _{x_{d}})\). The Carleson measure characterization of the Campanato space \(\Lambda_{d(1/p-1)}^{L}\) as (cf. [6]).
Proposition 4
Let
\(\frac{d}{d+1}< p\leq1\). Then, for any
\(f\in L_{\mathrm{loc}}^{2}(\mathbb{R}^{d})\), we have:
(a)
If
\(f \in\Lambda_{d(1/p-1)}^{L}\), then
\(D^{L}_{t}f\in T_{2}^{p,\infty}\); moreover, we have
$$ \bigl\Vert D^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}} \leq C \Vert f \Vert _{\Lambda_{d(1/p-1)}^{L}}. $$
(b)
Conversely, if
\(f\in L^{1}((1+|x|)^{-(d+1)}\,dx)\)
and
\(D^{L}_{t}f\in T_{2}^{p,\infty}\), then
\(f \in\Lambda_{d(1/p-1)}^{L}\)
and
$$ \Vert f \Vert _{\Lambda_{d(1/p-1)}^{L}} \leq C \bigl\Vert D^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}}. $$
Definition 2
Let \(\alpha>0\), we will say a function f of \(\Lambda_{\alpha}^{L}\) is in \(\lambda_{\alpha}^{L}\) if it satisfies \(\gamma_{1}(f)=\gamma_{2}(f)=\gamma_{3}(f)=0\), where
$$\begin{aligned}& \gamma_{1}(f)=\lim_{r\rightarrow0}\sup_{B\subset\mathbb{R}^{d} r_{B}< r} \frac{1}{|B|^{\alpha/d+1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f-f(B, V) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}; \\& \gamma_{2}(f)=\lim_{r\rightarrow\infty}\sup_{B\subset\mathbb{R}^{d} r_{B}\geq r} \frac{1}{|B|^{\alpha/d+1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f-f(B, V) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}; \\& \gamma_{3}(f)=\lim_{r\rightarrow\infty}\sup_{B\subset B(0,r)^{c}} \frac{1}{|B|^{\alpha/d+1/2}} \biggl( \int_{\widehat{B}} \bigl\vert f-f(B, V) \bigr\vert ^{2} \frac{dx\,dt}{t} \biggr)^{1/2}. \end{aligned}$$
The dual space of \(\lambda_{d(1/p-1)}^{L}\) is \(B_{L}^{p}(\mathbb{R}^{d})\), which is the completeness of \(H_{L}^{p}(\mathbb{R}^{d})\) (cf. [14]).
We can give a Carleson measure characterization of \(\lambda_{d(1/p-1)}^{L}\) as follows (see [14]).
Proposition 5
Let
\(\frac{d}{d+1}< p\leq1\). Then, for any
\(f\in L_{\mathrm{loc}}^{2}(\mathbb{R}^{d})\), we have:
(a)
If
\(f \in\lambda_{d(1/p-1)}^{L}\), then
\(D^{L}_{t}f\in T_{2}^{p,\infty}\); moreover, we have
$$ \bigl\Vert D^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}} \leq C \Vert f \Vert _{\lambda_{d(1/p-1)}^{L}}. $$
(b)
Conversely, if
\(f\in L^{1}((1+|x|)^{-(d+1)}\,dx)\)
and
\(D^{L}_{t}f\in T_{2}^{p,\infty}\), then
\(f \in\lambda_{d(1/p-1)}^{L}\)
and
$$ \Vert f \Vert _{\lambda_{d(1/p-1)}^{L}} \leq C \bigl\Vert D^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}}. $$
Let \(A_{j}=\partial_{x_{j}}+x_{j}\) and \(A_{-j}=\partial_{x_{j}}-x_{j}\) for \(j=1,2,\ldots, d\). Then
Therefore, in the harmonic analysis associated with L, the operators \(A_{j}\) play the role of the classical partial derivatives \(\partial _{x_{j}}\) in the Euclidean harmonic analysis (see [2, 11, 18]). Now, it is natural to consider the derivatives \(A_{i}\) other than \(\partial_{x_{j}}\). In [13], the author defined the Lusin area integral operator by \(A_{j}\) and characterized the Hardy space \(H_{L}^{1}(\mathbb {R}^{d})\). As a continuous study of the function spaces associated with L, in this paper we will define the Carleson measure by \(A_{j}\) and characterize the dual spaces and predual spaces of \(H_{L}^{p}(\mathbb {R}^{d})\). Moreover, let \(Q_{t}^{L}f(x)=t|\tilde{\nabla} P_{t}^{L}f |(x)\), where \(\tilde {\nabla}=(\partial_{t},A_{-1},\ldots,A_{-d}, A_{1},\ldots,A_{d})\). Then the main results of this paper can be stated as follows.
$$L= \sum_{j=1}^{d} A_{j}A_{-j}+A_{-j}A_{j}. $$
Theorem 1
Let
\(\frac{d}{d+1}< p\leq1\). Then, for every
\(f\in L_{\mathrm{loc}}^{2}(\mathbb{R}^{d})\), we have:
(a)
If
\(f \in\Lambda_{d(1/p-1)}^{L}\), then
\(Q^{L}_{t}f\in T_{2}^{p,\infty}\); moreover, we have
$$ \bigl\Vert Q^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}} \leq C \Vert f \Vert _{\Lambda_{d(1/p-1)}^{L}}. $$
(b)
Conversely, if
\(f\in L^{1}((1+|x|)^{-(d+1)}\,dx)\)
and
\(Q^{L}_{t}f\in T_{2}^{p,\infty}\), then
\(f \in\Lambda_{d(1/p-1)}^{L}\)
and
$$ \Vert f \Vert _{\Lambda_{d(1/p-1)}^{L}} \leq C \bigl\Vert Q^{L}_{t}f \bigr\Vert _{T_{2}^{p,\infty}}. $$
Remark 2
In [8], the authors characterize the case \(p=0\), i.e., \(\mathit{BMO}_{L}\), by the heat semigroup with the classical derivatives. In [15], the authors characterize the space \(\mathit{BMO}_{L}\) by the Poisson semigroup with the classical derivatives. In this paper, we will use the new derivatives \(A_{j}\) of the Poisson semigroup to characterize the space \(\Lambda_{d(1/p-1)}^{L}\) for \(\frac{d}{d+1}< p\leq1\).
Theorem 2
Let
\(\frac{d}{d+1}< p\leq1\). Then, for any
\(f\in L_{\mathrm{loc}}^{2}(\mathbb{R}^{d})\), we have:
(a)
If
\(f \in\lambda_{d(1/p-1)}^{L}\), then
\(Q^{L}_{t}f\in T_{2, 0}^{p,\infty}\); moreover, we have
$$ \bigl\Vert Q^{L}_{t}f \bigr\Vert _{T_{2, 0}^{p,\infty}} \leq C \Vert f \Vert _{\lambda_{d(1/p-1)}^{L}}. $$
(b)
Conversely, if
\(f\in L^{1}((1+|x|)^{-(d+1)}\,dx)\)
and
\(Q^{L}_{t}f\in T_{2, 0}^{p,\infty}\), then
\(f \in\lambda_{d(1/p-1)}^{L}\)
and
$$ \Vert f \Vert _{\lambda_{d(1/p-1)}^{L}} \leq C \bigl\Vert Q^{L}_{t}f \bigr\Vert _{T_{2, 0}^{p,\infty}}. $$
The paper is organized as follows. In Sect. 2, we give some estimates of the kernels. In Sect. 3, we give the proof of Theorem 1. The proofs of Theorem 2 will be given in Sect. 4.
Throughout the article, we will use A and C to denote the positive constants, which are independent of the main parameters and may be different at each occurrence. By \(B_{1} \sim B_{2}\), we mean that there exists a constant \(C>1\) such that \(\frac{1}{C} \leq \frac{B_{1}}{B_{2}}\leq C\).
2 Estimates of the kernels
In this section, we give some estimates of the kernels, which we will use in the sequel.
The proofs of these estimates can be found in [9].
Lemma 1
(a)
For every
\(N\in\mathbb{N}\), there is a constant
\(C_{N}>0\)
such that
$$ 0\leq K_{t}^{L}(x,y)\leq C_{N}t^{-\frac{d}{2}}e^{-(5t)^{-1}|x-y|^{2}} \biggl(1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)} \biggr)^{-N}. $$
(2)
(b)
There exists
\(C>0\)
such that, for every
\(N>0\), there is a constant
\(C_{N}>0\)
so that, for all
\(|h|\leq \sqrt{t}\),
$$ \bigl\vert K_{t}^{L}(x+h,y)-K_{t}^{L}(x,y) \bigr\vert \leq C_{N}\frac{|h|}{\sqrt{t}}t^{-\frac{d}{2}} e^{-At^{-1}|x-y|^{2}} \biggl(1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt {t}}{\rho(y)} \biggr)^{-N}. $$
(3)
By subordination formula, we can give the following estimates about the Poisson kernel.
Lemma 2
(a)
For every
N, there is a constant
\(C_{N}>0\), \(A>0\)
such that
$$ 0\leq P_{t}^{L}(x,y)\leq C_{N} \frac{t}{(t^{2}+A|x-y|^{2})^{(d+1)/2}} \biggl(1+\frac{t}{\rho(x)}+\frac{t}{\rho(y)} \biggr)^{-N}. $$
(4)
(b)
Let
\(|h|<\frac{|x-y|}{2}\). Then, for any
\(N>0\), there exist constants
\(C>0\), \(C_{N}>0\)
such that
$$ \bigl\vert P_{t}^{L}(x+h,y)-P_{t}^{L}(x,y) \bigr\vert \leq C_{N}\frac{|h|}{\sqrt{t}}\frac{t}{(t^{2}+A|x-y|^{2})^{(d+1)/2}} \biggl(1+ \frac{t}{\rho(x)}+\frac{t}{\rho(y)} \biggr)^{-N}. $$
(5)
Duong et al. [6] proved the following estimates about the kernel \(D_{t}^{L}(x,y)\).
Lemma 3
There exist constants
C
such that, for every
N, there is a constant
\(C_{N}>0\), so that
$$\begin{aligned} (\mathrm{a})&\quad \bigl\vert D_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-d} e^{-Ct^{-2} \vert x-y \vert ^{2}}\biggl(1+ \frac{t}{\rho(x)}+\frac{t}{\rho (y)}\biggr)^{-N} ; \\ (\mathrm{b})&\quad \bigl\vert D_{t}^{L}(x+h,y)-D_{t}^{L}(x,y) \bigr\vert \leq C_{k,N}\biggl(\frac{ \vert h \vert }{t}\biggr)^{\delta'}t^{-d} e^{-Ct^{-2} \vert x-y \vert ^{2}}\biggl(1+\frac{t}{\rho(x)}+\frac{t}{\rho (y)} \biggr)^{-N}, \\ &\qquad \textit{for all } \vert h \vert \leq t; \\ (\mathrm{c})&\quad \biggl\vert \int_{\mathbb {R}^{d}}D_{t}^{L}(x,y)\,dy \biggr\vert \leq C_{N}\frac{t/\rho(x)}{(1+t/\rho(x))^{N}}. \end{aligned}$$
Let \(t=\frac{1}{2}\ln\frac{1+s}{1-s}\), \(s\in(0,1)\). Then
The following estimations are very important for the proofs of the main result in this paper.
$$ K_{t}^{L}(x,y)= \biggl(\frac{1-s^{2}}{4\pi s} \biggr)^{d/2}\exp{ \biggl(-\frac{1}{4}\biggl(s|x+y|^{2}+ \frac{1}{s}|x-y|^{2}\biggr) \biggr)}\doteq K_{s}(x,y). $$
(6)
Lemma 4
There is
\(C>0\)
for
\(N\in\mathbb {N}\)
and
\(|x-x'|\leq\frac{|x-y|}{2}\), any
\(j=-1,\ldots,-d,1,\ldots,d\), we can find
\(C_{N}>0\)
such that
$$\begin{aligned} (\mathrm{a})&\quad \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-\frac {d}{2}}\exp{ \biggl(- \frac{ \vert x-y \vert ^{2}}{C t} \biggr)}\biggl(1+\frac{\sqrt {t}}{\rho(x)}+\frac{\sqrt{t}}{\rho(y)} \biggr)^{-N} ; \\ (\mathrm{b})&\quad \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y)- \sqrt{t}A_{j}K_{t}^{L}\bigl(x',y \bigr) \bigr\vert \\ &\qquad \leq C_{N}\frac{ \vert x-x' \vert }{t}t^{-\frac{d}{2}}\exp{ \biggl(- \frac{ \vert x-y \vert ^{2}}{C t} \biggr)}\biggl(1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt{t}}{\rho (y)} \biggr)^{-N}; \\ (\mathrm{c})&\quad \biggl\vert \int_{\mathbb {R}^{d}}\sqrt{t}A_{j}K_{t}^{L}(x,y) \,dy \biggr\vert \leq C\frac{t/\rho(x)}{(1+t/\rho(x))^{N}}. \end{aligned}$$
Proof
By
and \(t=\frac{1}{2}\ln\frac{1+s}{1-s}\sim s\), \(s\rightarrow0^{+}\), for \(s\in(0,\frac{1}{2}]\), we have
If \(x\cdot y\leq0\), then \(|x|\leq|x-y|\). So
If \(x\cdot y\geq0\), then \(|x|\leq|x+y|\). So
Therefore,
Since
we get \((\frac{1-s^{2}}{4\pi s} )^{d/2}\leq t^{-d}\) for \(s\in[\frac{1}{2}, 1)\).
$$\begin{aligned} \bigl\vert A_{j}K_{t}^{L}(x,y) \bigr\vert =& \biggl\vert \frac{\partial}{\partial x_{j}}K_{t}^{L}(x,y)+x_{j}K_{t}^{L}(x,y) \biggr\vert \\ \leq& \biggl\vert \frac{\partial}{\partial x_{j}}K_{t}^{L}(x,y) \biggr\vert + \bigl\vert x_{j}K_{t}^{L}(x,y) \bigr\vert \doteq I_{1}+I_{2}, \end{aligned}$$
$$\begin{aligned} I_{2} \le&C|x_{j}|s^{-\frac{d}{2}}\exp{ \biggl(- \frac {1}{4}s|x+y|^{2} \biggr)}\exp{ \biggl(-\frac{1}{4} \frac {|x-y|^{2}}{s} \biggr)} \\ \leq&C|x|s^{-\frac{d}{2}}\exp{ \biggl(-\frac{1}{4}s|x+y|^{2} \biggr)}\exp{ \biggl(-\frac{1}{4}\frac{|x-y|^{2}}{s} \biggr)}. \end{aligned}$$
$$\begin{aligned} I_{2} \le&Cs^{-\frac{d}{2}}|x-y|\exp{ \biggl(-\frac{1}{4} \frac {|x-y|^{2}}{s} \biggr)}\leq C s^{-\frac{d-1}{2}}\exp{ \biggl(-\frac {|x-y|^{2}}{8s} \biggr)} \\ \leq&C t^{-\frac{d-1}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. \end{aligned}$$
$$\begin{aligned} I_{2} \le&Cs^{-\frac{d}{2}}|x+y|\exp{ \biggl(-\frac {1}{4}s|x+y|^{2} \biggr)}\exp{ \biggl(-\frac{1}{4}\frac {|x-y|^{2}}{s} \biggr)} \\ \leq& C s^{-\frac{d+1}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{4s} \biggr)} \leq C t^{-\frac{d+1}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{4t} \biggr)}. \end{aligned}$$
$$ \vert \sqrt{t}I_{2} \vert \leq C(1+t)t^{-\frac{d}{2}} \exp{ \biggl(-\frac {|x-y|^{2}}{8t} \biggr)}\leq Ct^{-\frac{d}{2}}\exp{ \biggl(- \frac {|x-y|^{2}}{8t} \biggr)}. $$
(7)
$$\lim_{t\rightarrow\infty}t^{2} \biggl(1- \biggl(\frac {e^{2t}-1}{e^{2t}+1} \biggr)^{2} \biggr)=0, $$
When \(s\in[\frac{1}{2},1)\), we get \(t=\frac{1}{2}\ln\frac {1+s}{1-s}> s\). Therefore
Then
By (6), we get
and
Therefore, when \(s\in(0,\frac{1}{2}]\), we have
When \(s\in[\frac{1}{2},1)\), we have
Then
By (7)–(9), we get
Similar to the proof of (10), for any \(N>0\), we can prove
and
Since \(\rho(x)=\frac{1}{1+|x|}\), we get \(\frac{\sqrt{t}}{\rho(x)}=\sqrt {t}(1+|x|)\). Then, for \(N>0\),
Since x and y are symmetric, we also have
Then (a) follows from (10)–(12).
$$\begin{aligned} I_{2} \leq&C|x_{j}|\exp{ \biggl(-\frac{1}{4} \biggl(s|x+y|^{2}+\frac {|x-y|^{2}}{s}\biggr) \biggr)} \\ \leq&Ct^{-d}|x|\exp{ \biggl(-\frac{1}{4}\biggl(s|x+y|^{2}+ \frac {|x-y|^{2}}{s}\biggr) \biggr)} \\ \leq&Ct^{-d}\bigl( \vert x+y \vert + \vert x-y \vert \bigr) \exp{ \biggl(-\frac{1}{4}\biggl(s|x+y|^{2}+\frac {|x-y|^{2}}{s} \biggr) \biggr)} \\ \leq&Ct^{-d}\exp{ \biggl(-\frac{|x-y|^{2}}{8s} \biggr)} \\ \leq& Ct^{-d}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. \end{aligned}$$
$$ \vert \sqrt{t}I_{2} \vert \leq Ct^{-d+\frac{1}{2}} \exp{ \biggl(-\frac {|x-y|^{2}}{8t} \biggr)}\leq Ct^{-\frac{d}{2}}\exp{ \biggl(- \frac {|x-y|^{2}}{8t} \biggr)}. $$
(8)
$$\frac{\partial}{\partial x_{j}}K_{s}(x,y)=-\frac{1}{2}\biggl(s(x_{j}+y_{j})+ \frac {1}{s}(x_{j}-y_{j})\biggr)K_{s}(x,y) $$
$$ I_{1}\leq C\biggl(s|x_{j}+y_{j}|+ \frac{1}{s}|x_{j}-y_{j}|\biggr)K_{s}(x,y) \leq C\biggl(s|x+y|+\frac {1}{s}|x-y|\biggr)K_{s}(x,y). $$
$$I_{1}\leq C s^{-\frac{d}{2}}(1+s)\exp{ \biggl(-\frac{|x-y|^{2}}{8s} \biggr)}\leq Ct^{-\frac{d}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. $$
$$I_{1}\leq C t^{-d}\exp{ \biggl(-\frac{|x-y|^{2}}{8s} \biggr)} \leq Ct^{-d}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. $$
$$ \biggl\vert \sqrt{t}\frac{\partial}{\partial x_{j}}K_{t}^{L}(x,y) \biggr\vert \leq Ct^{-\frac{d}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. $$
(9)
$$ \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y) \bigr\vert \leq Ct^{-\frac{d}{2}}\exp{ \biggl(-\frac{|x-y|^{2}}{8t} \biggr)}. $$
(10)
$$\bigl(\sqrt{t} \vert x \vert \bigr)^{N} \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-\frac{d}{2}}\exp { \biggl(-\frac{ \vert x-y \vert ^{2}}{8t} \biggr)} $$
$$t^{N} \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-\frac{d}{2}}\exp{ \biggl(-\frac { \vert x-y \vert ^{2}}{8t} \biggr)}. $$
$$ \biggl(\frac{\sqrt{t}}{\rho(x)} \biggr)^{N} \bigl\vert \sqrt {t}A_{j}K_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-\frac{d}{2}}\exp{ \biggl(-\frac { \vert x-y \vert ^{2}}{8t} \biggr)}. $$
(11)
$$ \biggl(\frac{\sqrt{t}}{\rho(y)} \biggr)^{N} \bigl\vert tA_{j}K_{t}^{L}(x,y) \bigr\vert \leq C_{N}t^{-\frac{d}{2}}\exp{ \biggl(-\frac{ \vert x-y \vert ^{2}}{8t} \biggr)}. $$
(12)
(b) Note that
For \(J_{2}\), let
where \(\alpha(s,z,y)=s|z+y|^{2}+\frac{1}{s}|z-y|^{2}\).
$$\begin{aligned}& \bigl\vert \sqrt{t}A_{j}K_{t}^{L} \bigl(x',y\bigr)-\sqrt{t}A_{j}K_{t}^{L}(x,y) \bigr\vert \\& \quad \leq \biggl\vert \sqrt{t}\frac{\partial}{\partial x_{j}}K_{t}^{L} \bigl(x',y\bigr)-\sqrt {t}\frac{\partial}{\partial x_{j}}K_{t}^{L}(x,y) \biggr\vert + \bigl\vert \sqrt{t}x'_{j}K_{t}^{L} \bigl(x',y\bigr)-\sqrt{t}x_{j}K_{t}^{L}(x,y) \bigr\vert \\& \quad \doteq J_{1}+J_{2}. \end{aligned}$$
$$ \varphi(z)=\varphi_{y,s}(z)=z_{j}\exp{ \biggl(- \frac{1}{4}\alpha (s,z,y) \biggr)}, $$
Then
Therefore
Let \(\theta=\lambda x+(1-\lambda)x'\), \(0<\lambda<1\). Then
When \(|x-x'|\leq\frac{|x-y|}{2}\), we can get \(|\theta-y|\sim|x-y|\). Therefore, there exists \(A>0\) such that
For \(J_{1}\),
Let
Then
Therefore, similar to the proofs of (13) and (14), we can prove
and
Inequalities (13) and (15) show
Then, similar to the proof of (a), we have
$$\frac{\partial\varphi}{\partial z_{k}}(z)= \biggl(\delta_{jk}-\frac {s}{2}z_{j}(z_{k}+y_{k})- \frac{1}{2s}z_{j}(z_{k}-y_{k}) \biggr)\exp{ \biggl(-\frac {1}{4}\alpha(s,z,y) \biggr)}. $$
$$\begin{aligned} \biggl\vert \frac{\partial\varphi}{\partial z_{k}}(z) \biggr\vert \leq &C\biggl(1+s \vert z \vert \vert z+y \vert +\frac{1}{s} \vert z \vert \vert z-y \vert \biggr)\exp{ \biggl(-\frac{1}{4}\alpha (s,z,y) \biggr)} \\ \leq&C\biggl(1+s^{1/2} \vert z \vert +\frac{1}{s^{1/2}} \vert z \vert \biggr)\exp{ \biggl(-\frac {1}{8}\alpha(s,z,y) \biggr)} \\ \leq&C\biggl(1+s^{1/2}\bigl( \vert z-y \vert + \vert z+y \vert \bigr)+\frac{1}{s^{1/2}}\bigl( \vert z-y \vert + \vert z+y \vert \bigr) \biggr)\exp { \biggl(-\frac{1}{8}\alpha(s,z,y) \biggr)} \\ \leq&C\biggl(1+s+\frac{1}{s}\biggr)\exp{ \biggl(-\frac{1}{16s} \vert z-y \vert ^{2} \biggr)} \\ \leq&Cs^{-1}\exp{ \biggl(-\frac{1}{16s} \vert z-y \vert ^{2} \biggr)}. \end{aligned}$$
(13)
$$\begin{aligned} J_{2} \leq&Ct^{-d/2} \bigl\vert x'_{j}K_{s} \bigl(x',y\bigr)-x_{j}K_{s}(x,y) \bigr\vert \\ \leq&Ct^{-d/2} \bigl\vert x-x' \bigr\vert \sup _{\theta} \bigl\vert \nabla\varphi(\theta) \bigr\vert \\ \leq&Ct^{-d/2}\frac{|x-x'|}{s}|\sup_{\theta}\exp{ \biggl(-\frac {|\theta-y|^{2}}{16s} \biggr)} \\ \leq&Ct^{-d/2}\frac{|x-x'|}{t}|\sup_{\theta}\exp{ \biggl(-\frac {|\theta-y|^{2}}{16t} \biggr)}. \end{aligned}$$
$$ J_{2}\leq Ct^{-d/2}\frac{|x-x'|}{t}\exp{ \biggl(-\frac {|x-y|^{2}}{At} \biggr)}. $$
(14)
$$\begin{aligned} J_{1} =& \biggl\vert \sqrt{t}\frac{\partial}{\partial x_{j}}K_{t}^{L} \bigl(x',y\bigr)-\sqrt {t}\frac{\partial}{\partial x_{j}}K_{t}^{L}(x,y) \biggr\vert \\ =&\sqrt{t} \biggl\vert \frac{\partial}{\partial x_{j}}K_{s} \bigl(x',y\bigr)-\frac {\partial}{\partial x_{j}}K_{s}(x,y) \biggr\vert \\ =&\sqrt{t} \biggl\vert \biggl(s(x_{j}+y_{j})+ \frac{1}{s}(x_{j}-y_{j})\biggr)\exp{ \biggl(- \frac{1}{4}\alpha(s,x,y) \biggr)} \\ &{}-\biggl(s\bigl(x'_{j}+y_{j}\bigr)+ \frac{1}{s}\bigl(x'_{j}-y_{j}\bigr) \biggr)\exp{ \biggl(-\frac{1}{4}\alpha \bigl(s,x',y\bigr) \biggr)} \biggr\vert . \end{aligned}$$
$$\psi(z)=\psi_{y,s}(z)=\biggl(s(z_{j}+y_{j})+ \frac{1}{s}(z_{j}-y_{j})\biggr)\exp{ \biggl(- \frac{1}{4}\alpha(s,z,y) \biggr)}. $$
$$\begin{aligned} \frac{\partial\psi}{\partial z_{k}}(z) =&\biggl[\biggl(s+\frac{1}{s}\biggr)\delta _{jk}-\frac{1}{2}\biggl(s(z_{j}+y_{j})+ \frac{1}{s}(z_{j}-y_{j})\biggr) \\ &{}s(z_{k}+y_{k})+\frac{1}{s}(z_{k}-y_{k}) \biggr]\exp{ \biggl(-\frac{1}{4}\alpha (s,z,y) \biggr)}. \end{aligned}$$
$$\biggl\vert \frac{\partial\psi}{\partial z_{k}}(z) \biggr\vert \leq Cs^{-1}\exp { \biggl(-\frac{1}{4}\alpha(s,z,y) \biggr)} $$
$$\begin{aligned} J_{1} \leq&C\sup_{\theta}\bigl\vert \nabla\psi( \theta) \bigr\vert \bigl\vert x-x' \bigr\vert \\ \leq&Ct^{-d/2}\frac{|x-x'|}{t}|\exp{ \biggl(-\frac {|x-y|^{2}}{At} \biggr)}. \end{aligned}$$
(15)
$$ \bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y)- \sqrt{t}A_{j}K_{t}^{L}\bigl(x',y \bigr) \bigr\vert \leq C_{N}\frac{|x-x'|}{t}t^{-\frac{d}{2}}\exp{ \biggl(-\frac {|x-y|^{2}}{At} \biggr)}. $$
$$\bigl\vert \sqrt{t}A_{j}K_{t}^{L}(x,y)- \sqrt{t}A_{j}K_{t}^{L}\bigl(x',y \bigr) \bigr\vert \leq C_{N}\frac{|x-x'|}{t}t^{-\frac{d}{2}}\exp{ \biggl(-\frac {|x-y|^{2}}{At} \biggr)} \biggl(1+\frac{\sqrt{t}}{\rho(x)}+\frac{\sqrt {t}}{\rho(y)} \biggr)^{-N}. $$
(c) Noting that
The proof of part I can be found in Lemma 3.9 of [6]. For part II, since \(|x_{j}|\leq1+|x|=\frac{1}{\rho(x)}\) and Lemma 1, we get
Therefore, part (c) holds and this completes the proof of Proposition 4. □
$$\begin{aligned}& \biggl\vert \int_{\mathbb {R}^{d}}\sqrt{t}A_{j}K_{t}^{L}(x,y) \,dy \biggr\vert \\& \quad \leq \biggl\vert \int_{\mathbb {R}^{d}}\sqrt{t}\partial _{x_{j}}K_{t}^{L}(x,y) \,dy \biggr\vert + \biggl\vert \int_{\mathbb {R}^{d}}\sqrt {t}x_{j}K_{t}^{L}(x,y) \,dy \biggr\vert \\& \quad \doteq I+\mathit{II}. \end{aligned}$$
$$\mathit{II}\leq\frac{\sqrt{t}}{\rho(x)} \int_{\mathbb {R}^{d}}\sqrt {t} \bigl\vert K_{t}^{L}(x,y) \bigr\vert \,dy\leq\frac{\frac{\sqrt{t}}{\rho(x)}}{ (1+\frac{\sqrt{t}}{\rho(x)} )^{N}}. $$
Lemma 4 and the subordination formula give the following.
Lemma 5
There is
\(C>0\)
for
\(N\in\mathbb {N}\)
and
\(|x-x'|\leq\frac{|x-y|}{2}\), we can find
\(C_{N}>0\)
such that
$$\begin{aligned} (\mathrm{a})&\quad \bigl\vert Q_{t}^{L}(x,y) \bigr\vert \leq C_{N}\frac{t}{(t^{2}+A \vert x-y \vert ^{2})^{(d+1)/2}} \biggl(1+\frac{t}{\rho(x)}+ \frac{t}{\rho(y)}\biggr)^{-N}; \\ (\mathrm{b})&\quad \bigl\vert Q_{t}^{L}(x,y)-Q_{t}^{L} \bigl(x',y\bigr) \bigr\vert \leq C_{N} \biggl( \frac{ \vert x-x' \vert }{t} \biggr)\frac{t}{(t^{2}+A \vert x-y \vert ^{2})^{(d+1)/2}} \biggl(1+\frac{t}{\rho(x)}+ \frac{t}{\rho(y)}\biggr)^{-N}; \\ (\mathrm{c})&\quad \biggl\vert \int_{\mathbb {R}^{d}}Q_{t}^{L}(x,y)\,dy \biggr\vert \leq C_{N}\frac{t/\rho(x)}{(1+t/\rho(x))^{N}}. \end{aligned}$$
3 Carleson measure characterization of \(\Lambda_{\alpha}^{L}\)
Let \(s_{L}\) denote the Littlewood–Paley g-function associated with L, i.e.,
and \(A_{L}\) denote the Lusin area integral associated with L, i.e.,
Then we can prove the following.
$$s_{L}f(x)= \biggl( \int_{0}^{\infty} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac {dt}{t} \biggr)^{1/2}, $$
$$A_{L}f(x)= \biggl( \int_{0}^{\infty} \int_{\Gamma(x)} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dy\,dt}{t} \biggr)^{1/2}. $$
Lemma 6
The operators
\(s_{L}\)
and
\(A_{L}\)
are isometries on
\(L^{2}(\mathbb{R}^{d})\)
up to constant factors. Exactly,
$$\|s_{L}f\|_{L^{2}}=\frac{1}{2}\|f\|_{L^{2}}, \qquad \|A_{L}f\|_{L^{2}}=C_{d}\|f\|_{L^{2}}. $$
The proof of Lemma 6 is standard, we omit it.
Let \(F(x,t)=Q_{t}^{L}f(x)\) and \(G(x,t)=Q_{t}^{L}g(x)\). Then we have the following lemma.
Lemma 7
If
\(g\in L^{1} ((1+|x|)^{-(d+1)}\,dx )\)
and
f
is an
\(H_{L}^{p, \infty}\)-atom, then
$$\frac{1}{4} \int_{\mathbb{R}^{d}}f(x)\overline{g(x)}\,dx= \int _{\mathbb{R}_{+}^{d+1}}F(x,t)G(x,t)\frac{dx\,dt}{t}. $$
Lemma 8
There exists
\(C>0\)
such that, for any
\(H_{L}^{p, \infty}\)-atom
\(a(x)\), we have
\(\|A_{L}a\|_{L^{p}}\leq C\).
Now we can give the proof of Theorem 1.
Proof of Theorem 1
Let \(f \in \Lambda_{d(1/p-1)}^{L}\), then \(f\in L^{1} ((1+|x|)^{-(d+1)}\,dx )\). By Lemma 5(a), we know
is absolutely convergent. To prove the assertion (a), we need to prove that, for any ball \(B=B(x_{0},r)\),
Set \(B_{k}=B(x_{0},2^{k}r)\) and
By Lemma 6, we have
Note that
Therefore
For \(x \in B(x_{0},r)\), by Lemma 5(a) and (18),
In the last step of the above, we use the facts \(\frac{d}{d+1}< p\leq1\) to get \(d(1/p-1)-1<0\).
$$ Q^{L}_{t} f(x)= \int_{\mathbb{R}^{d}} Q^{L}_{t}(x,y) f(y) \,dy $$
$$ \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} f(x) \bigr\vert ^{2} \frac{dx \,dt}{t} \leq C \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}}. $$
(16)
$$ f = \bigl( f - f(B_{1}) \bigr) \chi_{{B_{1}}}+ \bigl( f - f(B_{1}) \bigr) \chi_{{B^{c}_{1}}}+ f(B_{1})= \widetilde{f}_{1} + \widetilde{f}_{2} + f(B_{1}). $$
$$\begin{aligned} \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \widetilde{f}_{1}(x) \bigr\vert ^{2} \frac{dx \,dt}{t} \leq&\frac{1}{|B|^{2/p-1}} \int_{B} \bigl\vert s_{L} \widetilde{f}_{1}(x) \bigr\vert ^{2} \,dx \\ =& \frac{1}{4|B|^{2/p-1}} \| \widetilde{f}_{1} \|^{2}_{L^{2}} = \frac{1}{4|B|^{2/p-1}} \int_{B_{1}} \bigl\vert f(g)- f(B_{1}) \bigr\vert ^{2} \,dx \\ \leq& C \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}}. \end{aligned}$$
(17)
$$\begin{aligned} \bigl\vert f(B_{2})- f(B_{1}) \bigr\vert \leq& 2^{d}\frac{1}{ \vert B_{2} \vert } \int_{B_{2}} \bigl\vert f(x)-f(B_{2}) \bigr\vert \,dx \\ \leq&2^{d}\frac{1}{ \vert B_{2} \vert ^{1/2}} \biggl( \int_{B_{2}} \bigl\vert f(x)-f(B_{2}) \bigr\vert ^{2}\,dx \biggr)^{1/2} \\ =&2^{d} \vert B_{2} \vert ^{1/p-1} \frac{1}{ \vert B_{2} \vert ^{1/p-1/2}} \biggl( \int_{B_{2}} \bigl\vert f(x)-f(B_{2}) \bigr\vert ^{2}\,dx \biggr)^{1/2} \\ \leq&2^{d} \vert B_{2} \vert ^{1/p-1}\| f \|_{\Lambda_{d(1/p-1)}^{L}}. \end{aligned}$$
$$ \bigl\vert f(B_{k+1})- f(B_{1}) \bigr\vert \leq C k |B_{k+1}|^{1/p-1} \| f \|_{\Lambda_{d(1/p-1)}^{L}}. $$
(18)
$$\begin{aligned} \bigl\vert Q^{L}_{t} \widetilde{f}_{2}(x) \bigr\vert \leq& C \int_{\mathbb{R}^{d}} \frac{t}{(t^{2}+C \vert x-y \vert ^{2})^{(d+1)/2}} \bigl\vert \widetilde{f}_{2}(y) \bigr\vert \,dy \\ \leq& C \int_{(B_{1})^{c}} \frac{t}{ \vert x_{0}-y \vert ^{(d+1)}} \bigl\vert f(y)-f(B_{1}) \bigr\vert \,dy \\ \leq& C \sum^{\infty}_{k=1} \frac{t}{(2^{k} r)^{d+1}} \biggl( \int_{B_{k+1} \setminus B_{k}} \bigl\vert f(y) - f(B_{k+1}) \bigr\vert \,dy + \bigl(2^{k} r\bigr)^{d} \bigl\vert f(B_{k+1})- f(B_{1}) \bigr\vert \biggr) \\ \leq& C \frac{t}{r^{1-d(1/p-1)}} \sum^{\infty}_{k=1} 2^{k({d(1/p-1)}-1)}(1+k) \| f \|_{\Lambda_{d(1/p-1)}^{L}} \\ \leq& C \frac{t}{r^{1-d(1/p-1)}} \| f \|_{\Lambda_{d(1/p-1)}^{L}}. \end{aligned}$$
Thus we have
It remains to estimate the constant term. Assume first that \(r< \rho(x_{0})\). Taking \(k_{0}\) such that \(2^{k_{0}}r< \rho(x_{0}) \leq 2^{k_{0}+1}r\), we have
Note that \(\rho(x) \sim\rho(x_{0})>r\) for any \(x \in B(x_{0},r)\), by using Lemma 5(c), we get
In the last step of the above, we use the fact \(d(1/p-1)<1\). For \(r \geq\rho(x_{0})\), we have \(| f(B_{1}) | \leq C |B_{1}|^{1/p-1} \| f \|_{\Lambda_{d(1/p-1)}^{L}}\).
$$ \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \widetilde{f}_{2}(x) \bigr\vert ^{2} \frac{dx \,dt}{t} \leq C \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}}. $$
(19)
$$\begin{aligned} \bigl\vert f(B_{1}) \bigr\vert \leq& \bigl\vert f(B_{k_{0}+1})- f(B_{1}) \bigr\vert + \bigl\vert f(B_{k_{0}+1}) \bigr\vert \\ \leq& C k_{0} \vert B_{k_{0}+1} \vert ^{1/p-1}\| f \|_{\Lambda_{d(1/p-1)}^{L}} + \vert B_{k_{0}+1} \vert ^{1/p-1}\| f \|_{\Lambda_{d(1/p-1)}^{L}} \\ \leq& C \biggl( 1+ \log_{2} \frac{\rho(x_{0})}{r} \biggr) \vert B_{k_{0}+1} \vert ^{1/p-1}\| f \|_{\Lambda_{d(1/p-1)}^{L}} . \end{aligned}$$
$$\begin{aligned}& \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \bigl( f(B_{1}) \mathbf{1} \bigr) (x) \bigr\vert ^{2} \frac{dx \,dt}{t} \\& \quad = \frac{ | f(B_{1}) |^{2}}{|B|^{2/p-1}} \int_{\widehat{B}} \biggl\vert \int_{\mathbb{R}^{d}} Q^{L}_{t}(x,y) \,dy \biggr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq \frac{C | f(B_{1}) |^{2}}{|B|^{2/p-1}} \int_{\widehat{B}} \biggl( \frac{t}{\rho(x_{0})} \biggr)^{2} \frac{dx \,dt}{t} \\& \quad \leq C \frac{|B_{k_{0}+1}|^{2/p-2}}{|B|^{2/p-2}} \biggl( 1+ \log_{2}\frac{\rho(x_{0})}{r} \biggr)^{2} \biggl( \frac{r}{\rho(x_{0})} \biggr)^{2} \|f \|^{2}_{\Lambda_{d(1/p-1)}^{L}} \\& \quad = C \biggl( 1+ \log_{2} \frac{\rho(x_{0})}{r} \biggr)^{2} \biggl( \frac{r}{\rho(x_{0})} \biggr)^{2-2d(1/p-1)} \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}} \\& \quad \leq C \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}}. \end{aligned}$$
(20)
Note that \(\rho(x) \leq Cr\) for any \(x \in B(x_{0},r)\), again by Lemma 5(c), we get
Then (17) follows from (18)–(21). This proves part (a).
$$\begin{aligned}& \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \bigl( f(B_{1})\mathbf{1} \bigr) (x) \bigr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq\frac{ |f(B_{1}) |^{2}}{|B|^{2/p-1}} \int^{\infty}_{0} \int_{B} \biggl\vert \int_{\mathbb{R}^{d}} Q^{L}_{t}(x,y) \,dy \biggr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq\frac{C | f(B_{1}) |^{2}}{|B|^{2/p-1}} \biggl( \int_{B} \int^{\rho(x)}_{0} \biggl( \frac{t}{\rho(x)} \biggr)^{2} \frac{dt}{t} \,dx + \int_{B} \int^{\infty}_{\rho(x)} \biggl( \frac{t}{\rho(x)} \biggr)^{-2} \frac{dt}{t} \,dx \biggr) \\& \quad \leq C \| f \|^{2}_{\Lambda_{d(1/p-1)}^{L}}. \end{aligned}$$
(21)
Let \(f\in L^{1} ((1+|x|)^{-(d+1)}\,dx )\) and \(Q_{t}^{L}f(x)\in T_{2}^{p,\infty}\). We want to prove that \(f\in \Lambda_{d(1/p-1)}^{L}\). By \(\Lambda_{d(1/p-1)}^{L}\) is the dual space of \(H_{L}^{p}(\mathbb {R}^{d})\), it is sufficient to prove that
defined on finite linear combinations of \(H_{L}^{p,\infty}\)-atoms satisfies the estimate
By Lemma 7, Lemma 8, and Proposition 3, we get
This gives the proof of part (b) and then Theorem 1 is proved. □
$$H_{L}^{p}\ni g\mapsto\mathcal{L}_{f}(g):= \int_{\mathbb{R}^{d}}f(x)g(x)\,dx $$
$$\bigl\vert \mathcal{L}_{f}(g) \bigr\vert \leq C \bigl\Vert Q_{t}^{L}f \bigr\Vert _{T_{2}^{p, \infty}}\|g \|_{H_{L}^{p}}. $$
$$\begin{aligned} \bigl\vert \mathcal{L}_{f}(g) \bigr\vert =& \biggl\vert \int_{\mathbb{R}^{d}}f(x)g(x)\, dx \biggr\vert \\ =&4 \biggl\vert \int_{\mathbb{R}^{d+1}_{+}}Q_{t}^{L}f(x)Q_{t}^{L}g(x) \frac{dx \,dt}{t} \biggr\vert \\ \leq&C \bigl\Vert Q_{t}^{L}f \bigr\Vert _{T_{2}^{p,\infty}} \bigl\Vert Q_{t}^{L}g \bigr\Vert _{T_{2}^{p}} \\ \leq&C \bigl\Vert Q_{t}^{L}f \bigr\Vert _{T_{2}^{p,\infty}} \Vert g \Vert _{H_{L}^{p}}. \end{aligned}$$
4 The predual space of Hardy space \(H_{L}^{p}(\mathbb{R}^{d})\)
In this section, we give a Carleson measure characterization of the space \(\lambda_{d(1/p-1)}^{L}(\mathbb{R}^{d})\).
Proof of Theorem 2
Let \(f\in\lambda_{d(1/p-1)}^{L}\), then \(f\in \Lambda_{d(1/p-1)}^{L}\). By Theorem 1, we know \(f\in L^{1} ((1+|x|)^{-(d+1)}\,dx )\). To prove \(Q_{t}^{L} f\in T_{2, 0}^{p,\infty}\), we first prove that there exists a constant \(C>0\) such that, for any ball \(B=B(x_{0}, r)\), we have
where
We first assume (22) holds, then we show that \(Q_{t}^{k} f\in T_{2, 0}^{p,\infty}\). In fact, as \(f\in\lambda_{d(1/p-1)}^{L}\), we have \(f\in \Lambda_{d(1/p-1)}^{L}\) and there exists a constant \(C>0\) such that
Then, for any \(k\in\mathbb{N}\), we have
By (22), we have
We can take \(k_{0}\) large enough such that \(2^{-k_{0}/2}\|f\|_{\Lambda_{d(1/p-1)}^{L}}^{2}\) is small. This proves that \(\|Q_{t}^{L} f\|_{T_{2}^{p,\infty}}<\infty\) and \(\eta_{1}(f)=\eta_{2}(f)=\eta_{3}(f)=0\) follows from (23). Therefore \(Q_{t}^{L} f\in T_{2,0}^{p,\infty}\).
$$ \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q_{t}^{L} f(x) \bigr\vert ^{2}\frac{dx\,dt}{t}\leq \sum _{k=1}^{\infty}2^{-k(1-d(1/p-1))}\beta_{k}(f, B), $$
(22)
$$\beta_{k}(f,B)=\sup_{B'\subset B_{k+1}}\frac{1}{|B'|^{2/p-1}} \int_{B'} \bigl\vert f(y)-f\bigl(B'\bigr) \bigr\vert ^{2} \,dy. $$
$$\beta_{k}(f,B)\leq C \|f\|_{\Lambda_{d(1/p-1)}^{L}}. $$
$$ \lim_{a\rightarrow0}\sup_{B\subset\mathbb{R}^{d} r_{B}\leq a}\beta _{k}(f,B)=\lim_{a\rightarrow\infty}\sup_{B\subset\mathbb{R}^{d} r_{B}\geq a} \beta_{k}(f,B)=\lim_{a\rightarrow\infty}\sup_{B\subset \mathbb{R}^{d} B\subset B(0,a)^{c}} \beta_{k}(f,B)=0. $$
(23)
$$\begin{aligned}& \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q_{t}^{L} f(x) \bigr\vert ^{2}\frac{dx\,dt}{t} \\& \quad \leq C\sum_{k=1}^{k_{0}}2^{-k(1-d(1/p-1))} \beta_{k}(f,B)+C\sum_{k=k_{0}}^{\infty} 2^{-k(1-d(1/p-1))}\|f\|_{\Lambda_{d(1/p-1)}^{L}}^{2} \\& \quad \leq C\sum_{k=1}^{k_{0}}2^{-k(1-d(1/p-1))} \beta_{k}(f,B)+C2^{-k_{0}(1-d(1/p-1))} \|f\|_{\Lambda_{d(1/p-1)}^{L}}^{2}. \end{aligned}$$
Now we give the proof of (22). Set \(B_{k}=B(x_{0}, 2^{k}r)\) and
By Lemma 6, we have
By
and Lemma 5(a), for \(x \in B(x_{0},r)\),
Therefore
It remains to estimate the constant term. Assume first that \(r< \rho(x_{0})\). Taking \(k_{0}\) such that \(2^{k_{0}}r< \rho(x_{0}) \leq 2^{k_{0}+1}r\), we have
Note that \(\rho(x) \sim\rho(x_{0})>r\) for any \(x \in B(x_{0},r)\), by Lemma 5(c), we get
For \(r \geq\rho(x_{0})\), we have
Note that \(\rho(x) \leq Cr\) for any \(x \in B(x_{0},r)\), again by Lemma 5(c),
Then (22) follows from (24)–(27).
$$f= \bigl(f-f(B_{1}) \bigr)\chi_{B_{1}}+ \bigl(f-f(B_{1}) \bigr)\chi _{(B_{1})^{c}}+f(B_{1})=\widetilde{f}_{1} + \widetilde{f}_{2} + f(B_{1}). $$
$$\begin{aligned} \frac{1}{ \vert B \vert ^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \widetilde{f}_{1}(x) \bigr\vert ^{2} \frac{dx \,dt}{t} \leq& \frac{1}{ \vert B \vert ^{2/p-1}} \int_{B} \bigl\vert s_{L} \widetilde{f}_{1}(x) \bigr\vert ^{2} \,dx \\ =& \frac{1}{4 \vert B \vert ^{2/p-1}} \int_{B_{1}} \bigl\vert f(x)- f(B_{1}) \bigr\vert ^{2} \,dx \leq C \beta_{1}(f,B). \end{aligned}$$
(24)
$$\bigl\vert f(B_{k+1})- f(B_{1}) \bigr\vert \leq C \sum _{i=2}^{k+1} \vert B_{i} \vert ^{1/p-1} \frac{1}{ \vert B_{i} \vert ^{1/p-1/2}} \biggl( \int_{B_{i}} \bigl\vert f(x)-f(B_{i}) \bigr\vert ^{2}\,dx \biggr)^{1/2} $$
$$\begin{aligned} \bigl\vert Q^{L}_{t} \widetilde{f}_{2}(x) \bigr\vert \leq& C \int_{\mathbb{R}^{d}} \frac{t}{ \vert x_{0}-y \vert ^{(d+1)}} \bigl\vert \widetilde {f}_{2}(y) \bigr\vert \,dy \\ \leq& C \int_{(B_{1})^{c}} \frac{t}{ \vert x_{0}-y \vert ^{(d+1)}} \bigl\vert f(y)-f(B_{1}) \bigr\vert \,dy \\ \leq& C \sum^{\infty}_{k=1} \frac{t}{(2^{k} r)^{d+1}} \biggl( \int_{B_{k+1} \setminus B_{k}} \bigl\vert f(y) - f(B_{k+1}) \bigr\vert \,dy \\ &{} + \bigl(2^{k} r\bigr)^{d} \bigl\vert f(B_{k+1})- f(B_{1}) \bigr\vert \biggr) \\ \leq& C \frac{t}{r^{1-d(1/p-1)}} \sum^{\infty}_{k=1} 2^{k(d(1/p-1)-1)}\biggl(\frac{1}{ \vert B_{k+1} \vert ^{1/p-1/2}}\biggl( \int _{B_{k+1}} \bigl\vert f-f(B_{k+1}) \bigr\vert ^{2} \,dy\biggr)^{1/2} \\ &{} +\sum_{i=2}^{k+1}\frac{1}{ \vert B_{i} \vert ^{1/p-1/2}} \biggl( \int _{B_{i}} \bigl\vert f-f(B_{i}) \bigr\vert ^{2} \,dy\biggr)^{1/2}\biggr) \\ \leq& C\frac{t}{r^{1-d(1/p-1)}}\sum^{\infty}_{k=1} 2^{k(d(1/p-1)-1)}(1+k)\beta_{k}(f,B)^{1/2}. \end{aligned}$$
$$ \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \widetilde{f}_{2}(x) \bigr\vert ^{2} \frac{dx \,dt}{t} \leq C \sum^{\infty}_{k=1} 2^{k(d(1/p-1)-1)} \beta_{k}(f,B). $$
(25)
$$\begin{aligned} \bigl\vert f(B_{1}) \bigr\vert \leq& \bigl\vert f(B_{k_{0}+1})- f(B_{1}) \bigr\vert + \bigl\vert f(B_{k_{0}+1}) \bigr\vert \\ \leq&C\sum_{i=2}^{k_{0}+1} \vert B_{i} \vert ^{1/p-1}\frac {1}{ \vert B_{i} \vert ^{1/p-1/2}} \biggl( \int_{B_{i}} \bigl\vert f-f(B_{i}) \bigr\vert ^{2} \,dy \biggr)^{1/2} \\ &{} + \vert B_{k_{0}+1} \vert ^{1/p-1}\frac{1}{ \vert B_{k_{0}+1} \vert ^{1/p-1/2}} \biggl( \int _{B_{k_{0}+1}} \vert f \vert ^{2} \,dy \biggr)^{1/2} \\ \leq& C \vert B_{k_{0}+1} \vert ^{1/p-1}(k_{0}+1) \beta_{k_{0}}^{1/2}(f,B). \end{aligned}$$
$$\begin{aligned}& \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \bigl( f(B_{1}) \mathbf{1} \bigr) (x) \bigr\vert ^{2} \frac{dx \,dt}{t} \\& \quad = \frac{ | f(B_{1}) |^{2}}{|B|^{2/p-1}} \int_{\widehat{B}} \biggl\vert \int_{\mathbb{R}^{d}} Q^{L}_{t}(x,y) \,dy \biggr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq \frac{C | f(B_{1}) |^{2}}{|B|^{2/p-1}} \int_{\widehat{B}} \biggl( \frac{t}{\rho(x_{0})} \biggr)^{2} \frac{dx \,dt}{t} \\& \quad \leq C \frac{|B_{k_{0}+1}|^{2/p-2}}{|B|^{2/p-2}} ( 1+k_{0})^{2} \biggl( \frac{r}{\rho(x_{0})} \biggr)^{2}\beta_{k_{0}}(f,B) \\& \quad \leq C 2^{-2k_{0}(1-d(1/p-1))}(1+k_{0})^{2} \beta_{k_{0}}(f,B) \\& \quad \leq C 2^{-k_{0}(1-d(1/p-1))}\beta_{k_{0}}(f,B). \end{aligned}$$
(26)
$$\bigl\vert f(B_{1}) \bigr\vert \leq C |B_{1}|^{1/p-1} \frac{1}{|B_{1}|^{1/p-1/2}} \biggl( \int_{B_{1}} \bigl\vert f-f(B_{1}) \bigr\vert ^{2} \,dy \biggr)^{1/2}. $$
$$\begin{aligned}& \frac{1}{|B|^{2/p-1}} \int_{\widehat{B}} \bigl\vert Q^{L}_{t} \bigl( f(B_{1})\mathbf{1} \bigr) (x) \bigr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq\frac{ |f(B_{1}) |^{2}}{|B|^{2/p-1}} \int^{\infty}_{0} \int_{B} \biggl\vert \int_{\mathbb{R}^{d}} Q^{L}_{t}(x,y) \,dy \biggr\vert ^{2} \frac{dx \,dt}{t} \\& \quad \leq\frac{C | f(B_{1}) |^{2}}{|B|^{2/p-1}} \biggl( \int_{B} \int^{\rho(x)}_{0} \biggl( \frac{t}{\rho(x)} \biggr)^{2} \frac{dt}{t} \,dx + \int_{B} \int^{\infty}_{\rho(x)} \biggl( \frac{t}{\rho(x)} \biggr)^{-2} \frac{dt}{t} \,dx \biggr) \\& \quad \leq C \frac{|B_{1}|^{2/p-1}}{|B|^{2/p-1}}\beta_{1}(f,B)\leq 2^{d(1/p-1)-1} \beta_{1}(f,B). \end{aligned}$$
(27)
For the reverse, by Theorem 1, we get \(f\in \Lambda_{d(1/p-1)}^{L}\) from \(Q_{t}^{L}f\in T_{2,0}^{p,\infty}\). For any ball \(B=B(x_{0},r)\),
Let \(G(x)= (g(x)-g(B) )\chi_{B}\). Then, by Lemma 7, we obtain
By Hölder’s inequality and Lemma 6, we have
Now, we estimate \(E_{k}\). By Hölder’s inequality again, we have that
where
and
When \(r<\rho(x_{0})\), then \(\int_{B}g(x)-g(B)\,dx=0\). Therefore, by Lemma 5(b),
Therefore
It follows that
When \(r\geq\rho(x_{0})\), we have \(\rho(y)\leq C r\) for \(y\in B(x_{0}, r)\). Then, by Lemma 5(a),
Then we can get
By (28) and (29), we know
where
Then we can get \(\gamma_{1}(f)=\gamma_{2}(f)=\gamma_{3}(f)=0\) as the proof of the first part of this theorem. Therefore \(f\in\lambda_{\alpha}^{L}\) and the proof of Theorem 2 is completed. □
$$\begin{aligned} \biggl( \int_{B} \bigl\vert f(x)-f(B) \bigr\vert ^{2} \,dx\biggr)^{1/2} =&\sup_{\operatorname{supp} g\subset B, \|g\|_{L^{2}(B)}\leq 1} \biggl\vert \int_{B}\bigl(f(x)-f(B)\bigr)g(x)\,dx \biggr\vert \\ =&\sup_{\operatorname{supp} g\subset B, \|g\|_{L^{2}(B)}\leq 1} \biggl\vert \int_{B}f(x) \bigl(g(x)-g(B)\bigr)\,dx \biggr\vert . \end{aligned}$$
$$\begin{aligned} \biggl\vert \int_{\mathbb{R}^{d}}f(x)G(x)\,dx \biggr\vert =&4 \biggl\vert \int_{\mathbb{R}^{d+1}_{+}}Q_{t}^{L}f(x) \bigl(Q_{t}^{L}g(x)-Q_{t}^{L}g(B) (x) \bigr)\frac{dx \,dt}{t} \biggr\vert \\ \leq&C \int_{\widehat{B_{2}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert \bigl\vert Q_{t}^{L}G(x) \bigr\vert \frac{dx \,dt}{t} \\ &{}+\sum_{k=2}^{\infty} \int_{\widehat{B_{k+1}}\setminus \widehat{B_{k}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert \bigl\vert Q_{t}^{L}G(x) \bigr\vert \frac{dx \,dt}{t} \\ =&E_{1}+\sum_{k=2}^{\infty}E_{k}. \end{aligned}$$
$$\begin{aligned} E_{1} \leq& \biggl( \int_{\widehat{B_{2}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2} \biggl\Vert \biggl( \int_{0}^{\infty}\bigl\vert Q_{t}^{L} \bigl(g-g(B) \bigr) (x) \bigr\vert ^{2}\frac{dt}{t} \biggr)^{1/2} \biggr\Vert _{L^{2}(B_{2})} \\ \leq&C \biggl( \int_{\widehat{B_{2}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2}. \end{aligned}$$
(28)
$$E_{k}\leq F_{k}\cdot I_{k}, $$
$$F_{k}= \biggl( \int_{\widehat{B_{k+1}}\setminus\widehat{B_{k}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx\,dt}{t} \biggr)^{1/2} $$
$$I_{k}= \biggl( \int_{\widehat{B_{k+1}}\setminus\widehat{B_{k}}} \bigl\vert Q_{t}^{L}g(x)-Q_{t}^{L}g(B) (x) \bigr\vert ^{2}\frac{dx\,dt}{t} \biggr)^{1/2}. $$
$$\begin{aligned} \bigl\vert Q_{t}^{L}g(x)-Q_{t}^{L}g(B) (x) \bigr\vert =& \biggl\vert \int_{B} \bigl(Q_{t}^{L}(x,y)-Q_{t}^{L}(x,x_{0}) \bigr) \bigl(g(y)-g(B)\bigr)\,dy \biggr\vert \\ \leq&C \int_{B}\frac{t}{(t+|x-y|)^{d+1}}\frac{|x_{0}-y|}{t} \bigl\vert g(y)-g(B) \bigr\vert \,dy \\ \leq&C \int_{B}\frac{t}{(2^{k}r)^{d+1}}\frac{r}{t} \bigl\vert g(y)-g(B) \bigr\vert \,dy \\ \leq&C\frac{t}{(2^{k}r)^{d+1}}\frac{r}{t}\|g\|_{L^{1}(B)}\leq C |B|^{1/2}\frac{t}{(2^{k}r)^{d+1}}\frac{r}{t}. \end{aligned}$$
$$ I_{k}^{2}\leq C|B| \int_{0}^{2^{k+1}r} \int_{B_{k+1}}\frac {t^{2}}{(2^{k}r)^{2d+2}} \biggl(\frac{r}{t} \biggr)^{2} \frac{dx \,dt}{t} \leq C|B|\frac{1}{(2^{k}r)^{d}}2^{-2k}. $$
$$E_{k}\leq C|B|^{1/2}|B_{k}|^{-1/2}2^{-k} \biggl( \int_{\widehat{B_{k+1}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2}. $$
$$\begin{aligned} \bigl\vert Q_{t}^{L}g(x)-Q_{t}^{L}g(B) (x) \bigr\vert =& \biggl\vert \int_{B}Q_{t}^{L}(x,y)g(y)\, dy \biggr\vert \\ \leq&C \int_{B}\frac{t}{(2^{k}r)^{d+1}}\frac{\rho(y)}{t} \bigl\vert g(y) \bigr\vert \,dy \\ \leq&C\frac{t}{(2^{k}r)^{d+1}}\frac{r}{t}\|g\|_{L^{1}(B)}\leq C |B|^{1/2}\frac{t}{(2^{k}r)^{d+1}}\frac{r}{t}. \end{aligned}$$
$$ E_{k}\leq C|B|^{1/2}|B_{k}|^{-1/2}2^{-k} \biggl( \int_{\widehat{B_{k+1}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx\,dt}{t} \biggr)^{1/2}. $$
(29)
$$\begin{aligned}& \frac{1}{ \vert B \vert ^{1/p-1/2}} \biggl( \int_{B} \bigl\vert f(x)-f(B) \bigr\vert ^{2}\, dx \biggr)^{1/2} \\& \quad \leq \frac{C}{ \vert B \vert ^{1/p-1}}\sum_{k=1}^{\infty}2^{-k} \vert B_{k} \vert ^{-1/2} \biggl( \int_{\widehat{B_{k}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2} \\& \quad = C\sum_{k=1}^{\infty}2^{-k} \frac{ \vert B_{k} \vert ^{1/p-1}}{ \vert B \vert ^{1/p-1}}\frac {1}{ \vert B_{k} \vert ^{1/p-1/2}} \biggl( \int_{\widehat{B_{k}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2} \\& \quad \leq C\sum_{k=1}^{\infty}2^{-k(1-\alpha)} \sigma_{k}(f,B), \end{aligned}$$
$$\sigma_{k}(f,B)=\frac{1}{|B_{k}|^{1/p-1/2}} \biggl( \int_{\widehat {B_{k}}} \bigl\vert Q_{t}^{L}f(x) \bigr\vert ^{2}\frac{dx \,dt}{t} \biggr)^{1/2}. $$
5 Conclusions
This paper defines a new version of Carleson measure associated with Hermite operator, which is adapted to the operator L. Then, we characterize the dual spaces and predual spaces of the Hardy spaces \(H_{L}^{p}(\mathbb {R}^{d})\) associated with L. The main results of this paper are the central problems in harmonic analysis, which can be used in PED or geometry widely.
Competing interests
The authors declare that they have no competing interests.
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