1 Introduction
Consider the generalized Schrödinger operator
where μ is a nonnegative Radon measure on \(\mathbb{R}^{n}\). Throughout this paper we assume that μ satisfies the following conditions: there exist positive constants \(C_{0}\), \(C_{1}\), and δ such that
and
for all \(x\in \mathbb{R}^{n}\) and \(0< r< R\), where \(B(x,r)\) denotes the open ball centered at x with radius r. Condition (1.2) may be regarded as scale-invariant Kato-condition, and (1.3) says that the measure μ is doubling on balls satisfying \(\mu (B(x,r)) \geq c r^{n-2}\).
$$ \mathcal{L}=-\Delta +\mu\quad \text{in }\mathbb{R}^{n}, n\geq 3, $$
(1.1)
$$ \mu \bigl(B(x,r) \bigr)\leq C_{0} \biggl( \frac{r}{R} \biggr)^{n-2+\delta }\mu \bigl(B(x,R) \bigr) $$
(1.2)
$$ \mu \bigl(B(x,2r) \bigr)\leq C_{1} \bigl\{ \mu \bigl(B(x,r) \bigr)+r^{n-2} \bigr\} $$
(1.3)
Hardy spaces are widely used various fields of analysis and partial differential equations. Let Δ be the Laplace operator on \(\mathbb{R}^{n}\). It is well known that \(H^{1}(\mathbb{R}^{n})\) can be characterized by the maximal function \(\sup_{t>0}|e^{-t\Delta }f(x)|\). See Stein [14]. In a sense, \(H^{1}(\mathbb{R}^{n})\) can be seen as the Hardy space associated with the operator −Δ. Let \(\mathcal{L}\) be a general differential operator, such as second order elliptic self-adjoint operators in divergence form, degenerate Schrödinger operators with nonnegative potential, Schrödinger operators with nonnegative potential, and so on. In recent years, the Hardy spaces associated with \(\mathcal{L}\) have become one of hot issues in harmonic analysis, see [2, 4‐10] and the references therein.
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Let \(\mathcal{L}\) be a generalized Schrödinger operator. Denote by \(\{T_{t}\}_{t>0}:=\{e^{-t\mathcal{L}}\}_{t>0}\) the heat semigroup generated by \(-\mathcal{L}\). The kernel of \(\{T_{t}\}\) is denoted by \(K_{t}^{\mathcal{L}}(\cdot ,\cdot )\), that is,
The maximal function associated with \(\{T_{t}\}\) is defined as
In [15], Wu and Yan introduced the following Hardy type space associated with \(\mathcal{L}\).
$$ T_{t}f(x)= \int _{\mathbb{R}^{n}}{}K_{t}^{\mathcal{L}}(x,y)f(y)\,d\mu (y). $$
$$ \mathcal{M}_{\mathcal{L}}(f) (x):=\sup_{t>0} \bigl\vert e^{-t\mathcal{L}}f(x) \bigr\vert \in L^{1} \bigl(\mathbb{R}^{n} \bigr). $$
Definition 1.1
A Hardy type space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) related to \(\mathcal{L}\) is defined as the set of all functions in \(f\in L^{1}( \mathbb{R}^{n})\) satisfying \(\mathcal{M}_{\mathcal{L}}(f)\in L^{1}( \mathbb{R}^{n})\). The norm of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) is defined as \(\|f\|_{H^{1}_{\mathcal{L}}}:=\|\mathcal{M}_{\mathcal{L}}(f) \|_{L^{1}}\).
Let \(\mathcal{L}=-\Delta \). \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) goes back to the classical Hardy space \(H^{1}(\mathbb{R}^{n})\). For a linear operator T, one of the methods to derive the \(H^{1}\)-boundedness is the so-called “atomic-molecular” method. In recent years, several authors used this method to investigate the boundedness on Hardy spaces associated with operators, see [3, 11, 13]. In Sect. 3.1, via a class of \((1, q)\)-type atoms associated with \(\mathcal{L}\), we obtain the corresponding atomic characterization of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\), see Sect. 3.1. Further, in Sect. 3.2, we introduce the \((p,q,\varepsilon )\)-moleculars associated with \(\mathcal{L}\) and establish the molecular decomposition of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\), see Theorem 3.6. In Sect. 4, let \(R_{\mathcal{L}}\) and \(\mathcal{L}^{i\gamma }\) denote the Riesz transforms and the imaginary power associated with \(\mathcal{L}\), i.e.,
By the aid of the regularities of the integral kernels, we can apply Theorems 3.3 & 3.6 to derive the \(H^{1}_{ \mathcal{L}}\)-boundedness of \(R_{\mathcal{L}}\) and \(\mathcal{L}^{i \gamma }\), see Theorems 4.4 & 4.6, respectively.
$$ \textstyle\begin{cases} R_{\mathcal{L}}:=\nabla (-\Delta +\mu )^{-1/2}; \\ \mathcal{L}^{\gamma }:=(-\Delta +\mu )^{i\gamma }. \end{cases} $$
Throughout this article, we will use c 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 \({1}/{C} \leq {B_{1}}/{B_{2}}\leq C\).
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2 Preliminaries
2.1 Generalized Schrödinger operators
Let μ be a Radon measure satisfying conditions (1.2) & (1.3). The auxiliary function \(m(x, \mu )\) is defined by
We begin by recalling some basic properties of the function \(m(x,\mu )\).
$$ \frac{1}{m(x,\mu )}=:\sup \biggl\{ r>0: \frac{\mu (B(x,r))}{r^{n-2}} \leq C_{1} \biggr\} . $$
Lemma 2.1
([12, Proposition 1.8 & Remark 1.9])
Suppose that
μ
satisfies (1.2) & (1.3). Then
(i)
\(0< m(x,\mu )<\infty \)
for every
\(x\in \mathbb{R}^{n}\).
(ii)
If
\(r=m(x,\mu )^{-1}\), then
\(r^{n-2}\leq \mu (B(x,r)) \leq C_{1}r^{n-2}\).
(iii)
If
\(|x-y|\leq Cm(x,\mu )^{-1}\), then
\(m(x,\mu ) \approx m(y,\mu )\).
(iv)
There exist constants
\(c, C>0\)
such that, for
\(x,y\in \mathbb{R}^{n}\),
with
\(k_{0}=C_{2}/\delta >0\)
and
\(C_{2}=\log _{2}(C_{1}+2^{n-2})\).
$$ \frac{cm(y,\mu )}{\{1+ \vert x-y \vert m(y,\mu )\}^{k_{0}/(1+k_{0})}}\leq m(x, \mu )\leq Cm(y,\mu ) \bigl\{ 1+ \vert x-y \vert m(y,\mu ) \bigr\} ^{k_{0}} $$
With the modified Agmon metric \(ds^{2}=m(x,\mu )\{dx^{2}_{1}+\cdots +dx _{n}^{2}\}\), the distance function \(d(x,y,\mu )\) is given by
where \(\gamma :[0,1]\rightarrow \mathbb{R}^{n}\) is absolutely continuous and \(\gamma (0)=x\), \(\gamma (1)=y\).
$$ d(x,y,\mu )=\inf_{\gamma } \int ^{1}_{0}m \bigl(\gamma (\tau ),\mu \bigr) \bigl\vert \gamma '( \tau ) \bigr\vert \,d\tau , $$
(2.1)
A parabolic-type distance function associated with \(m(x,\mu )\) is defined by
where \(\gamma (\tau )=(\gamma _{1}(\tau ),\ldots ,\gamma _{n}(\tau ))=( \tilde{\gamma }(\tau ),\gamma _{n+1}(\tau )):[0,1]\rightarrow \mathbb{R}^{n}\times \mathbb{R}_{+}\) is absolutely continuous with \(\gamma (0)=(x,0)\), \(\gamma (1)=(y,\sqrt{t})\).
$$ d_{\mu }(x,y,t)=\inf_{\gamma } \int ^{1}_{0}m \bigl(\tilde{\gamma }(\tau ), \mu \bigr)\max \bigl\{ \bigl\vert (\tilde{\gamma })'(\tau ) \bigr\vert , \bigl\vert (\gamma _{n+1})'(\tau ) \bigr\vert \bigr\} \,d \tau , $$
(2.2)
Lemma 2.2
([15, Lemma 2.2])
For the distance function
\(d(x,y,\mu )\)
in (2.1), we have
(i)
For every
\(x,y, z\in \mathbb{R}^{n}\),
$$ d(x,y,\mu )\leq d(x,z,\mu )+d(z,y,\mu ). $$
(ii)
There are two positive constants
c
and
C
such that, for any
\(x,y\in \mathbb{R}^{n}\),
$$ c \bigl\{ \bigl\{ 1+ \vert x-y \vert m(x,\mu ) \bigr\} ^{1/(k_{0}+1)}-1 \bigr\} \leq d(x,y,\mu ) \leq C \bigl\{ 1+ \vert x-y \vert m(x,\mu ) \bigr\} ^{k_{0}+1}. $$
Lemma 2.3
([15, Lemma 2.3])
For the distance function
\(d_{\mu }(x,y,t)\), there exist two positive constants
c
and
C
such that, for any
\(x,y\in \mathbb{R}^{n}\), \(x\neq y\), and
\(t>0\),
$$ \textstyle\begin{cases} d_{\mu }(x,y,t)\geq c \{\{1+\max \{ \vert x-y \vert ,\sqrt{t}\}m(x,\mu )\} ^{1/(k_{0}+1)}-1 \}; \\ d_{\mu }(x,y,t)\leq C \{1+\max \{ \vert x-y \vert , \sqrt{t}\}m(x,\mu ) \}^{k_{0}+1}. \end{cases} $$
It follows from (1.2), (1.3), and Lemma 2.1 that there exists a constant \(C>0\) such that, for every \(x\in \mathbb{R}^{n}\),
see [15, (2.1)]. The above estimate implies the following.
$$ \mu \bigl(B(x,r) \bigr)\leq \textstyle\begin{cases} C (rm(x,\mu ) )^{\delta }r^{n-2},& r< m(x,\mu )^{-1}; \\ C (rm(x,\mu ) )^{C _{2}}m(x,\mu )^{2-n},& r< m(x,\mu )^{-1}, \end{cases} $$
(2.3)
Lemma 2.4
([15, (2.2)])
For every nonnegative Schwarz function
ω,
$$ \int _{\mathbb{R}^{n}}t^{-n/2}\omega \biggl(\frac{x-y}{\sqrt{t}} \biggr)\,d\mu (y) \leq \textstyle\begin{cases} Ct^{-1} ( \sqrt{t}m(x,\mu ) )^{\delta },& t< m(x,\mu )^{-2}; \\ Ct ^{-1} (\sqrt{t}m(x,\mu ) )^{C_{2}-n+2},& t\geq m(x,\mu )^{-2}. \end{cases} $$
(2.4)
2.2 Function spaces associated with \(\mathcal{L}\)
In order to characterize \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\), Wu and Yan [15] introduced the following \(H^{1}_{\mathcal{L}}\)-atoms. For \(j\in \mathbb{Z}\), define the sets \(\mathcal{B}_{j}\) as
Since \(0< m(x,\mu )<\infty \), we have \(\mathbb{R}^{n}=\bigcup_{j\in \mathbb{Z}}\mathcal{B}_{j}\).
$$ \mathcal{B}_{j}= \bigl\{ x: 2^{j/2}\leq m(x,\mu )< 2^{(j+1)/2} \bigr\} . $$
Definition 2.5
A function a is a \((1,\infty )\)-atom for \(H^{1}_{\mathcal{L}}( \mathbb{R}^{n})\) associated with a ball \(B(x_{0},r)\) if The atomic norm of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) is defined by \(\|f\|_{\mathcal{L}\text{-}\mathrm{atom}}:=\inf \{\sum_{j}|\lambda _{j}|\}\), where the infimum is taken over all decompositions \(f=\sum_{j}\lambda _{j}a_{j}\), where \(\{a_{j}\}\) is a sequence of \((1,\infty )\)-atoms and \(\{\lambda _{j}\}\) is a sequence of scalars.
(i)
\(\operatorname{supp} a\subset B(x_{0},r)\);
(ii)
\(\|a\|_{L^{\infty }}\leq |B(x_{0},r)|^{-1}\);
(iii)
if \(x_{0}\in \mathcal{B}_{j}\), then \(r\leq 2^{1-j/2}\);
(iv)
if \(x_{0}\in \mathcal{ B}_{j}\) and \(r\leq 2^{-1-j/2}\), then \(\int a(x)\,dx=0\).
One of the main results of [15] is the following proposition.
Proposition 2.6
([15, Theorem 1.2])
Assume that
μ
is a nonnegative Radon measure on
\(\mathbb{R}^{n}\)
satisfying (1.2) & (1.3) for some
\(\delta >0\). Then the norms
\(\|f\|_{H^{1}_{\mathcal{L}}}\)
and
\(\|f\|_{\mathcal{L}\text{-}\mathrm{atom}}\)
are equivalent, that is, there exists a constant
\(C>0\)
such that
$$ \frac{1}{C} \Vert f \Vert _{H^{1}_{\mathcal{L}}}\leq \Vert f \Vert _{H_{\mathcal{L}}^{1}\text{-}\mathrm{atom}} \leq C \Vert f \Vert _{H^{1}_{\mathcal{L}}}. $$
At the end of this section, we state some regularity estimates for the kernel \(K^{{\mathcal{L}}}_{t}(\cdot ,\cdot )\).
Proposition 2.7
([15, Lemma 3.7])
(i)
(ii)
For every
\(0<\delta '<\delta _{0}=\min \{\alpha , \delta , \nu \}\), there exists a constant
C
such that, for every
\(N^{\prime }>0\), there exists a constant
\(C>0\)
such that, for
\(|h|<\sqrt{t}\), we have
$$ \bigl\vert K^{\mathcal{L}}_{t}(x+h,y)-K^{\mathcal{L}}_{t}(x,y) \bigr\vert \leq C_{N^{\prime }} \biggl(\frac{ \vert h \vert }{\sqrt{t}} \biggr)^{\delta '}\frac{1}{t^{n/2}}e^{-c \vert x-y \vert ^{2}/t} \frac{C_{N}}{\{1+\sqrt{t}m(x,\mu )+\sqrt{t}m(y,\mu )\}^{N^{\prime }}}. $$
3 Molecular characterization of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\)
3.1 The \((1,q)\)-atom decomposition
Now we introduce a new type of atoms.
Definition 3.1
A function a is a \((1,q)\)-atom of \(H^{1}_{\mathcal{L}}(\mathbb{R} ^{n})\) if
(i)
\(\operatorname{supp}a\subset B(x_{0},r)\);
(ii)
\(\|a\|_{q}\leq |B(x_{0},r)|^{1/q-1}\);
(iii)
if \(r\leq \rho (x_{0})\), then \(\int a(x)\,dx=0\).
Theorem 3.2
Any
\((1,\infty )\)-atom is a
\((1,q)\)-atom.
Proof
In fact, by Hölder’s inequality,
□
$$\begin{aligned} \Vert a \Vert _{q} \leq \Vert a \Vert _{\infty } \bigl\vert B(x_{0},r) \bigr\vert ^{1/q}\leq \bigl\vert B(x_{0},r) \bigr\vert ^{1/q-1}. \end{aligned}$$
Theorem 3.3
Let
\(\mathcal{L}=-\triangle +\mu \)
be a generalized Schrödinger operator, where
\(\mu \neq 0\)
is a nonnegative Radon measure on
\({\mathbb{R}}^{n}\)
satisfying (1.2) & (1.3) for some
\(\delta >0\). Then
\(f\in {H^{1}_{\mathcal{L}}}(\mathbb{R}^{n})\)
if and only if
\(f=\sum_{j}\lambda _{j}a_{j}\), where
\(\{{a_{j}}\}\)
are
\((1,q)\)-atoms and
\(\{{\lambda _{j}}\}\)
are scalars.
Proof
Because an \((1,\infty )\)-atom is also an \((1,q)\)-type atom, we only need to prove that there exists a constant c such that, for any \((1,q)\)-atom a, \({\|{\mathcal{M}_{\mathcal{L}}}(a)\|_{1}}\leq c\). Suppose that a is a \((1,q)\)-atom supported in \(B(x_{0},r)\). We write \({\|{\mathcal{M}_{\mathcal{L}}}(a)\|_{1}}\leq I_{1}+I_{2}\), where
$$ \textstyle\begin{cases} I_{1}:=\int _{B(x_{0},4r)} \vert {\mathcal{M}_{\mathcal{L}}}a(x) \vert \,dx; \\ I_{2}:=\int _{B^{c}(x_{0},4r)} \vert {\mathcal{M}_{\mathcal{L}}}a(x) \vert \,dx. \end{cases} $$
By Hölder’s inequality and the \(L^{q}\)-boundedness of \(\mathcal{M}_{\mathcal{L}}\), we can get
$$\begin{aligned} I_{1} \leq &{ \Vert {\mathcal{M}_{\mathcal{L}}}a \Vert _{q}} \bigl\vert B(x_{0},r) \bigr\vert ^{1- {1}/{q}} \leq {C \Vert a \Vert _{q}} \bigl\vert B(x_{0},r) \bigr\vert ^{1-{1}/{q}}\leq C. \end{aligned}$$
The estimation of \(I_{2}\) is divided into two cases.
Case 1:
\(1/{m(x_{0},\mu )}\leq r\leq 1/{4 m(x_{0},\mu )}\). For this case, by (i) of Lemma 2.7, we have
If \(y \in B(x_{0},r)\) and \(|x-x_{0}|>4r\), then \(|y-x_{0}|\leq |x-x _{0}|/4\) and \(|y-x|\geq 3|x-x_{0}|/4\). We can apply Lemma 2.1 to obtain
which gives
where in the last inequality we have used the fact that \(1\leq r m(x _{0},\mu )\leq 4\).
$$\begin{aligned} {\mathcal{M}_{\mathcal{L}}}(a) (x) \leq &c\sup_{t>0} { \int _{B(x _{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/t}} { \bigl(1+m(x,\mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c\sup_{t>0} { \int _{B(x_{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/t}} { \bigl(1+{ \vert x-y \vert ^{2}}/{\sqrt{t}} \bigr)^{-n-N}} { \bigl(1+m(x,\mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy. \end{aligned}$$
$$\begin{aligned} \bigl\vert {\mathcal{M}_{\mathcal{L}}}(a) (x) \bigr\vert \leq & c\sup _{t>0}\frac{1}{t ^{n/2}}{ \int _{B(x_{0},r)}} { \bigl({ \vert x-x_{0} \vert }/{ \sqrt{t}} \bigr)^{-n-N}} { \bigl(m(x, \mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c{ \vert x-x_{0} \vert ^{-n-N}} { \bigl[m(x,\mu ) \bigr]^{-N}} { \int _{B(x_{0},r)}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c{ \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}} \bigl[m(x_{0},\mu ) \bigr]^{-{N}/( {{k_{0}}+1})}, \end{aligned}$$
$$\begin{aligned} { \int _{ \vert x-x_{0} \vert >4r}} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \,dx \leq &c{ \int _{ \vert x-x _{0} \vert >4r}} { \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}} \bigl[m(x_{0},\mu ) \bigr]^{-{N}/( {{k_{0}}+1})}\,dx \\ \leq &c{ \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}}r^{-{N}/({{k_{0}}+1})} \leq C, \end{aligned}$$
Case2:
\(r<1/m(x_{0},\mu )\). By Proposition 2.7 and the symmetry of \(K^{\mathcal{L}}_{t}(\cdot ,\cdot )\), we have
Notice that \(|y-x_{0}|< r\), \(|x-x_{0}|>4r\Rightarrow |x-y|\geq {3|x-x _{0}|}/{4}\). By the canceling condition of a, we can get
which gives
□
$$\begin{aligned} \bigl\vert K^{\mathcal{L}}_{t}(x,y+h)-K^{\mathcal{L}}_{t}(x,y) \bigr\vert \leq C_{N}{ \bigl( { \vert h \vert }/{\sqrt{t}} \bigr)^{\delta '}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/ct}} { \bigl\{ {1+ \sqrt{t}m(x,\mu )+\sqrt{t}m(y,\mu ) \bigr\} }^{-N}}. \end{aligned}$$
$$\begin{aligned} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \leq &\sup _{t>0} \biggl\vert { \int _{B(x_{0},r)}} \bigl[K^{\mathcal{L}}_{t}(x,y)-K^{\mathcal{L}}_{t}(x,x _{0}) \bigr]a(y)\,dy \biggr\vert \\ \leq &c\sup_{t>0} { \int _{B(x_{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/ct}} { \bigl({ \vert y-x_{0} \vert }/{\sqrt{t}} \bigr)^{\delta '}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c\sup_{t>0} \biggl\{ {t^{-n/2}} { \int _{B(x_{0},r)}} { \bigl(1+ { \vert x-y \vert }/{\sqrt{t}} \bigr)^{-n-\delta '}} { \vert y-x_{0} \vert ^{\delta '}}/{t^{\delta '/2}} \bigl\vert a(y) \bigr\vert \,dy \biggr\} \\ \leq &{cr^{\delta '}} { \vert x-x_{0} \vert ^{-n-\delta '}}, \end{aligned}$$
$$\begin{aligned} { \int _{ \vert x-x_{0} \vert \geq 4r}} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \,dx \leq &c { \int _{ \vert x-x_{0} \vert \geq 4r}} {r^{\delta '}} { \vert x-x_{0} \vert ^{-n-\delta '}}\,dx \leq C. \end{aligned}$$
3.2 Molecular characterization of \(H^{1}_{\mathcal{L}}( \mathbb{R}^{n})\)
Now we introduce the molecular of \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\).
Definition 3.4
Let \(1\leq q\leq \infty \), \(\varepsilon >0\), \(b=1-{1}/{q}+\varepsilon \). An \(L^{q}\)-function M is called a \((1,q,\varepsilon )\)-molecular centered at \(x_{0}\) if
(i)
\(|x|^{nb}M(x)\in L^{q}(\mathbb{R}^{n})_{j}\);
(ii)
\(\|M\|^{{\varepsilon }/{b}}_{q}\||x-x_{0}|^{nb}M(\cdot ) \|^{1-{\varepsilon }/{b}}_{q}\leq 1\);
(iii)
if \(x_{0}\in B_{k}\) and \(\|M\|_{q}^{\{n({1}/{q}-1)\}^{-1}} \leq m(x_{0},\mu )^{-1}\), \(\int M(x)\,dx=0\).
Lemma 3.5
If
a
is a
\((1,q)\)-atom supported on
\(B(x_{0},r)\), a
is also a
\((1,q,\varepsilon )\)-molecular centered at
\(x_{0}\).
Proof
Recall that \(\|a\|_{q}\leq |B(x_{0},r)|^{{1}/{q}-1}\). It is easy to see that
which indicates that \(|\cdot -x_{0}|^{nb}a\in L^{q}(\mathbb{R}^{n})\) with \(\||\cdot -x_{0}|^{nb}a\|_{q}\leq |B(x_{0},r)|^{q}\). Moreover, for \(b=1-{1}/{q}+\varepsilon \),
$$\begin{aligned} \int _{\mathbb{R}^{n}} \bigl\vert \vert x-x_{0} \vert ^{nb}a(x) \bigr\vert ^{q}\,dx \leq & \bigl\vert B(x_{0},r) \bigr\vert ^{bq+1-q}, \end{aligned}$$
$$\begin{aligned} \Vert a \Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}a(\cdot ) \bigr\Vert ^{1- {\varepsilon }/{b}}_{q} \leq \bigl\vert B(x_{0},r) \bigr\vert ^{({1}/{q}-1)({\varepsilon }/ {b})} \bigl\vert B(x_{0},r) \bigr\vert ^{\varepsilon (1-{\varepsilon }/{b})}\leq 1. \end{aligned}$$
We only need to verify the canceling condition, i.e., \(\|a\|^{1/\{n( {1}/{q}-1)\}}_{q}\leq m(x_{0},\mu )^{-1}\). Denote by \(\omega _{n}\) the volume of the unit ball in \(\mathbb{R}^{n}\). It is clear that \(\omega _{n}>1\) and \(\|a\|_{q}\leq \omega ^{({1}/{q}-1)}_{n}r^{n({1}/ {q}-1)}\leq r^{n({1}/{q}-1)}\), equivalently,
By the canceling condition of \((1,q)\)-atoms, we can see that \(\int _{\mathbb{R}^{n}}a(x)\,dx=0\). So a is a \((1,q,\varepsilon )\)-molecular centered at \(x_{0}\). □
$$ r\leq \Vert a \Vert ^{1/\{n({1}/{q}-1)\}}_{q}\leq m(x_{0}, \mu )^{-1}. $$
Theorem 3.6
Let
\(1\leq q\leq \infty \), \(\varepsilon >0\), \(b=1-{1}/{q}+\varepsilon \). Then
\(f\in H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\)
if and only if
\(f=\sum_{j}\lambda _{j}M_{j}\), where
\(\{M_{j}\}\)
are
\((1,q, \varepsilon )\)-moleculars and
\(\{\lambda _{j}\}\)
are scalars with
\(\inf \sum_{j}|\lambda _{j}|\sim \|f\|_{H^{1}_{\mathcal{L}}}\), where the infimum is taken over all decompositions.
Proof
We have known that any \((1,q)\)-type atom is also a \((1,q,\varepsilon )\)-type molecular. By Theorem 3.3, if \(f\in H^{1}_{ \mathcal{L}}(\mathbb{R}^{n})\), then there exist a sequence of \((1,q)\)-type atoms \(\{a_{j}\}\) and a sequence of scalars \(\{\lambda _{j}\}\) such that \(f=\sum_{j}\lambda _{j}a_{j}\). This means that f can be represented as a linear combination of \((1,q,\varepsilon )\)-moleculars. Conversely, we only need to verify that, for any \((1,q,\varepsilon )\)-molecular, \(\|M\|_{H^{1}_{ \mathcal{L}}}\leq C\). For simplicity, denote
Without loss of generality, we assume that \(N_{\mathcal{L}}(M)=1\) and \(q=2\). Write \(\sigma =\|M\|^{1/\{n({1}/{2}-1)\}}_{2}\). Let
Denote by \(\psi _{k}\) the characteristic function \(\chi _{E_{k}}(x)\) and write \(M(x)=\sum_{k}M_{k}(x)\), where \(M_{k}:=M(x)\psi _{k}(x)\).
$$ N_{\mathcal{L}}(M)=: \Vert M \Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}M( \cdot ) \bigr\Vert ^{1-{\varepsilon }/{b}}_{q}. $$
$$ \textstyle\begin{cases} E_{0}=\{x: \vert x-x_{0} \vert \leq \sigma \}; \\ E_{k}=\{x:2^{k-1}\sigma < \vert x-x_{0} \vert \leq 2^{k}\sigma \},\quad k\in N; \\ B_{k}=\{x: \vert x-x_{0} \vert \leq 2^{k}\sigma \},\quad k=0,1,2,\ldots . \end{cases} $$
Case 1:
\(\sigma \leq 1/ m(x_{0},\mu )\). Then \(\|M\|^{1/\{n( {1}/{2}-1)\}}_{2}\leq m(x_{0},\mu )^{-1}\) and \(\int _{\mathbb{R}^{n}} M(x)\,dx=0\). The proof is similar to the classical case, and we omit it.
Case 2:
\(\sigma >1/m(x_{0},\mu )\). For this case, \(\||\cdot -x _{0}|^{n({1}/{2}+\varepsilon )}M(\cdot )\|^{1-{\varepsilon }/{b}}_{2}= \|M\|^{-{\varepsilon }/{b}}_{2}\). Denote by σ the term \(\|M\|^{1/\{n({1}/{2}-1)\}}_{2}\). Then \(\||\cdot -x_{0}|^{n({1}/{2}+ \varepsilon )}M(\cdot )\|_{2}=\sigma ^{n\varepsilon }\) and
which implies that \(\|M_{0}\|_{2}\leq |B_{0}|^{-{1}/{2}}\).
$$\begin{aligned} \frac{1}{ \vert B_{0} \vert } \int _{\mathbb{R}^{n}} \bigl\vert M_{0}(x) \bigr\vert ^{2}\,dx \leq &\frac{1}{ \sigma ^{n}} \Vert M \Vert ^{2}_{2}=\frac{1}{\sigma ^{2n}}, \end{aligned}$$
For the term \(M_{k}\), we have
that is, \(\|M\|_{2}\leq C|B_{k}|^{-{1}/{2}}2^{-k\varepsilon n}\). Let \(a_{k}(x)=\lambda ^{-1}_{k}M_{k}(x)\), \(k=0,1,2,\ldots \) , where \(\lambda _{k}=2^{-2k\varepsilon n}\) and \(a_{k}\), \(k\in \mathbb{Z}_{+}\), are \((1,2)\)-atoms. Hence \(M(x)=\sum_{k}\lambda _{k}a_{k}(x)= \sum_{k}M_{k}(x)\) and \(\sum_{k}|\lambda _{k}|=C\sum_{k}2^{-2k\varepsilon n}<\infty \). Repeating the procedure of [3, Theorem 4], we can prove that \(M\in H^{1}_{\mathcal{L}}( \mathbb{R}^{n})\). We omit the details, and this completes the proof of Theorem 3.6. □
$$\begin{aligned} \frac{1}{ \vert B_{k} \vert } \int _{\mathbb{R}^{n}} \bigl\vert M_{k}(x) \bigr\vert ^{2}\,dx \leq &\frac{1}{(2^{k-1} \sigma )^{n}} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}M_{k}(\cdot ) \bigr\Vert ^{2}_{2} \bigl(2^{k-1} \sigma \bigr)^{-n(1+2\varepsilon )} \\ \leq & \bigl(2^{k-1}\sigma \bigr)^{-2n-2\varepsilon n}\sigma ^{2n\varepsilon } \\ \leq &C_{n,\varepsilon } \bigl(2^{k}\sigma \bigr)^{-2n}2^{-2k\varepsilon }, \end{aligned}$$
4 Operators on the Hardy type space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\)
4.1 The \(H^{1}_{\mathcal{L}}\)-boundedness of \({\mathcal{L}} ^{i\gamma }\)
Let \(q_{t}(\cdot ,\cdot )\) denote the kernel of \(e^{-t\mathcal{L}}-e ^{-t(-\Delta )}\). We have
The following estimate was obtained by Wu and Yan [15].
$$ q_{t}(x,y)=h_{t}(x-y)-K^{\mathcal{L}}_{t}(x-y)= \int ^{t}_{0} \int _{\mathbb{R}^{n}}K^{\mathcal{L}}_{s}(x,t)h_{t-s}(z-y) \,d\mu (z)\,ds. $$
Lemma 4.1
([15, Lemma 3.6])
(i)
There exist constants
C
and
c
such that, for every
\(x,y\in \mathbb{R}^{n}\)
and
\(t>0\),
$$ q_{t}(x,y)\leq \textstyle\begin{cases} C( \sqrt{t}m(x, \mu ))^{\delta }t^{-{n}/{2}}e^{-{|x-y|^{2}}/{ct}} , & \sqrt{t}\leq m(x, \mu )^{-1}; \\ C( \sqrt{t}m(y, \mu ))^{\delta }t^{-{n}/{2}}e^{-{|x-y|^{2}}/{ct}} , & \sqrt{t}\leq m(y, \mu )^{-1}; \\ h_{t}(x-y), & \textit{elsewhere}. \end{cases} $$
(ii)
For every
\(0<\delta '<\min \{1,\delta \}\)
and
\(C>0\), there exist constants
\(C'\)
and
c
such that, for every
\(h,x,y\in \mathbb{R}^{n}\), \(|h|\leq {|x-y|}/{4}\), \(|h|\leq C m(y, \mu )^{-1}\), we have
$$ \bigl\vert q_{t}(x,y+h)-q_{t}(x,y) \bigr\vert \leq C' \bigl( \vert h \vert m(x, \mu ) \bigr)^{\delta '}t^{-{n}/ {2}}e^{-{ \vert x-y \vert ^{2}}/{ct}}. $$
By the functional calculus, we can see that the kernel of \((-\Delta )^{i \gamma }-\mathcal{L}^{i\gamma }\) can be expressed as
$$ g(x,y):= \int ^{\infty }_{0}t^{-i\gamma }q_{t}(x,y) \frac{dt}{t}. $$
(4.1)
Lemma 4.2
Let
\(\mathcal{L}=-\Delta +\mu \)
be a generalized Schrödinger operator, where
\(\mu \neq 0\)
is a nonnegative Radon measure on
\(\mathbb{R}^{n}\)
satisfying (1.2) & (1.3) for some
\(\delta >0\).
(i)
If
\(y\in B(x_{0}, r)\), then
$$ \bigl\vert g(x,y) \bigr\vert \leq C m(x_{0}, \mu )^{\delta }{ \vert x-y \vert ^{\delta -n}}. $$
(4.2)
(ii)
There exists
\(0<\delta '<\delta \)
such that
$$ \bigl\vert g(x,y)-g(x,x_{0}) \bigr\vert \leq {C} { \vert x-y \vert ^{-n}} \bigl( \vert y-x_{0} \vert m(x,\mu ) \bigr) ^{\delta '}. $$
Proof
Case 1:
\(\sqrt{t}\leq 1/m(y, \mu )\). Because \(y\in B\), then \(|y-x_{0}|< r< {1}/{m(x_{0}, \mu )}\), \(m(y, \mu )\thicksim m(x_{0}, \mu )\). By (i) of Lemma 4.1, we can get
$$\begin{aligned} \bigl\vert g(x,y) \bigr\vert \leq &Cm(x_{0}, \mu )^{\delta } \int ^{\infty }_{0}t^{-{n}/ {2}+{\delta }/{2}-1}e^{-{ \vert x-y \vert ^{2}}/{ct}}\,dt \leq C m(x_{0}, \mu )^{ \delta }{ \vert x-y \vert ^{\delta -n}}. \end{aligned}$$
Case 2:
\(\sqrt{t}>1/m(y, \mu )\). For this case, \(\sqrt{t}m(y, \mu )>1\). Using Lemma 4.1 again, we can deduce that
$$\begin{aligned} \bigl\vert g(x,y) \bigr\vert \leq &C \int ^{\infty }_{0}t^{-{n}/{2}}e^{-{ \vert x-y \vert ^{2}}/{ct}} \bigl( \sqrt{t}m(y, \mu ) \bigr)^{\delta }\frac{dt}{t} \leq C m(x_{0}, \mu )^{ \delta }{ \vert x-y \vert ^{\delta -n}}. \end{aligned}$$
(ii). It follows from (4.1) that
By (ii) of Lemma 4.1 and a direct computation, we get
This completes the proof of Lemma 4.2. □
$$\begin{aligned} \bigl\vert g(x,y)-g(x,x_{0}) \bigr\vert =& \biggl\vert \int ^{\infty }_{0}t^{-i\gamma }q_{t}(x,y) {dt}/{t}- \int ^{\infty }_{0}t^{-i\gamma }q_{t}(x,x_{0}) \frac{dt}{t} \biggr\vert \\ \leq & \int ^{\infty }_{0}t^{-i\gamma } \bigl\vert q_{t}(x,y)-q_{t}(x,x_{0}) \bigr\vert \frac{dt}{t}. \end{aligned}$$
$$\begin{aligned} \bigl\vert g(x,y)-g(x,x_{0}) \bigr\vert \leq & \biggl\vert \int ^{\infty }_{0}t^{-i\gamma } \bigl[ \vert y-x _{0} \vert m(x,\mu ) \bigr]^{\delta '}t^{-{n}/{2}}e^{-{c \vert x-y \vert ^{2}}/{t}} \frac{dt}{t} \biggr\vert \\ \leq &C \bigl[ \vert y-x_{0} \vert m(x,\mu ) \bigr]^{\delta '} \int ^{\infty }_{0}t^{-{n}/ {2}}e^{-{c \vert x-y \vert ^{2}}/{t}} \frac{dt}{t} \\ \leq & C{ \vert x-y \vert ^{-n}} \bigl[ \vert y-x_{0} \vert m(x,\mu ) \bigr]^{\delta '}. \end{aligned}$$
We recall that an operator T taking \(\mathcal{C}^{\infty }( \mathbb{R}^{n})\) into \(L^{1}_{\mathrm{loc}}(\mathbb{R}^{n})\) is called a Calderón–Zygmund operator if
(a)
T extends to a bounded operator on \(L^{2}(\mathbb{R}^{n},dx)\);
(b)
there exists a kernel K such that, for every \(f\in L^{1} _{c}(\mathbb{R}^{n},dx)\),
$$ Tf(x)= \int _{\mathbb{R}^{n}}K(x,y)f(y)\,dy\quad \text{a.e. on }\{\operatorname{supp} f\} ^{c}; $$
(c)
the kernel K satisfies
$$ \textstyle\begin{cases} \vert K(x,y) \vert \leq {c}/{ \vert x-y \vert ^{n}}; \\ \vert K(x+h,y)-K(x,y) \vert \leq {c \vert h \vert ^{\delta }}/{ \vert x-y \vert ^{n+\delta }}; \\ \vert K(x,y+h)-K(x,y) \vert \leq {c \vert h \vert ^{\delta }}/{ \vert x-y \vert ^{n+\delta }}. \end{cases} $$
(4.3)
In [12], Shen proved the following result.
Theorem 4.3
Now we prove the \(H^{1}_{\mathcal{L}}\)-boundedness of \(\mathcal{L} ^{i\gamma }\).
Theorem 4.4
Proof
We only need to prove that, for any \((1,\infty )\) atom a, \(\mathcal{L}^{i\gamma }(a)\) is a \((1,q, \varepsilon )\)-molecular and \(\|\mathcal{L}^{i\gamma }(a)\|_{H^{1}_{\mathcal{L}}}\leq C\). Let a be a \((1,\infty )\) atom supported on \(B(x_{0},r)\). Then \(\|a\|_{\infty } \leq {1}/{|B(x_{0},r)|}\). If \(r< m(x_{0},\mu )^{-1}\), \(\int a(x)\,dx=0\). Set \(B^{\sharp }=B(x_{0}, {2}/{m(x_{0},\mu )})\) and \(B^{*}=B(x_{0},2r)\). We divide the proof into three parts.
Part I:
\(\mathcal{L}^{i\gamma }a\in L^{q}(\mathbb{R}^{n}) \& |x|^{nb}L ^{i\gamma }(a)\in L^{q}(\mathbb{R}^{n})\).
By the \(L^{q}\)-boundedness of \(\mathcal{L}^{i\gamma }\) and Minkowski’s inequality, \(\||x|^{nb}\mathcal{L}^{i\gamma }a\|_{q}\leq S_{1}+S_{2}\), where
$$\begin{aligned} \bigl\Vert \vert x \vert ^{nb}\mathcal{L}^{i\gamma }a \bigr\Vert _{q} \leq & \bigl\Vert x_{B^{*}} \vert \cdot \vert ^{nb} \mathcal{L}^{i\gamma }a \bigr\Vert _{q}+ \bigl\Vert x_{({B^{*}})^{c}} \vert \cdot \vert ^{nb} \mathcal{L}^{i\gamma }a \bigr\Vert _{q} \\ =& \biggl( \int _{B^{*}} \vert x \vert ^{qnb} \bigl\vert \mathcal{L}^{i\gamma }(a) \bigr\vert ^{q}\,dx \biggr)^{ {1}/{q}}+ \biggl( \int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \bigl\vert \mathcal{L}^{i\gamma }(a) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \\ \leq & \bigl(r+ \vert x_{0} \vert \bigr)^{nb} \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}+ \biggl[ \int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \biggl\vert \int _{\mathbb{R}^{n}}K^{\mathcal{L}} _{\gamma }(x,y)a(y)\,dy \biggr\vert ^{q}\,dx \biggr]^{{1}/{q}}. \end{aligned}$$
$$ \textstyle\begin{cases} S_{1}:=(r+ \vert x_{0} \vert )^{nb} \Vert a \Vert _{q}; \\ S_{2}:=\int _{B(x_{0},r)} \vert a(y) \vert [\int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \vert K ^{\mathcal{L}}_{\gamma }(x,y) \vert ^{q}\,dx ]^{{1}/{q}}. \end{cases} $$
Because \(\|a\|_{q}\leq |B(x_{0},r)|^{{1}/{q}-1}\), then
$$ S_{1}= \bigl(r+ \vert x_{0} \vert \bigr)^{nb} \Vert a \Vert _{q}\leq \bigl(r+ \vert x_{0} \vert \bigr)^{nb}r^{n({1}/{q}-1)}. $$
For the term \(S_{2}\), recall that
$$ K^{\mathcal{L}}_{\gamma }(x,y)= \int ^{\infty }_{0}t^{-i\gamma }K^{ \mathcal{L}}_{t}(x,y) \frac{dt}{t}. $$
Proposition 2.7 implies that
which gives
For \(x\in B\) and \(y\in ({B^{*}})^{c}\), we can see that \(|x-y|\geq {|x-x_{0}|}/{2}\). Notice that for \(y\in B(x_{0},r)\), \(|x_{0}-y|< r\) and
Then, via a direct computation, we have
$$\begin{aligned} \bigl\vert K^{\mathcal{L}}_{\gamma }(x,y) \bigr\vert \leq &C_{N}{t^{-{n}/{2}}}\frac{e ^{-c{ \vert x-y \vert ^{2}}/{t}}}{\{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{N}} \frac{dt}{t} \\ \leq &\frac{C_{N}}{\{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{N}} \frac{1}{ \vert x-y \vert ^{n}}, \end{aligned}$$
$$\begin{aligned} S_{2}\leq C \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \int _{({B^{*}})^{c}} \vert x \vert ^{qnb}\frac{1}{ \{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{qN}} \frac{dx}{ \vert x-y \vert ^{qn}} \biggr]^{ {1}/{q}}\,dy. \end{aligned}$$
$$ m(y,\mu )^{N}\geq \biggl[\frac{cm(x_{0},\mu )}{\{1+ \vert x_{0}-y \vert m(x_{0}, \mu )\}^{{k_{0}}/{(k_{0}+1)}}} \biggr]^{N}. $$
$$\begin{aligned} S_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl\{ \int _{({B^{*}})^{c}}\frac{ \vert x \vert ^{qnb}\,dx}{ \vert x-x_{0} \vert ^{(n+N)q}} \biggr\} ^{ {1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl\{ \biggl[ \int _{({B^{*}})^{c}}\frac{1}{ \vert x-x_{0} \vert ^{(n+N)q-qnb}}\,dx \biggr]^{{1}/ {q}} \\ &{}+ \biggl[ \int _{({B^{*}})^{c}} \frac{ \vert x_{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr]^{{1}/{q}} \biggr\} \,dy \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \bigl\{ r^{n\varepsilon -N}+ \vert x _{0} \vert ^{nb}r^{-n-N+{n}/{q}} \bigr\} \,dy \\ \leq &C\frac{\{1+r m(x_{0},\mu )\}^{k_{0}N/(k_{0}+1)}}{m(x_{0}, \mu )^{N}} \bigl\{ r^{n\varepsilon -N}+ \vert x_{0} \vert ^{nb}r^{-n-N+{n}/{q}} \bigr\} \biggl( \int _{B} \bigl\vert a(y) \bigr\vert \,dy \biggr)< \infty . \end{aligned}$$
Part II:
\(N_{\mathcal{L}}(\mathcal{L}^{i\gamma }a)=\| \mathcal{L}^{i\gamma }a\|^{{\varepsilon }/{b}}_{q}\||\cdot -x_{0}|^{nb} \mathcal{L}^{i\gamma }a\|^{1-{\varepsilon }/{b}}_{q}\leq C\)
.
Case I:
\(r\geq \rho (x_{0})\). Because \(x\in ({B^{*}})^{c}\) and \(y\in B\),
Because \(\mathcal{L}^{i\gamma }\) is bounded on \(L^{q}(\mathbb{R}^{n})\), \(q>1\), then
$$ \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}\leq \biggl( \int _{B^{*}} \bigl\vert \mathcal{L}^{i \gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}+ \biggl( \int _{({B^{*}})^{c}} \bigl\vert \mathcal{L}^{i\gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}. $$
$$ \biggl( \int _{B^{*}} \bigl\vert \mathcal{L}^{i\gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}} \leq \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}\leq C \Vert a \Vert _{q}\leq \vert B \vert ^{{1}/ {q}-1}. $$
For \(y\in B\) and \(x\in (B^{\ast })^{c}\), \(|x-y|\geq |x-x_{0}|/2\). By Theorem 4.3, we can get
$$\begin{aligned} \biggl( \int _{({B^{*}})^{c}} \bigl\vert \mathcal{L}^{i\gamma }a(x) \bigr\vert ^{q}\,dx \biggr)^{ {1}/{q}} \leq & \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{({B^{*}})^{c}} \bigl\vert K^{ \mathcal{L}}_{\gamma }(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq & \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{({B^{*}})^{c}}{ \vert x-y \vert ^{-nq}}\,dx \biggr)^{ {1}/{q}}\,dy \\ \leq & \int _{B} \bigl\vert a(y) \bigr\vert r^{{n}/{q}-n} \,dy\leq C r^{{n}/{q}-n}. \end{aligned}$$
Because \(q>1 \& r\geq \rho (x_{0})\), the above estimates indicate that
which means that such \(\mathcal{L}^{i\gamma }a\) need not satisfy the canceling condition.
$$ \bigl\Vert \mathcal{L}^{i\gamma }a(x) \bigr\Vert ^{1/\{{n}/{q}-n\}}\geq \rho (x_{0})^{( {n}/{q}-n)/\{{n}/{q}-n\}}=\rho (x_{0}), $$
On the other hand, we write \(\||\cdot -x_{0}|^{nb}\mathcal{L}^{i \gamma }a\|_{q}\leq I_{1}+I_{2}\), where
$$ \textstyle\begin{cases} I_{1}:= \Vert \chi _{B^{*}} \vert \cdot -x_{0} \vert ^{nb}\mathcal{L}^{i\gamma }a \Vert _{q}; \\ I_{2}:= \Vert \chi _{({B^{*}})^{c}} \vert \cdot -x_{0} \vert ^{nb}\mathcal{L}^{i\gamma }a \Vert _{q}. \end{cases} $$
For \(I_{1}\), by the \(L^{q}\)-boundedness of \(\mathcal{L}^{i\gamma }\) and the fact that \(\varepsilon -b={1}/{q}-1\), we have
$$\begin{aligned} I_{1} \leq &C \vert B \vert ^{b} \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q} \leq C \vert B \vert ^{\varepsilon }. \end{aligned}$$
For \(I_{2}\), because
we can use Lemma 2.1 and the fact that \(r\geq {1}/{m(x_{0}, \mu )}\) to obtain
The estimates for \(I_{1}\) and \(I_{2}\) imply that
$$ \bigl\vert K^{\mathcal{L}}_{\gamma }(x,y) \bigr\vert \leq \frac{C_{N}}{\{1+ \vert x-y \vert [m(x, \mu )+m(y,\mu )]\}^{N}}\frac{1}{ \vert x-y \vert ^{n}}, $$
$$\begin{aligned} I_{2} \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \biggl( \int _{({B^{*}})^{c}}\frac{ \vert x-x _{0} \vert ^{nbq}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl\{ \frac{1}{m(x_{0},\mu )}+ \vert y-x_{0} \vert \biggr\} ^{N}r^{nb+ {n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{N}r^{nb+{n}/{q}-n-N} \,dy \\ \leq &C \vert B \vert ^{\varepsilon }. \end{aligned}$$
$$\begin{aligned} N_{\mathcal{L}} \bigl(\mathcal{L}^{i\gamma }a \bigr) =& \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert \mathcal{L}^{i\gamma }a \bigr\Vert ^{1-{\varepsilon }/{b}}_{q}\leq C. \end{aligned}$$
Case 2:
\(r<{1}/{m(x_{0},\mu )}\). For this case, the atom a has the canceling property. There exists a positive integer m such that \({2^{-m-1}}/{m(x_{0},\mu )}\leq r< {2^{-m}}/{m(x_{0},\mu )}\). Let \(B^{\sharp }=B(x_{0}, {2}/{m(x_{0},\mu )})\) and \(B^{\ast }=B(x_{0},2r)\). We write
We will prove that \((\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a\) and \((-\Delta )^{i\gamma }a\) are both moleculars. For \(r<{1}/{m(x_{0}, \mu )}\), any \((1,q)\)-atom is a classical atom. By Alverez–Milman [1], \((-\Delta )^{i\gamma }a\) is a \((1,q, \varepsilon )\)-molecular. Hence, \((-\Delta )^{i\gamma }a\) is a molecular of \(H^{1}_{\mathcal{L}}( \mathbb{R}^{n})\). We write \(\|(\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a\|_{q}\leq I_{1}+I_{2}+I_{3}\), where
$$ \mathcal{L}^{i\gamma }a= \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a+(- \Delta )^{i\gamma }a. $$
$$ \textstyle\begin{cases} I_{1}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a\chi _{B^{ \ast }} \Vert _{q}; \\ I_{2}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a \chi _{B^{\sharp }\backslash B^{\ast }} \Vert _{q}; \\ I_{3}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a \chi _{(B^{\sharp })^{c}} \Vert _{q}. \end{cases} $$
We first estimate the term \(I_{1}\). Because \(\delta \in (0,n)\), then \({n}/{q}-n+\delta >0\). Estimate (4.2) implies that
$$\begin{aligned} I_{1} \leq &C \int _{B} \biggl( \int _{B^{\ast }} \bigl\vert g(x,y) \bigr\vert ^{q} \,dx \biggr)^{{1}/ {q}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert m(x_{0}, \mu )^{\delta } \biggl( \int _{B^{\ast }} { \vert x-y \vert ^{-q(n-\delta )}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert m(x_{0}, \mu )^{\delta }r^{{(n-qn+q\delta )}/ {q}}\,dy \\ \leq &Cm(x_{0}, \mu )^{\delta } \bigl[{2^{m}m(x_{0}, \mu )} \bigr]^{- {n}/{q}+n-\delta } \\ \leq &C{m(x_{0}, \mu )}^{{n}-n/{q}}. \end{aligned}$$
Now we deal with \(I_{3}\). If \(x\in (B^{\sharp })^{c}\) and \(y\in B\), then \(|y-x|\sim |x-x_{0}|\). By the canceling property of a, we have
Then, since \(r m(x_{0}, \mu )<1\), we obtain that
$$\begin{aligned} \bigl\vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr](a) (x) \bigr\vert \leq & \biggl\vert \int _{B} \bigl[ K^{\mathcal{L}}_{\gamma }(x,y)-K_{\gamma }(x,y) \bigr]a(y)\,dy \biggr\vert \\ \leq & \biggl\vert \int _{B} \bigl[ K^{\mathcal{L}}_{\gamma }(x,x_{0})-K^{ \mathcal{L}}_{\gamma }(x,y) \bigr]a(y)\,dy \biggr\vert \\ &{} + \biggl\vert \int _{B} \bigl[K_{\gamma }(x,y)-K _{\gamma }(x,x_{0}) \bigr]a(y)\,dy \biggr\vert \\ \leq & \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \frac{ \vert y-x_{0} \vert ^{\delta }}{ \vert x-x_{0} \vert ^{n+ \delta }}+\frac{ \vert y-x_{0} \vert }{ \vert x-x_{0} \vert ^{n+1}} \biggr]\,dy \\ \leq &C \biggl(\frac{r^{\delta }}{ \vert x-x_{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x _{0} \vert ^{n+1}} \biggr). \end{aligned}$$
$$\begin{aligned} I_{3} \leq &C \biggl[ \int _{(B^{\sharp })^{c}} \biggl(\frac{r^{\delta }}{ \vert x-x _{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x_{0} \vert ^{n+1}} \biggr)^{q}\,dx \biggr]^{{1}/ {q}} \\ \leq &C \biggl\{ \biggl[ \int _{(B^{\sharp })^{c}}\frac{r^{q\delta }}{ \vert x-x _{0} \vert ^{q(n+\delta )}}\,dx \biggr]^{{1}/{q}} + \biggl[ \int _{(B^{\sharp })^{c}}\frac{r ^{q}}{ \vert x-x_{0} \vert ^{q(n+1)}}\,dx \biggr]^{{1}/{q}} \biggr\} \\ \leq &Cr^{\delta }m(x_{0}, \mu )^{\delta +n-{n}/{q}}+r m(x_{0}, \mu )^{\delta +1-{n}/{q}} \\ \leq &Cm(x_{0}, \mu )^{n-{n}/{q}}. \end{aligned}$$
At last, we estimate \(I_{2}\). For this case, \(x\in B^{\sharp }\backslash B^{\ast }\), then \(2r<|x-x_{0}|<{2}/{m(x_{0}, \mu )}\) and \(|x-y|\sim |x-x _{0}|\). Applying (4.2) and the canceling property of a again, we get
which implies that
where in the last inequality we have used the fact that \(q<{n}/{(n- \delta ')}\).
$$\begin{aligned} \bigl\vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a(x) \bigr\vert \leq &C \int _{B} { \bigl[ \vert y-x_{0} \vert m(x_{0}, \mu ) \bigr]^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &C \bigl[m(x_{0}, \mu ) \bigr]^{\delta '}{r^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \int _{B} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &C \bigl[m(x_{0}, \mu ) \bigr]^{\delta '}{r^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \\ \leq &C{2^{-m\delta '}} { \vert x-x_{0} \vert ^{-n}}, \end{aligned}$$
$$\begin{aligned} I_{2} \leq &C \biggl[ \int _{B^{\sharp }\backslash B^{\ast }} \bigl\vert \bigl[\mathcal{L} ^{i\gamma }-(- \Delta )^{i\gamma } \bigr]a(x) \bigr\vert ^{q}\,dx \biggr]^{{1}/{q}} \\ \leq &C \biggl( \int _{B^{\sharp }\backslash B^{\ast }}{2^{-m\delta 'q}} { \vert x-x_{0} \vert ^{-nq}}\,dx \biggr)^{{1}/{q}} \\ \leq &C{m(x_{0}, \mu )}^{{n}-n/{q}}, \end{aligned}$$
Finally, it follows from the estimates for \(I_{i}\), \(i=1,2,3\), that
On the other hand, the \(L^{q}\)-boundedness of \(\mathcal{L}^{i\gamma }-(- \Delta )^{i\gamma }\) gives
This means that for this case, \((\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a\) need not satisfy the canceling condition.
$$ \bigl\Vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr\Vert ^{1/({n}/{q}-n)} _{q}\geq {m(x_{0}, \mu )}^{-1}. $$
$$ \bigl\Vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr\Vert _{q}\leq \Vert a \Vert _{q}= \biggl( \int _{B} \bigl\vert a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}\leq r^{{n}/{q}-n}. $$
Part III: There exists a constant
C
such that, for any
\((1,\infty )\)-atom, uniformly,
We write \(b=1-{1}/{q}+\varepsilon \), then \(\varepsilon -b={1}/{q}-1\). We have proved that
Now we split: \(\||\cdot -x_{0}|^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a\|_{q}\leq I_{1}+I_{2}\), where
$$ {N}_{\mathcal{L}} \bigl( \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr) \leq C. $$
$$\begin{aligned} \bigl\Vert \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert _{q} \leq &C\rho (x _{0})^{{n}/{q}-n}\leq \rho (x_{0})^{n(\varepsilon -b)}. \end{aligned}$$
$$ \textstyle\begin{cases} I_{1}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a \Vert _{L^{q}(B^{\sharp })}; \\ I_{2}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a \Vert _{L^{q}((B^{\sharp })^{c})}. \end{cases} $$
For \(I_{1}\), because \(B^{\sharp }=(x_{0},2\rho (x_{0}))\),
$$\begin{aligned} I_{1} \leq &C\rho (x_{0})^{nb} \biggl[ \int _{B^{\sharp }} \bigl\vert \bigl(\mathcal{L} ^{i\gamma }-(- \Delta )^{i\gamma } \bigr)a(x) \bigr\vert ^{q} \biggr]^{{1}/{q}} \leq C \rho (x_{0})^{nb}\rho (x_{0})^{n(\varepsilon -b)}\leq C\rho (x_{0})^{n \varepsilon }. \end{aligned}$$
For \(I_{2}\), we further split \(I_{2}\) into \(I_{2,1}+I_{2,2}\), where
Notice that \(\varepsilon <{\delta }/{n}\) and \(nb-(n+\delta )+{n}/ {q}<0\). By Theorem 4.3, we have
$$ \textstyle\begin{cases} I_{2,1}:= \Vert \vert \cdot -x_{0} \vert ^{nb}\mathcal{L}^{i\gamma }a \Vert _{L^{q}((B^{ \sharp })^{c})}; \\ I_{2,2}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(-\Delta )^{i\gamma }a \Vert _{L^{q}((B^{ \sharp })^{c})}. \end{cases} $$
$$\begin{aligned} I_{2,1} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb}| \int _{B}{ \vert y-x_{0} \vert ^{q\delta }} { \vert x-x_{0} \vert ^{-q(n+\delta )}}\,dx \biggr]^{ {1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert y-x_{0} \vert ^{\delta } \biggl[ \int _{(B^{\sharp })^{c}} \vert x-x _{0} \vert ^{qnb-q(n+\delta )}\,dx \biggr]^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{\delta }\rho (x_{0})^{nb-(n+\delta )+{n}/ {q}}\,dy \\ \leq &C \rho (x_{0})^{n\varepsilon }. \end{aligned}$$
For \(I_{2,2}\), similarly, we have
where we have used the fact that \(0<\varepsilon <\min \{{\delta }/ {n},{1}/{n}\}\). Finally, we get
and, hence,
Finally, we have proved that, for any \((1,\infty )\)-atom, \(\mathcal{L}^{i\gamma }a\) is a \((1,q,\varepsilon )\)-molecular or the linear combination of finite \((1,q,\varepsilon )\)-moleculars. □
$$\begin{aligned} I_{2,2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert x-x_{0} \vert \biggl( \int _{ \vert x-x_{0} \vert \geq 2\rho (x_{0})}{ \vert x-x_{0} \vert ^{q(nb-n-1)}} \vert x-x_{0} \vert ^{n-1}\,d \vert x-x _{0} \vert \biggr)^{{1}/{q}}\,dy \\ \leq & C \int _{B} \bigl\vert a(y) \bigr\vert r\rho (x_{0})^{nb-(n+1)+{n}/{q}}\,dy \\ \leq &C \rho (x_{0})^{n\varepsilon }, \end{aligned}$$
$$ \bigl\Vert \vert \cdot -x_{0} \vert ^{nb} \bigl( \mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert _{q}\leq \rho (x_{0})^{n\varepsilon }, $$
$$\begin{aligned}& \bigl\Vert \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert ^{{\varepsilon }/ {b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb} \bigl(\mathcal{L}^{i\gamma }-(- \Delta )^{i \gamma } \bigr)a \bigr\Vert ^{(1-{\varepsilon }/{b})}_{q} \\& \quad \leq C\rho (x_{0})^{n( \varepsilon -b){\varepsilon }/{b}}\rho (x_{0})^{n\varepsilon (1-{\varepsilon }/ {b})} \leq C. \end{aligned}$$
4.2 The \(H^{1}_{\mathcal{L}}\)-boundedness of Riesz transforms \(R_{\mathcal{L}}\)
In this section, we prove that Riesz transforms \(R_{\mathcal{L}}\) are bounded on \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\). The Riesz transforms associated with \(\mathcal{L}\) are defined as
where \((-\Delta +\mu )^{-{1}/{2}}=\frac{1}{\pi }\int ^{\infty }_{0} \lambda ^{-{1}/{2}}(-\Delta +\mu +\lambda )^{-1}\,d\lambda \). Shen proved the following estimate of \(R_{\mathcal{L}}\). Assume that μ satisfies (1.2) & (1.3) for some \(\delta >1\). Then \(\nabla (- \Delta +\mu )^{-{1}/{2}}\) is a Calderón–Zygmund operator. Precisely,
where
In [12], Shen proved the following results, see [12, (7.20), (7.26), (7.29)], respectively.
$$ R_{\mathcal{L}}:=\nabla (-\Delta +\mu )^{-1/2}, $$
$$ \nabla (-\Delta +\mu )^{-{1}/{2}}f(x)= \int _{\mathbb{R}^{n}}R_{ \mathcal{L}}(x,y)f(y)\,dy, $$
$$ R_{\mathcal{L}}(x,y):=\frac{1}{\pi } \int ^{\infty }_{0}\lambda ^{-{1}/ {2}}\nabla _{x}\varGamma _{\mu +\lambda }(x,y)\,d\lambda . $$
Lemma 4.5
The kernel
\(R_{\mathcal{L}}(\cdot ,\cdot )\)
satisfies the following estimates:
$$ \textstyle\begin{cases} (1) \quad \vert R_{\mathcal{L}}(x,y) \vert \leq {Ce^{-cd(x,y,\mu )}}{ \vert x-y \vert ^{-n}}; \\ (2) \quad \vert R_{\mathcal{L}}(x+h,y)-R_{\mathcal{L}}(x,y) \vert \leq C ({ \vert h \vert }/ { \vert x-y \vert } )^{\delta -1}{ \vert x-y \vert ^{-n}}; \\ (3) \quad \vert R_{\mathcal{L}}(x,y+h)-R_{\mathcal{L}}(x,y) \vert \leq C ({ \vert h \vert }/ { \vert x-y \vert } )^{\delta _{1}}{ \vert x-y \vert ^{-n}},\quad \delta _{1}\in (0,1). \end{cases} $$
Theorem 4.6
Proof
Similar to Theorem 4.4, the proof of this theorem is divided into three parts.
Part I:
\(\|{|\cdot |}^{nb}R_{\mathcal{L}}a\|_{q}<\infty \), uniformly. For any atom a and \(B^{\ast }=B(x_{0},2r)\), we write \(\|{|\cdot |}^{nb}R_{\mathcal{L}}a\|_{q}\leq I_{1}+I_{2}\), where \(I_{1}:=\|{|\cdot |}^{nb}R_{\mathcal{L}}a\|_{L^{q}(B^{\ast })}\) and \(I_{2}:=\|{|\cdot |}^{nb}R_{\mathcal{L}}a\|_{L^{q}((B^{\ast })^{c})}\).
By the \(L^{q}\)-boundedness of \(R_{\mathcal{L}}\), we have
By Lemma 4.5, for any positive \(N>0\),
On the other hand, for \(y\in B\) and \(x\in (B^{\ast })^{c}\), \(|x-y|\geq {|x-x_{0}|}/{2}\). We can obtain that
Because \(y\in B(x_{0},r)\),
which implies that \(\|{|\cdot |}^{nb}a\|_{L^{q}((B^{\ast })^{c})}< \infty \).
$$\begin{aligned} I_{1} \leq &r^{nb} \biggl( \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}a(x) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\leq r^{nb} \Vert a \Vert _{q} \leq r^{nb} \bigl\vert B(x_{0},r) \bigr\vert ^{{1}/ {q}-1} \leq r^{n\varepsilon }. \end{aligned}$$
$$\begin{aligned} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert \leq &C{ \vert x-y \vert ^{-n-N} \bigl[m(y,\mu ) \bigr]^{-N}}. \end{aligned}$$
$$\begin{aligned} I_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\ast })^{c}} \vert x \vert ^{qnb} \bigl\vert R_{ \mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}} \vert x \vert ^{qnb} \frac{1}{ \vert x-y \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \biggl\{ \int _{B} \bigl\vert a(y) \bigr\vert \frac{ \vert x_{0} \vert ^{nb}}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}}\frac{dx}{ \vert x-x_{0} \vert ^{(n+N)q}} \biggr)^{{1}/{q}}\,dy \\ &{}+ \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}}\frac{ \vert x-x _{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \biggr\} \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \bigl\{ r^{n\epsilon -N}+ \vert x _{0} \vert ^{-n-N+{n}/{q}} \bigr\} \,dy. \end{aligned}$$
$$ m(y,\mu ) \geq \frac{Cm(x_{0},\mu )}{\{1+ \vert x-x_{0} \vert m(x_{0},\mu )\}^{ {k_{0}}/{(k_{0}+1)}}}, $$
Part II:
\(N_{\mathcal{L}}(R_{\mathcal{L}}(a))\leq C\). We divide the proof into two cases.
Case1:
\(r\geq {1}/{m(x_{0},\mu )}\). By the boundedness of the Riesz transform \(R_{\mathcal{L}}\), we have
By Lemma 4.5, we can get
which means that \(\|R_{\mathcal{L}}a\|^{1/({n}/{q}-n)}_{q}\geq \rho (x _{0})\), i.e., \(R_{\mathcal{L}}a\) does not need the canceling condition for this case. Now we split \(\||\cdot -x_{0}|^{nb}R_{\mathcal{L}}a\| _{q}\leq I_{1}+I_{2}\), where
It is easy to see that
$$\begin{aligned} \Vert R_{\mathcal{L}}a \Vert _{q} \leq &C \bigl\{ \Vert R_{\mathcal{L}}a \Vert _{q}+ \Vert \chi _{(B^{\ast })^{c}}R_{\mathcal{L}}a \Vert _{q} \bigr\} \\ \leq &C \bigl\{ \Vert a \Vert _{q}+ \Vert \chi _{(B^{\ast })^{c}}R_{\mathcal{L}}a \Vert _{q} \bigr\} \\ \leq &C \biggl\{ \bigl\vert B(x_{0},r) \bigr\vert ^{{1}/{q}-1}+ \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\,dy \biggr\} . \end{aligned}$$
$$\begin{aligned} \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\,dy \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int ^{\infty }_{2r}s^{n-qn-1}\,ds \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \,dy \times r^{{n}/{q}-n} \\ \leq &C\rho (x_{0})^{{n}/{q}-n}, \end{aligned}$$
$$ \textstyle\begin{cases} I_{1}:= (\int _{B^{\ast }} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ I_{2}:= (\int _{(B^{\ast })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
$$\begin{aligned} I_{1}\leq r^{nb} \biggl( \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}a(x) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \leq r^{nb} \Vert R_{\mathcal{L}}a \Vert _{q} \leq r^{nb} \Vert a \Vert _{q} \leq r^{n\varepsilon }. \end{aligned}$$
For \(I_{2}\), by Minkowski’s inequality,
which gives \(N_{\mathcal{L}}(R_{\mathcal{L}}a)=\|R_{\mathcal{L}}a\| ^{{\varepsilon }/{b}}_{q}\||\cdot -x_{0}|^{nb}R_{\mathcal{L}}a\|^{1- {\varepsilon }/{b}}_{q}\leq C\).
$$\begin{aligned} I_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert {m(y,\mu )^{-N}} \biggl( \int _{B^{\ast }}{ \vert x-x _{0} \vert ^{qnb-q(n+N)}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert {m(y,\mu )^{-N}}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \bigl\{ {m(x_{0},\mu )^{-1}}+ \vert y-x_{0} \vert \bigr\} ^{N}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{N}r^{nb+{n}/{q}-n-N} \,dy \\ \leq &C \vert B \vert ^{(\varepsilon -b){\varepsilon }/{b}} \vert B \vert ^{\varepsilon (1- {\varepsilon }/{b})}, \end{aligned}$$
Case 2: \(r\leq \rho (x_{0})\). Let \(B^{\sharp }=B(x_{0},2\rho (x_{0}))\) and \(B^{\ast }=B(x_{0},2r)\). So \(R_{\mathcal{L}}a=R_{0}a+(R_{ \mathcal{L}}-R_{0})a\), where \(R_{0}:=\nabla (-\Delta )^{-1/2}\). For any a with the canceling condition, \(R_{0}a\) is a molecular. We only need to deal with \((R_{\mathcal{L}}-R_{0})a\). Split \(\|(R_{\mathcal{L}}-R _{0})a\|_{q}\leq I_{1}+I_{2}+I_{3}\), where
$$ \textstyle\begin{cases} I_{1}:= (\int _{B^{\ast }} \vert (R_{\mathcal{L}}-R_{0})a(x) \vert ^{q}\,dx )^{ {1}/{q}}; \\ I_{2}:= (\int _{B^{\sharp } \backslash B^{\ast }} \vert (R_{\mathcal{L}}-R _{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ I_{3}:= (\int _{(B^{\sharp })^{c}} \vert (R_{\mathcal{L}}-R_{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
We first estimate \(I_{3}\). For \(x\in (B^{\sharp })^{c}\) and \(y\in B\), \(|x-y|\sim |x-x_{0}|\). Denote by \(R_{0}(\cdot ,\cdot )\) the kernel of \(\nabla (-\Delta )^{-1/2}\). We can get
It follows from the above estimate that
For the estimates of \(I_{1} \& I_{2}\), we need the following estimate:
For \(r={|x-y|}/{2}\),
We get \(I_{1}\leq \int _{B}|a(y)|A_{1}(y)\,dy\), where
Due to (4.4), we further obtain \(A_{1}\leq U_{1}+U_{2}\), where
$$\begin{aligned}& \bigl\vert (R_{\mathcal{L}}-R_{0}) (a) (x) \bigr\vert \\& \quad \leq C \biggl\{ \int _{B} \bigl\vert R_{ \mathcal{L}}(x,y)-R_{\mathcal{L}}(x,x_{0}) \bigr\vert \bigl\vert a(y) \bigr\vert \,dy+ \int _{B} \bigl\vert R_{0}(x,y)-R _{0}(x,x_{0}) \bigr\vert \bigl\vert a(y) \bigr\vert \,dy \biggr\} \\& \quad \leq C \Vert a \Vert _{\infty } \biggl( \int _{B}\frac{ \vert y-x_{0} \vert ^{\delta }}{ \vert x-x _{0} \vert ^{n+\delta }}\,dy+ \int _{B}\frac{ \vert y-x_{0} \vert }{ \vert x-x_{0} \vert ^{n+1}}\,dy \biggr) \\& \quad \leq C \Vert a \Vert _{\infty } \vert B \vert \biggl( \frac{r^{\delta }}{ \vert x-x_{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x_{0} \vert ^{n+1}} \biggr). \end{aligned}$$
$$\begin{aligned} I_{3} \leq &C \biggl\{ \biggl( \int _{(B^{\sharp })^{c}}\frac{r^{q\delta }}{ \vert x-x _{0} \vert ^{q(n+\delta )}}\,dx \biggr)^{{1}/{q}} + \biggl( \int _{(B^{\sharp })^{c}}\frac{r ^{q}}{ \vert x-x_{0} \vert ^{q(n+1)}}\,dx \biggr)^{{1}/{q}} \biggr\} \\ \leq &C \bigl\{ r^{\delta }m(x_{0},\mu )^{n+\delta -{n}/{q}}+r \bigl[m(x_{0}, \mu ) \bigr]^{n+1-{n}/{q}} \bigr\} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}. \end{aligned}$$
$$ \bigl\vert R_{\mathcal{L}}(y,x)-R_{0}(y,x) \bigr\vert \leq C \biggl\{ \frac{1}{r^{n-1}} \int _{B(y,r)}\frac{d\mu (z)}{ \vert z-y \vert ^{n-1}}+\frac{(r m(x,\mu ))^{\delta }}{r^{n}} \biggr\} . $$
(4.4)
$$ \bigl\vert R_{\mathcal{L}}(y,x)-R_{0}(y,x) \bigr\vert \leq C \biggl\{ \frac{1}{ \vert x-y \vert ^{n-1}} \int _{B(y, { \vert x-y \vert }/{2})}\frac{d\mu (z)}{ \vert z-y \vert ^{n-1}} +\frac{( \vert x-y \vert m(x, \mu ))^{\delta }}{ \vert x-y \vert ^{n}} \biggr\} . $$
$$ A_{1}:= \biggl\{ \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}(x,y)-R_{0}(x,y) \bigr\vert ^{q}\,dx \biggr\} ^{{1}/{q}}. $$
$$ \textstyle\begin{cases} U_{1}:= \{\int _{B^{\sharp }}{ \vert x-y \vert ^{q(1-n)}} ( \int _{B(x,{ \vert x-y \vert }/{2})}{ \vert z-x \vert ^{1-n}}\,d\mu (z) )^{q}\,dx \}^{{1}/ {q}}; \\ U_{2}:= \{\int _{B^{\sharp }}{( \vert x-y \vert m(y,\mu ))^{q\delta }}{ \vert x-y \vert ^{-qn}}\,dx \}^{{1}/{q}}. \end{cases} $$
For \(U_{2}\), if \(y\in B\), then \(|y-x_{0}|< r<2\rho (x_{0})\) and \(m(y,\mu )\sim m(x_{0},\mu )\). On the other hand, because \(x\in B^{ \sharp }\), then \(|x-y|<{3}/{m(x_{0},\mu )}\). We can get
$$\begin{aligned} U_{2} \leq & Cm(x_{0},\mu )^{\delta } \biggl( \int _{B^{\sharp }}{ \vert x-y \vert ^{q \delta -qn}}\,dx \biggr)^{{1}/{q}}\leq C m(x_{0},\mu )^{n-{n}/{q}}. \end{aligned}$$
Now we estimate the term \(U_{1}\). Let \(T_{j}=B(y,{2^{j+2}}/{ m(x_{0}, \mu )})\). If \(y\in B\) and \(x\in B^{\sharp }\), by the triangle inequality, it is easy to see that \(B^{\sharp }\subset B(y,{4}/{ m(x _{0},\mu )})\). Also, for \(x\in T_{j+1}\setminus T_{j}\), \(|x-y|\geq {2^{j+2}}/{ m(x_{0},\mu )}\). On the other hand, \(B(x,{|x-y|}/{2}) \subset B(y, {3|x-y|}/{2})\). Then
Notice that
A direct computation gives
which implies that
The estimate for \(I_{2}\) is similar. Then we obtain \(\|(R_{ \mathcal{L}}-R_{0})a\|^{1/\{{n}/{q}-n\}}_{q}\geq {C}/{m(x_{0},\mu )}\), which means \((R_{\mathcal{L}}-R_{0})a\) does not need the canceling condition. What is left to prove is the norm \(\||\cdot -x_{0}|^{nb}(R _{\mathcal{L}}-R_{0})a\|_{q}\). We write \(\||\cdot -x_{0}|^{nb}(R_{ \mathcal{L}}-R_{0})a\|_{q}\leq E_{1}+E_{2}\), where
$$\begin{aligned} U_{1} \leq & C\sum^{0}_{j=-\infty } \biggl( \int _{T_{j+1}\setminus T_{j}} \frac{1}{ \vert x-y \vert ^{q(n-1)}} \biggl( \int _{B(x,{ \vert x-y \vert }/{2})} \frac{d\mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq & C\sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1} \biggl( \int _{T_{j+1}\setminus T_{j}} \biggl( \int _{B(y,{ \vert x-y \vert }/{2})}\frac{d \mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C\sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1} \biggl( \int _{T_{j+1}\setminus T_{j}} \biggl( \int _{B(y,{2^{j+2}}/{m(x_{0},\mu )})}\frac{d\mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C \sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1}\frac{ \mu (3T_{j+1})}{2^{(j+2)(n-{n}/{q}-1)}} \bigl[m(x_{0}, \mu ) \bigr]^{n-{n}/{q}-1}. \end{aligned}$$
$$ \mu (T_{j+1})=\mu \bigl(B \bigl(y,{2^{j+2}}/{m(x_{0}, \mu )} \bigr) \bigr)\leq \bigl(2^{j+2} \bigr)^{n-2+ \delta }{m(x_{0}, \mu )}^{2-n}. $$
$$\begin{aligned} U_{1} \leq &C \sum^{0}_{j=-\infty } \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}2^{j(n-2+ \delta ')}{2^{-j(2n-{1}/{q}-2)}} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}\sum ^{0}_{j=-\infty }2^{j(n-2+ \delta '-2n+{n}/{q}+2)} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}, \end{aligned}$$
$$\begin{aligned} I_{1}\leq C \int _{B} \bigl\vert a(y) \bigr\vert A_{1}(y) \,dy \leq C \Vert a \Vert _{1} \bigl[m(x_{0},\mu ) \bigr]^{n- {n}/{q}} \leq C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}. \end{aligned}$$
$$ \textstyle\begin{cases} E_{1}:= (\int _{B^{\sharp }} \vert x-x_{0} \vert ^{qnb} \vert (R_{\mathcal{L}}-R_{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ E_{2}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert (R_{\mathcal{L}}-R _{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
By the \(L^{p}\)-boundedness of \(R_{\mathcal{L}}\) and \(R_{0}\), we get
For the term \(E_{2}\), we have \(E_{2}\leq E_{2,1}+E_{2,2}\), where
A direct computation gives
For \(E_{2,2}\), because \(R_{0}\) is a Calderón–Zygmund operator, the kernel \(K_{0}(\cdot ,\cdot )\) satisfies
We can get
Finally, we obtain that
□
$$\begin{aligned} E_{1} \leq &C{m(x_{0},\mu )}^{-nb} \bigl\Vert (R_{\mathcal{L}}-R_{0})a \bigr\Vert _{q} \leq C{m(x_{0},\mu )}^{-nb} \Vert a \Vert _{q} \leq C{m(x_{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
$$ \textstyle\begin{cases} E_{2,1}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ E_{2,2}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{0}a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
$$\begin{aligned} E_{2,1} \leq & C \int _{B} \bigl\vert a(y) \bigr\vert \,dy \biggl( \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \bigl\vert R _{\mathcal{L}}(x,y)-R_{\mathcal{L}}(x,x_{0}) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert y-x_{0} \vert ^{\delta } \biggl( \int _{(B^{\sharp })^{c}}\frac{ \vert x-x _{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+\delta )q}}\,dx \biggr)^{{1}/{q}}\,dy \leq C{m(x _{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
$$ \bigl\vert R_{0}(x,y)-R_{0}(x,x_{0}) \bigr\vert \leq C{ \vert y-x_{0} \vert } { \vert x-x_{0} \vert ^{-n-1}}. $$
$$\begin{aligned} E_{2,2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{q(nb-n)}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C r^{n+1-{n}/{q}} \biggl( \int ^{\infty }_{{2}/{m(x_{0},\mu )}}s ^{-(q-1)n-q+nbq-1}\,ds \biggr)^{{1}/{q}} \leq C{m(x_{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
$$\begin{aligned} N_{L} \bigl((R_{\mathcal{L}}-R_{0})a \bigr) \leq &C{m(x_{0},\mu )}^{(n/{q}-{n})( {\varepsilon }/{b})}{m(x_{0},\mu )}^{n\varepsilon ({\varepsilon }/ {b}-1)}\leq C. \end{aligned}$$
Acknowledgements
Some part of this manuscript was completed when the second author was visiting Mudanjiang Normal University. The authors would like to express their deepest thanks to Professor Pu Zhang for his discussion on this topic.
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