Advertisement

Riesz transforms on the Hardy space associated with generalized Schrödinger operators

  • Yixin Wang
  • Pengtao LiEmail author
Open Access
Research
  • 64 Downloads

Abstract

Let \(\mathcal{L}=-\Delta +\mu \) be a generalized Schrödinger operator, where the measure μ is a nonnegative Radon measure. In this paper, we establish the molecular characterization of the Hardy type space \(H^{1}_{\mathcal{L}}(\mathbb{R}^{n})\) associated with \(\mathcal{L}\). As applications, we obtain the \(H^{1}_{\mathcal{L}}\)-boundedness of Riesz transforms and the imaginary power related to \(\mathcal{L}\).

Keywords

Generalized Schrödinger operator Hardy type space Riesz transform Molecular decomposition 

MSC

42B35 47B38 42B35 

1 Introduction

Consider the generalized Schrödinger operator
$$ \mathcal{L}=-\Delta +\mu\quad \text{in }\mathbb{R}^{n}, n\geq 3, $$
(1.1)
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
$$ \mu \bigl(B(x,r) \bigr)\leq C_{0} \biggl( \frac{r}{R} \biggr)^{n-2+\delta }\mu \bigl(B(x,R) \bigr) $$
(1.2)
and
$$ \mu \bigl(B(x,2r) \bigr)\leq C_{1} \bigl\{ \mu \bigl(B(x,r) \bigr)+r^{n-2} \bigr\} $$
(1.3)
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 \(\mu (B(x,r)) \geq c r^{n-2}\).

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, 5, 6, 7, 8, 9, 10] and the references therein.

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,
$$ T_{t}f(x)= \int _{\mathbb{R}^{n}}{}K_{t}^{\mathcal{L}}(x,y)f(y)\,d\mu (y). $$
The maximal function associated with \(\{T_{t}\}\) is defined as
$$ \mathcal{M}_{\mathcal{L}}(f) (x):=\sup_{t>0} \bigl\vert e^{-t\mathcal{L}}f(x) \bigr\vert \in L^{1} \bigl(\mathbb{R}^{n} \bigr). $$
In [15], Wu and Yan introduced the following Hardy type space associated with \(\mathcal{L}\).

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.,
$$ \textstyle\begin{cases} R_{\mathcal{L}}:=\nabla (-\Delta +\mu )^{-1/2}; \\ \mathcal{L}^{\gamma }:=(-\Delta +\mu )^{i\gamma }. \end{cases} $$
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.

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 \({1}/{C} \leq {B_{1}}/{B_{2}}\leq C\).

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
$$ \frac{1}{m(x,\mu )}=:\sup \biggl\{ r>0: \frac{\mu (B(x,r))}{r^{n-2}} \leq C_{1} \biggr\} . $$
We begin by recalling some basic properties of the function \(m(x,\mu )\).

Lemma 2.1

([12, Proposition 1.8 & Remark 1.9])

Suppose thatμsatisfies (1.2) & (1.3). Then
  1. (i)

    \(0< m(x,\mu )<\infty \)for every\(x\in \mathbb{R}^{n}\).

     
  2. (ii)

    If\(r=m(x,\mu )^{-1}\), then\(r^{n-2}\leq \mu (B(x,r)) \leq C_{1}r^{n-2}\).

     
  3. (iii)

    If\(|x-y|\leq Cm(x,\mu )^{-1}\), then\(m(x,\mu ) \approx m(y,\mu )\).

     
  4. (iv)
    There exist constants\(c, C>0\)such that, for\(x,y\in \mathbb{R}^{n}\),
    $$ \frac{cm(y,\mu )}{\{1+ \vert x-y \vert m(y,\mu )\}^{k_{0}/(1+k_{0})}}\leq m(x, \mu )\leq Cm(y,\mu ) \bigl\{ 1+ \vert x-y \vert m(y,\mu ) \bigr\} ^{k_{0}} $$
    with\(k_{0}=C_{2}/\delta >0\)and\(C_{2}=\log _{2}(C_{1}+2^{n-2})\).
     
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
$$ d(x,y,\mu )=\inf_{\gamma } \int ^{1}_{0}m \bigl(\gamma (\tau ),\mu \bigr) \bigl\vert \gamma '( \tau ) \bigr\vert \,d\tau , $$
(2.1)
where \(\gamma :[0,1]\rightarrow \mathbb{R}^{n}\) is absolutely continuous and \(\gamma (0)=x\), \(\gamma (1)=y\).
A parabolic-type distance function associated with \(m(x,\mu )\) is defined by
$$ d_{\mu }(x,y,t)=\inf_{\gamma } \int ^{1}_{0}m \bigl(\tilde{\gamma }(\tau ), \mu \bigr)\max \bigl\{ \bigl\vert (\tilde{\gamma })'(\tau ) \bigr\vert , \bigl\vert (\gamma _{n+1})'(\tau ) \bigr\vert \bigr\} \,d \tau , $$
(2.2)
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 \(\gamma (0)=(x,0)\), \(\gamma (1)=(y,\sqrt{t})\).

Lemma 2.2

([15, Lemma 2.2])

For the distance function\(d(x,y,\mu )\)in (2.1), we have
  1. (i)
    For every\(x,y, z\in \mathbb{R}^{n}\),
    $$ d(x,y,\mu )\leq d(x,z,\mu )+d(z,y,\mu ). $$
     
  2. (ii)
    There are two positive constantscandCsuch 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 constantscandCsuch 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 \(x\in \mathbb{R}^{n}\),
$$ \mu \bigl(B(x,r) \bigr)\leq \textstyle\begin{cases} C (rm(x,\mu ) )^{\delta }r^{n-2},& r< m(x,\mu )^{-1}; \\ C (rm(x,\mu ) )^{C _{2}}m(x,\mu )^{2-n},& r< m(x,\mu )^{-1}, \end{cases} $$
(2.3)
see [15, (2.1)]. The above estimate implies the following.

Lemma 2.4

([15, (2.2)])

For every nonnegative Schwarz functionω,
$$ \int _{\mathbb{R}^{n}}t^{-n/2}\omega \biggl(\frac{x-y}{\sqrt{t}} \biggr)\,d\mu (y) \leq \textstyle\begin{cases} Ct^{-1} ( \sqrt{t}m(x,\mu ) )^{\delta },& t< m(x,\mu )^{-2}; \\ Ct ^{-1} (\sqrt{t}m(x,\mu ) )^{C_{2}-n+2},& t\geq m(x,\mu )^{-2}. \end{cases} $$
(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
$$ \mathcal{B}_{j}= \bigl\{ x: 2^{j/2}\leq m(x,\mu )< 2^{(j+1)/2} \bigr\} . $$
Since \(0< m(x,\mu )<\infty \), we have \(\mathbb{R}^{n}=\bigcup_{j\in \mathbb{Z}}\mathcal{B}_{j}\).

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
  1. (i)

    \(\operatorname{supp} a\subset B(x_{0},r)\);

     
  2. (ii)

    \(\|a\|_{L^{\infty }}\leq |B(x_{0},r)|^{-1}\);

     
  3. (iii)

    if \(x_{0}\in \mathcal{B}_{j}\), then \(r\leq 2^{1-j/2}\);

     
  4. (iv)

    if \(x_{0}\in \mathcal{ B}_{j}\) and \(r\leq 2^{-1-j/2}\), then \(\int a(x)\,dx=0\).

     
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.

One of the main results of [15] is the following proposition.

Proposition 2.6

([15, Theorem 1.2])

Assume thatμis 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])

  1. (i)
    There exist positive constantsCandcdepending only onnand constants\(C_{0}\), \(C_{1}\)andδin (1.2) & (1.3) such that
    $$ 0\leq K^{{\mathcal{L}}}_{t}(x,y)\leq C h_{t}(x-y)e^{-c d_{\mu }(x,y,t)}. $$
     
  2. (ii)
    For every\(0<\delta '<\delta _{0}=\min \{\alpha , \delta , \nu \}\), there exists a constantCsuch 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
  1. (i)

    \(\operatorname{supp}a\subset B(x_{0},r)\);

     
  2. (ii)

    \(\|a\|_{q}\leq |B(x_{0},r)|^{1/q-1}\);

     
  3. (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
$$\begin{aligned} {\mathcal{M}_{\mathcal{L}}}(a) (x) \leq &c\sup_{t>0} { \int _{B(x _{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/t}} { \bigl(1+m(x,\mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c\sup_{t>0} { \int _{B(x_{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/t}} { \bigl(1+{ \vert x-y \vert ^{2}}/{\sqrt{t}} \bigr)^{-n-N}} { \bigl(1+m(x,\mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy. \end{aligned}$$
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
$$\begin{aligned} \bigl\vert {\mathcal{M}_{\mathcal{L}}}(a) (x) \bigr\vert \leq & c\sup _{t>0}\frac{1}{t ^{n/2}}{ \int _{B(x_{0},r)}} { \bigl({ \vert x-x_{0} \vert }/{ \sqrt{t}} \bigr)^{-n-N}} { \bigl(m(x, \mu )\sqrt{t} \bigr)^{-N}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c{ \vert x-x_{0} \vert ^{-n-N}} { \bigl[m(x,\mu ) \bigr]^{-N}} { \int _{B(x_{0},r)}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c{ \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}} \bigl[m(x_{0},\mu ) \bigr]^{-{N}/( {{k_{0}}+1})}, \end{aligned}$$
which gives
$$\begin{aligned} { \int _{ \vert x-x_{0} \vert >4r}} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \,dx \leq &c{ \int _{ \vert x-x _{0} \vert >4r}} { \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}} \bigl[m(x_{0},\mu ) \bigr]^{-{N}/( {{k_{0}}+1})}\,dx \\ \leq &c{ \vert x-x_{0} \vert ^{-n-{N}/{({k_{0}}+1})}}r^{-{N}/({{k_{0}}+1})} \leq C, \end{aligned}$$
where in the last inequality we have used the fact that \(1\leq r m(x _{0},\mu )\leq 4\).
Case2:\(r<1/m(x_{0},\mu )\). By Proposition 2.7 and the symmetry of \(K^{\mathcal{L}}_{t}(\cdot ,\cdot )\), we have
$$\begin{aligned} \bigl\vert K^{\mathcal{L}}_{t}(x,y+h)-K^{\mathcal{L}}_{t}(x,y) \bigr\vert \leq C_{N}{ \bigl( { \vert h \vert }/{\sqrt{t}} \bigr)^{\delta '}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/ct}} { \bigl\{ {1+ \sqrt{t}m(x,\mu )+\sqrt{t}m(y,\mu ) \bigr\} }^{-N}}. \end{aligned}$$
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
$$\begin{aligned} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \leq &\sup _{t>0} \biggl\vert { \int _{B(x_{0},r)}} \bigl[K^{\mathcal{L}}_{t}(x,y)-K^{\mathcal{L}}_{t}(x,x _{0}) \bigr]a(y)\,dy \biggr\vert \\ \leq &c\sup_{t>0} { \int _{B(x_{0},r)}} {t^{-n/2}} {e^{-{ \vert x-y \vert ^{2}}/ct}} { \bigl({ \vert y-x_{0} \vert }/{\sqrt{t}} \bigr)^{\delta '}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &c\sup_{t>0} \biggl\{ {t^{-n/2}} { \int _{B(x_{0},r)}} { \bigl(1+ { \vert x-y \vert }/{\sqrt{t}} \bigr)^{-n-\delta '}} { \vert y-x_{0} \vert ^{\delta '}}/{t^{\delta '/2}} \bigl\vert a(y) \bigr\vert \,dy \biggr\} \\ \leq &{cr^{\delta '}} { \vert x-x_{0} \vert ^{-n-\delta '}}, \end{aligned}$$
which gives
$$\begin{aligned} { \int _{ \vert x-x_{0} \vert \geq 4r}} \bigl\vert {\mathcal{M}_{\mathcal{L}}}a(x) \bigr\vert \,dx \leq &c { \int _{ \vert x-x_{0} \vert \geq 4r}} {r^{\delta '}} { \vert x-x_{0} \vert ^{-n-\delta '}}\,dx \leq C. \end{aligned}$$
 □

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
  1. (i)

    \(|x|^{nb}M(x)\in L^{q}(\mathbb{R}^{n})_{j}\);

     
  2. (ii)

    \(\|M\|^{{\varepsilon }/{b}}_{q}\||x-x_{0}|^{nb}M(\cdot ) \|^{1-{\varepsilon }/{b}}_{q}\leq 1\);

     
  3. (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

Ifais a\((1,q)\)-atom supported on\(B(x_{0},r)\), ais 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
$$\begin{aligned} \int _{\mathbb{R}^{n}} \bigl\vert \vert x-x_{0} \vert ^{nb}a(x) \bigr\vert ^{q}\,dx \leq & \bigl\vert B(x_{0},r) \bigr\vert ^{bq+1-q}, \end{aligned}$$
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 \(b=1-{1}/{q}+\varepsilon \),
$$\begin{aligned} \Vert a \Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}a(\cdot ) \bigr\Vert ^{1- {\varepsilon }/{b}}_{q} \leq \bigl\vert B(x_{0},r) \bigr\vert ^{({1}/{q}-1)({\varepsilon }/ {b})} \bigl\vert B(x_{0},r) \bigr\vert ^{\varepsilon (1-{\varepsilon }/{b})}\leq 1. \end{aligned}$$
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,
$$ r\leq \Vert a \Vert ^{1/\{n({1}/{q}-1)\}}_{q}\leq m(x_{0}, \mu )^{-1}. $$
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 \(x_{0}\). □

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
$$ N_{\mathcal{L}}(M)=: \Vert M \Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}M( \cdot ) \bigr\Vert ^{1-{\varepsilon }/{b}}_{q}. $$
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
$$ \textstyle\begin{cases} E_{0}=\{x: \vert x-x_{0} \vert \leq \sigma \}; \\ E_{k}=\{x:2^{k-1}\sigma < \vert x-x_{0} \vert \leq 2^{k}\sigma \},\quad k\in N; \\ B_{k}=\{x: \vert x-x_{0} \vert \leq 2^{k}\sigma \},\quad k=0,1,2,\ldots . \end{cases} $$
Denote by \(\psi _{k}\) the characteristic function \(\chi _{E_{k}}(x)\) and write \(M(x)=\sum_{k}M_{k}(x)\), where \(M_{k}:=M(x)\psi _{k}(x)\).

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
$$\begin{aligned} \frac{1}{ \vert B_{0} \vert } \int _{\mathbb{R}^{n}} \bigl\vert M_{0}(x) \bigr\vert ^{2}\,dx \leq &\frac{1}{ \sigma ^{n}} \Vert M \Vert ^{2}_{2}=\frac{1}{\sigma ^{2n}}, \end{aligned}$$
which implies that \(\|M_{0}\|_{2}\leq |B_{0}|^{-{1}/{2}}\).
For the term \(M_{k}\), we have
$$\begin{aligned} \frac{1}{ \vert B_{k} \vert } \int _{\mathbb{R}^{n}} \bigl\vert M_{k}(x) \bigr\vert ^{2}\,dx \leq &\frac{1}{(2^{k-1} \sigma )^{n}} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb}M_{k}(\cdot ) \bigr\Vert ^{2}_{2} \bigl(2^{k-1} \sigma \bigr)^{-n(1+2\varepsilon )} \\ \leq & \bigl(2^{k-1}\sigma \bigr)^{-2n-2\varepsilon n}\sigma ^{2n\varepsilon } \\ \leq &C_{n,\varepsilon } \bigl(2^{k}\sigma \bigr)^{-2n}2^{-2k\varepsilon }, \end{aligned}$$
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. □

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
$$ q_{t}(x,y)=h_{t}(x-y)-K^{\mathcal{L}}_{t}(x-y)= \int ^{t}_{0} \int _{\mathbb{R}^{n}}K^{\mathcal{L}}_{s}(x,t)h_{t-s}(z-y) \,d\mu (z)\,ds. $$
The following estimate was obtained by Wu and Yan [15].

Lemma 4.1

([15, Lemma 3.6])

  1. (i)
    There exist constantsCandcsuch that, for every\(x,y\in \mathbb{R}^{n}\)and\(t>0\),
    $$ q_{t}(x,y)\leq \textstyle\begin{cases} C( \sqrt{t}m(x, \mu ))^{\delta }t^{-{n}/{2}}e^{-{|x-y|^{2}}/{ct}} , & \sqrt{t}\leq m(x, \mu )^{-1}; \\ C( \sqrt{t}m(y, \mu ))^{\delta }t^{-{n}/{2}}e^{-{|x-y|^{2}}/{ct}} , & \sqrt{t}\leq m(y, \mu )^{-1}; \\ h_{t}(x-y), & \textit{elsewhere}. \end{cases} $$
     
  2. (ii)
    For every\(0<\delta '<\min \{1,\delta \}\)and\(C>0\), there exist constants\(C'\)andcsuch 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\(\delta >0\).
  1. (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)
     
  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

(i). In fact, we can deduce (4.2) from Lemma 4.1. Precisely,

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
$$\begin{aligned} \bigl\vert g(x,y)-g(x,x_{0}) \bigr\vert =& \biggl\vert \int ^{\infty }_{0}t^{-i\gamma }q_{t}(x,y) {dt}/{t}- \int ^{\infty }_{0}t^{-i\gamma }q_{t}(x,x_{0}) \frac{dt}{t} \biggr\vert \\ \leq & \int ^{\infty }_{0}t^{-i\gamma } \bigl\vert q_{t}(x,y)-q_{t}(x,x_{0}) \bigr\vert \frac{dt}{t}. \end{aligned}$$
By (ii) of Lemma 4.1 and a direct computation, we get
$$\begin{aligned} \bigl\vert g(x,y)-g(x,x_{0}) \bigr\vert \leq & \biggl\vert \int ^{\infty }_{0}t^{-i\gamma } \bigl[ \vert y-x _{0} \vert m(x,\mu ) \bigr]^{\delta '}t^{-{n}/{2}}e^{-{c \vert x-y \vert ^{2}}/{t}} \frac{dt}{t} \biggr\vert \\ \leq &C \bigl[ \vert y-x_{0} \vert m(x,\mu ) \bigr]^{\delta '} \int ^{\infty }_{0}t^{-{n}/ {2}}e^{-{c \vert x-y \vert ^{2}}/{t}} \frac{dt}{t} \\ \leq & C{ \vert x-y \vert ^{-n}} \bigl[ \vert y-x_{0} \vert m(x,\mu ) \bigr]^{\delta '}. \end{aligned}$$
This completes the proof of Lemma 4.2. □
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
  1. (a)

    T extends to a bounded operator on \(L^{2}(\mathbb{R}^{n},dx)\);

     
  2. (b)
    there exists a kernel K such that, for every \(f\in L^{1} _{c}(\mathbb{R}^{n},dx)\),
    $$ Tf(x)= \int _{\mathbb{R}^{n}}K(x,y)f(y)\,dy\quad \text{a.e. on }\{\operatorname{supp} f\} ^{c}; $$
     
  3. (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

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\(\delta >0\). Then, for\(\gamma \in \mathbb{R}^{n}\), \(\mathcal{L}^{i\gamma }\)is a Calderón–Zygmund operator.

Now we prove the \(H^{1}_{\mathcal{L}}\)-boundedness of \(\mathcal{L} ^{i\gamma }\).

Theorem 4.4

Letμbe a nonnegative Radon measure in\(\mathbb{R}^{n}\), \(n \geq 3\). Suppose thatμsatisfies conditions (1.2) & (1.3) for some\(\delta >0\). Then, for\(\gamma \in \mathbb{R}^{n}\), \(\mathcal{L}^{i\gamma }\)is bounded on\(H^{1}_{\mathcal{L}}( \mathbb{R}^{n})\).

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:\(\mathcal{L}^{i\gamma }a\in L^{q}(\mathbb{R}^{n}) \& |x|^{nb}L ^{i\gamma }(a)\in L^{q}(\mathbb{R}^{n})\).
$$\begin{aligned} \bigl\Vert \vert x \vert ^{nb}\mathcal{L}^{i\gamma }a \bigr\Vert _{q} \leq & \bigl\Vert x_{B^{*}} \vert \cdot \vert ^{nb} \mathcal{L}^{i\gamma }a \bigr\Vert _{q}+ \bigl\Vert x_{({B^{*}})^{c}} \vert \cdot \vert ^{nb} \mathcal{L}^{i\gamma }a \bigr\Vert _{q} \\ =& \biggl( \int _{B^{*}} \vert x \vert ^{qnb} \bigl\vert \mathcal{L}^{i\gamma }(a) \bigr\vert ^{q}\,dx \biggr)^{ {1}/{q}}+ \biggl( \int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \bigl\vert \mathcal{L}^{i\gamma }(a) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \\ \leq & \bigl(r+ \vert x_{0} \vert \bigr)^{nb} \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}+ \biggl[ \int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \biggl\vert \int _{\mathbb{R}^{n}}K^{\mathcal{L}} _{\gamma }(x,y)a(y)\,dy \biggr\vert ^{q}\,dx \biggr]^{{1}/{q}}. \end{aligned}$$
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
$$ \textstyle\begin{cases} S_{1}:=(r+ \vert x_{0} \vert )^{nb} \Vert a \Vert _{q}; \\ S_{2}:=\int _{B(x_{0},r)} \vert a(y) \vert [\int _{({B^{*}})^{c}} \vert x \vert ^{qnb} \vert K ^{\mathcal{L}}_{\gamma }(x,y) \vert ^{q}\,dx ]^{{1}/{q}}. \end{cases} $$
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
$$\begin{aligned} \bigl\vert K^{\mathcal{L}}_{\gamma }(x,y) \bigr\vert \leq &C_{N}{t^{-{n}/{2}}}\frac{e ^{-c{ \vert x-y \vert ^{2}}/{t}}}{\{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{N}} \frac{dt}{t} \\ \leq &\frac{C_{N}}{\{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{N}} \frac{1}{ \vert x-y \vert ^{n}}, \end{aligned}$$
which gives
$$\begin{aligned} S_{2}\leq C \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \int _{({B^{*}})^{c}} \vert x \vert ^{qnb}\frac{1}{ \{1+ \vert x-y \vert [m(x,\mu )+m(y,\mu )]\}^{qN}} \frac{dx}{ \vert x-y \vert ^{qn}} \biggr]^{ {1}/{q}}\,dy. \end{aligned}$$
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
$$ m(y,\mu )^{N}\geq \biggl[\frac{cm(x_{0},\mu )}{\{1+ \vert x_{0}-y \vert m(x_{0}, \mu )\}^{{k_{0}}/{(k_{0}+1)}}} \biggr]^{N}. $$
Then, via a direct computation, we have
$$\begin{aligned} S_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl\{ \int _{({B^{*}})^{c}}\frac{ \vert x \vert ^{qnb}\,dx}{ \vert x-x_{0} \vert ^{(n+N)q}} \biggr\} ^{ {1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl\{ \biggl[ \int _{({B^{*}})^{c}}\frac{1}{ \vert x-x_{0} \vert ^{(n+N)q-qnb}}\,dx \biggr]^{{1}/ {q}} \\ &{}+ \biggl[ \int _{({B^{*}})^{c}} \frac{ \vert x_{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr]^{{1}/{q}} \biggr\} \,dy \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \bigl\{ r^{n\varepsilon -N}+ \vert x _{0} \vert ^{nb}r^{-n-N+{n}/{q}} \bigr\} \,dy \\ \leq &C\frac{\{1+r m(x_{0},\mu )\}^{k_{0}N/(k_{0}+1)}}{m(x_{0}, \mu )^{N}} \bigl\{ r^{n\varepsilon -N}+ \vert x_{0} \vert ^{nb}r^{-n-N+{n}/{q}} \bigr\} \biggl( \int _{B} \bigl\vert a(y) \bigr\vert \,dy \biggr)< \infty . \end{aligned}$$

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 \(y\in B\),
$$ \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}\leq \biggl( \int _{B^{*}} \bigl\vert \mathcal{L}^{i \gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}+ \biggl( \int _{({B^{*}})^{c}} \bigl\vert \mathcal{L}^{i\gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}. $$
Because \(\mathcal{L}^{i\gamma }\) is bounded on \(L^{q}(\mathbb{R}^{n})\), \(q>1\), then
$$ \biggl( \int _{B^{*}} \bigl\vert \mathcal{L}^{i\gamma }a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}} \leq \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert _{q}\leq C \Vert a \Vert _{q}\leq \vert B \vert ^{{1}/ {q}-1}. $$
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
$$ \bigl\Vert \mathcal{L}^{i\gamma }a(x) \bigr\Vert ^{1/\{{n}/{q}-n\}}\geq \rho (x_{0})^{( {n}/{q}-n)/\{{n}/{q}-n\}}=\rho (x_{0}), $$
which means that such \(\mathcal{L}^{i\gamma }a\) need not satisfy the canceling condition.
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
$$ \bigl\vert K^{\mathcal{L}}_{\gamma }(x,y) \bigr\vert \leq \frac{C_{N}}{\{1+ \vert x-y \vert [m(x, \mu )+m(y,\mu )]\}^{N}}\frac{1}{ \vert x-y \vert ^{n}}, $$
we can use Lemma 2.1 and the fact that \(r\geq {1}/{m(x_{0}, \mu )}\) to obtain
$$\begin{aligned} I_{2} \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \biggl( \int _{({B^{*}})^{c}}\frac{ \vert x-x _{0} \vert ^{nbq}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl\{ \frac{1}{m(x_{0},\mu )}+ \vert y-x_{0} \vert \biggr\} ^{N}r^{nb+ {n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{N}r^{nb+{n}/{q}-n-N} \,dy \\ \leq &C \vert B \vert ^{\varepsilon }. \end{aligned}$$
The estimates for \(I_{1}\) and \(I_{2}\) imply that
$$\begin{aligned} N_{\mathcal{L}} \bigl(\mathcal{L}^{i\gamma }a \bigr) =& \bigl\Vert \mathcal{L}^{i\gamma }a \bigr\Vert ^{{\varepsilon }/{b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert \mathcal{L}^{i\gamma }a \bigr\Vert ^{1-{\varepsilon }/{b}}_{q}\leq C. \end{aligned}$$
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
$$ \mathcal{L}^{i\gamma }a= \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a+(- \Delta )^{i\gamma }a. $$
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
$$ \textstyle\begin{cases} I_{1}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a\chi _{B^{ \ast }} \Vert _{q}; \\ I_{2}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a \chi _{B^{\sharp }\backslash B^{\ast }} \Vert _{q}; \\ I_{3}:= \Vert (\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a \chi _{(B^{\sharp })^{c}} \Vert _{q}. \end{cases} $$
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
$$\begin{aligned} \bigl\vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr](a) (x) \bigr\vert \leq & \biggl\vert \int _{B} \bigl[ K^{\mathcal{L}}_{\gamma }(x,y)-K_{\gamma }(x,y) \bigr]a(y)\,dy \biggr\vert \\ \leq & \biggl\vert \int _{B} \bigl[ K^{\mathcal{L}}_{\gamma }(x,x_{0})-K^{ \mathcal{L}}_{\gamma }(x,y) \bigr]a(y)\,dy \biggr\vert \\ &{} + \biggl\vert \int _{B} \bigl[K_{\gamma }(x,y)-K _{\gamma }(x,x_{0}) \bigr]a(y)\,dy \biggr\vert \\ \leq & \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \frac{ \vert y-x_{0} \vert ^{\delta }}{ \vert x-x_{0} \vert ^{n+ \delta }}+\frac{ \vert y-x_{0} \vert }{ \vert x-x_{0} \vert ^{n+1}} \biggr]\,dy \\ \leq &C \biggl(\frac{r^{\delta }}{ \vert x-x_{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x _{0} \vert ^{n+1}} \biggr). \end{aligned}$$
Then, since \(r m(x_{0}, \mu )<1\), we obtain that
$$\begin{aligned} I_{3} \leq &C \biggl[ \int _{(B^{\sharp })^{c}} \biggl(\frac{r^{\delta }}{ \vert x-x _{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x_{0} \vert ^{n+1}} \biggr)^{q}\,dx \biggr]^{{1}/ {q}} \\ \leq &C \biggl\{ \biggl[ \int _{(B^{\sharp })^{c}}\frac{r^{q\delta }}{ \vert x-x _{0} \vert ^{q(n+\delta )}}\,dx \biggr]^{{1}/{q}} + \biggl[ \int _{(B^{\sharp })^{c}}\frac{r ^{q}}{ \vert x-x_{0} \vert ^{q(n+1)}}\,dx \biggr]^{{1}/{q}} \biggr\} \\ \leq &Cr^{\delta }m(x_{0}, \mu )^{\delta +n-{n}/{q}}+r m(x_{0}, \mu )^{\delta +1-{n}/{q}} \\ \leq &Cm(x_{0}, \mu )^{n-{n}/{q}}. \end{aligned}$$
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
$$\begin{aligned} \bigl\vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a(x) \bigr\vert \leq &C \int _{B} { \bigl[ \vert y-x_{0} \vert m(x_{0}, \mu ) \bigr]^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &C \bigl[m(x_{0}, \mu ) \bigr]^{\delta '}{r^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \int _{B} \bigl\vert a(y) \bigr\vert \,dy \\ \leq &C \bigl[m(x_{0}, \mu ) \bigr]^{\delta '}{r^{\delta '}} { \vert x-x_{0} \vert ^{-n}} \\ \leq &C{2^{-m\delta '}} { \vert x-x_{0} \vert ^{-n}}, \end{aligned}$$
which implies that
$$\begin{aligned} I_{2} \leq &C \biggl[ \int _{B^{\sharp }\backslash B^{\ast }} \bigl\vert \bigl[\mathcal{L} ^{i\gamma }-(- \Delta )^{i\gamma } \bigr]a(x) \bigr\vert ^{q}\,dx \biggr]^{{1}/{q}} \\ \leq &C \biggl( \int _{B^{\sharp }\backslash B^{\ast }}{2^{-m\delta 'q}} { \vert x-x_{0} \vert ^{-nq}}\,dx \biggr)^{{1}/{q}} \\ \leq &C{m(x_{0}, \mu )}^{{n}-n/{q}}, \end{aligned}$$
where in the last inequality we have used the fact that \(q<{n}/{(n- \delta ')}\).
Finally, it follows from the estimates for \(I_{i}\), \(i=1,2,3\), that
$$ \bigl\Vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr\Vert ^{1/({n}/{q}-n)} _{q}\geq {m(x_{0}, \mu )}^{-1}. $$
On the other hand, the \(L^{q}\)-boundedness of \(\mathcal{L}^{i\gamma }-(- \Delta )^{i\gamma }\) gives
$$ \bigl\Vert \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr\Vert _{q}\leq \Vert a \Vert _{q}= \biggl( \int _{B} \bigl\vert a(y) \bigr\vert ^{q}\,dy \biggr)^{{1}/{q}}\leq r^{{n}/{q}-n}. $$
This means that for this case, \((\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a\) need not satisfy the canceling condition.
Part III: There exists a constantCsuch that, for any\((1,\infty )\)-atom, uniformly,
$$ {N}_{\mathcal{L}} \bigl( \bigl[\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr]a \bigr) \leq C. $$
We write \(b=1-{1}/{q}+\varepsilon \), then \(\varepsilon -b={1}/{q}-1\). We have proved that
$$\begin{aligned} \bigl\Vert \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert _{q} \leq &C\rho (x _{0})^{{n}/{q}-n}\leq \rho (x_{0})^{n(\varepsilon -b)}. \end{aligned}$$
Now we split: \(\||\cdot -x_{0}|^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma })a\|_{q}\leq I_{1}+I_{2}\), where
$$ \textstyle\begin{cases} I_{1}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a \Vert _{L^{q}(B^{\sharp })}; \\ I_{2}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(\mathcal{L}^{i\gamma }-(-\Delta )^{i \gamma })a \Vert _{L^{q}((B^{\sharp })^{c})}. \end{cases} $$
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
$$ \textstyle\begin{cases} I_{2,1}:= \Vert \vert \cdot -x_{0} \vert ^{nb}\mathcal{L}^{i\gamma }a \Vert _{L^{q}((B^{ \sharp })^{c})}; \\ I_{2,2}:= \Vert \vert \cdot -x_{0} \vert ^{nb}(-\Delta )^{i\gamma }a \Vert _{L^{q}((B^{ \sharp })^{c})}. \end{cases} $$
Notice that \(\varepsilon <{\delta }/{n}\) and \(nb-(n+\delta )+{n}/ {q}<0\). By Theorem 4.3, we have
$$\begin{aligned} I_{2,1} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl[ \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb}| \int _{B}{ \vert y-x_{0} \vert ^{q\delta }} { \vert x-x_{0} \vert ^{-q(n+\delta )}}\,dx \biggr]^{ {1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert y-x_{0} \vert ^{\delta } \biggl[ \int _{(B^{\sharp })^{c}} \vert x-x _{0} \vert ^{qnb-q(n+\delta )}\,dx \biggr]^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{\delta }\rho (x_{0})^{nb-(n+\delta )+{n}/ {q}}\,dy \\ \leq &C \rho (x_{0})^{n\varepsilon }. \end{aligned}$$
For \(I_{2,2}\), similarly, we have
$$\begin{aligned} I_{2,2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert x-x_{0} \vert \biggl( \int _{ \vert x-x_{0} \vert \geq 2\rho (x_{0})}{ \vert x-x_{0} \vert ^{q(nb-n-1)}} \vert x-x_{0} \vert ^{n-1}\,d \vert x-x _{0} \vert \biggr)^{{1}/{q}}\,dy \\ \leq & C \int _{B} \bigl\vert a(y) \bigr\vert r\rho (x_{0})^{nb-(n+1)+{n}/{q}}\,dy \\ \leq &C \rho (x_{0})^{n\varepsilon }, \end{aligned}$$
where we have used the fact that \(0<\varepsilon <\min \{{\delta }/ {n},{1}/{n}\}\). Finally, we get
$$ \bigl\Vert \vert \cdot -x_{0} \vert ^{nb} \bigl( \mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert _{q}\leq \rho (x_{0})^{n\varepsilon }, $$
and, hence,
$$\begin{aligned}& \bigl\Vert \bigl(\mathcal{L}^{i\gamma }-(-\Delta )^{i\gamma } \bigr)a \bigr\Vert ^{{\varepsilon }/ {b}}_{q} \bigl\Vert \vert \cdot -x_{0} \vert ^{nb} \bigl(\mathcal{L}^{i\gamma }-(- \Delta )^{i \gamma } \bigr)a \bigr\Vert ^{(1-{\varepsilon }/{b})}_{q} \\& \quad \leq C\rho (x_{0})^{n( \varepsilon -b){\varepsilon }/{b}}\rho (x_{0})^{n\varepsilon (1-{\varepsilon }/ {b})} \leq C. \end{aligned}$$
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. □

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
$$ R_{\mathcal{L}}:=\nabla (-\Delta +\mu )^{-1/2}, $$
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,
$$ \nabla (-\Delta +\mu )^{-{1}/{2}}f(x)= \int _{\mathbb{R}^{n}}R_{ \mathcal{L}}(x,y)f(y)\,dy, $$
where
$$ R_{\mathcal{L}}(x,y):=\frac{1}{\pi } \int ^{\infty }_{0}\lambda ^{-{1}/ {2}}\nabla _{x}\varGamma _{\mu +\lambda }(x,y)\,d\lambda . $$
In [12], Shen proved the following results, see [12, (7.20), (7.26), (7.29)], respectively.

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

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\(\delta >0\). The Riesz transform\(R_{\mathcal{L}}\)is bounded on\(H^{1}_{ \mathcal{L}}(\mathbb{R}^{n})\).

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
$$\begin{aligned} I_{1} \leq &r^{nb} \biggl( \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}a(x) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\leq r^{nb} \Vert a \Vert _{q} \leq r^{nb} \bigl\vert B(x_{0},r) \bigr\vert ^{{1}/ {q}-1} \leq r^{n\varepsilon }. \end{aligned}$$
By Lemma 4.5, for any positive \(N>0\),
$$\begin{aligned} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert \leq &C{ \vert x-y \vert ^{-n-N} \bigl[m(y,\mu ) \bigr]^{-N}}. \end{aligned}$$
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
$$\begin{aligned} I_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\ast })^{c}} \vert x \vert ^{qnb} \bigl\vert R_{ \mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}} \vert x \vert ^{qnb} \frac{1}{ \vert x-y \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \biggl\{ \int _{B} \bigl\vert a(y) \bigr\vert \frac{ \vert x_{0} \vert ^{nb}}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}}\frac{dx}{ \vert x-x_{0} \vert ^{(n+N)q}} \biggr)^{{1}/{q}}\,dy \\ &{}+ \int _{B} \bigl\vert a(y) \bigr\vert \frac{1}{m(y,\mu )^{N}} \biggl( \int _{(B^{\ast })^{c}}\frac{ \vert x-x _{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+N)q}}\,dx \biggr)^{{1}/{q}}\,dy \biggr\} \\ \leq &C \int _{B}\frac{ \vert a(y) \vert }{m(y,\mu )^{N}} \bigl\{ r^{n\epsilon -N}+ \vert x _{0} \vert ^{-n-N+{n}/{q}} \bigr\} \,dy. \end{aligned}$$
Because \(y\in B(x_{0},r)\),
$$ m(y,\mu ) \geq \frac{Cm(x_{0},\mu )}{\{1+ \vert x-x_{0} \vert m(x_{0},\mu )\}^{ {k_{0}}/{(k_{0}+1)}}}, $$
which implies that \(\|{|\cdot |}^{nb}a\|_{L^{q}((B^{\ast })^{c})}< \infty \).

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
$$\begin{aligned} \Vert R_{\mathcal{L}}a \Vert _{q} \leq &C \bigl\{ \Vert R_{\mathcal{L}}a \Vert _{q}+ \Vert \chi _{(B^{\ast })^{c}}R_{\mathcal{L}}a \Vert _{q} \bigr\} \\ \leq &C \bigl\{ \Vert a \Vert _{q}+ \Vert \chi _{(B^{\ast })^{c}}R_{\mathcal{L}}a \Vert _{q} \bigr\} \\ \leq &C \biggl\{ \bigl\vert B(x_{0},r) \bigr\vert ^{{1}/{q}-1}+ \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\,dy \biggr\} . \end{aligned}$$
By Lemma 4.5, we can get
$$\begin{aligned} \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \bigl\vert R_{\mathcal{L}}(x,y) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}}\,dy \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int ^{\infty }_{2r}s^{n-qn-1}\,ds \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \,dy \times r^{{n}/{q}-n} \\ \leq &C\rho (x_{0})^{{n}/{q}-n}, \end{aligned}$$
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
$$ \textstyle\begin{cases} I_{1}:= (\int _{B^{\ast }} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ I_{2}:= (\int _{(B^{\ast })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
It is easy to see that
$$\begin{aligned} I_{1}\leq r^{nb} \biggl( \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}a(x) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \leq r^{nb} \Vert R_{\mathcal{L}}a \Vert _{q} \leq r^{nb} \Vert a \Vert _{q} \leq r^{n\varepsilon }. \end{aligned}$$
For \(I_{2}\), by Minkowski’s inequality,
$$\begin{aligned} I_{2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert {m(y,\mu )^{-N}} \biggl( \int _{B^{\ast }}{ \vert x-x _{0} \vert ^{qnb-q(n+N)}}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert {m(y,\mu )^{-N}}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \bigl\{ {m(x_{0},\mu )^{-1}}+ \vert y-x_{0} \vert \bigr\} ^{N}r^{nb+{n}/{q}-n-N}\,dy \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert r^{N}r^{nb+{n}/{q}-n-N} \,dy \\ \leq &C \vert B \vert ^{(\varepsilon -b){\varepsilon }/{b}} \vert B \vert ^{\varepsilon (1- {\varepsilon }/{b})}, \end{aligned}$$
which gives \(N_{\mathcal{L}}(R_{\mathcal{L}}a)=\|R_{\mathcal{L}}a\| ^{{\varepsilon }/{b}}_{q}\||\cdot -x_{0}|^{nb}R_{\mathcal{L}}a\|^{1- {\varepsilon }/{b}}_{q}\leq C\).
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
$$\begin{aligned}& \bigl\vert (R_{\mathcal{L}}-R_{0}) (a) (x) \bigr\vert \\& \quad \leq C \biggl\{ \int _{B} \bigl\vert R_{ \mathcal{L}}(x,y)-R_{\mathcal{L}}(x,x_{0}) \bigr\vert \bigl\vert a(y) \bigr\vert \,dy+ \int _{B} \bigl\vert R_{0}(x,y)-R _{0}(x,x_{0}) \bigr\vert \bigl\vert a(y) \bigr\vert \,dy \biggr\} \\& \quad \leq C \Vert a \Vert _{\infty } \biggl( \int _{B}\frac{ \vert y-x_{0} \vert ^{\delta }}{ \vert x-x _{0} \vert ^{n+\delta }}\,dy+ \int _{B}\frac{ \vert y-x_{0} \vert }{ \vert x-x_{0} \vert ^{n+1}}\,dy \biggr) \\& \quad \leq C \Vert a \Vert _{\infty } \vert B \vert \biggl( \frac{r^{\delta }}{ \vert x-x_{0} \vert ^{n+\delta }}+\frac{r}{ \vert x-x_{0} \vert ^{n+1}} \biggr). \end{aligned}$$
It follows from the above estimate that
$$\begin{aligned} I_{3} \leq &C \biggl\{ \biggl( \int _{(B^{\sharp })^{c}}\frac{r^{q\delta }}{ \vert x-x _{0} \vert ^{q(n+\delta )}}\,dx \biggr)^{{1}/{q}} + \biggl( \int _{(B^{\sharp })^{c}}\frac{r ^{q}}{ \vert x-x_{0} \vert ^{q(n+1)}}\,dx \biggr)^{{1}/{q}} \biggr\} \\ \leq &C \bigl\{ r^{\delta }m(x_{0},\mu )^{n+\delta -{n}/{q}}+r \bigl[m(x_{0}, \mu ) \bigr]^{n+1-{n}/{q}} \bigr\} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}. \end{aligned}$$
For the estimates of \(I_{1} \& I_{2}\), we need the following estimate:
$$ \bigl\vert R_{\mathcal{L}}(y,x)-R_{0}(y,x) \bigr\vert \leq C \biggl\{ \frac{1}{r^{n-1}} \int _{B(y,r)}\frac{d\mu (z)}{ \vert z-y \vert ^{n-1}}+\frac{(r m(x,\mu ))^{\delta }}{r^{n}} \biggr\} . $$
(4.4)
For \(r={|x-y|}/{2}\),
$$ \bigl\vert R_{\mathcal{L}}(y,x)-R_{0}(y,x) \bigr\vert \leq C \biggl\{ \frac{1}{ \vert x-y \vert ^{n-1}} \int _{B(y, { \vert x-y \vert }/{2})}\frac{d\mu (z)}{ \vert z-y \vert ^{n-1}} +\frac{( \vert x-y \vert m(x, \mu ))^{\delta }}{ \vert x-y \vert ^{n}} \biggr\} . $$
We get \(I_{1}\leq \int _{B}|a(y)|A_{1}(y)\,dy\), where
$$ A_{1}:= \biggl\{ \int _{B^{\ast }} \bigl\vert R_{\mathcal{L}}(x,y)-R_{0}(x,y) \bigr\vert ^{q}\,dx \biggr\} ^{{1}/{q}}. $$
Due to (4.4), we further obtain \(A_{1}\leq U_{1}+U_{2}\), where
$$ \textstyle\begin{cases} U_{1}:= \{\int _{B^{\sharp }}{ \vert x-y \vert ^{q(1-n)}} ( \int _{B(x,{ \vert x-y \vert }/{2})}{ \vert z-x \vert ^{1-n}}\,d\mu (z) )^{q}\,dx \}^{{1}/ {q}}; \\ U_{2}:= \{\int _{B^{\sharp }}{( \vert x-y \vert m(y,\mu ))^{q\delta }}{ \vert x-y \vert ^{-qn}}\,dx \}^{{1}/{q}}. \end{cases} $$
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
$$\begin{aligned} U_{1} \leq & C\sum^{0}_{j=-\infty } \biggl( \int _{T_{j+1}\setminus T_{j}} \frac{1}{ \vert x-y \vert ^{q(n-1)}} \biggl( \int _{B(x,{ \vert x-y \vert }/{2})} \frac{d\mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq & C\sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1} \biggl( \int _{T_{j+1}\setminus T_{j}} \biggl( \int _{B(y,{ \vert x-y \vert }/{2})}\frac{d \mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C\sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1} \biggl( \int _{T_{j+1}\setminus T_{j}} \biggl( \int _{B(y,{2^{j+2}}/{m(x_{0},\mu )})}\frac{d\mu (z)}{ \vert z-x \vert ^{n-1}} \biggr)^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C \sum^{0}_{j=-\infty } \biggl[ \frac{m(x_{0},\mu )}{2^{j}} \biggr]^{n-1}\frac{ \mu (3T_{j+1})}{2^{(j+2)(n-{n}/{q}-1)}} \bigl[m(x_{0}, \mu ) \bigr]^{n-{n}/{q}-1}. \end{aligned}$$
Notice that
$$ \mu (T_{j+1})=\mu \bigl(B \bigl(y,{2^{j+2}}/{m(x_{0}, \mu )} \bigr) \bigr)\leq \bigl(2^{j+2} \bigr)^{n-2+ \delta }{m(x_{0}, \mu )}^{2-n}. $$
A direct computation gives
$$\begin{aligned} U_{1} \leq &C \sum^{0}_{j=-\infty } \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}2^{j(n-2+ \delta ')}{2^{-j(2n-{1}/{q}-2)}} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}\sum ^{0}_{j=-\infty }2^{j(n-2+ \delta '-2n+{n}/{q}+2)} \\ \leq &C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}, \end{aligned}$$
which implies that
$$\begin{aligned} I_{1}\leq C \int _{B} \bigl\vert a(y) \bigr\vert A_{1}(y) \,dy \leq C \Vert a \Vert _{1} \bigl[m(x_{0},\mu ) \bigr]^{n- {n}/{q}} \leq C \bigl[m(x_{0},\mu ) \bigr]^{n-{n}/{q}}. \end{aligned}$$
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
$$ \textstyle\begin{cases} E_{1}:= (\int _{B^{\sharp }} \vert x-x_{0} \vert ^{qnb} \vert (R_{\mathcal{L}}-R_{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ E_{2}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert (R_{\mathcal{L}}-R _{0})a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
By the \(L^{p}\)-boundedness of \(R_{\mathcal{L}}\) and \(R_{0}\), we get
$$\begin{aligned} E_{1} \leq &C{m(x_{0},\mu )}^{-nb} \bigl\Vert (R_{\mathcal{L}}-R_{0})a \bigr\Vert _{q} \leq C{m(x_{0},\mu )}^{-nb} \Vert a \Vert _{q} \leq C{m(x_{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
For the term \(E_{2}\), we have \(E_{2}\leq E_{2,1}+E_{2,2}\), where
$$ \textstyle\begin{cases} E_{2,1}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{\mathcal{L}}a(x) \vert ^{q}\,dx )^{{1}/{q}}; \\ E_{2,2}:= (\int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \vert R_{0}a(x) \vert ^{q}\,dx )^{{1}/{q}}. \end{cases} $$
A direct computation gives
$$\begin{aligned} E_{2,1} \leq & C \int _{B} \bigl\vert a(y) \bigr\vert \,dy \biggl( \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{qnb} \bigl\vert R _{\mathcal{L}}(x,y)-R_{\mathcal{L}}(x,x_{0}) \bigr\vert ^{q}\,dx \biggr)^{{1}/{q}} \\ \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \vert y-x_{0} \vert ^{\delta } \biggl( \int _{(B^{\sharp })^{c}}\frac{ \vert x-x _{0} \vert ^{qnb}}{ \vert x-x_{0} \vert ^{(n+\delta )q}}\,dx \biggr)^{{1}/{q}}\,dy \leq C{m(x _{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
For \(E_{2,2}\), because \(R_{0}\) is a Calderón–Zygmund operator, the kernel \(K_{0}(\cdot ,\cdot )\) satisfies
$$ \bigl\vert R_{0}(x,y)-R_{0}(x,x_{0}) \bigr\vert \leq C{ \vert y-x_{0} \vert } { \vert x-x_{0} \vert ^{-n-1}}. $$
We can get
$$\begin{aligned} E_{2,2} \leq &C \int _{B} \bigl\vert a(y) \bigr\vert \biggl( \int _{(B^{\sharp })^{c}} \vert x-x_{0} \vert ^{q(nb-n)}\,dx \biggr)^{{1}/{q}}\,dy \\ \leq &C r^{n+1-{n}/{q}} \biggl( \int ^{\infty }_{{2}/{m(x_{0},\mu )}}s ^{-(q-1)n-q+nbq-1}\,ds \biggr)^{{1}/{q}} \leq C{m(x_{0},\mu )}^{-n\varepsilon }. \end{aligned}$$
Finally, we obtain that
$$\begin{aligned} N_{L} \bigl((R_{\mathcal{L}}-R_{0})a \bigr) \leq &C{m(x_{0},\mu )}^{(n/{q}-{n})( {\varepsilon }/{b})}{m(x_{0},\mu )}^{n\varepsilon ({\varepsilon }/ {b}-1)}\leq C. \end{aligned}$$
 □

Notes

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.

Availability of data and materials

We declare that the materials described in the manuscript, including all relevant raw data, will be freely available to any scientist wishing to use them for non-commercial purposes, without breaching participant confidentiality.

Authors’ contributions

Both authors conceived of the study, participated in its design and coordination, drafted the manuscript, and participated in the sequence alignment. All authors read and approved the final manuscript.

Funding

Project supported by the National Natural Science Foundation of China under grants No. 11871293 & No. 11571217; Shandong Natural Science Foundation of China (No. ZR2017JL008, No. ZR2016AM05); University Science and Technology Projects of Shandong Province (No. J15LI15).

Competing interests

The authors declare that they have no competing interests.

References

  1. 1.
    Alvarez, J., Milman, M.: \(H^{p}\) continuty properties of Calderón–Zygmund operators. J. Math. Anal. Appl. 118, 65–79 (1986) CrossRefGoogle Scholar
  2. 2.
    Cao, J., Yang, Da.: Hardy spaces \(H^{p}_{L}(\mathbb{R}^{n})\) associated with operators satisfying k-Davies–Gaffney estimates. Sci. China Math. 55, 1403–1440 (2012) MathSciNetCrossRefGoogle Scholar
  3. 3.
    Dong, J., Huang, J., Liu, H.: Boundedness of singular integrals on Hardy type spaces associated with Schrödinger operators. J. Funct. Spaces 2015, Article ID 409215 (2015) zbMATHGoogle Scholar
  4. 4.
    Duong, X., Yan, L.: Duality of Hardy and BMO spaces associated with operators with heat kernel bounds. J. Am. Math. Soc. 18, 943–973 (2005) MathSciNetCrossRefGoogle Scholar
  5. 5.
    Dziubański, J.: Note on \(H^{1}\) spaces related to degenerate Schrödinger operators. Ill. J. Math. 49, 1271–1297 (2005) CrossRefGoogle Scholar
  6. 6.
    Dziubański, J., Zienkiewicz, J.: Hardy space \(H^{1}\) associated to Schrödinger operator with potential satisfying reverse Hölder inequality. Rev. Mat. Iberoam. 15, 279–296 (1999) CrossRefGoogle Scholar
  7. 7.
    Hofmann, S., Lu, G., Mitrea, D., Mitrea, M., Yan, L.: Hardy spaces associated to non-negative self-adjoint operators satisfying Davies–Gaffney estimates. Mem. Am. Math. Soc. 214 no. 1007 (2011) MathSciNetzbMATHGoogle Scholar
  8. 8.
    Jiang, R., Yang, Da.: Orlicz–Hardy spaces associated with operators. Sci. China Ser. A 52, 1042–1080 (2009) MathSciNetCrossRefGoogle Scholar
  9. 9.
    Jiang, R., Yang, Da.: Orlicz-Hardy spaces associated with operators satisfying Davies–Gaffney estimates. Commun. Contemp. Math. 13, 331–373 (2011) MathSciNetCrossRefGoogle Scholar
  10. 10.
    Lin, C., Liu, H., Liu, Y.: Hardy spaces associated with Schrödinger operators on the Heisenberg group. arXiv:1106.4960 [math.AP]
  11. 11.
    Liu, Y., Qi, S.: Endpoint estimates of Riesz transforms associated with generalized Schrödinger operators. Can. Math. Bull. 61, 787–801 (2018) CrossRefGoogle Scholar
  12. 12.
    Shen, Z.: On fundamental solutions of generalized Schrödinger operators. J. Funct. Anal. 167, 521–564 (1999) MathSciNetCrossRefGoogle Scholar
  13. 13.
    Song, L., Yan, L.: Riesz transforms associated to Schrödinger operators on weighted Hardy spaces. J. Funct. Anal. 259, 1466–1490 (2010) MathSciNetCrossRefGoogle Scholar
  14. 14.
    Stein, E.M.: Harmonic Analysis: Real Variable Methods, Orthogonality and Oscillatory Integrals. Princeton Math. Serises, vol. 43. Princeton University Press, Princeton (1993) zbMATHGoogle Scholar
  15. 15.
    Wu, L., Yan, L.: Heat kernel, upper bounds and Hardy spaces associated to the generalized Schrödinger operators. J. Funct. Anal. 270, 3709–3749 (2016) MathSciNetCrossRefGoogle Scholar

Copyright information

© The Author(s) 2019

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  1. 1.College of Mathematics and StatisticsQingdao UniversityQingdaoChina

Personalised recommendations