Boundedness of operators generated by fractional semigroups associated with Schrödinger operators on Campanato type spaces via T1 theorem

Abstract

Let ${\mathcal {L}}=-\varDelta +V$ be a Schrödinger operator, where the nonnegative potential V belongs to the reverse Hölder class $B_{q}$. By the aid of the subordinative formula, we estimate the regularities of the fractional heat semigroup, $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0},$ associated with ${\mathcal {L}}$. As an application, we obtain the BMO$^{\gamma }_{{\mathcal {L}}}$-boundedness of the maximal function, and the Littlewood–Paley g-functions associated with ${\mathcal {L}}$ via T1 theorem, respectively.

1 Introduction

In the research of harmonic analysis and partial differential equations, the maximal operators and Littlewood–Paley g-functions paly an important role and were investigated by many mathematicians extensively. For any integrable function f on ${\mathbb {R}}^{n}$, the Hardy–Littlewood maximal operator is defined as

$$\begin{aligned} M(f)(x):=\sup _{Q}\frac{1}{|Q|}\int _{Q}|f(y)|{\mathrm{d}}y, \end{aligned}$$

where the supremum is taken over all cubes $Q\subset {\mathbb {R}}^{n}$. For $f\in {\mathrm{BMO}}({\mathbb {R}}^{n})$, Bennett–DeVore–Sharpley proved in [1] that M(f) is either infinite or belongs to BMO$({\mathbb {R}}^{n})$. The boundedness result in [1] can be extended to other maximal operators. For example, let $-\varDelta$ be the Laplace operator: $\varDelta =\sum ^{n}_{i=1}\frac{\partial ^{2}}{\partial x_{i}^{2}}$. Denote by $M_{\varDelta }$ and g the maximal operator and Littlewood–Paley g-function generated by the heat semigroup $\{e^{-t(-\varDelta )}\}_{t>0}$, respectively, i.e.,

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle&M_{\varDelta }(f)(x):=\sup _{t>0}|e^{-t(-\varDelta )}(f)(x)|;\\&g(f)(x):=\left( \int _{0}^{\infty }|e^{-t(-\varDelta )}(f)(x)|^{2}\frac{{\mathrm{d}}t}{t^{n+1}}\right) ^{1/2}. \end{aligned}\right. \end{aligned}$$

(1)

Due to the mean value on “large” cubes may be infinite, the BMO$({\mathbb {R}}^{n})$-boundedness of $M_{\varDelta }$ or g holds if $M_{\varDelta }(f)<\infty$ or $g(f)<\infty$ for $f\in {\mathrm{BMO}}({\mathbb {R}}^{n})$.

However, if the Laplacian $-\varDelta$ is replaced by other second-order differential operators, the situation becomes different. Consider the Schrödinger ${\mathcal {L}}=-\varDelta +V$ in ${\mathbb {R}}^{n},\ \ n\ge 3,$ where V is a nonnegative potential belonging to the reverse Hölder class $B_{q}$ for some $q>n/2$. Here a nonnegative potential V is said to belong to $B_{q}$ if there exists $C>0$ such that for every ball B,

$$\begin{aligned} \left( \frac{1}{|B|}\int _{B}V^{q}(x){\mathrm{d}}x\right) ^{1/q}\le \frac{C}{|B|}\int _{B}V(x){\mathrm{d}}x. \end{aligned}$$

In [1], the authors pointed out that for $f\in {\mathrm{BMO}}({\mathbb {R}}^{n})$ a supremum of averages of f over “large” cubes may be infinite, see [1, page 610]. In 2005, Dziubański et al.  [9] proved the square functions associated with Schrödinger operators are bounded on the BMO type space BMO$_{{\mathcal {L}}}({\mathbb {R}}^{n})$ related with ${\mathcal {L}}$ which is distinguished from the case of $BMO({\mathbb {R}}^{n})$. See also [13, 18] for similar results in the setting of Heisenberg groups and stratified Lie groups.

Let $H=-\varDelta +|x|^{2}$ be the harmonic oscillator. In [2], Betancor et al. introduced a T1 criterion for Calderón–Zygmund operators related to H on the BMO type space BMO$_{H}({\mathbb {R}}^{n})$. Later, Ma et al. [15] generalized the T1 criterion to the case of Campanato type spaces BMO$_{{\mathcal {L}}}^{\gamma }({\mathbb {R}}^{n})$ related with ${\mathcal {L}}$. As applications, the authors in [15] proved that the maximal operators associated with the heat semigroup $\{e^{-t{\mathcal {L}}}\}_{t>0}$ and with the generalized Poisson operators $\{P^{\sigma }_{t}\}_{t>0}(0<\sigma <1)$, the Littlewood–Paley g-functions given in terms of the heat and the Poisson semigroups are bounded on BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$.

Notice that for $\sigma \in (0,1)$, the generalized Poisson operator $\{P^{\sigma }_{t}\}_{t>0}$ is expressed as

$$\begin{aligned} P^{\sigma }_{t}(f)(x):=\frac{t^{2\sigma }}{4^{\sigma }\varGamma (\sigma )}\int ^{\infty }_{0}e^{-\frac{t^{2}}{4r}} e^{-r{\mathcal {L}}}(f)\frac{{\mathrm{d}}r}{r^{1+\sigma }}. \end{aligned}$$

(2)

Specially, for $\sigma =1/2$, $\{P^{1/2}_{t}\}_{t>0}$ is corresponding to the Poisson semigroup $\{e^{-t{\mathcal {L}}^{1/2}}\}_{t>0}$ associated with ${\mathcal {L}}$. The main purpose of this paper is to derive the pointwise estimate and regularity properties of the fractional heat semigroup $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0}$, $\alpha >0$, to prove the boundedness of the maximal function and the Littlewood–Paley g-functions generated by $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0}$ on BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$, $0<\gamma <\min \{2\alpha ,\delta _{0},1\}$, via T1 theorem, respectively.

When ${\mathcal {L}}=-\varDelta$, the kernels of the fractional heat semigroup $\{e^{-t(-\varDelta )^{\alpha }}\}_{t>0}$ can be defined via the Fourier transform, i.e.,

$$\begin{aligned} K_{\alpha , t}(x)=(2\pi )^{-n/2}\int _{{\mathbb {R}}^n}e^{ix\cdot \xi -t|\xi |^{2\alpha }}{\mathrm{d}}\xi . \end{aligned}$$

(3)

For ${\mathcal {L}}=-\varDelta +V$ with $V>0$, the kernels of fractional heat semigroups $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0}$, $\alpha \in (0,1)$, can not be defined via (3). However, for $\alpha >0$, the subordinative formula (cf. [10]) indicates that

$$\begin{aligned} K^{{\mathcal {L}}}_{\alpha ,t}(x,y)=\int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)K^{{\mathcal {L}}}_{s}(x,y){\mathrm{d}}s, \end{aligned}$$

(4)

where $\eta ^{\alpha }_{t}(\cdot )$ is a continuous function on $(0,\infty )$ satisfying (19) below. In [12], the identity (4) was applied to estimate $K^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$ via the heat kernel $K^{{\mathcal {L}}}_{t}(\cdot ,\cdot )$, see Proposition 2. Specially, for $\alpha =1/2$, the estimates of $K^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$ goes back to those of the Poisson kernel $P_{t}^{{\mathcal {L}}}(\cdot ,\cdot )$, see [5, Lemma 3.9].

We point out that, compared with the case of $\{P^{\sigma }_{t}\}_{t>0}$, some new regularity estimates should be introduced to prove the BMO$^{\gamma }_{{\mathcal {L}}}$-boundedness of the maximal function and Littlewood–Paley g-functions generated by $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0}$. Let $E=L^{\infty }((0,\infty ), {\mathrm{d}}t)$. It follows from (2) and the Minkowski integral inequality that

$$\begin{aligned} \Vert P^{\sigma }_{t}(f)\Vert _{E}\le C_{\sigma }\int ^{\infty }_{0}t^{2\sigma }e^{-\frac{t^{2}}{4r}}\Vert e^{-r{\mathcal {L}}}(f)\Vert _{E}\frac{{\mathrm{d}}r}{r^{1+\sigma }}. \end{aligned}$$

The fact that

$$\begin{aligned} \int ^{\infty }_{0}t^{2\sigma }e^{-\frac{t^{2}}{4r}}\frac{{\mathrm{d}}r}{r^{1+\sigma }}<\infty \end{aligned}$$

ensures that the BMO$^{\gamma }_{{\mathcal {L}}}$-boundedness of the maximal function $\sup _{t>0}|P^{\sigma }_{t}f(x)|$ can be deduced from that of the heat maximal function $\sup _{t>0}|e^{-t{\mathcal {L}}}f(x)|$, see [15, Proposition 4.7]. However, we can see from the identity (4) that this method is not applicable to the case $\{e^{-t{\mathcal {L}}^{\alpha }}\}_{t>0}$.

In this paper, we get the following results:

• In Sect. 3.1, let $\displaystyle \nabla _{x}=\left( {\partial }/{\partial x_{1}},{\partial }/{\partial x_{2}},\ldots ,{\partial }/{\partial x_{n}}\right)$. By the perturbation theory for semigroups of operators, we deduce the pointwise estimates and the Hölder type estimates of the kernels:

  $$\begin{aligned} \left\{ \begin{aligned}&\left| \nabla _{x}(K_{t}(x-y)-K^{{\mathcal {L}}}_{t}(x,y))\right| ,\\&\left| t^{m}\partial ^{m}_{t}(K_{t}(x-y)-K^{{\mathcal {L}}}_{t}(x,y))\right| ,\\ \end{aligned}\right. \end{aligned}$$

  see Lemmas 8, 10 and Proposition 1, respectively.

• In Sect. 3.2, we use (3) to obtain the corresponding estimates for

  $$\begin{aligned} \left\{ \begin{aligned}&\left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right| ,\\&\left| t^{1/2\alpha }\nabla _{x}(K_{\alpha ,t}(x-y)-K^{{\mathcal {L}}}_{\alpha , t}(x,y))\right| ,\\&\left| t^{m}\partial ^{m}_{t}(K_{\alpha ,t}(x-y)-K^{{\mathcal {L}}}_{\alpha , t}(x,y))\right| ,\\ \end{aligned}\right. \end{aligned}$$

  see Propositions 5–10, respectively.

• In Sect. 4, as applications of the regularity estimates obtained in Sect. 3, we use the T1 criterion established in [15] to prove the boundedness on Campanato type spaces BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$, $0<\gamma <\min \{2\alpha ,\delta _{0},1\}$, of the following maximal operator and g-functions:

  $$\begin{aligned} \left\{ \begin{aligned}&M^{\alpha }_{{\mathcal {L}}}f(x):=\sup _{t>0}|e^{-t{\mathcal {L}}^{\alpha }}f(x)|;\\&g^{{\mathcal {L}}}_{\alpha }(f)(x):=\left( \int ^{\infty }_{0}|D^{{\mathcal {L}},m}_{\alpha ,t}(f)(x)|^{2}\frac{{\mathrm{d}}t}{t}\right) ^{1/2};\\&{\widetilde{g}}^{{\mathcal {L}}}_{\alpha }f(x):=\left( \int ^{\infty }_{0}|{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(f)(x)|^{2}\frac{{\mathrm{d}}t}{t}\right) ^{1/2}, \end{aligned} \right. \end{aligned}$$

  see Theorems 3–5, respectively, where $D^{{\mathcal {L}},m}_{\alpha ,t}$ and ${\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}$ are the operators with the integral kernels

  $$\begin{aligned} {\left\{ \begin{array}{ll} &{}D^{{\mathcal {L}},m}_{\alpha ,t}(x,y):=t^{m}\partial _{t}^{m}K^{{\mathcal {L}}}_{\alpha ,t}(x,y),\\ &{}{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y):=t^{1/2\alpha }\nabla _{x}K^{{\mathcal {L}}}_{\alpha ,t}(x,y), \end{array}\right. } \end{aligned}$$

  (5)

  respectively.

Notations We will use c and C to denote the positive constants, which are independent of 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 $C^{-1}\le B_{1}/B_{2}\le C.$

2 Preliminaries

2.1 Schrödinger operators and function spaces

In this paper, let $\delta _{0}=2- n/q$. At first, we list some properties of the potential V which will be used in the sequel.

Lemma 1

[16, Lemma1.2]

• (i) For $0<r<R<\infty$,

  $$\begin{aligned} \frac{1}{r^{n-2}}\int _{B(x,r)}V(y){\mathrm{d}}y\le C\left( \frac{r}{R}\right) ^{\delta }\frac{1}{R^{n-2}}\int _{B(x,R)}V(y){\mathrm{d}}y. \end{aligned}$$

• (ii) $r^{2-n}\int _{B(x,r)}V(y){\mathrm{d}}y=1$ if $r=\rho (x)$. $r\sim \rho (x)$ if and only if $r^{2-n}\int _{B(x,r)}V(y){\mathrm{d}}y\sim 1.$

Lemma 2

[16, Lemma 1.4]

• (i) There exist $C>0$ and $k_{0}\ge 1$ such that for all $x,y\in {\mathbb {R}}^{n}$,

  $$\begin{aligned} C^{-1}\rho (x)\left( 1+|x-y|/\rho (x)\right) ^{-k_{0}}\le \rho (y)\le C\rho (x)\left( 1+|x-y|/\rho (x)\right) ^ {k_{0}/(1+k_{0})}. \end{aligned}$$

  In particular, $\rho (y)\sim \rho (x)$ if $|x-y|<C\rho (x)$.

• (ii) There exists $l_{0}>1$ such that

  $$\begin{aligned} \int _{B(x,R)}\frac{V(y)}{|x-y|^{n-2}}{\mathrm{d}}y\le \frac{C}{R^{n-2}}\int _{B(x,R)}V(y){\mathrm{d}}y\le C\left( 1+ \frac{R}{\rho (x)}\right) ^{l_{0}}. \end{aligned}$$

Lemma 3

[7, Corollary 2.8] For every nonnegative Schwarz function $\omega$, there exist $\delta >0$ and $C>0$ such that

$$\begin{aligned} \int _{{\mathbb {R}}^{n}}t^{-n/2}\omega (|x-y|/\sqrt{t})V(y){\mathrm{d}}y \le {\left\{ \begin{array}{ll} Ct^{-1}(\sqrt{t}/\rho (x))^{\delta }, &{} t< \rho (x)^{2}; \\ Ct^{-1}(\sqrt{t}/\rho (x))^{l_{0}}, &{} t\ge {\rho (x)^{2}}, \end{array}\right. } \end{aligned}$$

where $l_{0}$ is the constant given in Lemma 2.

It is well known that the classical Hardy space $H^{1}({\mathbb {R}}^{n})$ can be defined via the maximal function $\sup _{t>0}|e^{-t(-\varDelta )}f(x)|$ (cf. [17]). In this sense, we can say that the Hardy space $H^{1}({\mathbb {R}}^{n})$ is the Hardy space associated with $-\varDelta$. Since 1990s, the theory of Hardy spaces associated with operators on ${\mathbb {R}}^{n}$ has been investigated extensively. In [6], Dziubański and Zienkiewicz introduced the Hardy space $H^{1}_{{\mathcal {L}}}({\mathbb {R}}^{n})$ related to Schrödinger operators ${\mathcal {L}}$ and obtained the atomic characterization and the Riesz transform characterization of $H^{1}_{{\mathcal {L}}}({\mathbb {R}}^{n})$ via local Hardy spaces. By the aid of Campanato type spaces, the spaces $H^{p}_{{\mathcal {L}}}({\mathbb {R}}^{n})\ (0<p\le 1)$ were introduced by Dziubański and Zienkiewicz [7]. In recent years, the results of [6, 7] have been extended to other second-ordered differential operators, and various function spaces associated to operators have been established. For further information, we refer the reader to [3, 18,19,20] and the references therein.

For a Schrödinger operator ${\mathcal {L}}$, let $\{e^{-t{\mathcal {L}}}\}_{t>0}$ be the heat semigroup generated by ${\mathcal {L}}$ and denote by $K^{{\mathcal {L}}}_{t}(\cdot ,\cdot )$ the integral kernel of $e^{-t{\mathcal {L}}}$. Because the potential $V\ge 0$, the Feynman-Kac formula implies that

$$\begin{aligned} 0\le K^{{\mathcal {L}}}_{t}(x,y)\le K_{t}(x-y):=(4\pi t)^{-n/2}e^{-{|x-y|^{2}}/{(4t)}}. \end{aligned}$$

The Hardy type spaces $H^{p}_{{\mathcal {L}}}({\mathbb {R}}^{n})$, $0<p\le 1$, are defined as follows (cf. [7]):

Definition 1

For $0<p\le 1$, the Hardy type space $H^{p}_{{\mathcal {L}}}({\mathbb {R}}^{n})$ is defined as the completion of the space of compactly supported $L^{1}({\mathbb {R}}^{n})$-functions such that the maximal function

$$\begin{aligned} M_{{\mathcal {L}}}(f)(x):=\sup _{t>0}|e^{-t{\mathcal {L}}}(f)(x)| \end{aligned}$$

belongs to $L^{p}({\mathbb {R}}^{n})$. The quasi-norm in $H^{p}_{{\mathcal {L}}}({\mathbb {R}}^{n})$ is defined as $\Vert f\Vert _{H^{p}_{{\mathcal {L}}}}:=\Vert M_{{\mathcal {L}}}(f)\Vert _{L^{p}}.$

Let f be a locally integrable function on ${\mathbb {R}}^{n}$ and $B=B(x,r)$ be a ball. Denote by $f_{B}$ the mean of f on B, i.e., $f_{B}=|B|^{-1}\int _{B}f(y){\mathrm{d}}y$. Let

$$\begin{aligned} f(B,V):={\left\{ \begin{array}{ll} f_{B}, &{} r<\rho (x);\\ 0, &{} r\ge \rho (x), \end{array}\right. } \end{aligned}$$

where the auxiliary function $\rho (\cdot )$ is defined as

$$\begin{aligned} \rho (x):=\sup \left\{ r>0:\frac{1}{r^{n-2}}\int _{B(x,y)}V(y){\mathrm{d}}y\le 1\right\} . \end{aligned}$$

Definition 2

Let $0<\gamma \le 1$. The Campanato type space BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$ is defined as the set of all locally integrable functions f satisfying

$$\begin{aligned} \Vert f\Vert _{{\mathrm{BMO}}^{\gamma }_{{\mathcal {L}}}}:=\sup _{B\subset {\mathbb {R}}^{n}}\left\{ \frac{1}{|B|^{1+\gamma /n}}\int _{B}|f(x)-f(B,V)|{\mathrm{d}}x\right\} <\infty . \end{aligned}$$

The dual space of $H^{n/(n+\gamma )}_{{\mathcal {L}}}({\mathbb {R}}^{n})$, $0\le \gamma < 1$, is the Campanato type space BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$ (cf. [14, Theorem 4.5]).

2.2 The T1 criterion on Campanato type spaces

We denote by $L^{p}_{c}({\mathbb {R}}^{n})$ the set of functions $f\in L^{p}({\mathbb {R}}^{n})$, $1\le p\le \infty$, whose support $\text { supp }(f)$ is a compact subset of ${\mathbb {R}}^{n}$.

Definition 3

Let $0\le \beta <n$, $1<p\le q<\infty$ with $1/q=1/p-\beta /n$. Let T be a bounded linear operator from $L^{p}({\mathbb {R}}^{n})$ into $L^{q}({\mathbb {R}}^{n})$ such that

$$\begin{aligned} Tf(x)=\int _{{\mathbb {R}}^{n}}K(x,y)f(y){\mathrm{d}}y,\ \ \ f\in L^{p}_{c}({\mathbb {R}}^{n})\ {\mathrm{and}}\ a.e.\ x\notin \text {supp}(f). \end{aligned}$$

We shall say that T is a $\beta$-Schrödinger–Calderón–Zygmund operator with regularity exponent $\delta >0$ if there exists a constant $C>0$ such that

• (i) $\displaystyle |K(x,y)|\le \frac{C}{|x-y|^{n-\beta }}\left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N}$ for all $N>0$ and $x\ne y$;

• (ii) $\displaystyle |K(x,y)-K(x,z)|+|K(y,x)-K(z,x)|\le C\frac{|y-z|^{\delta }}{|x-y|^{n-\beta +\delta }}$ when $|x-y|>2|y-z|$.

The following T1 type criterions on Campanato type spaces were established by Ma et al. [15].

Theorem 1

[15, Theorem1.1] Let T be a $\beta$-Schrödinger–Calderón–Zygmund operator, $\beta \ge 0$, $0< \beta +\gamma <\min \{1,\delta \}$, with smoothness exponent $\delta$. Then T is a bounded operator from BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$ into BMO$^{\gamma +\beta }_{{\mathcal {L}}}({\mathbb {R}}^{n})$ if and only if there exists a constant C such that

$$\begin{aligned} \left( \frac{\rho (x)}{r}\right) ^{\gamma }\frac{1}{|B|^{1+\beta /n}}\int _{B}|T1(y)-(T1)_{B}|{\mathrm{d}}y\le C \end{aligned}$$

for every ball B(x, r), $x\in {\mathbb {R}}^{n}$ and $0<r\le \rho (x)/2$.

When $\gamma =0$, the authors in [15] also proved

Theorem 2

[15, Theorem1.2] Let T be a $\beta$-Schrödinger–Calderón–Zygmund operator, $0\le \beta <\min \{1,\delta \}$, with smoothness exponent $\delta$. Then T is a bounded operator from BMO$_{{\mathcal {L}}}({\mathbb {R}}^{n})$ into BMO$^{\beta }_{{\mathcal {L}}}({\mathbb {R}}^{n})$ if and only if there exists a constant C such that

$$\begin{aligned} \log \left( \frac{\rho (x)}{r}\right) \frac{1}{|B|^{1+\beta /n}}\int _{B}|T1(y)-(T1)_{B}|{\mathrm{d}}y\le C \end{aligned}$$

for every ball B(x, r), $x\in {\mathbb {R}}^{n}$ and $0<r\le \rho (x)/2$.

Lemma 4

[15, Remark 4.1] Theorems 1and 2can be also stated in a vector-valued setting. If Tf takes values in a Banach space ${\mathbb {B}}$ and the absolute values in the conditions are replaced by the norm in ${\mathbb {B}}$, then both results hold.

3 Regularity estimates

3.1 Regularities of heat kernels

By the fundamental solutions of Schrödinger operators, Dziubański and Zienkiewicz proved that the heat kernel $K^{{\mathcal {L}}}_{t}(\cdot ,\cdot )$ satisfies the following estimates, see also [11].

Lemma 5

• (i) ([6, Theorem 2.11]) For any $N>0$, there exist constants $C_{N},c>0$ such that

  $$\begin{aligned} |K^{{\mathcal {L}}}_{t}(x,y)|\le C_{N}t^{-n/2}e^{-c|x-y|^{2}/t}\left( 1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) ([8, Theorem 4.11]) Assume that $0<\delta \le \min \{1,\delta _{0}\}.$ For any $N>0$, there exist constants $C_{N},c>0$ such that for all $|h|<\sqrt{t}$,

  $$\begin{aligned} \left| K^{{\mathcal {L}}}_{t}(x+h,y)-K^{{\mathcal {L}}}_{t}(x,y)\right| \le C_{N}\left( \frac{|h|}{\sqrt{t}}\right) ^{\delta }t^{-n/2}e^{-c|x-y|^{2}/t}\left( 1+\frac{\sqrt{t}}{\rho (x)}+ \frac{\sqrt{t}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

Lemma 6

[7, Proposition 2.16] There exist constants $C,c>0$ such that for $x,y\in {\mathbb {R}}^{n}$ and $t>0,$

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{t}(x,y)-K_{t}(x-y)\right| \le C \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta _{0}}t^{-n/2}e^{-c|x-y|^{2}/t}. \end{aligned}$$

In [5], under the assumption that $V\in B_{q}, q>n$, Duong et al. obtained the following regularity estimate for the kernel $K^{{\mathcal {L}}}_{t}(\cdot ,\cdot )$.

Lemma 7

[5, Lemma 3.8] Suppose that $V\in B_{q}$ for some $q>n$. For any $N>0$, there exist constants $C >0$ and $c>0$ such that for all $x,y\in {\mathbb {R}}^n$ and $t>0,$

$$\begin{aligned} | \nabla _x { K}^{{\mathcal {L}}}_t(x,y)| \le C t^{-(n+1)/2}e^{-c{{|x-y|}^2}/{ t}}\left( 1+\frac{\sqrt{t}}{\rho (x)} +\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

(6)

By the perturbation theory for semigroups of operators,

$$\begin{aligned} K_{t}(x-y)-K_{t}^{{\mathcal {L}}}(x,y)= & {} \int ^{t}_{0}\int _{{\mathbb {R}}^{n}}K_{t-s}(w-x)V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s \nonumber \\= & {} \int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}K_{t-s}(w-x)V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s\nonumber \\&+ \int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}K_{s}(w-x)V(w)K^{{\mathcal {L}}}_{t-s}(w,y){\mathrm{d}}w{\mathrm{d}}s. \end{aligned}$$

(7)

Similar to [7, Proposition 2.16], we can prove the following lemma.

Lemma 8

Suppose that $V\in B_{q}$ for some $q>n$. There exist constants $C,c>0$ such that

$$\begin{aligned} \left| \nabla _{x}K_{t}(x-y)-\nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)\right| \le Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\min \left\{ \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta _{0}},\ \left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}\right\} . \end{aligned}$$

Proof

If $t\ge \rho (y)^{2}$, it is easy to see that

$$\begin{aligned} \left| \nabla _{x}K_{t}(x-y)-\nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)\right| \le Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}. \end{aligned}$$

If $t<\rho (y)^{2}$, by (7), we get

$$\begin{aligned} \left| \nabla _{x}K_{t}(x-y)-\nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)\right|= & {} \left| \int ^{t}_{0}\int _{{\mathbb {R}}^{n}}\nabla _{x}K_{t-s}(x-w)V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s\right| \nonumber \\\le & {} I_{1}+I_{2}, \end{aligned}$$

(8)

where

$$\begin{aligned} \left\{ \begin{aligned} I_{1}&:= \int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}|\nabla _{x}K_{t-s}(x-w)|V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s;\\ I_{2}&:=\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}|\nabla _{x}K_{s}(x-w)|V(w)K^{{\mathcal {L}}}_{t-s}(w,y){\mathrm{d}}w{\mathrm{d}}s. \end{aligned}\right. \end{aligned}$$

For $I_{1}$, it follows from Lemmas 7 and 3 that

$$\begin{aligned} I_{1}= & {} \int ^{t/2}_{0}\int _{|w-y|<|x-y|/2}|\nabla _{x}K_{t-s}(x-w)|V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s\\&+\int ^{t/2}_{0}\int _{|w-y|\ge |x-y|/2}|\nabla _{x}K_{t-s}(x-w)|V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s \\\le \, & {} Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\int ^{t/2}_{0}\int _{|w-y|<|x-y|/2}V(w)s^{-n/2}e^{-c|w-y|^{2}/s}{\mathrm{d}}w{\mathrm{d}}s\\&+ Ct^{-(n+1)/2}\int ^{t/2}_{0}\int _{|w-y|\ge |x-y|/2}V(w)s^{-n/2}e^{-c(|w-y|^{2}+|x-y|^{2})/s}{\mathrm{d}}w{\mathrm{d}}s \\\le \, & {} Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\int ^{t/2}_{0}\frac{1}{s}\left( \frac{\sqrt{s}}{\rho (y)}\right) ^{\delta _{0}}{\mathrm{d}}s \\= \, & {} Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}. \end{aligned}$$

Similar to $I_{1}$, for the term $I_{2}$, we can obtain

$$\begin{aligned} I_{2}\le Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

It follows from (i) of Lemma 2 that

$$\begin{aligned} \frac{\sqrt{t}}{\rho (x)}\le C\left( 1+\frac{|x-y|}{\sqrt{t}}\frac{\sqrt{t}}{\rho (y)}\right) ^{l_{0}}\frac{\sqrt{t}}{\rho (y)}\le C_{\varepsilon }e^{\varepsilon |x-y|^{2}/t}\frac{\sqrt{t}}{\rho (y)}, \end{aligned}$$

where $\varepsilon >0$ is an arbitrary small constant. Hence

$$\begin{aligned} I_{2}\le Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}. \end{aligned}$$

$\square$

Lemma 9

[7, Proposition 2.17] Let $0<\delta <\min \{1,\delta _{0}\}$. For every $C'>0$ there exists a constant C such that for every $z,x,y\in {\mathbb {R}}^{n}$, $|y-z|\le |x-y|/4$, $|y-z|\le C'\rho (x)$ we have

$$\begin{aligned} \left| \left( K^{{\mathcal {L}}}_{t}(x,y)-K_{t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{t}(x,z)-K_{t}(x-z)\right) \right| \le C\left( \frac{|y-z|}{\rho (y)}\right) ^{\delta }t^{-n/2}e^{-c|x-y|^{2}/t}. \end{aligned}$$

Lemma 10

Suppose that $V\in B_{q}$ for some $q>n$. Let $\delta _{1}=1-n/q$ and $0<\delta '<\delta _{1}$. For every $C'>0$ there exists a constant C such that for every $u,x,y\in {\mathbb {R}}^{n}$, $|u|\le |x-y|/4$, $|u|\le C'\rho (x)$ we have

$$\begin{aligned}&\left| \left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)-\nabla _{x}K_{t}(x-y)\right) -\left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)-\nabla _{x}K_{t}(x+u-y)\right) \right| \\&\quad \le Ct^{-(n+1)/2}e^{-c|x-y|^{2}/t}\left( \frac{|u|}{\rho (y)}\right) ^{\delta '}. \end{aligned}$$

Proof

We prove this lemma by the same argument as Lemma 8. It is enough to verify that

$$\begin{aligned}&\left| \left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)-\nabla _{x}K_{t}(x-y)\right) -\left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)-\nabla _{x}K_{t}(x+u-y)\right) \right| \\&\quad \le C_{\epsilon }t^{-(n+1)/2}e^{\epsilon |x-y|^{2}/t}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}},\nonumber \end{aligned}$$

(9)

where $\epsilon >0$ is an arbitrary small constant. In fact, under the condition $|u|<|x-y|/4$, it is easy to see that $|x-y|\sim |x+u-y|$. We can deduce from Lemma 7 that

$$\begin{aligned}&\left| \left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)-\nabla _{x}K_{t}(x-y)\right) -\left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)-\nabla _{x}K_{t}(x+u-y)\right) \right| \\&\quad \le \left| \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)\right| +\left| \nabla _{x}K_{t}(x-y)\right| +\left| \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)\right| +\left| \nabla _{x}K_{t}(x+u-y)\right| \nonumber \\&\quad \le C t^{-(n+1)/2}e^{-c{{|x-y|}^2}/t}.\nonumber \end{aligned}$$

(10)

Then, for $\delta '\in (0,\delta _{1})$, it follows from (9) and (10) that

$$\begin{aligned}&\left| \left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)-\nabla _{x}K_{t}(x-y)\right) -\left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)-\nabla _{x}K_{t}(x+u-y)\right) \right| \\&\quad \le C \left\{ t^{-(n+1)/2}e^{\epsilon |x-y|^{2}/t}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\right\} ^{\delta '/\delta _{1}} \left\{ t^{-(n+1)/2}e^{-c{{|x-y|}^2}/t}\right\} ^{1-\delta '/\delta _{1}}, \end{aligned}$$

which gives the desired estimate.

Now we prove (9). Since the case for $|u|\ge \rho (y)$ is trivial, we may assume $|u|<\rho (y)$. If $t\le 2|u|^{2}$, the required estimate follows from Lemma 8. Hence we consider the case $t>2|u|^{2}$ only. Recall that for the classical heat kernel $K_{t}(\cdot )$, it holds

$$\begin{aligned} \left| \nabla ^{2}_{x}K_{t}(x)\right| \le Ct^{-n/2-1}e^{-c|x|^{2}/t}. \end{aligned}$$

A direct computation gives

$$\begin{aligned} \left| \nabla _{x}K_{t}(x+u)-\nabla _{x}K_{t}(x)\right| \le C|u|t^{-n/2-1}, \end{aligned}$$

(11)

and for $|u|\le |x|/2$,

$$\begin{aligned} \left| \nabla _{x}K_{t}(x+u)-\nabla _{x}K_{t}(x)\right| \le C|u|t^{-n/2-1}e^{-c|x|^{2}/t}. \end{aligned}$$

(12)

Similar to (8), we split

$$\begin{aligned} \left| \left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x,y)-\nabla _{x}K_{t}(x-y)\right) -\left( \nabla _{x}K^{{\mathcal {L}}}_{t}(x+u,y)-\nabla _{x}K_{t}(x+u-y)\right) \right| \le J_{1}+J_{2}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{aligned} J_{1}&:=\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left| \nabla _{x}K_{t-s}(w-(x+u))-\nabla _{x}K_{t-s}(w-x)\right| V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s;\\ J_{2}&:=\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left| \nabla _{x}K_{s}(w-(x+u))-\nabla _{x}K_{s}(w-x)\right| V(w)K^{{\mathcal {L}}}_{t-s}(w,y){\mathrm{d}}w{\mathrm{d}}s. \end{aligned}\right. \end{aligned}$$

For $J_{1}$, if $t<2\rho (y)^{2}$, using Lemma 3 and (11), we get

$$\begin{aligned} J_{1} &\le C|u|t^{-n/2-1}\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}V(w)s^{-n/2}e^{-c|y-w|^{2}/s}{\mathrm{d}}w{\mathrm{d}}s \\ & \le C|u|t^{-n/2-1}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}} \\ & \le Ct^{-(n+1)/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}. \end{aligned}$$

If $t\ge 2\rho (y)^{2}$, applying Lemmas 3 and  5, we have

$$\begin{aligned} J_{1}& \le C|u|t^{-n/2-1}\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}V(w)s^{-n/2}e^{-c|y-w|^{2}/s}\left( 1+\frac{\sqrt{s}}{\rho (y)}\right) ^{-N}{\mathrm{d}}w{\mathrm{d}}s \\& \le C|u|t^{-n/2-1}\left\{ \int ^{\rho (y)^{2}}_{0}\frac{1}{s}\left( \frac{\sqrt{s}}{\rho (y)}\right) ^{\delta _{0}}ds+\int ^{t/2}_{\rho (y)^{2}} \frac{1}{s}\left( \frac{\sqrt{s}}{\rho (y)}\right) ^{l_{0}-N}{\mathrm{d}}s\right\} \\& \le C|u|t^{-n/2-1} \\& \le Ct^{-(n+1)/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}, \end{aligned}$$

where N is chosen large enough satisfying $N>l_{0}$.

To estimate $J_{2}$, we use Lemma 5 and write $J_{2}\le C(J_{2,1}+J_{2,2}+J_{2,3})$, where

$$\begin{aligned} \left\{ \begin{aligned} J_{2,1}&:=t^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{|u|^{2}}_{0}\int _{{\mathbb {R}}^{n}}\left| \nabla _{x}K_{s}(w-(x+u))-\nabla _{x}K_{s} (w-x)\right| V(w){\mathrm{d}}w{\mathrm{d}}s;\\ J_{2,2}&:=t^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{|u|^{2}}\int _{|w-x|<2|u|}\left| \nabla _{x}K_{s}(w-(x+u))-\nabla _{x}K_{s} (w-x)\right| V(w){\mathrm{d}}w{\mathrm{d}}s;\\ J_{2,3}&:=t^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{|u|^{2}}\int _{|w-x|\ge 2|u|}\left| \nabla _{x}K_{s}(w-(x+u))-\nabla _{x}K_{s} (w-x)\right| V(w){\mathrm{d}}w{\mathrm{d}}s. \end{aligned}\right. \end{aligned}$$

Notice that $\rho (x+u)\sim \rho (x)$ as $|u|\le \rho (x)$. It holds

$$\begin{aligned} J_{2,1}& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{|u|^{2}}_{0}\frac{1}{s^{3/2}}\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _{0}}{\mathrm{d}}s \\& \le Ct^{-n/2}|u|^{-1}\left( \frac{|u|}{\rho (y)}\right) ^{-1}\left( \frac{\sqrt{t}}{|u|}\right) ^{-1}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1} \left( \frac{|u|}{\rho (y)}\right) ^{\delta _{0}}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}} \\& \le Ct^{-(n+1)/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}}, \end{aligned}$$

where in the last inequality we have used the fact that $\delta _{0}=2-n/q$. By Lemmas 1 and 2, we apply (11) to get

$$\begin{aligned} J_{2,2}& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{|u|^{2}}\int _{|w-x|<2|u|}|u|s^{-n/2-1}V(w){\mathrm{d}}w{\mathrm{d}}s \\& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{|u|^{2}}|u|^{n-1}s^{-n/2-1}\left( \frac{|u|}{\rho (x)}\right) ^{\delta _{0}}{\mathrm{d}}s \\& \le Ct^{-n/2}|u|^{-1}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{0}}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}} \\& \le Ct^{-(n+1)/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

For $J_{2,3}$, if $t\le 2\rho (x)^{2}$, it can be deduced from Lemma 3 and (12) that

$$\begin{aligned} J_{2,3}& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{|u|^{2}}\int _{|w-x|\ge 2|u|}|u|s^{-n/2-1}e^{-c|w-x|^{2}/s}V(w){\mathrm{d}}w{\mathrm{d}}s \\& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}|u|\int ^{t/2}_{|u|^{2}}\frac{1}{s^{2}}\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _{0}}{\mathrm{d}}s \\& \le Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}|u|^{-1}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{0}} \left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}} \\& \le Ct^{-(n+1)/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

If $t>2\rho (x)^{2}$, then

$$\begin{aligned} J_{2,3}\le \, & {} Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{\rho (x)^{2}}_{|u|^{2}}\int _{|w-x|\ge 2|u|}\nonumber \\&\left| \nabla _{x}K_{s}(w-(x+u))-\nabla _{x}K_{s}(w-x)\right| V(w){\mathrm{d}}w{\mathrm{d}}s\nonumber \\&+ Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{\rho (x)^{2}}\int _{|w-x|\ge 2|u|}|u|s^{-n/2-1}e^{-c|w-x|^{2}/s}V(w){\mathrm{d}}w{\mathrm{d}}s\nonumber \\\le \, & {} Ct^{-(n+1)/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}}\nonumber \\&+Ct^{-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int ^{t/2}_{\rho (x)^{2}}\frac{|u|}{s}\frac{1}{s}\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{l_{0}} {\mathrm{d}}s \nonumber \\\le \, & {} Ct^{-(n+1)/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta _{1}}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+1}\left( \frac{\rho (y)}{\rho (x)}\right) ^{\delta _{0}}\nonumber \\&+Ct^{-(n+1)/2}\frac{|u|}{\rho (y)}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N-1+l_{0}}\left( \frac{\rho (y)}{\rho (x)}\right) ^{l_{0}}, \end{aligned}$$

(13)

where in (13) we have used the estimate obtained for $t\le 2\rho (x)^{2}$ and Lemma 3 for $s\ge \rho (x)^{2}$.

By Lemma 2,

$$\begin{aligned} \frac{\rho (y)}{\rho (x)}\le C\left( 1+\frac{|x-y|}{\sqrt{t}}\frac{\sqrt{t}}{\rho (y)}\right) ^{m_{0}}\le C_{\epsilon }e^{\epsilon |x-y|^{2}/t} \left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{m_{0}}, \end{aligned}$$

where $\epsilon >0$ is an arbitrary small constant. Choosing N large enough in the estimates of $J_{2,1},J_{2,2}$ and $J_{2,3}$, we obtain (9) and hence Lemma 10 is proved.

$\square$

We can obtain the following estimates, which generalize [4, Lemmas 3.7 and 3.8]. We also refer to [13, (57)] for the case $m=1$ in the setting of Heisenberg groups.

Proposition 1

• (i) There exist constants $C,c>0$ such that

  $$\begin{aligned} \left| t^{m}\partial _{t}^{m}K^{{\mathcal {L}}}_{t}(x,y)-t^{m}\partial _{t}^{m}K_{t}(x-y)\right| \le Ct^{-n/2}e^{-c|x-y|^{2}/t}\min \left\{ \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta _{0}}, \left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}\right\} . \end{aligned}$$

• (ii) Let $0<\delta <\min \{1,\delta _{0}\}$. For every $C'>0$ there exist constants C and c such that for every $z,x,y\in {\mathbb {R}}^{n}$, $|y-z|\le |x-y|/4$, $|y-z|\le C'\rho (x)$ we have

  $$\begin{aligned}&\left| \left( t^{m}\partial _{t}^{m}K^{{\mathcal {L}}}_{t}(x,y)-t^{m}\partial _{t}^{m}K_{t}(x-y)\right) -\left( t^{m}\partial _{t}^{m}K^{{\mathcal {L}}}_{t}(x,z)-t^{m}\partial _{t}^{m}K_{t}(x-z)\right) \right| \\&\quad \le C\left( \frac{|y-z|}{\rho (y)}\right) ^{\delta }t^{-n/2}e^{-c|x-y|^{2}/t}. \end{aligned}$$

Proof

For $t>0$ and $m\in {\mathbb {Z}}_{+}$, define $Q^{{\mathcal {L}}}_{t,m}(x,y):= t^{m}\partial _{t}^{m}K^{{\mathcal {L}}}_{t}(x,y)$ and $Q_{t,m}(x-y):= t^{m}\partial _{t}^{m}K_{t}(x-y).$ The proof of (i) is similar to [18, Lemmas 4.10 and 4.11], so we omit the details.

For (ii), by (7), we get

$$\begin{aligned}&\left( K_{t}(x+u-y)-K^{{\mathcal {L}}}_{t}(x+u,y)\right) -\left( K_{t}(x-y)-K_{t}^{{\mathcal {L}}}(x,y)\right) \\&= \int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left( K_{t-s}(w-(x+u))-K_{t-s}(w-x)\right) V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s\\&\quad + \int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left( K_{s}(w-(x+u))-K_{s}(w-x)\right) V(w)K^{{\mathcal {L}}}_{t-s}(w,y){\mathrm{d}}w{\mathrm{d}}s. \end{aligned}$$

Similar to [18, Proposition 4.8], we can use a direct calculus to deduce:

• (1) If m is even with $m\ge 2$, there exists a sequence of coefficients $\{C_{m,j}\}_{m\ge 2,2\le j\le m/2}$ such that

  $$\begin{aligned}&t^{m}\frac{{\mathrm{d}}^{m}}{{\mathrm{d}}t^{m}}\left\{ \left( K_{t}(x+u-y)-K^{{\mathcal {L}}}_{t}(x+u,y)\right) -\left( K_{t}(x-y)-K_{t}^{{\mathcal {L}}}(x,y)\right) \right\} \nonumber \\&\quad = \frac{m+1}{2}(E_{1}+E_{2}) +\sum ^{m/2}_{j=2}(C_{m-1,j-1}+C_{m-1,j})\left( E^{j}_{3,1}+E_{3,2}^{j}\right) +E_{4}+E_{5}.\nonumber \\ \end{aligned}$$

  (14)

• (2) If m is odd with $m\ge 3$, there exists a sequence of coefficients $\{C_{m,j}\}_{m\ge 3,2\le j\le [m/2]}$ such that

  $$\begin{aligned}&t^{m}\frac{{\mathrm{d}}^{m}}{{\mathrm{d}}t^{m}}\left\{ \left( K_{t}(x+u-y)-K^{{\mathcal {L}}}_{t}(x+u,y)\right) -\left( K_{t}(x-y)-K_{t}^{{\mathcal {L}}}(x,y)\right) \right\} \nonumber \\&\quad =\frac{m+1}{2}(E_{1}+E_{2})+ E_{4}-E_{5} +\sum ^{[m/2]}_{j=2}(C_{m-1,j-1}+C_{m-1,j})\left( E^{j}_{3,1}+E^{j}_{3,2}\right) \nonumber \\&\qquad +2C_{m,[m/2]}\frac{{\mathrm{d}}^{[m/2]}}{{\mathrm{d}}t^{[m/2]}}\left( K_{t/2}(w-(x+u))-K_{t/2}(w-x)\right) V(w)\frac{{\mathrm{d}}^{[m/2]}}{{\mathrm{d}}t^{[m/2]}}K^{{\mathcal {L}}}_{t/2}(w,y).\nonumber \\ \end{aligned}$$

  (15)

Here in the above (14) and (15),

$$\begin{aligned} \left\{ \begin{aligned} E_{1}:&=\int _{{\mathbb {R}}^{n}}t\left\{ t^{m-1}\frac{{\mathrm{d}}^{m-1}}{{\mathrm{d}}t^{m-1}}\left( K_{t/2}(w-x-u)-K_{t/2}(w-x)\right) \right\} V(w)K^{{\mathcal {L}}}_{t/2}(w,y){\mathrm{d}}w;\\ E_{2}:&=\int _{{\mathbb {R}}^{n}}t\left( K_{t/2}(w-x-u))-K_{t/2}(w-x)\right) V(w)\left( t^{m-1}\frac{{\mathrm{d}}^{m-1}}{{\mathrm{d}}t^{m-1}}K^{{\mathcal {L}}}_{t/2}(w,y)\right) {\mathrm{d}}w;\\ E_{3,1}^{j}:&=\int _{{\mathbb {R}}^{n}} t\left\{ t^{m-j}\frac{{\mathrm{d}}^{m-j}}{{\mathrm{d}}t^{m-j}}\left( K_{t/2}(w-x-u)-K_{t/2}(w-x)\right) \right\} V(w)\left( t^{j-1}\frac{{\mathrm{d}}^{j-1}}{{\mathrm{d}}t^{j-1}}K^{{\mathcal {L}}}_{t/2}(w,y)\right) {\mathrm{d}}w;\\ E_{3,2}^{j}:&=\int _{{\mathbb {R}}^{n}} t\left\{ t^{j-1}\frac{{\mathrm{d}}^{j-1}}{{\mathrm{d}}t^{j-1}}\left( K_{t/2}(w-x-u)-K_{t/2}(w-x)\right) \right\} V(w)\left( t^{m-j}\frac{{\mathrm{d}}^{m-j}}{{\mathrm{d}}t^{m-j}}K^{{\mathcal {L}}}_{t/2}(w,y)\right) {\mathrm{d}}w;\\ E_{4}:&=\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left( t^{m}\frac{{\mathrm{d}}^{m}}{{\mathrm{d}}t^{m}}(K_{t-s}(w-x-u)-K_{t-s}(w-x)\right) V(w)K^{{\mathcal {L}}}_{s}(w,y){\mathrm{d}}w{\mathrm{d}}s;\\ E_{5}:&=\int ^{t/2}_{0}\int _{{\mathbb {R}}^{n}}\left( K_{s}(w-x-u)-K_{s}(w,x)\right) V(w)\left( t^{m}\frac{{\mathrm{d}}^{m}}{{\mathrm{d}}t^{m}}K^{{\mathcal {L}}}_{t-s}(w,y)\right) {\mathrm{d}}w{\mathrm{d}}s. \end{aligned}\right. \end{aligned}$$

Below, for the sake of simplicity, we only estimate $E_{1}, E_{4}, E_{5}$. The estimations for $E_{2}$, $E_{3,1}^{j}$, $E^{j}_{3,2}$ are similar, and so we omit the details. By the mean value theorem, we know that there exist constants C, c such that

$$\begin{aligned} \left| Q_{t,m}(x+u)-Q_{t,m}(x)\right| \le \left\{ \begin{aligned}&C|u|t^{-(n+1)/2},&\quad \forall \ x, u \in {\mathbb {R}}^{n},\ t\in (0,\infty );\\&C|u|t^{-(n+1)/2}e^{-c|x|^{2}/t},&\quad |u|\le |x|/2,\ t>0. \end{aligned}\right. \end{aligned}$$

(16)

We divide $E_{1}$ as $E_{1}\le E_{1,1}+E_{1,2}$, where

$$\begin{aligned} \left\{ \begin{aligned} E_{1,1}&:=t\int _{|w-x|<2u}\left| Q_{t/2,m-1}(w-(x+u))-Q_{t/2,m-1}(w-x)\right| V(w)K^{{\mathcal {L}}}_{t/2}(w,y){\mathrm{d}}w;\\ E_{1,2}&:=t\int _{|w-x|\ge 2u}\left| Q_{t/2,m-1}(w-(x+u))-Q_{t/2,m-1}(w-x)\right| V(w)K^{{\mathcal {L}}}_{t/2}(w,y){\mathrm{d}}w. \end{aligned}\right. \end{aligned}$$

If $t<2\rho (y)^{2}$, for $E_{1,1}$, By (16), Lemma 3 (i) and Lemma 5, we obtain

$$\begin{aligned} E_{1,1} & \le Ct^{1-n/2}\int _{|w-x|<2|u|}|u|t^{-(n+1)/2}V(w)e^{-c|w-y|^{2}/t}{\mathrm{d}}w \\ &\le Ct^{1-n/2}|u|\frac{1}{t^{3/2}}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}} \\ &= Ct^{-n/2}\frac{|u|}{\sqrt{t}}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}}\le Ct^{-n/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta }. \end{aligned}$$

Similar to $E_{1,1}$, by (16) again, we get

$$\begin{aligned} E_{1,2} & \le Ct^{1-n/2}\int _{|w-x|\ge 2|u|}|u|t^{-(n+1)/2}e^{-c|w-y|^{2}/t}V(w){\mathrm{d}}w \\ & \le Ct^{1-n/2}\frac{|u|}{t^{3/2}}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{\delta _{0}} \\ & \le Ct^{-n/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta }. \end{aligned}$$

If $t>2\rho (y)^{2}$, for $E_{1,1}$, by (16), Lemma 3 (ii) and Lemma 5, we obtain

$$\begin{aligned} E_{1,1} & \le Ct^{1-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}\int _{|w-x|<2|u|}|u|t^{-(n+1)/2}V(w)e^{-c|w-y|^{2}/t}{\mathrm{d}}w \\ & \le Ct^{1-n/2}\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N}|u|\frac{1}{t^{3/2}}\left( \frac{\sqrt{t}}{\rho (y)}\right) ^{l_{0}} \\ & \le Ct^{-n/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta }\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+l_{0}}. \end{aligned}$$

Similar to $E_{1,1}$, we can also choose N large enough such that

$$\begin{aligned} E_{1,2}\le Ct^{-n/2}\left( \frac{|u|}{\rho (y)}\right) ^{\delta }\left( 1+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N+l_{0}}. \end{aligned}$$

For $E_{4}$ and $E_{5}$, similar to [13, Lemma 10], by Lemma 5 and (16), we can obtain

$$\begin{aligned} |E_{4}+E_{5}|\le C_{\epsilon }t^{-n/2}e^{\epsilon |x-y|^{2}/t}\left( \frac{|u|}{\rho (y)}\right) ^{\delta }, \end{aligned}$$

(17)

where $\epsilon >0$ is an arbitrary small constant. Hence Proposition 1 is proved. $\square$

3.2 Fractional heat kernels associated with ${\mathcal {L}}$

In the following, we will derive some regularity estimates for the fractional heat kernels related with ${\mathcal {L}}$. For $\alpha \in (0,1)$, the fractional power of ${\mathcal {L}}$, denoted by ${\mathcal {L}}^{\alpha }$, is defined as

$$\begin{aligned} {\mathcal {L}}^{\alpha }:=\frac{1}{\varGamma (-\alpha )}\int ^{\infty }_{0}\left( e^{-t\sqrt{{\mathcal {L}}}}f(x)-f(x)\right) \frac{{\mathrm{d}}t}{t^{1+2\alpha }}\quad \forall \ f\in L^{2}({\mathbb {R}}^{n}). \end{aligned}$$

(18)

We use the subordinative formula to express the integral kernel $K^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$ of $e^{-t{\mathcal {L}}^{\alpha }}$ as (cf. [10])

$$\begin{aligned} K^{{\mathcal {L}}}_{\alpha ,t}(x,y)=\int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)K^{{\mathcal {L}}}_{s}(x,y){\mathrm{d}}s, \end{aligned}$$

where $\eta ^{\alpha }_{t}(\cdot )$ satisfies

$$\begin{aligned} \left\{ \begin{aligned}&\eta ^{\alpha }_{t}(s)=1/t^{1/\alpha }\eta _{1}^{\alpha }(s/t^{1/\alpha });\\&\eta ^{\alpha }_{t}(s)\le t/s^{1+\alpha }\quad \forall \ s,t>0;\\&\int ^{\infty }_{0}s^{-r}\eta ^{\alpha }_{1}(s){\mathrm{d}}s<\infty , \ \ r>0;\\&\eta ^{\alpha }_{t}(s)\simeq t/s^{1+\alpha }\ \ \forall \ s\ge t^{1/\alpha }>0. \end{aligned}\right. \end{aligned}$$

(19)

By the subordinative formula (4) and Lemma 5, Li et al. [12] proved the following estimates for $K^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$.

Proposition 2

[12, Propositions 3.1 and 3.2] Let $0<\alpha <1$.

• (i) For any $N>0$, there exists a constant $C_{N}>0$ such that

  $$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right| \le \frac{C_{N}t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}\left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) Let $0<\delta \le \min \{1,\delta _{0}\}$. For any $N>0$, there exists a constant $C_{N}>0$ such that for all $|h|\le \sqrt{t^{1/\alpha }}$,

  $$\begin{aligned}&\left| K^{{\mathcal {L}}}_{\alpha ,t}(x+h,y)-K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right| \nonumber \\&\quad \le C_{N}\left( \frac{|h|}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }\frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x}+ \frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

For the kernels ${\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$ and $D^{{\mathcal {L}},m}_{\alpha ,t}(\cdot ,\cdot ), m\in {\mathbb {Z}}_{+}, t>0$, defined by (5), the following regularity estimates were obtained by Li et al. [12].

Proposition 3

[12, Proposition 3.3] Let $0<\alpha <1$.

• (i) For every N, there is a constant $C_{N}>0$ such that

  $$\begin{aligned} \left| D_{\alpha ,t}^{{\mathcal {L}},m}(x,y)\right| \le \frac{C_{N}t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}\left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) Let $0<\delta <\min \{2\alpha ,\delta _{0},1\}$. For every $N>0$, there exists a constant $C_{N}>0$ such that, for all $|h|<\sqrt{t^{1/\alpha }}$,

  $$\begin{aligned}&\left| D^{{\mathcal {L}},m}_{\alpha ,t}(x+h,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\right| \nonumber \\&\quad \le C_{N}\left( \frac{h}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }\frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (iii) There exists a constant $C_{N}>0$ such that

  $$\begin{aligned} \left| \int _{{\mathbb {R}}^{n}}D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)dy\right| \le \frac{C_{N}(\sqrt{t^{1/\alpha }}/\rho (x))^{\delta }}{(1+\sqrt{t^{1/\alpha }}/\rho (x))^{N}}. \end{aligned}$$

Proposition 4

[12, Propositions 3.6, 3.9 and 3.10] Suppose that and $V\in B_{q}$ for some $q>n$.

• (i) Let $\alpha \in (0,1)$. For every N, there is a constant $C_{N}>0$ such that

  $$\begin{aligned} \left| {\widetilde{D}}_{\alpha ,t}^{{\mathcal {L}}}(x,y)\right| \le \frac{C_{N}t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}\left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) Let $\alpha \in (0,1)$ and $\delta _{1}=1-n/q$. For every $N>0$, there exists a constant $C_{N}>0$ such that for all $|h|<\sqrt{t^{1/\alpha }}$,

  $$\begin{aligned}&\left| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x+h,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right| \nonumber \\&\quad \le C_{N}\left( \frac{h}{\sqrt{t^{1/\alpha }}}\right) ^{\delta _{1}}\frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (iii) Let $\alpha \in (0, 1/2-n/2q)$. There exists a constant $C_{N}>0$ such that

  $$\begin{aligned} \left| \int _{{\mathbb {R}}^{n}}{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y){\mathrm{d}}y\right| \le C_{N}\min \left\{ \left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{1+2\alpha },\ \left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{-N}\right\} . \end{aligned}$$

To establish the BMO$^{\gamma }_{L}$-boundedness of operators via T1 type theorem, we need the following propositions.

Proposition 5

There exists a constant $C>0$ such that

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right| \le \left\{ \begin{aligned}&C\left( \frac{|x-y|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\le |x-y|;\\&C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\ge |x-y|. \end{aligned}\right. \end{aligned}$$

Proof

By the subordinative formula (4) and Lemma 6, we obtain

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right|\le & {} \int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)\left| K^{{\mathcal {L}}}_{s}(x,y)-K_{s}(x-y)\right| {\mathrm{d}}s \\\le & {} C\int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _{0}}s^{-n/2}e^{-|x-y|^{2}/s}{\mathrm{d}}s. \end{aligned}$$

On the one hand, letting $s=t^{1/\alpha }u$, we can get

$$\begin{aligned}&\left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right| \nonumber \\&\quad \le C\int ^{\infty }_{0}\frac{t}{s^{1+\alpha }}\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _{0}}s^{-n/2}e^{-|x-y|^{2}/s}{\mathrm{d}}s \\&\quad \le C\int ^{\infty }_{0}\frac{t}{(t^{1/\alpha }u)^{1+\alpha }}\left( \frac{\sqrt{t^{1/\alpha }u}}{\rho (x)}\right) ^{\delta _{0}}(t^{1/\alpha }u)^{-n/2} e^{-c|x-y|^{2}/(t^{1/\alpha }u)}t^{1/\alpha }{\mathrm{d}}u \\&\quad \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}t^{-{n}/{(2\alpha )}}\int ^{\infty }_{0}u^{-1-\alpha -n/2+\delta _{0}/2}e^{-|x-y|^{2}/(t^{1/\alpha }u)} {\mathrm{d}}u. \end{aligned}$$

Applying the change of variables: ${|x-y|^{2}}/{(t^{1/\alpha }u)}=r^{2}$, we deduce that

$$\begin{aligned}&\left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right| \nonumber \\&\quad \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}t^{-n/(2\alpha )}\int ^{\infty }_{0} \left( \frac{|x-y|^{2}}{t^{1/\alpha }r^{2}}\right) ^{-1-\alpha +\delta _{0}/2-n/2}e^{-r^{2}}\frac{|x-y|^{2}}{t^{1/\alpha }r^{3}}{\mathrm{d}}r \\&\quad \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}|x-y|^{-2\alpha +\delta _{0}-n}t^{1-\delta _{0}/(2\alpha )}\int ^{\infty }_{0} e^{-r^{2}}r^{2\alpha -\delta _{0}+n+1}{\mathrm{d}}r \\&\quad \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t^{1-\delta _{0}/(2\alpha )}}{|x-y|^{2\alpha +n-\delta _{0}}}. \end{aligned}$$

On the other hand, taking $\tau =s/t^{1/\alpha }$, we obtain

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right|\le & {} C\int ^{\infty }_{0}\left( \frac{\sqrt{s}}{\rho (x)}\right) ^{\delta _{0}}s^{-n/2}t^{-1/\alpha }\eta ^{\alpha }_{1}(s/t^{1/\alpha }){\mathrm{d}}s \\\le & {} C\int ^{\infty }_{0}\left( \frac{\sqrt{t^{1/\alpha }\tau }}{\rho (x)}\right) ^{\delta _{0}}(t^{1/\alpha }\tau )^{-n/2}\eta ^{\alpha }_{1}(\tau ){\mathrm{d}}\tau \\\le & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}t^{-n/2\alpha }\int ^{\infty }_{0}\eta ^{\alpha }_{1}(\tau )\tau ^{\delta _{0}/2-n/2}{\mathrm{d}}\tau \\\le & {} C \left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}t^{-n/2\alpha }. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\le |x-y|$, then

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right|\le & {} C\left( \frac{|x-y|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{|x-y|^{2\alpha +n}} \\\le & {} C\left( \frac{|x-y|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{2\alpha +n}}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}>|x-y|$, we can see that

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right|\le & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{t^{n/(2\alpha )+1}} \\\le & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{2\alpha +n}}. \end{aligned}$$

$\square$

Let ${\widetilde{D}}_{\alpha ,t}(\cdot )=t^{1/(2\alpha )}\nabla _{x}e^{-t(-\varDelta )^{\alpha }}(\cdot ).$ Similar to the proof of Proposition 5, by (4) and Lemma 8, we have

Proposition 6

There exists a constant $C>0$ such that

$$\begin{aligned} \left| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}_{\alpha ,t}(x-y)\right| \le \left\{ \begin{aligned}&C\left( \frac{|x-y|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\le |x-y|;\\&C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\ge |x-y|. \end{aligned}\right. \end{aligned}$$

Let $D^{m}_{\alpha ,t}(\cdot )=t^{m}\partial ^{m}_{t}e^{-t(-\varDelta )^{\alpha }}(\cdot ).$ We have

Proposition 7

There exists a constant $C>0$ such that

$$\begin{aligned} \left| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{m}_{\alpha ,t}(x-y)\right| \le \left\{ \begin{aligned}&C\left( \frac{|x-y|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\le |x-y|;\\&C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(|x-y|+\sqrt{t^{1/\alpha }})^{n+2\alpha }},\ \ \ \sqrt{t^{1/\alpha }}\ge |x-y|. \end{aligned}\right. \end{aligned}$$

Proof

The proposition can be proved by Proposition 1 and (4). Since the argument is similar to that of Proposition 5, the details is omitted. $\square$

Proposition 8

Let $0<\delta <\min \{2\alpha ,\delta _{0}\}$. For every $C'>0$ there exists a constant C such that for every $z,x,y\in {\mathbb {R}}^{n}$, $|y-z|\le |x-y|/4$, $|y-z|\le C'\rho (y)$ we have

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \nonumber \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta } \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}. \end{aligned}$$

Proof

By the subordinative formula (4), we can use Lemma 9 to deduce

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le \int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)\left| \left( K^{{\mathcal {L}}}_{s}(x,y)-K_{s}(x-y)\right) -\left( K^{{\mathcal {L}}}_{s}(x,z)-K_{s}(x-z)\right) \right| {\mathrm{d}}s \\&\quad \le C\int ^{\infty }_{0}\eta ^{\alpha }_{t}(s)\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }s^{-n/2}e^{-|x-y|^{2}/s}{\mathrm{d}}s. \end{aligned}$$

On the one hand, taking $s=t^{1/\alpha }u$, we obtain

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le \int ^{\infty }_{0}\frac{Ct}{s^{1+\alpha }}\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }s^{-n/2}e^{-|x-y|^{2}/s}{\mathrm{d}}s \\&\quad \le C\int ^{\infty }_{0}\frac{t}{(t^{1/\alpha }u)^{1+\alpha }}\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }(t^{1/\alpha }u)^{-n/2} e^{-|x-y|^{2}/(t^{1/\alpha }u)}t^{1/\alpha }{\mathrm{d}}u \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }t^{-n/(2\alpha )}\int ^{\infty }_{0}e^{-|x-y|^{2}/(t^{1/\alpha }u)}u^{-1-\alpha -n/2}{\mathrm{d}}u. \end{aligned}$$

Let $\displaystyle \frac{|x-y|^{2}}{(t^{1/\alpha }u)}=r$. We can see that

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }t^{-n/(2\alpha )}\int ^{\infty }_{0}e^{-r}\left( \frac{|x-y|^{2}}{t^{1/\alpha }r}\right) ^{-1-\alpha -n/2} \frac{|x-y|^{2}}{t^{1/\alpha }r^{2}}{\mathrm{d}}r \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }\frac{t}{|x-y|^{2\alpha +n}}. \end{aligned}$$

On the other hand, letting $\displaystyle \frac{s}{t^{1/\alpha }}=\tau$, we obtain

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le \int ^{\infty }_{0}\frac{1}{t^{1/\alpha }}\eta ^{\alpha }_{1}(s/t^{1/\alpha })\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }s^{-n/2}{\mathrm{d}}s \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }\int ^{\infty }_{0}\eta ^{\alpha }_{1}(\tau )(t^{1/\alpha }\tau )^{-n/2}{\mathrm{d}}\tau \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }t^{-n/(2\alpha )}. \end{aligned}$$

Case 1: $\sqrt{t^{1/\alpha }}\le |x-y|$. We obtain

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta } \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}. \end{aligned}$$

Case 2: $\sqrt{t^{1/\alpha }}>|x-y|$. We can see that

$$\begin{aligned}&\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K_{\alpha ,t}(x-y)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(x,z)-K_{\alpha ,t}(x-z)\right) \right| \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta } \frac{t}{t^{n/(2\alpha )+1}}\le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta }\frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}. \end{aligned}$$

$\square$

Proposition 9

Let $0<\delta <\min \{2\alpha ,\delta _{0}\}$. For every $C'>0$ there exists a constant C such that for every $z,x,y\in {\mathbb {R}}^{n}$, $|y-z|\le |x-y|/4$, $|y-z|\le C'\rho (y)$ we have

$$\begin{aligned}&\left| \left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{m}_{\alpha ,t}(x-y)\right) -\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)-D^{m}_{\alpha ,t}(x-z)\right) \right| \nonumber \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta } \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}. \end{aligned}$$

Proof

The proof is similar to that of Proposition 8, so we omit the details. $\square$

Proposition 10

Suppose that $V\in B_{q}$ for some $q>n$. Let $0<\delta '<\delta _{1}:=1-n/q$. For every $C'>0$ there exists a constant C such that for every $z,x,y\in {\mathbb {R}}^{n}$, $|y-z|\le |x-y|/4$, $|y-z|\le C'\rho (y)$ we have

$$\begin{aligned}&\left| \left( {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}_{\alpha ,t}(x-y)\right) - \left( {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)-{\widetilde{D}}_{\alpha ,t}(x-z)\right) \right| \nonumber \\&\quad \le C\left( \frac{|y-z|}{\rho (x)}\right) ^{\delta '} \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}. \end{aligned}$$

Proof

The proof is similar to that of Proposition 8, so we omit the details. $\square$

4 BMO$^{\gamma }_{{\mathcal {L}}}$-boundedness via T1 theorem

4.1 Maximal operators for fractional heat semigroups

Definition 4

Let $0<\gamma \le 1$. The Campanato type space BMO$^{\gamma }_{{\mathcal {L}},L^{\infty }((0,\infty ),{\mathrm{d}}t)}({\mathbb {R}}^{n})$ is defined as the set of all locally integrable functions f satisfying

$$\begin{aligned} \Vert f\Vert _{{\mathrm{BMO}}^{\gamma }_{{\mathcal {L}},L^{\infty }((0,\infty ),{\mathrm{d}}t)}}:=\sup _{B\subset {\mathbb {R}}^{n}} \left\{ \frac{1}{|B|^{1+\gamma /n}}\int _{B}\Vert f(x,t)-f(B,V)\Vert _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}x\right\} <\infty . \end{aligned}$$

To prove that $M^{\alpha }_{{\mathcal {L}}}$ is bounded from BMO$_{{\mathcal {L}}}^{\gamma }({\mathbb {R}}^{n})$, $0<\gamma <\min \{2\alpha ,\delta _{0},1 \}$, into itself, we give a vector-valued interpretation of the operator and apply Lemma 4. Indeed, it is clear that $M^{\alpha }_{{\mathcal {L}}}f=\Vert e^{-t{\mathcal {L}}^{\alpha }}f\Vert _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}$. Hence, it is enough to show that the operator $\varLambda (f):=\{e^{-t{\mathcal {L}}^{\alpha }}f\}_{t>0}$ is bounded from BMO$^{\gamma }_{{\mathcal {L}}}$ into BMO$_{{\mathcal {L}},L^{\infty }((0,\infty ),{\mathrm{d}}t)}^{\gamma }$.

By the spectral theorem, $\varLambda$ is bounded from $L^{2}({\mathbb {R}}^{n})$ into $L^{2}_{L^{\infty }((0,\infty ),{\mathrm{d}}t)}({\mathbb {R}}^{n})$. The desired result can be then deduced from the following theorem.

Theorem 3

Assume that the potential $V\in B_{q}$ with $q>n/2$. Let $x,y,z\in {\mathbb {R}}^{n}$.

• (i) For any $N>0$, there exists a constant $C_{N}$ such that

  $$\begin{aligned} \left\| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}\le \frac{C_{N}}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) For $|x-y|>2|y-z|$ and any $0<\delta <\min \{2\alpha ,\delta _{0}\}$, there exists a constant $C>0$ such that

  $$\begin{aligned}&\left\| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K^{{\mathcal {L}}}_{\alpha ,t}(x,z)\left\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}+ \right\| K^{{\mathcal {L}}}_{\alpha ,t}(y,x)-K^{{\mathcal {L}}}_{\alpha ,t}(z,x)\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}\nonumber \\&\quad \le \frac{C|y-z|^{\delta }}{|x-y|^{n+\delta }}. \end{aligned}$$

  (20)

• (iii) There exists a constant C such that for every ball $B=B(x,r)$ with $0<r\le \rho (x)/2$,

  $$\begin{aligned} \log \left( \frac{\rho (x)}{r}\right) \frac{1}{|B|}\int _{B}\left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-(e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}y\le C, \end{aligned}$$

  and, if $\gamma <\min \{2\alpha ,1,\delta _{0}\}$ then

  $$\begin{aligned} \left( \frac{\rho (x)}{r}\right) ^{\gamma }\frac{1}{|B|}\int _{B}\left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-(e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}y\le C. \end{aligned}$$

Proof

For (i), from (i) of Proposition 2, we can get

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right| \le C_{N} \min \left\{ \frac{t^{1+N/\alpha }}{|x-y|^{n+2\alpha +2N}},\ t^{-n/(2\alpha )}\right\} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}>|x-y|$, then

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right|\le & {} \frac{C_{N}}{(\sqrt{t^{1/\alpha }})^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N} \\\le & {} \frac{C}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\le |x-y|$, we obtain

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right|\le & {} \frac{C_{N}t^{1+N/\alpha }}{|x-y|^{n+2\alpha +2N}}\left( \frac{\sqrt{t^{1/\alpha }}}{|x-y|}\right) ^{-N} \left( \frac{|x-y|}{\sqrt{t^{1/\alpha }}}+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N} \\\le & {} \frac{C_{N}(\sqrt{t^{1/\alpha }})^{2\alpha +N}}{|x-y|^{n+2\alpha +N}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N} \\\le & {} \frac{C_{N}}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N}. \end{aligned}$$

For (ii), from (ii) of Proposition 2, we obtain

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right| \le C_{N}\min \left\{ \frac{t^{1+N/\alpha }|y-z|^{\delta }}{|x-y|^{2\alpha +n+2N+\delta }},\ t^{-n/(2\alpha )}\right\} \left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}>|x-y|$, then

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right|\le & {} \frac{C}{|x-y|^{n}}\left( \frac{|y-z|}{|x-y|}\right) ^{\delta } \le \frac{C|y-z|^{\delta }}{|x-y|^{n+\delta }}. \end{aligned}$$

If $|x-y|>\sqrt{t^{1/\alpha }}$, we can also get

$$\begin{aligned} \left| K^{{\mathcal {L}}}_{\alpha ,t}(x,y)-K^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right|\le & {} \frac{C|x-y|^{2\alpha +2N}}{|x-y|^{2\alpha +n+2N+\delta }}|y-z|^{\delta } \le \frac{C|y-z|^{\delta }}{|x-y|^{n+\delta }}. \end{aligned}$$

The symmetry of the kernel $K^{L}_{\alpha ,t}(\cdot ,\cdot )$ gives the conclusion of (ii).

For (iii), letting $B=B(x,r)$ with $0<r\le \rho (x)/2$, the triangle inequality gives

$$\begin{aligned} \left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-(e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}\le \frac{1}{|B|}\int _{B}\left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}z. \end{aligned}$$

We estimate $\Vert e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\Vert _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}$. Because $y,z\in B$, $\rho (y)\sim \rho (z)\sim \rho (x).$ By Proposition 5, we split $|e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)|\le S_{1}+S_{2}$, where

$$\begin{aligned} \left\{ \begin{aligned} S_{1}:= \int _{{\mathbb {R}}^{n}}\left| K^{{\mathcal {L}}}_{\alpha ,t}(y,w)-K_{\alpha ,t}(y,w)\right| {\mathrm{d}}w;\\ S_{2}:=\int _{{\mathbb {R}}^{n}}\left| K^{{\mathcal {L}}}_{\alpha ,t}(z,w)-K_{\alpha ,t}(z,w)\right| {\mathrm{d}}w. \end{aligned}\right. \end{aligned}$$

For $S_{1}$, if $|y-w|\le \sqrt{t^{1/\alpha }}$, we obtain

$$\begin{aligned} S_{1} \le C\int _{{\mathbb {R}}^{n}}\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(\sqrt{t^{1/\alpha }}+|y-w|)^{n+2\alpha }}{\mathrm{d}}w \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

If $|y-w|>\sqrt{t^{1/\alpha }}$, we can see that

$$\begin{aligned} S_{1}& \le C\int _{{\mathbb {R}}^{n}}\left( \frac{|y-w|}{\rho (x)}\right) ^{\delta _{0}}\frac{t}{(\sqrt{t^{1/\alpha }}+|y-w|)^{n+2\alpha }}{\mathrm{d}}w \\& \le \frac{C}{\rho (x)^{\delta _{0}}}\int ^{\infty }_{0}\frac{|y-w|^{\delta _{0}+n-1}t}{(|y-w|+\sqrt{t^{1/\alpha }})^{n+2\alpha }}{\mathrm{d}}|y-w| \\& \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\int ^{\infty }_{0}u^{\delta _{0}+n-1}(1+u)^{-n-2\alpha }{\mathrm{d}}u \\& \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}, \end{aligned}$$

where we have chosen $\delta _{0}<2\alpha$ since $\delta _{0}=2-n/q$, $q>n/2$. The proof of the term $S_{2}$ is similar to that of the term $S_{1}$, so we omit it. Then we can get

$$\begin{aligned} \left| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right| \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}, \end{aligned}$$

which shows that if $\sqrt{t^{1/\alpha }}\le 2r$,

$$\begin{aligned} \left| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right| \le C\left( \frac{r}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}>2r$, then $|y-z|\le 2r<\sqrt{t^{1/\alpha }}$. Hence, Proposition 2 implies that for $0<\delta <\delta _{0}$,

$$\begin{aligned}&\left| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right| \le \int _{{\mathbb {R}}^{n}}\left| K^{{\mathcal {L}}}_{\alpha ,t}(y,w)-K^{{\mathcal {L}}}_{\alpha ,t}(z,w)\right| {\mathrm{d}}w \nonumber \\&\quad \le C\int _{{\mathbb {R}}^{n}}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }\frac{t}{(\sqrt{t^{1/\alpha }}+|y-w|)^{n+2\alpha }}{\mathrm{d}}w \le C\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }\le C \left( \frac{r}{\sqrt{t^{1/\alpha }}}\right) ^{\delta }. \end{aligned}$$

(21)

Therefore, if $\sqrt{t^{1/\alpha }}>\rho (x)$, (21) gives

$$\begin{aligned} |e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)|\le C\left( \frac{r}{\rho (x)}\right) ^{\delta }. \end{aligned}$$

When $2r<\sqrt{t^{1/\alpha }}<\rho (x)$, we have $|e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)| = {\mathrm{I}}+{\mathrm{II}}+{\mathrm{III}},$ where

$$\begin{aligned} \left\{ \begin{aligned} {\mathrm{I}}&:=\int _{|w-y|>c\rho (y)>4|y-z|}\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(y,w)-K_{\alpha ,t}(y,w)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(z,w)-K_{\alpha ,t}(z,w)\right) \right| {\mathrm{d}}w;\\ {\mathrm{II}}&:=\int _{4|y-z|<|w-y|<c\rho (y)}\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(y,w)-K_{\alpha ,t}(y,w)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(z,w)-K_{\alpha ,t}(z,w)\right) \right| {\mathrm{d}}w;\\ {\mathrm{III}}&:= \int _{|w-y|<4|y-z|}\left| \left( K^{{\mathcal {L}}}_{\alpha ,t}(y,w)-K_{\alpha ,t}(y,w)\right) -\left( K^{{\mathcal {L}}}_{\alpha ,t}(z,w)-K_{\alpha ,t}(z,w)\right) \right| {\mathrm{d}}w. \end{aligned}\right. \end{aligned}$$

Notice that the estimate (20) is valid for the classical fractional heat kernel. For I, by (20), we can get

$$\begin{aligned} {\mathrm{I}}\le C\int _{|w-y|>c\rho (y)>4|y-z|}\frac{|y-z|^{\delta }}{|w-y|^{n+\delta }}{\mathrm{d}}w\le C\left( \frac{r}{\rho (x)}\right) ^{\delta }. \end{aligned}$$

For II, we apply Proposition 8 and the fact that $\rho (w)\sim \rho (y)$ in the region of integration to deduce that

$$\begin{aligned} {\mathrm{II}} \le C|y-z|^{\delta }\int _{4|y-z|<|w-y|<c\rho (y)}\frac{tdw}{\rho (w)^{\delta }(\sqrt{t^{1/\alpha }}+|w-y|)^{n+2\alpha }} \le C\left( \frac{r}{\rho (x)}\right) ^{\delta }. \end{aligned}$$

For III, since $|y-z|\le 2r<\sqrt{t^{1/\alpha }},$ we have $|w-y|<C\sqrt{t^{1/\alpha }}$. For $n-\delta _{0}>0$, by Proposition 5, we obtain

$$\begin{aligned} {\mathrm{III}}\le \, & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\left( \int _{|w-y|<4|y-z|}\frac{t{\mathrm{d}}w}{(\sqrt{t^{1/\alpha }}+|w-y|)^{n+2\alpha }}\right. \nonumber \\&\left. + \int _{|w-z|\le 5|y-z|}\frac{t{\mathrm{d}}w}{(\sqrt{t^{1/\alpha }}+|w-z|)^{n+2\alpha }}\right) \\\le \, & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\int ^{5|y-z|/\sqrt{t^{1/\alpha }}}_{0}u^{n-1}{\mathrm{d}}u \le C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x)}\right) ^{\delta _{0}}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{n} \\\le \, & {} \frac{Cr^{n}}{\rho (x)^{\delta _{0}}(\sqrt{t^{1/\alpha }})^{n-\delta _{0}}}\le \frac{Cr^{n}}{\rho (x)^{\delta _{0}}r^{n-\delta _{0}}}=C\left( \frac{r}{\rho (x)}\right) ^{\delta _{0}}. \end{aligned}$$

Thus, when $2r<\sqrt{t^{1/\alpha }}<\rho (x)$,

$$\begin{aligned} \left| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right| \le C\left( \frac{r}{\rho (x)}\right) ^{\delta }. \end{aligned}$$

Combining the above estimates, we can get

$$\begin{aligned} \left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}\le C\left( \frac{r}{\rho (x)}\right) ^{\delta }. \end{aligned}$$

(22)

Therefore, it holds

$$\begin{aligned}&\log \left( \frac{\rho (x)}{r}\right) \frac{1}{|B|}\int _{B}\left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-(e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}y\nonumber \\&\quad \le C\left( \frac{r}{\rho (x)}\right) ^{\delta }\log \left( \frac{\rho (x)}{r}\right) \le C, \end{aligned}$$

which is the first conclusion of (iii).

For the second estimate of (iii), take $\delta \in (\gamma ,\ \min \{2\alpha ,\ 2-n/q\})$. By (22), we have

$$\begin{aligned} \left( \frac{\rho (x)}{r}\right) ^{\gamma }\frac{1}{|B|}\int _{B}\left\| e^{-t{\mathcal {L}}^{\alpha }}1(y)-(e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{\infty }((0,\infty ),{\mathrm{d}}t)}{\mathrm{d}}y\le C\left( \frac{r}{\rho (x)}\right) ^{\delta -\gamma }\le C. \end{aligned}$$

$\square$

4.2 Boundedness of the Littlewood–Paley g-function $g^{{\mathcal {L}}}_{\alpha }$

Similar to Sect. 4.1, we introduce the following function space:

Definition 5

Let $0<\gamma \le 1$. The Campanato type space BMO$^{\gamma }_{ {\mathcal {L}},L^{2}((0,\infty ),{\mathrm{d}}t/t)}({\mathbb {R}}^{n})$ is defined as the set of all locally integrable functions f satisfying

$$\begin{aligned} \Vert f\Vert _{{\mathrm{BMO}}^{\gamma }_{{\mathcal {L}},L^{2}((0,\infty ),{\mathrm{d}}t)}}:=\sup _{B\subset {\mathbb {R}}^{n}} \left\{ \frac{1}{|B|^{1+\gamma /n}}\int _{B}\Vert f(x,t)-f(B,V)\Vert _{L^{2}((0,\infty ),{\mathrm{d}}t/t)}{\mathrm{d}}x\right\} <\infty . \end{aligned}$$

The functional calculus and the spectral theorem imply that $g^{{\mathcal {L}}}_{\alpha }$ is an isometry on $L^{2}({\mathbb {R}}^{n})$. As before, to get the boundedness of $g^{{\mathcal {L}}}_{\alpha }$ on BMO$^{\gamma }_{{\mathcal {L}}}({\mathbb {R}}^{n})$, $0<\gamma <\min \{2\alpha ,\delta _{0},1\}$, it is sufficient to prove the following result.

Theorem 4

Assume that the potential $V\in B_{q}$ with $q>n/2$. Let $x,y,z\in {\mathbb {R}}^{n}$ and $N>0$.

• (i) For any $N>0$, there exists a constant $C_{N}$ such that

  $$\begin{aligned} \left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\right\| _{L^{2}((0,\infty ),\frac{{\mathrm{d}}t}{t})}\le \frac{C_{N}}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N}. \end{aligned}$$

• (ii) If $|x-y|>2|y-z|$ and $0<\delta <\min \{2\alpha ,\delta _{0},1\}$, there exists a constant C such that

  $$\begin{aligned}&\left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right\| _{L^{2}((0,\infty ),\frac{{\mathrm{d}}t}{t})}+ \left\| D^{{\mathcal {L}},m}_{\alpha ,t}(y,x)-D^{{\mathcal {L}},m}_{\alpha ,t}(z,x)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\nonumber \\&\quad \le {{C}}\frac{|y-z|^{\delta }}{|x-y|^{n+\delta }}. \end{aligned}$$

  (23)

• (iii) There exists a constant C such that for every ball $B=B(x_{0},r)$ with $0<r\le \rho (x)/2$,

  $$\begin{aligned} \log \left( \frac{\rho (x)}{r}\right) \frac{1}{|B|}\int _{B}\left\| t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(y)- (t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }{\mathrm{d}}y\le C, \end{aligned}$$

  and, if $\gamma <\min \{2\alpha ,\delta _{0},1\}$, then

  $$\begin{aligned} \left( \frac{\rho (x)}{r}\right) ^{\gamma }\frac{1}{|B|}\int _{B}\left\| t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(y)-(t^{m}\partial _{t}^{m} e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }{\mathrm{d}}y\le C. \end{aligned}$$

Proof

For (i), from Proposition 3, we have

$$\begin{aligned} \left| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\right| \le C_{N}\min \left\{ \frac{t^{1+N/\alpha }}{|x-y|^{n+2\alpha +2N}},\ t^{-n/(2\alpha )} \right\} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\le |x-y|$, we obtain

$$\begin{aligned}&\left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\le |x-y|\}}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }^{2} \\&\quad \le C_{N}\int ^{|x-y|^{2\alpha }}_{0}\frac{t^{2+2N/\alpha }}{|x-y|^{2n+4\alpha +4N}} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-2N}\frac{{\mathrm{d}}t}{t} \\&\quad \le C_{N}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\int ^{|x-y|^{2\alpha }}_{0} \frac{(\sqrt{t^{1/\alpha }})^{4\alpha +4N}}{|x-y|^{2n+4\alpha +4N}}\left( \frac{\sqrt{t^{1/\alpha }}}{|x-y|}\right) ^{-2N}\frac{{\mathrm{d}}t}{t} \\&\quad \le \left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\frac{C_{N}}{|x-y|^{2n}}\int ^{|x-y|^{2\alpha }}_{0} \left( \frac{\sqrt{t^{1/\alpha }}}{|x-y|}\right) ^{4\alpha +2N}\frac{{\mathrm{d}}t}{t}. \end{aligned}$$

Let ${\sqrt{t^{1/\alpha }}}/{|x-y|}=u$. We can see that

$$\begin{aligned}&\left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\le |x-y|\}}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }^{2}\nonumber \\&\quad \le \left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\frac{C_{N}}{|x-y|^{2n}}\int ^{1}_{0}u^{4\alpha +2N-1}{\mathrm{d}}u\\&\quad \le \frac{C_{N}}{|x-y|^{2n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\ge |x-y|$, we can get

$$\begin{aligned}&\left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\ge |x-y|\}}\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\\&\quad \le C_{N}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\int ^{\infty }_{|x-y|^{2\alpha }}t^{-n/\alpha -1}{\mathrm{d}}t \\&\quad \le \frac{C_{N}}{|x-y|^{2n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}. \end{aligned}$$

For (ii), by Proposition 3, we have

$$\begin{aligned} \left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le & {} \int ^{\infty }_{0}\frac{Ct}{(\sqrt{t^{1/\alpha }}+|x-y|)^{2n+4\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }{\mathrm{d}}t \\\le & {} C|y-z|^{2\delta }\int ^{\infty }_{0}\frac{(\sqrt{t^{1/\alpha }})^{2\alpha -2\delta }}{(\sqrt{t^{1/\alpha }}+|x-y|)^{2n+4\alpha }}{\mathrm{d}}t. \end{aligned}$$

Let ${\sqrt{t^{1/\alpha }}}/{|x-y|}=u$. We obtain

$$\begin{aligned} \left\| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le & {} C|y-z|^{2\delta }|x-y|^{-2n-2\delta }\int ^{\infty }_{0}\frac{u^{4\alpha -2\delta -1}}{(1+u)^{2n+4\alpha }}{\mathrm{d}}u \\\le & {} \frac{C|y-z|^{2\delta }}{|x-y|^{2n+2\delta }}. \end{aligned}$$

The symmetry of the kernel $D^{L,m}_{\alpha ,t}(\cdot ,\cdot )$ gives the conclusion of (ii).

For (iii), let us fix $y,z\in B=B(x_{0},r)$, $0<r\le \rho (x_{0})/2$. Similar to Theorem 3, we must handle

$$\begin{aligned} \left\| t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }. \end{aligned}$$

We can write

$$\begin{aligned} \left\| t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }= & {} M_{1}+M_{2}+M_{3}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{aligned} M_{1}&:=\int ^{(2r)^{2\alpha }}_{0}\left| \int _{{\mathbb {R}}^{n}}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t};\\ M_{2}&:=\int ^{\rho (x_{0})^{2\alpha }}_{(2r)^{2\alpha }}\left| \int _{{\mathbb {R}}^{n}}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t};\\ M_{3}&:=\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left| \int _{{\mathbb {R}}^{n}}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t}. \end{aligned}\right. \end{aligned}$$

Since $y,z\in B\subset B(x_{0},\rho (x_{0}))$, it follows that $\rho (y)\sim \rho (x_{0})\sim \rho (z)$. By Proposition 3 (iii),

$$\begin{aligned} M_{1}\le C\int ^{(2r)^{2\alpha }}_{0}\frac{(\sqrt{t^{1/\alpha }}/\rho (x_{0}))^{2\delta }}{(1+\sqrt{t^{1/\alpha }}/\rho (x_{0}))^{2N}}\frac{{\mathrm{d}}t}{t} \le C\int ^{(2r)^{2\alpha }}_{0}\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x_{0})}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}=C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }. \end{aligned}$$

Also, by Proposition 3(ii),

$$\begin{aligned} M_{3}& \le C\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }\left| \int _{{\mathbb {R}}^{n}} \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}{\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t} \\& \le C\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}\le C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }. \end{aligned}$$

It remains to estimate the term $M_{2}$. In this case, $|y-z|\le 2r\le \sqrt{t^{1/\alpha }}\le \rho (x_{0})$. Then we can use the methods in Theorem 3 to obtain

$$\begin{aligned} M_{2} =\int ^{\rho (x_{0})^{2\alpha }}_{(2r)^{2\alpha }}\left| M_{2,1}+M_{2,2}+M_{2,3}\right| ^{2}\frac{{\mathrm{d}}t}{t}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{aligned} M_{2,1}&:=\int _{|x-y|>c\rho (y)>4|y-z|}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{m}_{\alpha ,t}(x-y)\right) -\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)-D^{m}_{\alpha ,t}(x-z)\right) {\mathrm{d}}x;\\ M_{2,2}&:=\int _{4|y-z|<|x-y|<c\rho (y)}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{m}_{\alpha ,t}(x-y)\right) -\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)-D^{m}_{\alpha ,t}(x-z)\right) {\mathrm{d}}x;\\ M_{2,3}&:=\int _{|x-y|<4|y-z|}\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{m}_{\alpha ,t}(x-y)\right) -\left( D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)-D^{m}_{\alpha ,t}(x-z)\right) {\mathrm{d}}x. \end{aligned}\right. \end{aligned}$$

For $M_{2,1}$, similar to prove (23), we can also get

$$\begin{aligned} \left| D^{{\mathcal {L}},m}_{\alpha ,t}(x,y)-D^{{\mathcal {L}},m}_{\alpha ,t}(x,z)\right| \le \frac{C|y-z|^{\delta }}{|x-y|^{n+\delta }}, \end{aligned}$$

which is valid to $D^{m}_{\alpha ,t}(\cdot )$. So we obtain

$$\begin{aligned} |M_{2,1}|\le C\int _{|x-y|>c\rho (y)>4|y-z|}\frac{|y-z|^{\delta }}{|x-y|^{n+\delta }}{\mathrm{d}}x\le C\left( \frac{r}{\rho (x_{0})}\right) ^{\delta }. \end{aligned}$$

For $M_{2,2}$, by Proposition 9 and the fact that $\rho (x)\sim \rho (y)$ in the region of integration.

$$\begin{aligned} |M_{2,2}|\le & {} C|y-z|^{\delta }\int _{4|y-z|<|x-y|<c\rho (y)}\frac{t}{\rho (x)^{\delta }(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}{\mathrm{d}}x \le C\left( \frac{r}{\rho (x_{0})}\right) ^{\delta }. \end{aligned}$$

For $M_{2,3}$, since $|y-z|\le 2r<\sqrt{t^{1/\alpha }},$ we have $|x-y|<C\sqrt{t^{1/\alpha }}$. For $n-\delta _{0}>0$, by Proposition 7, we obtain

$$\begin{aligned} | M_{2,3}|\le & {} C\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x_{0})}\right) ^{\delta _{0}}\left( \int _{|x-y|<4|y-z|}\frac{t{\mathrm{d}}x}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}\right. \nonumber \\&\left. + \int _{|x-z|\le 5|y-z|}\frac{t{\mathrm{d}}x}{(\sqrt{t^{1/\alpha }}+|x-z|)^{n+2\alpha }}\right) \\\le & {} \frac{Cr^{n}}{\rho (x_{0})^{\delta _{0}}(\sqrt{t^{1/\alpha }})^{n-\delta _{0}}}\le C\left( \frac{r}{\rho (x_{0})}\right) ^{\delta _{0}}. \end{aligned}$$

The estimates for $M_{2,i}, i=1,2,3$, imply that

$$\begin{aligned} M_{2}\le \int ^{\rho (x_{0})^{2\alpha }}_{(2r)^{2\alpha }}\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}=C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta } \log \left( \frac{\rho (x_{0})}{r}\right) . \end{aligned}$$

Finally, we can get

$$\begin{aligned} \left\| t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{m}\partial _{t}^{m}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le C\left( \frac{r}{\rho (x_{0})}\right) ^{\delta }\left( \log (\frac{\rho (x_{0})}{r})\right) ^{1/2}. \end{aligned}$$

Thus (iii) readily follows. $\square$

4.3 Boundedness of Littlewood–Paley g-function ${\widetilde{g}}^{{\mathcal {L}}}_{\alpha }$

By the $L^{2}$-boundedness of Riesz transforms $\nabla _{x}{\mathcal {L}}^{-1/2}$, we can see that

$$\begin{aligned} \Vert {\widetilde{g}}^{{\mathcal {L}}}_{\alpha }f\Vert ^{2}_{L^{2}}\le C\int ^{\infty }_{0}\left( \int _{{\mathbb {R}}^{n}}|t^{1/2\alpha }{\mathcal {L}}^{1/2}e^{-t{\mathcal {L}}^{\alpha }}f(x)|^{2}{\mathrm{d}}x\right) \frac{{\mathrm{d}}t}{t}. \end{aligned}$$

Then by the spectral theorem, we know that ${\widetilde{g}}^{{\mathcal {L}}}_{\alpha }$ is bounded from $L^{2}({\mathbb {R}}^{n})$ into $L^{2}({\mathbb {R}}^{n})$.

Theorem 5

Assume that the potential $V\in B_{q}$ with $q>n$. Let $x,y,z\in {\mathbb {R}}^{n}$.

• (i) For any $N>0$, there exists a constant $C_{N}$ such that

  $$\begin{aligned} \left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le \frac{C_{N}}{|x-y|^{n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-N}; \end{aligned}$$

• (ii) Let $|x-y|>2|y-z|$ and $0<\delta <\min \{2\alpha ,\delta _{1},1\}$. There exists a constant C such that

  $$\begin{aligned}&\left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\\&\quad + \left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(y,x)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(z,x)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le {{C}}\frac{|y-z|^{\delta }}{|x-y|^{n+\delta }}; \end{aligned}$$

• (iii) There exists a constant C such that for every ball $B=B(x_{0},r)$ with $0<r\le \rho (x)/2$,

  $$\begin{aligned} \log \left( \frac{\rho (x)}{r}\right) \frac{1}{|B|}\int _{B}\left\| t^{1/(2\alpha )}\nabla _{x} e^{-t{\mathcal {L}}^{\alpha }}1(y)-(t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }{\mathrm{d}}y\le C, \end{aligned}$$

  and, if $\gamma <\min \{2\alpha ,\delta _{1},1\}$ then

  $$\begin{aligned} \left( \frac{\rho (x)}{r}\right) ^{\gamma }\frac{1}{|B|}\int _{B}\left\| t^{1/(2\alpha )}\nabla _{x} e^{-t{\mathcal {L}}^{\alpha }}1(y)-(t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1)_{B}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }{\mathrm{d}}y\le C. \end{aligned}$$

Proof

For (i), from Proposition 4, we have

$$\begin{aligned} \left| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\right| \le C_{N}\min \left\{ \frac{t^{1+N/\alpha +1/(2\alpha )}}{|x-y|^{n+2\alpha +2N+1}},\ t^{-n/(2\alpha )} \right\} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\le |x-y|$, we obtain

$$\begin{aligned}&\left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\le |x-y|\}}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }^{2}\\&\quad \le C_{N}\int ^{|x-y|^{2\alpha }}_{0}\frac{t^{2+2N/\alpha +1/\alpha }}{|x-y|^{2n+4\alpha +4N+2}} \left( 1+\frac{\sqrt{t^{1/\alpha }}}{\rho (x)}+\frac{\sqrt{t^{1/\alpha }}}{\rho (y)}\right) ^{-2N}\frac{{\mathrm{d}}t}{t} \\&\quad \le C_{N}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\int ^{|x-y|^{2\alpha }}_{0} \frac{(\sqrt{t^{1/\alpha }})^{4\alpha +4N+2}}{|x-y|^{2n+4\alpha +4N+2}}\left( \frac{\sqrt{t^{1/\alpha }}}{|x-y|}\right) ^{-2N}\frac{{\mathrm{d}}t}{t} \\&\quad \le \left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\frac{C_{N}}{|x-y|^{2n}}\int ^{|x-y|^{2\alpha }}_{0} \left( \frac{\sqrt{t^{1/\alpha }}}{|x-y|}\right) ^{4\alpha +2N+2}\frac{{\mathrm{d}}t}{t}. \end{aligned}$$

Let $\sqrt{t^{1/\alpha }}/|x-y|=u$. We can see that

$$\begin{aligned}&\left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\le |x-y|\}}\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }^{2}\\&\quad \le \left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\frac{C_{N}}{|x-y|^{2n}}\int ^{1}_{0}u^{4\alpha +2N+1}{\mathrm{d}}u\\&\quad \le \frac{C_{N}}{|x-y|^{2n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}. \end{aligned}$$

If $\sqrt{t^{1/\alpha }}\ge |x-y|$, we can get

$$\begin{aligned}&\left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)\cdot \chi _{\{t^{1/2\alpha }\ge |x-y|\}}\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\\&\quad \le C_{N}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}\int ^{\infty }_{|x-y|^{2\alpha }}t^{-n/\alpha -1}{\mathrm{d}}t \\&\quad \le \frac{C_{N}}{|x-y|^{2n}}\left( 1+\frac{|x-y|}{\rho (x)}+\frac{|x-y|}{\rho (y)}\right) ^{-2N}, \end{aligned}$$

which proves (i).

For (ii), by Proposition 4, we have

$$\begin{aligned} \left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le & {} \int ^{\infty }_{0}\frac{Ct}{(\sqrt{t^{1/\alpha }}+|x-y|)^{2n+4\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }{\mathrm{d}}t \\\le & {} C|y-z|^{2\delta }\int ^{\infty }_{0}\frac{(\sqrt{t^{1/\alpha }})^{2\alpha -2\delta }}{(\sqrt{t^{1/\alpha }}+|x-y|)^{2n+4\alpha }}{\mathrm{d}}t. \end{aligned}$$

Let ${\sqrt{t^{1/\alpha }}}/{|x-y|}=u$. We obtain

$$\begin{aligned} \left\| {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }\le & {} C|y-z|^{2\delta }|x-y|^{-2n-2\delta }\int ^{\infty }_{0}\frac{u^{4\alpha -2\delta -1}}{(1+u)^{2n+4\alpha }}{\mathrm{d}}u \\\le & {} \frac{C|y-z|^{2\delta }}{|x-y|^{2n+2\delta }}. \end{aligned}$$

The symmetry of the kernel $D^{{\mathcal {L}}}_{\alpha ,t}(\cdot ,\cdot )$ gives the desired conclusion of (ii).

For (iii), fix $y,z\in B=B(x_{0},r)$ with $0<r\le \rho (x_{0})/2$. Similar to Theorem 4, we need to deal with the term

$$\begin{aligned} \left\| t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| _{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) } \end{aligned}$$

first. Write

$$\begin{aligned} \left\| t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) }= & {} G_{1}+G_{2}+G_{3}, \end{aligned}$$

where

$$\begin{aligned} \left\{ \begin{aligned} G_{1}&:=\int ^{(2r)^{2\alpha }}_{0}\left| \int _{{\mathbb {R}}^{n}}\left( {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t};\\ G_{2}&:=\int ^{\rho (x_{0})^{2\alpha }}_{(2r)^{2\alpha }}\left| \int _{{\mathbb {R}}^{n}}\left( {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t};\\ G_{3}&:=\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left| \int _{{\mathbb {R}}^{n}}\left( {\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,y)-{\widetilde{D}}^{{\mathcal {L}}}_{\alpha ,t}(x,z)\right) {\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t}. \end{aligned}\right. \end{aligned}$$

Since $y,z\in B\subset B(x_{0},\rho (x_{0}))$, then $\rho (y)\sim \rho (x_{0})\sim \rho (z)$. It follows from Proposition 4 (iii) that

$$\begin{aligned} G_{1}\le C\int ^{(2r)^{2\alpha }}_{0}(\sqrt{t^{1/\alpha }}/\rho (x_{0}))^{1+2\alpha }\frac{{\mathrm{d}}t}{t} \le C\int ^{(2s)^{2\alpha }}_{0}\left( \frac{\sqrt{t^{1/\alpha }}}{\rho (x_{0})}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}=C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }. \end{aligned}$$

Also, we apply Proposition 4 (ii) to deduce that

$$\begin{aligned} G_{3}\le & {} C\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }\left| \int _{{\mathbb {R}}^{n}} \frac{t}{(\sqrt{t^{1/\alpha }}+|x-y|)^{n+2\alpha }}{\mathrm{d}}x\right| ^{2}\frac{{\mathrm{d}}t}{t} \\\le & {} C\int ^{\infty }_{\rho (x_{0})^{2\alpha }}\left( \frac{|y-z|}{\sqrt{t^{1/\alpha }}}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}\le C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }. \end{aligned}$$

Then for $G_{2}$, following the procedure of the treatment for $M_{2}$ in Theorem 4, we obtain

$$\begin{aligned} G_{2}\le \int ^{\rho (x_{0})^{2\alpha }}_{(2r)^{2\alpha }}\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta }\frac{{\mathrm{d}}t}{t}=C\left( \frac{r}{\rho (x_{0})}\right) ^{2\delta } \log \left( \frac{\rho (x_{0})}{r}\right) . \end{aligned}$$

From the above estimates, we can get

$$\begin{aligned} \left\| t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(y)-t^{1/(2\alpha )}\nabla _{x}e^{-t{\mathcal {L}}^{\alpha }}1(z)\right\| ^{2}_{L^{2}\left( (0,\infty ),\frac{{\mathrm{d}}t}{t}\right) } \le C\left( \frac{r}{\rho (x_{0})}\right) ^{\delta }\left( \log \left( \frac{\rho (x_{0})}{r}\right) \right) ^{1/2}. \end{aligned}$$

Thus (iii) readily follows. $\square$