New Atomic Decomposition for Besov Type Space $\dot{B}^0_{1,1}$ Associated with Schrödinger Type Operators

Abstract

Let $(X, d, \mu )$ be a space of homogeneous type. Let L be a nonnegative self-adjoint operator on $L^2(X)$ satisfying certain conditions on the heat kernel estimates which are motivated from the heat kernel of the Schrödinger operator on $\mathbb {R}^n$. The main aim of this paper is to prove a new atomic decomposition for the Besov space $\dot{B}^{0, L}_{1,1}(X)$ associated with the operator L. As a consequence, we prove the boundedness of the Riesz transform associated with L on the Besov space $\dot{B}^{0, L}_{1,1}(X)$.

1 Introduction

Let $(X,d, \mu )$ be a metric spaces endowed with a nonnegative Borel measure $\mu$. Denote $B(x,r):=\{y\in X: d(x,y)<r\}$. In this paper we assume that the measure satisfies the doubling property condition, i.e., there exists a constant $C_1>0$ such that

$$\begin{aligned} \mu (B(x,2r))\le C_1\mu (B(x,r)) \end{aligned}$$

(1)

for all $x\in X$ and $r>0$. This condition implies that there exist constants $C_2, D\ge 0$ such that

$$\begin{aligned} V(x,\lambda r)\le C_2\lambda ^D V(x,r) \end{aligned}$$

(2)

for all $x\in X, r>0$ and $\lambda \ge 1$. See [11].

We also assume further that $(X,d,\mu )$ satisfies the noncollapsing condition, i.e., there exists $c_0>0$ such that

$$\begin{aligned} V(x,1)\ge c_0 \end{aligned}$$

(3)

for all $x\in X$.

From now on, for any measurable subset $E\subset X$, we denote $V(E):=\mu (E)$. For all $x\in X$ and $r>0$, we also denote $V(x,r)=\mu (B(x,r))$.

Note that the classical Hardy space $H^1(X)$ is a suitable substitution for the space $L^1(X)$ when we work with Calderón–Zygmund operators but the classical Hardy space might not be suitable for the study of certain operators that lie beyond the Calderon Zygmund class. This observation highlights the need for the development of new function spaces that adapt well to these operators. In recent times, there has been a remarkable progress in the field of function spaces associated with operators, reflecting the growing interest in understanding the behaviour of these operators and their associated function spaces. See for example [1, 5, 15, 21, 23, 29] and the references therein.

Motivated by this ongoing research, we aim to study new atomic decomposition of Besov spaces associated to Schrödinger type operators. Throughout this paper, we assume that H is a non-negative self-adjoint operator on $L^2(X)$ which generates the analytic semigroup $\{e^{-tH}\}_{t>0}$. Denote by ${\widetilde{p}}_t(x,y)$ and ${\widetilde{q}}_t(x,y)$ the kernels of $e^{-tH}$ and $tH e^{-tH}$, respectively, we assume that the kernels ${\widetilde{p}}_t(x,y)$ satisfy the following conditions:

(H1):

There exist positive constants C and c such that

$$\begin{aligned} \displaystyle |{\widetilde{p}}_t(x,y)|+|{\widetilde{q}}_t(x,y)|\le \frac{C}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big ) \end{aligned}$$

for all $x,y\in X$ and $t>0$;

(H2):

There exist positive constants $\delta _1$, c and C such that

$$\begin{aligned}{} & {} |{\widetilde{p}}_t(x,y)-{\widetilde{p}}_t(\overline{x},y)|+|{\widetilde{q}}_t(x,y)-{\widetilde{q}}_t(\overline{x},y)|\\ {}{} & {} \le \frac{C}{V(x,\sqrt{t})}\Big [\frac{d(x,\overline{x})}{d(x,y)}\Big ]^{\delta _1}\exp \Big (-c\frac{d(x,y)^2}{t}\Big ) \end{aligned}$$

whenever $d(x,\overline{x})\le \sqrt{t}$ and $t>0$;

(H3):

$\displaystyle \int _X {\widetilde{p}}_t(x,y)d\mu (x)=1$ for $y\in X$.

In fact, the assumptions (H1) and (H2) can be assumed only for the kernel ${\widetilde{p}}_t(x,y)$ since the estimates in (H1) and (H2) for ${\widetilde{p}}_t(x,y)$ imply similar estimates for ${\widetilde{q}}_t(x,y)$. However, for the sake of simplicity, we make the assumptions for both ${\widetilde{p}}_t(x,y)$ and ${\widetilde{q}}_t(x,y)$.

Standard examples of operators which satisfy conditions (H1), (H2) and (H3) include the Laplacians $\Delta$ on the Euclidean spaces ${\mathbb {R}}^n$, the Laplace-Beltrami operators on non-compact Riemannian manifolds with doubling property, the Bessel operators on $(0, \infty )^n$, the sub-Laplacians on stratified Lie groups and certain degenerate elliptic operators on doubling spaces and domains.

Our motivation is to study the Schrödinger operator $L = H + V$ which is a non-negative self-adjoint operator on $L^2(X)$. Under suitable conditions, the potential V induces a critical function $\rho$ which appears on the upper bounds and regularity estimates of the heat kernels of L and its time derivative. We refer the reader to Sect. 2.1 for a general definition of critical functions and further details.

In this paper, without the assumption $L = H + V$, we assume that L is a non-negative self-adjoint operator on $L^2(X)$. Denote by $p_t(x,y)$ and $q_t(x,y)$ the kernels of $e^{-tL}$ and $tL e^{-tL}$, respectively. Suppose that $\rho$ is a critical function defined on X. See Sect. 2.1 for the precise definition of critical functions. We assume that the kernels $p_t(x,y)$ and $q_t(x,y)$ satisfy the following conditions:

• (L1) For all $N>0$, there exist positive constants c and C so that

  $$\begin{aligned} |p_t(x,y)|\le \frac{C}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )\Big (1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (y)}\Big )^{-N} \end{aligned}$$

  for all $x,y\in X$ and $t>0$;

• (L2) There is a positive constant $\delta _2$ so that for all $N>0$, there exist positive constants c and C which satisfy

  $$\begin{aligned}{} & {} |q_t(x,y)-q_t(\overline{x},y)|\le \frac{C}{V(x,\sqrt{t})}\Big [\frac{d(x,\overline{x})}{d(x,y)}\Big ]^{\delta _2}\\ {}{} & {} \exp \Big (-c\frac{d(x,y)^2}{t}\Big )\Big (1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (y)}\Big )^{-N} \end{aligned}$$

  whenever $d(x,\overline{x})\le \sqrt{t}$ and $t>0$;

• (L3) There is a positive constant $\delta _3$ such that

  $$\begin{aligned} |p_t(x,y)- & {} {\widetilde{p}}_t(x,y)|+|q_t(x,y)-{\widetilde{q}}_t(x,y)|\\\le & {} \frac{C}{V(x,\sqrt{t})}\Big (\frac{\sqrt{t}}{\sqrt{t}+\rho (x)}\Big )^{\delta _3}\exp \Big (-c\frac{d(x,y)^2}{t}\Big ) \end{aligned}$$

  for all $x,y\in X$ and $t>0$.

Remark 1.1

• (a) If we set $\delta =\min \{\delta _1,\delta _2,\delta _3\}$, then (H2), (L2) and (L3) are satisfied with the exponent $\delta$. For this reason, we might assume that $\delta _1=\delta _2=\delta _3 = \delta$.

• (b) Note that the condition (L1) implies that for all $N>0$, there exist positive constants c and C so that

  $$\begin{aligned} |q_t(x,y)|\le \frac{C}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )\Big (1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (y)}\Big )^{-N} \end{aligned}$$

  (4)

  for all $x,y\in X$ and $t>0$. Since the proof of (4) is standard, we leave it to the interested reader.

• (c) As mentioned above, an example of the pairs of operators (H, L) which satisfy our assumptions are the operators H mentioned above and $L = H +V$ for suitable potentials V. See Sect. 2.1, also [9, Section 6] and [34]. We remark that our work on the operator L in this paper only relies on the assumptions (L1), (L2), (L3) and does not use the representation $L = H +V$.

Our aim is to study the homogeneous Besov space $\dot{B}^{0,L}_{1,1}(X)$ associated with the operator L.

Definition 1.2

The homogeneous Besov space $\dot{B}^{0,L}_{1,1}(X)$ is defined as the set of $f\in L^1(X)$ such that

$$\begin{aligned} \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}:= \int _0^{\infty }\Vert tL e^{-tL}f\Vert _1\frac{dt}{t}<\infty . \end{aligned}$$

When $L=-\Delta$ the Laplacian on ${\mathbb {R}}^n$, the Besov space $\dot{B}^{0,L}_{1,1}({\mathbb {R}}^n)$ coincides with the classical Besov space $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)$. It is well known that the Besov space $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)$ is contained in the Hardy space $H^1({\mathbb {R}}^n)$ and is used in proving the dispersive estimates of the wave equations (see for example [3, 8, 13]) and the regularity of the Green functions on domains (see for example [20]). See also [17, 18, 24,25,26] and the references therein for further discussion on the Besov space type $\dot{B}^{0}_{1,1}$ and the Besov spaces on spaces of homogeneous type. It is worth noticing that in the definition above we define the Besov space a subset of $L^1(X)$. This is more advantageous than the approach using new distributions as in [5, 26].

We are interested in atomic decompositions of the Besov space $\dot{B}^{0,L}_{1,1}(X)$. Note that atomic decompositions of Besov spaces associated to non-negative self-adjoint operators satisfying Gaussian upper bounds were obtained in [5] for homogeneous Besov spaces and in [27] for inhomogeneous Besov spaces. Adapting ideas in [5, 27], we can define atoms for the Besov spaces $\dot{B}^{0,L}_{1,1}(X)$ as follows.

Definition 1.3

Let $M\in \mathbb {N}_+$. A function a is said to be an (L, M) atom if there exists a ball B so that

• (i) $a=L^{M} b$ with $b\in D(L^M)$, where $D(L^M)$ is the domain of $L^M$;

• (ii) $\textrm{supp} \,L^{k} b\subset B$, $k=0,\ldots , 2M$;

• (iii) $\displaystyle |L^{k} b(x)|\le r_B^{2(M-k)}V(B)^{-1}$, $k=0,\ldots , 2M,$ where $r_B$ denotes the radius of ball B..

Note that the atoms defined in [5, 27] are supported in balls associated to dyadic cubes. See Lemma 2.2 for the definition of dyadic cubes. In this paper, we do not need the dyadic cubes in Definition 1.3 and we are able to prove the following result.

Theorem 1.4

Let $M\in \mathbb {N}_+$. Assume $f\in \dot{B}^{0,L}_{1,1}(X)$. Then there exist a sequence of (L, M) atoms $\{a_j\}$ and a sequence of coefficients $\{\lambda _j\}\in \ell ^1$ so that

$$\begin{aligned} f=\sum _{j}\lambda _j a_j \ \ \text {in } L^1(X), \end{aligned}$$

and

$$\begin{aligned} \sum _{j}|\lambda _j| \lesssim \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}. \end{aligned}$$

Conversely, if

$$\begin{aligned} f=\sum _{j}\lambda _j a_j \ \ \text {in } L^1(X), \end{aligned}$$

where $\{a_j\}$ is a sequence of (L, M)-atoms and $\{\lambda _j\}\in \ell ^1$, then

$$\begin{aligned} \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}\lesssim \sum _{j}|\lambda _j|. \end{aligned}$$

The proof of Theorem 1.4 will be presented later. In comparison with the atomic decomposition in Theorems 4.2 and 4.3 in [5], the main difference is that in Theorem 1.4, the convergence used in the atomic decomposition is in $L^1(X)$ instead of in the space of new distributions associated with the operator L; moreover, Theorem 1.4 uses the atoms associated with balls rather than the dyadic cubes as in Theorems 4.2 and 4.3 in [5].

We now consider new atoms associated with the critical function $\rho$ which will be defined in Sect. 2.1. Note that the idea of the atomic decomposition associated to the critical functions was used in the setting of Hardy spaces. In [16], the atomic decomposition associated to the critical functions was studied for the Hardy spaces associated to Schrödinger operators with potential satisfying certain reverse Hölder inequality. Then the results were extended to encompass a broader scope, incorporating Schrödinger operators in various contexts such as stratified Lie groups and doubling manifolds. See for example [9, 34]. However, this is the first time the atomic decomposition associated to the critical functions was established for the Besov spaces.

Definition 1.5

Let $\epsilon >0$ and $\rho$ be a critical function. A function a is said to be an $(\epsilon ,\rho (\cdot ))$-atom if there exists a ball B such that

• (i) ${\text {supp}}\,a \subset B$;

• (ii) $|a(x)| \le V(B)^{-1}$;

• (iii) $|a(x)-a(y)| \le V(B)^{-1}\left( \dfrac{d(x,y)}{r_B}\right) ^\epsilon , \ \ x,y\in X$;

• (iv) $\displaystyle \int a(x)d\mu (x)=0$ if $r_B<\rho (x_B)$.

It is interesting that the atoms in Definition 1.5 depend on the critical function $\rho$ only. This type of atoms can be viewed as an extended version of the atoms used for the inhomogeneous Besov type. In fact, in the particular case $\rho =\textrm{constant}$, the atoms in Definition 1.5 turn out to be the atoms which characterize the inhomogeneous Besov spaces. See for example [26]. Our main result is the following theorem.

Theorem 1.6

If $f\in \dot{B}^{0,L}_{1,1}(X)$, then there exist a sequence of $(\epsilon ,\rho (\cdot ))$-atoms $\{a_j\}$ for some $\epsilon >0$ and a sequence of coefficients $\{\lambda _j\}\in \ell ^1$ so that

$$\begin{aligned} f=\sum _{j}\lambda _j a_j \ \ \text {in } L^1(X), \end{aligned}$$

and

$$\begin{aligned} \sum _{j}|\lambda _j| \lesssim \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}. \end{aligned}$$

Conversely, if

$$\begin{aligned} f=\sum _{j}\lambda _j a_j \ \ \text {in } L^1(X), \end{aligned}$$

where $\{a_j\}$ is a sequence of $(\epsilon ,\rho (\cdot ))$-atoms with $\epsilon >0$ and $\{\lambda _j\}\in \ell ^1$, then

$$\begin{aligned} \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}\lesssim \sum _{j}|\lambda _j|. \end{aligned}$$

The organization of the paper is as follows. In Sect. 2, we recall the definitions of critical functions and dyadic cubes, and prove some kernel estimates of the spectral multipliers of H. In Sect. 3, we will set up the theory of the inhomogeneous Besov space $B^0_{1,1}(X)$ including atomic decomposition results. The proofs of the main results will be given in Sect. 4. Finally, Sect. 5 is devoted in the proof of the boundedness of the Riesz transform associated with L in Besov spaces.

Throughout the paper, we always use C and c to denote positive constants that are independent of the main parameters involved but whose values may differ from line to line. We write $A\lesssim B$ if there is a universal constant C so that $A\le CB$ and $A\approx B$ if $A\lesssim B$ and $B\lesssim A$. Given a $\lambda > 0$ and a ball $B:=B(x,r)$, we write $\lambda B$ for the $\lambda$-dilated ball, which is the ball with the same center as B and with radius $\lambda r$. For each ball $B\subset X$, we set

$$\begin{aligned} S_0(B)=B \ \text {and} \ S_j(B) = 2^jB\backslash 2^{j-1}B \ \text {for} \ j\in \mathbb {N}. \end{aligned}$$

2 Preliminaries

2.1 Critical Functions

A function $\rho :X\rightarrow (0,\infty )$ is called a critical function if there exist positive constants $C_\rho$ and $k_0$ so that

$$\begin{aligned} \rho (y)\le C_\rho \rho (x)\left( 1 +\frac{d(x,y)}{\rho (x)}\right) ^{\frac{k_0}{k_0+1}} \end{aligned}$$

(5)

for all $x,y\in X$.

Note that the concept of critical functions was introduced in the setting of Schrödinger operators on $\mathbb {R}^D$ in [19] (see also [30]) and then was extended to the spaces of homogeneous type in [34].

A simple example of a critical function is $\rho \equiv 1$. Moreover, one of the most important classes of the critical functions is the one involving the weights satisfying the reverse Hölder inequality. Recall that a non-negative locally integrable function w is said to be in the reverse Hölder class $RH_q(X)$ with $q>1$ if there exists a constant $C>0$ so that

$$\begin{aligned} \Big (\frac{1}{V(B)}\int _B (w(x))^q d\mu (x)\Big )^{1/q}\le \frac{C}{V(B)}\int _B w(x)d\mu (x)\end{aligned}$$

for all balls $B\subset X$. Note that if $w\in RH_q(X)$ then w is a Muckenhoupt weight. See [32].

Now suppose $V\in RH_q(X)$ for some $q>\max \{1,D/2\}$ and, following [30, 34], set

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

(6)

Then it was proved in [30, 34] that $\rho$ is a critical function provided $q>\max \{1,D/2\}$. The following result will be useful in the sequel which is taken from Lemma 2.3 and Lemma 2.4 of [34].

Lemma 2.1

Let $\rho$ be a critical function on X. Then there exist a sequence of points $\{x_\alpha \}_{\alpha \in \mathcal {I}}\subset X$ and a family of functions $\{\psi _\alpha \}_{\alpha \in \mathcal {I}}$ satisfying the following:

• (i) $\displaystyle \bigcup _{\alpha \in \mathcal {I}} B(x_\alpha , \rho (x_\alpha )) = X$.

• (ii) For every $\lambda \ge 1$ there exist constants C and $N_1$ such that $\displaystyle \sum _{\alpha \in \mathcal {I}} 1_{B(x_\alpha , \lambda \rho (x_\alpha ))}\le C\lambda ^{N_1}$.

• (iii) $\textrm{supp}\, \psi _{\alpha }\subset B_\alpha :=B(x_\alpha , \epsilon _0\rho (x_\alpha ))$ and $0\le \psi _\alpha (x)\le 1$ for all $x\in X$, where $\epsilon _0$ is a fixed constant such that $C_\rho \epsilon _0 (1 +\epsilon _0)^{\frac{k_0}{k_0+1}}<1$.

• (iv) $\displaystyle |\psi _\alpha (x)-\psi _\alpha (y)|\le C d(x,y)/\rho (x_\alpha )$;

• (v) $\displaystyle \sum _{\alpha \in \mathcal {I}}\psi _\alpha (x)=1$ for all $x\in X$.

2.2 Dyadic Cubes

We now recall an important covering lemma in [10].

Lemma 2.2

There exists a collection of open sets $\{Q_\tau ^k\subset X: k\in \mathbb {Z}, \tau \in I_k\}$, where $I_k$ denotes certain (possibly finite) index set depending on k, and constants $\eta \in (0,1), a_0\in (0,1]$ and $\kappa _0\in (0,\infty )$ such that

• (i) $\mu (X\backslash \cup _\tau Q_\tau ^k)=0$ for all $k\in \mathbb {Z}$;

• (ii) if $i\ge k$, then either $Q_\tau ^i \subset Q_\beta ^k$ or $Q_\tau ^i \cap Q_\beta ^k=\emptyset$;

• (iii) for every $(k,\tau )$ and each $i<k$, there exists a unique $\tau '$ such $Q_\tau ^k\subset Q_{\tau '}^i$;

• (iv) the diameter $\textrm{diam}\,(Q_\tau ^k)\le \kappa _0 \eta ^k$;

• (v) each $Q_\tau ^k$ contains certain ball $B(x_{Q_\tau ^k}, a_0\eta ^k)$.

Remark 2.3

Since the constants $\eta$ and $a_0$ are not essential in the paper, without loss of generality, we may assume that $\eta =a_0=1/2$. We then fix a collection of open sets in Lemma 2.2 and denote this collection by $\mathcal {D}$. We call open sets in $\mathcal {D}$ the dyadic cubes in X and $x_{Q_\tau ^k}$ the center of the cube $Q_\tau ^k \in \mathcal {D}$. We also denote

$$\begin{aligned} \mathcal {D}_\nu :=\{Q_\tau ^{\nu +1} \in \mathcal {D}: \tau \in I_{\nu +1}\} \end{aligned}$$

for each $\nu \in \mathbb {Z}$. Then for $Q\in \mathcal {D}_\nu$, we have $B(x_Q, c_02^{-\nu })\subset Q\subset B(x_Q, \kappa _0 2^{-\nu })=:B_Q$, where $c_0$ is a constant independent of Q. For the sake of simplicity we might assume that $\kappa _0=1$.

2.3 Kernel Estimates

Denote by $E_H(\lambda )$ a spectral decomposition of H. Then by spectral theory, for any bounded Borel funtion $F:[0,\infty )\rightarrow \mathbb {C}$ we can define

$$\begin{aligned} F(H)=\int _0^\infty F(\lambda )dE_H(\lambda ) \end{aligned}$$

as a bounded operator on $L^2(X)$. It is well-known that the kernel ${\cos (t\sqrt{H})}(\cdot ,\cdot )$ of $\cos (t\sqrt{H})$ satisfies the finite propagation speed

$$\begin{aligned} \textrm{supp}\, {\cos (t\sqrt{H})}\subset \{(x,y)\in X\times X: d(x,y)\le {\tilde{c}}_0t\} \end{aligned}$$

(7)

for some ${\tilde{c}}_0>0$. See for example [31].

In what follows, without loss of generality we may assume that ${\tilde{c}}_0=1$.

We have the following useful lemma.

Lemma 2.4

([23]) Let $\phi \in C^\infty _0(\mathbb {R})$ be an even function with supp $\phi \subset (-1, 1)$ and $\displaystyle \int \phi =2\pi$. Denote by $\Phi$ the Fourier transform of $\phi$, i.e.,

$$\begin{aligned} \Phi (\xi ):= {\mathcal {F}} \phi (\xi ):= \frac{1}{2\pi }\int _{{\mathbb {R}}} e^{-ix\xi }\phi (x)dx. \end{aligned}$$

(8)

Then for every $k\in \mathbb {N}$, the operator $(t^2{H})^k\Phi (t\sqrt{H})$ is an integral operator with kernel denoted by $(t^2{H})^k\Phi (t\sqrt{H})(x,y)$ satisfying the following

$$\begin{aligned} \displaystyle \textrm{supp}\, (t^2{H})^k\Phi (t\sqrt{H})(\cdot , \cdot )\subset \{(x,y)\in X\times X: d(x,y)\le t\}, \end{aligned}$$

(9)

and

$$\begin{aligned} |(t^2{H})^k\Phi (t\sqrt{H})(x,y)|\le \frac{C}{V(x,t)} \end{aligned}$$

(10)

for all $t>0$ and $x,y\in X$.

Lemma 2.5

([7]) Let $\lambda >0$. Then we have:

• (a) For any $N>0$ and $s=N+2D + 1$, there exists $C=C(N)$ so that

  $$\begin{aligned} |{{F(\lambda \sqrt{H})}}(x,y)|\le \frac{C}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-N} \Vert F\Vert _{W^2_{s}} \end{aligned}$$

  (11)

  for all $x,y\in X$, and all functions F supported in [1/2, 2].

• (b) For any $N>0$ and $s=2(N +2D +1)$ there exists $C=C(N)$ so that

  $$\begin{aligned} |F(\lambda \sqrt{H})(x,y)|\le \frac{C}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-N} \Vert F\Vert _{W^\infty _{s}} \end{aligned}$$

  (12)

  for all $x,y\in X$, and for all functions F supported in [0, 2] with $F^{(2\nu +1)}(0)=0$ for all $\nu \in \mathbb {N}$. Here, $\Vert F\Vert _{W_s^q} =\Vert (I-d^2/dx^2)F\Vert _{q}$ for $s>0$ and $q\in [1,\infty ]$.

Lemma 2.6

Let $\lambda >0$. Then we have:

• (a) For any $N>0$ and $s=N+3D + 2$, there exists $C=C(N)$ so that

  $$\begin{aligned}{} & {} |{{F(\lambda \sqrt{H})}}(x,y)-{{F(\lambda \sqrt{H})}}(x,y')|\nonumber \\ {}{} & {} \le C\Big (\frac{d(y,y')}{\lambda }\Big )^\delta \frac{1}{ V(x,\lambda ) } \Big (1+\frac{d(x,y)}{\lambda }\Big )^{-N} \Vert F\Vert _{W^2_{s}} \end{aligned}$$

  (13)

  for all $x,y,y'\in X$ with $d(y,y')<\lambda$, and all functions supported in [1/2, 2].

• (b) For any $N>0$ and $s=2(N +3D +2)$ there exists $C=C(N)$ so that

  $$\begin{aligned}{} & {} |F(\lambda \sqrt{H})(x,y)-F(\lambda \sqrt{H})(x,y')|\nonumber \\ {}{} & {} \le C\Big (\frac{d(y,y')}{\lambda }\Big )^\delta \frac{1}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-N} \Vert F\Vert _{W^\infty _{s}} \end{aligned}$$

  (14)

  for all $x,y,y'\in X$ with $d(y,y')<\lambda$, and for all functions F supported in [0, 2] with $F^{(2\nu +1)}(0)=0$ for all $\nu \in \mathbb {N}$.

Proof

(a) We write $F(\lambda )= G(\lambda )e^{-\lambda ^2}$, where $G(\lambda ) = F(\lambda )e^{\lambda ^2}$. Then we have

$$\begin{aligned} \begin{aligned} F(\lambda \sqrt{H})(x,y) =\int _X G(\lambda \sqrt{H})(x,z){\widetilde{p}}_{\lambda ^2}(z,y)d\mu (z). \end{aligned} \end{aligned}$$

This, along with Lemma 2.5, (H2) and the fact $\Vert G\Vert _{W^2_s}\lesssim \Vert F\Vert _{W^2_s}$ for every $s>0$, yields that, for $x,y,y'\in X$ with $d(y,y')<\lambda$, $N>0$ and $s={\tilde{N}} +D + 1$ with ${\tilde{N}} = N+ D + 1$,

$$\begin{aligned} \begin{aligned} |F(\lambda \sqrt{H})(x,y)&-F(\lambda \sqrt{H})(x,y')|\\&\le \int _X |G(\lambda \sqrt{H})(x,z)||{\widetilde{p}}_{\lambda ^2}(z,y)-{\widetilde{p}}_{\lambda ^2}(z,y')|d\mu (z)\\&\lesssim \Vert G\Vert _{W^2_s}\Big [\frac{d(y,y')}{\lambda }\Big ]^{\delta }\int _X\frac{1}{ V(x,\lambda ) }\Big (1\\&\quad +\frac{d(x,z)}{\lambda }\Big )^{-{\tilde{N}}} \frac{1}{V(z,\lambda )}\exp \Big (-c\frac{d(y,z)^2}{\lambda ^2}\Big )d\mu (z)\\&\lesssim \Vert F\Vert _{W^2_s}\Big [\frac{d(y,y')}{\lambda }\Big ]^{\delta }\int _X\frac{1}{ V(x,\lambda ) }\Big (1\\&\quad +\frac{d(x,z)}{\lambda }\Big )^{-{\tilde{N}}} \frac{1}{V(z,\lambda )}\exp \Big (-c\frac{d(y,z)^2}{\lambda ^2}\Big )d\mu (z). \end{aligned} \end{aligned}$$

On the other hand,

$$\begin{aligned} \begin{aligned} \Big (1+\frac{d(x,z)}{\lambda }\Big )^{-{\tilde{N}}}\exp \Big (-c\frac{d(y,z)^2}{\lambda ^2}\Big )&\lesssim \Big (1+\frac{d(x,z)}{\lambda }\Big )^{-{\tilde{N}}}\Big (1\\&\quad +\frac{d(y,z)}{\lambda }\Big )^{-{\tilde{N}}}\exp \Big (-c\frac{d(y,z)^2}{2\lambda ^2}\Big )\\&\lesssim \Big (1+\frac{d(x,y)}{\lambda }\Big )^{-{\tilde{N}}}\exp \Big (-c\frac{d(y,z)^2}{2\lambda ^2}\Big ). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \int _X\frac{1}{ V(x,\lambda ) }\Big (1+\frac{d(x,z)}{\lambda }\Big )^{-{\tilde{N}}}&\frac{1}{V(z,\lambda )}\exp \Big (-c\frac{d(y,z)^2}{\lambda ^2}\Big )d\mu (z)\\&\lesssim \frac{1}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-{\tilde{N}}}\int _X\frac{1}{V(z,\lambda )}\\ {}&\quad \exp \Big (-c\frac{d(y,z)^2}{2\lambda ^2}\Big )d\mu (z)\\&\lesssim \frac{1}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-{\tilde{N}}}\\&\lesssim \frac{1}{ V(x,\lambda ) }\Big (1+\frac{d(x,y)}{\lambda }\Big )^{-N}, \end{aligned} \end{aligned}$$

which implies (13).

The estimate (14) can be proved similarly.

This completes our proof. $\square$

Lemma 2.7

Let $\varphi \in \mathcal {S}(\mathbb {R})$ be an even function. Then for any $N>0$ there exists $C_N$ such that

$$\begin{aligned} |{\varphi (t\sqrt{H})}(x,y)|\le \frac{C_N}{V(x,t)}\Big (1+\frac{d(x,y)}{t}\Big )^{-N}, \end{aligned}$$

(15)

and

$$\begin{aligned} |{\varphi (t\sqrt{H})}(x,y)-{\varphi (t\sqrt{H})}(x,y')|\le C_N\Big [\frac{d(y,y')}{t}\Big ]^{\delta }\frac{1}{V(x,t)}\Big (1+\frac{d(x,y)}{t}\Big )^{-N}\nonumber \\ \end{aligned}$$

(16)

for all $t>0$ and $x,y,y'\in X$ with $d(y,y')<t$.

Consequently, ${\varphi (t\sqrt{H})}$ is bounded on $L^1(X)$.

Proof

The inequality (15) was proved in [7]. Taking $N>D$, it follows that ${\varphi (t\sqrt{H})}$ is bounded on $L^1(X)$ since

$$\begin{aligned} \int _X\frac{1}{V(x,t)}\Big (1+\frac{d(x,y)}{t}\Big )^{-N}d\mu (x)\lesssim 1, \end{aligned}$$

as long as $N>D$.

We need only to prove (16).

Let $\psi _0\in C^\infty (\mathbb {R})$ supported in [0, 2] such that $\psi _0 = 1$ on [0, 1] and $0\le \psi _0\le 1$. Set $\psi (\lambda )=\psi _0(\lambda ) -\psi _0(2\lambda )$ and $\psi _j(\lambda )=\psi (2^{-j}\lambda )$ for $j\ge 1$. Then we have

$$\begin{aligned} \sum _{j\ge 0}\psi _j(\lambda )=1, \lambda >0. \end{aligned}$$

Hence,

$$\begin{aligned} \varphi (t\sqrt{H})=\sum _{j\ge 0}\psi _j(t\sqrt{H})\varphi (t\sqrt{H}). \end{aligned}$$

(17)

By (14), for $N>0$ we have

$$\begin{aligned}{} & {} |\psi _0(t\sqrt{H})\varphi (t\sqrt{H})(x,y)-\psi _0(t\sqrt{H})\varphi (t\sqrt{H})(x,y')|\nonumber \\{} & {} \lesssim \Big [\frac{d(y,y')}{t}\Big ]^{\delta } \frac{1}{V(x,t)}\Big (1+\frac{d(x,y)}{t}\Big )^{-N} \end{aligned}$$

(18)

for all $t>0$ and $x,y,y'\in X$ with $d(y,y')<t$.

Since supp $\psi \subset [1/2,2]$, using (11) and (13), we have, for $j\ge 1$, $t>0$, $x,y,y'\in X$ with $d(y,y')<t$,

$$\begin{aligned} \begin{aligned} |\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)&-\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y')|\\&\lesssim \Big [\frac{d(y,y')}{2^{-j}t}\Big ]^{\delta } \frac{1}{V(x,2^{-j}t)}\Big [\Big (1+\frac{d(x,y)}{2^{-j}t}\Big )^{-N}\\&\quad +\Big (1+\frac{d(x,y')}{2^{-j}t}\Big )^{-N}\Big ]\Vert h_j\Vert _{W^2_s}\\&\lesssim \Big [\frac{d(y,y')}{t}\Big ]^{\delta } \frac{2^{j(n+\delta )}}{V(x,t)}\Big [\Big (1+\frac{d(x,y)}{t}\Big )^{-N}\\&\quad +\Big (1+\frac{d(x,y')}{t}\Big )^{-N}\Big ]\Vert h_j\Vert _{W^2_s}\\&\lesssim \Big [\frac{d(y,y')}{t}\Big ]^{\delta } \frac{2^{j(n+\delta )}}{V(x,t)} \Big (1+\frac{d(x,y)}{t}\Big )^{-N} \Vert h_j\Vert _{W^2_s}, \end{aligned} \end{aligned}$$

where $s= N+3n + 2$ and $h_j(\lambda )=\psi (\lambda )\varphi (2^{-j}\lambda )$.

Since $\varphi \in {\mathcal {S}}(\mathbb {R})$, $\Vert h_j\Vert _{W^2_s}\le C_s 2^{-j(n+\delta +1)}$ for every $s>0$. As a consequence,

$$\begin{aligned} \begin{aligned}&|\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)-\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y')|\\&\lesssim 2^{-j} \Big [\frac{d(y,y')}{t}\Big ]^{\delta } \frac{1}{V(x,t)} \Big (1+\frac{d(x,y)}{t}\Big )^{-N}, \end{aligned} \end{aligned}$$

whenever $d(y,y')<t$.

This, along with (17) and (18), implies that for each $N>0$ there exists C such that

$$\begin{aligned} |\varphi (t\sqrt{H})(x,y)-\varphi (t\sqrt{H})(x,y')|\lesssim \Big [\frac{d(y,y')}{t}\Big ]^{\delta } \frac{1}{V(x,t)} \Big (1+\frac{d(x,y)}{t}\Big )^{-N} \end{aligned}$$

for all $t>0$ and $x,y,y'\in X$ with $d(y,y')<t$.

This completes the proof. $\square$

Remark 2.8

The results in Lemmas 2.5, 2.6 and 2.7 hold true if we replace H by L since we do not use the assumption (H3) in the proofs.

Lemma 2.9

Assume that $\varphi (\lambda )=\lambda ^2\phi (\lambda )$, where $\phi \in {\mathcal {S}}({\mathbb {R}})$ is an even function. Then we have

$$\begin{aligned} \int _X \varphi (t\sqrt{H})(x,y)d\mu (y)=\int _X \varphi (t\sqrt{H})(y,x)d\mu (y)=0 \end{aligned}$$

for all $x\in X$ and $t>0$.

Proof

Let $\psi _j$ be the function as in the proof of Lemma 2.7 for $j=0,1,2,\ldots$. Then we have

$$\begin{aligned} \varphi (t\sqrt{H})f=\sum _{j\ge 0}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})f \ \ \ \text {in } L^2(X) \end{aligned}$$

for $f\in L^2(X)$.

Let $B_R=B(x_0,R)$ for a fixed $x_0\in X$ and $R>0$. Taking $f=1_{B_R}$, then it follows that

$$\begin{aligned} \int _{B_R}\varphi (t\sqrt{H})(x,y)d\mu (y)=\sum _{j\ge 0}\int _{B_R}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y) \ \ \ \text {in } L^2(X).\nonumber \\ \end{aligned}$$

(19)

Arguing similarly to the proof of Lemma 2.7, we also yield that for any $N>n$ and $j=0,1,2,\ldots$,

$$\begin{aligned} |\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y) |\lesssim 2^{-j} \frac{1}{V(x,t)} \Big (1+\frac{d(x,y)}{t}\Big )^{-N}. \end{aligned}$$

Consequently,

$$\begin{aligned} \begin{aligned} \sum _{j\ge 0}\int _{B_R} |\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)| d\mu (y)&\lesssim \sum _{j\ge 0}2^{-j}\int _{X} \frac{1}{V(x,t)} \Big (1+\frac{d(x,y)}{t}\Big )^{-N}d\mu (y)\\&\lesssim 1. \end{aligned} \end{aligned}$$

(20)

This, together with (19), implies that

$$\begin{aligned} \int _{B_R}\varphi (t\sqrt{H})(x,y)d\mu (y)=\sum _{j\ge 0}\int _{B_R}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y) \end{aligned}$$

for $x\in X$.

Using (20), and letting $R\rightarrow \infty$, the above identity deduces that

$$\begin{aligned} \int _{X}\varphi (t\sqrt{H})(x,y)d\mu (y)=\sum _{j\ge 0}\int _{X}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y) \end{aligned}$$

for $x\in X$.

It now suffices to prove

$$\begin{aligned} \int _{X}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y)=0 \end{aligned}$$

for $x\in X$ and $j=0,1,2,\ldots$.

Indeed, since $\varphi (\lambda )=\lambda ^2\phi (\lambda )$, we have

$$\begin{aligned} \psi _j(t\sqrt{H})\varphi (t\sqrt{H})= G_{j,t}(H)\circ \big [t^2H e^{-t^2H}\big ], \end{aligned}$$

where $G_{j,t}(\lambda )= e^{t^2\lambda ^2}\psi _j(t\lambda )\phi (t\lambda )$.

Therefore, due to Lemma 2.5, the upper bound of ${\widetilde{q}}_t(x,y)$ and Fubini’s theorem,

$$\begin{aligned} \begin{aligned} \int _{X}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y)&= \int _X\int _X G_{j,t}(H)(x,z){\widetilde{q}}_{t^2}(z,y)d\mu (z)d\mu (y)\\&= \int _X G_{j,t}(H)(x,z)\int _X{\widetilde{q}}_{t^2}(z,y)d\mu (y)d\mu (z). \end{aligned} \end{aligned}$$

In addition, from the conservation property (H3), we immediately have

$$\begin{aligned} \int _X{\widetilde{q}}_{t^2}(z,y)d\mu (y) =0, \end{aligned}$$

which implies

$$\begin{aligned} \int _{X}\psi _j(t\sqrt{H})\varphi (t\sqrt{H})(x,y)d\mu (y)=0. \end{aligned}$$

This completes our proof. $\square$

3 Inhomogeneous Besov Spaces $B^0_{1,1}(X)$ and Atomic Decomposition

In this section, we will introduce the Besov space $B^0_{1,1}(X)$. Our approach relies on the function spaces associated to the “Laplace-like” operator. This is motivated from the classical case in which the classical Besov spaces can be viewed as Besov spaces associated with the Laplacian. In our setting, under the three conditions (H1), (H2) and (H3), the operator H satisfies important properties which are similar to the Laplacian on the Euclidean space.

3.1 Inhomogeneous Besov Spaces $B^0_{1,1}(X)$

Definition 3.1

The (inhomogeneous) Besov space $B^0_{1,1}(X)$ is defined as the set of $f\in L^1(X)$ such that

$$\begin{aligned} \Vert f\Vert _{B^{0}_{1,1}(X)}:=\Vert e^{-H}f\Vert _1+ \int _0^{1}\Vert tH e^{-tH}f\Vert _1\frac{dt}{t}<\infty . \end{aligned}$$

In the sequel we will show that the Besov space $B^0_{1,1}(X)$ is independent of the operator H. This is a reason why we do not include the operator H in the notation of the Besov space.

Lemma 3.2

The inhomogeneous Besov space $B^{0}_{1,1}(X)$ is complete.

In order to prove Lemma 3.2 we need the following technical lemmas.

Lemma 3.3

For each $1\le p<\infty$, the space $L^p(X)$ is dense in inhomogeneous Besov space $B^{0}_{1,1}(X)$. In fact, for each $f\in B^{0}_{1,1}(X)$ and each $1\le p<\infty$, there exists a sequence $\{f_k\}\subset L^1(X)\cap L^p(X)$ such that

$$\begin{aligned} \Vert f_k-f\Vert _1 + \Vert f_k-f\Vert _{B^{0}_{1,1}(X)} \rightarrow 0 \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

Proof

We first recall the following fact in [4]

$$\begin{aligned} \lim _{s\rightarrow 0} \Vert e^{-sH}f - f\Vert _1=0 \ \ \text {for } f\in L^1(X). \end{aligned}$$

(21)

Assume that $f\in B^{0}_{1,1}(X)$. It follows that $f\in L^1(X)$. For each $n\in {\mathbb {N}}$, define

$$\begin{aligned} f_k = e^{-H/k}f. \end{aligned}$$

From the Gaussian upper bound condition (H1) and (3),

$$\begin{aligned} \Vert e^{-H/k}f\Vert _p \lesssim c_0 k^{n/p'} \Vert f\Vert _1, \ \ \ p \in [1,\infty ), \end{aligned}$$

which implies $f_k\in L^p(X)$ for each $1\le p<\infty$.

Hence,

$$\begin{aligned} \begin{aligned} \Vert f-f_k\Vert _{B^{0}_{1,1}(X)}&=\Vert e^{-H}(f_k-f)\Vert _1+ \int _0^{1}\Vert tH e^{-tH}(f_k-f)\Vert _1\frac{dt}{t}. \end{aligned} \end{aligned}$$

By (21),

$$\begin{aligned} \begin{aligned} \Vert e^{-H}(f_k-f)\Vert _1&\lesssim \Vert f_k-f\Vert _1 = \Vert e^{-H/k}f-f\Vert _1 \rightarrow 0 \ \ \ \text {as } k\rightarrow \infty . \end{aligned} \end{aligned}$$

Similarly,

$$\begin{aligned} \Vert tLe^{-tH}(f_k-f)\Vert _1 \rightarrow 0 \ \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

On the other hand, since $e^{-sH}$ is bounded on $L^1(X)$, we have

$$\begin{aligned} \begin{aligned} \Vert tH e^{-tH}(f_k-f)\Vert _1&\le \Vert tH e^{-tH}f\Vert _1 + \Vert tHe^{-tH}f_k\Vert _1\\&=\Vert tH e^{-tH}f\Vert _1+\big \Vert e^{-H/k}\big (tH e^{-tH}f\big )\big \Vert _1\\&\lesssim \Vert tH e^{-tH}f\Vert _{1}. \end{aligned} \end{aligned}$$

In addition,

$$\begin{aligned} \int _0^1\Vert tH e^{-tH}f\Vert _{1}\frac{dt}{t}\le \Vert f\Vert _{B^{0}_{1,1}(X)}. \end{aligned}$$

By the Dominated Convergence Theorem,

$$\begin{aligned} \int _0^{1}\Vert tH e^{-tH}(f_k-f)\Vert _1\frac{dt}{t} \rightarrow 0 \ \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

It follows that

$$\begin{aligned} \Vert f-f_k\Vert _{B^{0}_{1,1}(X)} \rightarrow 0 \ \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

This, along with the fact that $f_k\in L^p(X)$ for each $n\in {\mathbb {N}}$ and $p\in [1,\infty )$, implies that $L^p(X)$ is dense in $B^{0}_{1,1}(X)$ for each $p\in [1,\infty )$.

This completes our proof. $\square$

Lemma 3.4

Let $\psi _0,\psi$ be even functions such that ${\text {supp}}\psi _0\subset \{\lambda : |\lambda |\le 2\}$ and ${\text {supp}}\psi \subset \{\lambda : 1/2\le |\lambda |\le 2\}$, and

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\lambda )=1, \ \ \ \lambda \in {\mathbb {R}}, \end{aligned}$$

where $\psi _j(\lambda )=\psi (2^{-j}\lambda ), \ j =1,2,\ldots$.

Then we have

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H})f=f \ \ \text {in } L^1(X) \end{aligned}$$

for $f\in B^{0}_{1,1}(X)$.

Proof

Let $f\in B^{0}_{1,1}(X)$. By Lemma 2.7, we have

$$\begin{aligned} \Vert \psi _0(\sqrt{H})f\Vert _1 = \Vert \psi _0(\sqrt{H})e^{H}(e^{-H}f)\Vert _1\lesssim \Vert e^{-H}f\Vert _1, \end{aligned}$$

and for $j\ge 1$,

$$\begin{aligned} \begin{aligned} \Vert \psi _j(\sqrt{H})f\Vert _1&= \Big \Vert \widetilde{\psi }_j(\sqrt{H})\big (2^{-2j}H e^{-2^{-2j}H}f\big )\Big \Vert _1\\&\lesssim \Vert 2^{-2j}H e^{-2^{-2j}H}f\Vert _1, \end{aligned} \end{aligned}$$

where $\widetilde{\psi }_j(\lambda ) = (2^{-2j}\lambda ^2)^{-1} e^{2^{-2j }\lambda ^2}\psi _j(\lambda )$.

Note that for $t\in [2^{-2j-2}, 2^{-2j}]$,

$$\begin{aligned} \begin{aligned} \Vert 2^{-2j}H e^{-2^{-2j}H}f\Vert _1&=\frac{2^{-2j}}{t}\Big \Vert e^{-(2^{-2j}-t)H}\big (tH e^{-tH}f\big )\Big \Vert _1\\&\lesssim \Vert tH e^{-tH}f\Vert _1, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \Vert \psi _j(\sqrt{H})f\Vert _1\lesssim \int _{2^{-2j-2}}^{2^{-2j}}\Vert tH e^{-tH}f\Vert _1\frac{dt}{t}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \sum _{j=0}^\infty \Vert \psi _j(\sqrt{H})f\Vert _1&\lesssim \Vert e^{-H}f\Vert _1 + \sum _{j\ge 1}\int _{2^{-2j-2}}^{2^{-2j}}\Vert tH e^{-tH}f\Vert _1\frac{dt}{t}\\&\lesssim \Vert e^{-H}f\Vert _1 + \sum _{j\ge 1}\int _{0}^{1}\Vert tH e^{-tH}f\Vert _1\frac{dt}{t}\\&\lesssim \Vert f\Vert _{B^{0}_{1,1}(X)}. \end{aligned} \end{aligned}$$

(22)

It follows that there exists $g\in L^1(X)$ such that

$$\begin{aligned} g= \sum _{j=0}^\infty \psi _j(\sqrt{H})f \ \ \text {in } L^1(X). \end{aligned}$$

If $f\in L^2(X)$, then by the spectral theory,

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H})f=f \ \ \text {in } L^2(X). \end{aligned}$$

Consequently, $f=g$ for a.e.. Hence,

$$\begin{aligned} f= \sum _{j=0}^\infty \psi _j(\sqrt{H})f \ \ \text {in } L^1(X). \end{aligned}$$

In general, for $f\in B^{0}_{1,1}(X)$, by Lemma 3.3 there exists a sequence $\{f_k\}\subset L^2(X)$ such that

$$\begin{aligned} \Vert f-f_k\Vert _1+\Vert f-f_k\Vert _{B^{0}_{1,1}} \rightarrow 0 \ \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

Similarly to (22),

$$\begin{aligned} \Big \Vert \sum _{j=0}^\infty \psi _j(\sqrt{H})(f_k-f) \Big \Vert _1 \lesssim \Vert f_k-f\Vert _{B^0_{1,1}(X)}. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} \lim _{k\rightarrow \infty } \Big \Vert \sum _{j=0}^\infty \psi _j(\sqrt{H})(f_k-f) \Big \Vert _1= 0. \end{aligned} \end{aligned}$$

(23)

We now write

$$\begin{aligned} \begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H})f =&\ \sum _{j=0}^\infty \psi _j(\sqrt{H})(f-f_k)+ \Big [\sum _{j=0}^\infty \psi _j(\sqrt{H})f_k -f_k\Big ] + \big [f_k-f\big ]+f. \end{aligned} \end{aligned}$$

From (23),

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H})(f-f_k) \rightarrow 0 \ \ \ \text {in } L^1(X) \text { as } k\rightarrow \infty . \end{aligned}$$

Since $f_k\in L^2(X)\cap B^0_{1,1}(X)$, we have proved that

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H}) f_k -f_k =0 \ \ \ \text {in } L^1(X) \text { for } k\in {\mathbb {N}}. \end{aligned}$$

In addition,

$$\begin{aligned} \Vert f_k-f\Vert _1 \rightarrow 0 \ \ \ \text { as } k\rightarrow \infty . \end{aligned}$$

Consequently,

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{L}) f = f \ \ \text {in } L^1(X) \end{aligned}$$

for all $f\in B^0_{1,1}(X)$.

This completes our proof. $\square$

Corollary 3.5

We have the following continuous embedding

$$\begin{aligned} B^{0}_{1,1}\hookrightarrow L^1(X). \end{aligned}$$

Proof

Let $\psi _0,\psi$ be even functions such that ${\text {supp}}\psi _0\subset \{\lambda : |\lambda |\le 2\}$ and ${\text {supp}}\psi \subset \{\lambda : 1/2\le |\lambda |\le 2\}$, and

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\lambda )=1, \ \ \ \lambda \in {\mathbb {R}}, \end{aligned}$$

where $\psi _j(\lambda )=\psi (2^{-j}\lambda ), \ j =1,2,\ldots$.

By Lemma 3.4,

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\sqrt{H})f=f \ \ \text {in } L^1(X) \end{aligned}$$

for $f\in B^{0}_{1,1}(X)$.

It follows that

$$\begin{aligned} \Vert f\Vert _1 \le \sum _{j=0}^\infty \Vert \psi _j(\sqrt{H})f\Vert _1. \end{aligned}$$

This, along with (22), implies that

$$\begin{aligned} \Vert f\Vert _1\lesssim \Vert f\Vert _{B^{0}_{1,1}}. \end{aligned}$$

This completes our proof. $\square$

We are now ready to prove Lemma 3.2.

Proof of Lemma 3.2

Assume that $\{f_k\}$ is a Cauchy sequence in $B^0_{1,1}(X)$. Hence, this is also a Cauchy sequence in $L^1(X)$ since $B^0_{1,1}(X)\hookrightarrow L^1(X)$. As a consequence, $f_k \rightarrow f \in L^1(X)$ for some $f\in L^1(X)$. On the other hand, we have

$$\begin{aligned} \Vert e^{-H}\Vert _{1\rightarrow 1} + \Vert tH e^{-tH}\Vert _{1\rightarrow 1}\lesssim 1 \end{aligned}$$

uniformly in $t>0$.

Therefore,

$$\begin{aligned} \Vert e^{-H}f_k\Vert _1 \rightarrow \Vert e^{-H}f\Vert _1 \ \ \text {as } k\rightarrow \infty , \end{aligned}$$

and

$$\begin{aligned} \Vert tH e^{-tH}f_k\Vert _1 \rightarrow \Vert tH e^{-tH}f\Vert _1 \ \ \text {as } k\rightarrow \infty . \end{aligned}$$

Since $\{f_k\}$ is a Cauchy sequence in $B^0_{1,1}(X)$, for any $\epsilon >0$ there exists N such that for $m, k\ge N$,

$$\begin{aligned} \Vert e^{-H}(f_k-f_m)\Vert _1+ \int _0^{1}\Vert tL e^{-tH}(f_k-f_m)\Vert _1\frac{dt}{t}<\epsilon . \end{aligned}$$

Fixing k, then using Fatou’s Lemma we have

$$\begin{aligned} \begin{aligned} \Vert e^{-H}(f_k-f)\Vert _1&+ \int _0^{1}\Vert tL e^{-tH}(f_k-f)\Vert _1\frac{dt}{t}\\&\le \lim _{m\rightarrow \infty }\Vert e^{-H}(f_k-f_m)\Vert _1+ \liminf _{m\rightarrow \infty }\int _0^{1}\Vert tL e^{-tH}(f_k-f_m)\Vert _1\frac{dt}{t}\\&<\epsilon . \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} f_k \rightarrow f \ \ \ \text {in } B^0_{1,1}(X). \end{aligned}$$

This completes our proof. $\square$

3.2 Atomic Decomposition

In order to establish atomic decomposition for the Besov space, we need another Calderón reproducing formula.

Proposition 3.6

Let $\varphi$ be as in Lemma 2.4. Let $\psi \in C^\infty _0(\mathbb {R})$ be an even function with supp $\psi \subset (-1, 1)$ and $\int \psi =2\pi$. Let $\Phi$ and $\Psi$ be the Fourier transforms of $\varphi$ and $\psi$, respectively. Then we have, for $f\in B^0_{1,1}(X)$,

$$\begin{aligned} \begin{aligned} f&=\Phi (2^{-2}\sqrt{H})\Psi (2^{-2}\sqrt{H})f-\int _0^{1/4} (t\sqrt{H})\Phi '(t\sqrt{H}) \Psi (t\sqrt{H})f\frac{dt}{t} \\&\quad - \int _0^{1/4} (t\sqrt{H})\Psi '(t\sqrt{H}) \Phi (t\sqrt{H})f\frac{dt}{t} \end{aligned} \end{aligned}$$

(24)

in $L^1(X)$.

Proof

Similarly to the proof of Lemma 3.4, it suffices to prove the proposition for $f\in L^2(X)\cap B^0_{1,1}(X)$. Observe that

$$\begin{aligned} \begin{aligned} \int _0^{1/4} (tz)(\Phi \Psi )^{'}(tz)\frac{dt}{t}&= \int _0^{z/4} (\Phi \Psi )^{'}(u) du\\&=\Phi (z/4)\Psi (z/4) - \Phi (0)\Psi (0)\\&=\Phi (z/4)\Psi (z/4) - 1, \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \begin{aligned} 1&= \Phi (z/4)\Psi (z/4)-\int _0^{1/4} (tz)(\Phi \Psi )^{'}(tz)\frac{dt}{t}\\&= \Phi (z/4)\Psi (z/4)-\int _0^{1/4} (tz)\Phi '(tz) \Psi (tz)\frac{dt}{t} - \int _0^{1/4} (tz)\Psi '(tz) \Phi (tz)\frac{dt}{t}. \end{aligned} \end{aligned}$$

This, along with spectral theory, yields

$$\begin{aligned} \begin{aligned} f&=\Phi (2^{-2}\sqrt{H})\Psi (2^{-2}\sqrt{H})f-\int _0^{1/4} (t\sqrt{H})\Phi '(t\sqrt{H}) \Psi (t\sqrt{H})f\frac{dt}{t} \\&\quad - \int _0^{1} (t\sqrt{H})\Psi '(t\sqrt{H}) \Phi (t\sqrt{H})f\frac{dt}{t} \end{aligned} \end{aligned}$$

(25)

in $L^2(X)$.

Set

$$\begin{aligned} {\mathcal {F}}(\sqrt{H})&= \Phi (\sqrt{H})\Psi (\sqrt{H})-\int _0^{1/4} (t\sqrt{H})\Phi '(t\sqrt{H}) \Psi (t\sqrt{H})\frac{dt}{t} \\&\quad - \int _0^{1/4} (t\sqrt{H})\Psi '(t\sqrt{H}) \Phi (t\sqrt{H})\frac{dt}{t}. \end{aligned}$$

Then, by Lemma 3.9 and Corollary 3.5,

$$\begin{aligned} \begin{aligned} \Vert {\mathcal {F}}( \sqrt{H})f \Vert _1&\lesssim \Vert f\Vert _1 + \Vert f\Vert _{B^0_{1,1}(X)}\\&\lesssim \Vert f\Vert _{B^0_{1,1}(X)}. \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} {\mathcal {F}}(\sqrt{H})f = g \ \ \ \text {in } L^1(X) \end{aligned}$$

for some $g\in L^1(X)$.

This, in combination with (25), implies that $f=g$ for a.e.. Therefore,

$$\begin{aligned} f= & {} \Phi (\sqrt{H})\Psi (\sqrt{H})f-\int _0^{1} (t\sqrt{H})\Phi '(t\sqrt{H}) \Psi (t\sqrt{H})f\frac{dt}{t} \\{} & {} - \int _0^{1} (t\sqrt{H})\Psi '(t\sqrt{H}) \Phi (t\sqrt{H})f\frac{dt}{t} \ \ \ \text {in } L^1(X) \end{aligned}$$

for $f\in L^2(X)\cap B^0_{1,1}(X)$.

This completes our proof. $\square$

For any bounded Borel function $\varphi$ defined on $[0,\infty )$. We now define, for $\lambda >0$,

$$\begin{aligned} \varphi ^*_{\lambda } (t\sqrt{H})f(x) =\sup _{y\in X}\frac{|\varphi (t\sqrt{H})f(y)|}{(1+ d(x,y)/t)^\lambda } \end{aligned}$$

for all $f\in L^1(X)$, $x\in X$ and $t>0$.

Definition 3.7

([27]) Let $(\varphi ,\varphi _0)$ be a pair of even functions in ${\mathcal {S}}({\mathbb {R}})$. We say that the pair $(\varphi ,\varphi _0)$ belongs to the class ${\mathcal {A}}({\mathbb {R}})$ if

$$\begin{aligned} |\varphi _0(\lambda )|>0 \ \text {for} \ \ |\lambda |< 4\epsilon , \ \ \ |\varphi (\lambda )|>0 \ \text {for} \ \ \epsilon /4< |\lambda |< 4\epsilon \end{aligned}$$

(26)

for some $\epsilon >0$, and

$$\begin{aligned} \lambda ^{-2}\varphi (\lambda ) \in {\mathcal {S}}([0,\infty )). \end{aligned}$$

Arguing similarly to the proof of Therem 1.2 in [28], we have:

Lemma 3.8

Let $(\varphi ,\varphi _0)$ be a pair of even functions in ${\mathcal {A}}({\mathbb {R}})$. Then, for $\lambda >2n$, we have

$$\begin{aligned} \Vert f\Vert _{B^{0}_{1,1}(X)}\sim \Vert (\varphi _0)^*_\lambda (\sqrt{H})f\Vert _1 +\sum _{j=1}^\infty \Vert \varphi ^*_\lambda (2^{-j}\sqrt{H}) f\Vert _1 \end{aligned}$$

for all $f\in B^{0}_{1,1}(X)$.

Lemma 3.9

Let $\varphi (\lambda )=\lambda ^2 \phi (\lambda )$ be an even function in ${\mathcal {S}}({\mathbb {R}})$. Then, for $\lambda >2n$, we have

$$\begin{aligned} \int _0^1 \Vert \varphi ^*_\lambda (t\sqrt{H})f\Vert \frac{dt}{t}\lesssim \Vert f\Vert _{B^{0}_{1,1}(X)} \end{aligned}$$

for all $f\in B^{0}_{1,1}(X)$.

Proof

Let $\psi _0,\psi$ be even functions such that ${\text {supp}}\psi _0\subset \{\lambda : |\lambda |\le 2\}$, ${\text {supp}}\psi \subset \{\lambda : 1/2\le |\lambda |\le 2\}$, and

$$\begin{aligned} \sum _{j=0}^\infty \psi _j(\lambda )=1, \ \ \ \lambda \in {\mathbb {R}}, \end{aligned}$$

where $\psi _j(\lambda )=\psi (2^{-j}\lambda ), \ j =1,2,\ldots$.

Then we have

$$\begin{aligned} \varphi (t\sqrt{H})f=\sum _{j=0}^\infty \psi _j(\sqrt{H})\varphi (t\sqrt{H})f \end{aligned}$$

(27)

for all $t\in (0,1)$.

Let $\lambda >0$ and let $t\in [2^{-j_0-1},2^{-j_0}]$ for some $j_0\ge 0$ and $M>\lambda /2$. We then have

$$\begin{aligned} \begin{aligned} \varphi (t\sqrt{H})f=\sum _{j = j_0 + 1}^\infty \varphi (t\sqrt{H})\psi _j(\sqrt{H})f+\sum _{0\le j\le j_0} \varphi (t\sqrt{H})\psi _j(\sqrt{H})f. \end{aligned} \end{aligned}$$

Set $\psi _{j,M}(\lambda )=(2^{-j}\lambda )^{-2M}\psi _j(\lambda )$. This, along with the fact that $\varphi (\lambda )=\lambda ^2 \phi (\lambda )$, yields

$$\begin{aligned} \begin{aligned} \varphi (t\sqrt{H})f=&\sum _{j=j_0+1}^\infty 2^{-2M(j-j_0)}(t\sqrt{H})^{2M}\varphi (t\sqrt{H})\psi _{j,M}(\sqrt{H})f\\&+\sum _{0\le j\le j_0} 2^{-2(j_0-j)}\phi (t\sqrt{H})(2^{-j}\sqrt{H})\psi _j(\sqrt{H})f. \end{aligned} \end{aligned}$$

By Lemma 2.7, for each $y\in X$ and $N>n$,

$$\begin{aligned} \begin{aligned} |\varphi (t\sqrt{H})f(y)|=&\sum _{j = j_0+1}^\infty 2^{-2M(j-j_0)}\int _X \frac{1}{V(y,t)}\Big (1+\frac{d(y,z)}{t}\Big )^{-N-\lambda }|\psi _{j,M}(\sqrt{H})f(z)|d\mu (z)\\&+\sum _{0\le j\le j_0} 2^{-2(j_0-j)}\int _X \frac{1}{V(y,t)}\Big (1+\frac{d(y,z)}{t}\Big )^{-N-\lambda }|\psi _j(\sqrt{H})f(z)|d\mu (z). \end{aligned} \end{aligned}$$

Using the inequality

$$\begin{aligned} \Big (1+\frac{d(x,y)}{t}\Big )^{-\lambda }\Big (1+\frac{d(y,z)}{t}\Big )^{-\lambda }\le \Big (1+\frac{d(x,z)}{t}\Big )^{-\lambda }, \end{aligned}$$

we obtain, for $x,y\in X$,

$$\begin{aligned} \begin{aligned} \frac{|\varphi (t\sqrt{H})f(y)|}{(1+d(x,y)/t)^\lambda }&\lesssim \sum _{j = j_0+1}^\infty 2^{-2M(j-j_0)}\int _X \frac{1}{V(y,t)}\Big (1+\frac{d(y,z)}{t}\Big )^{-N}\frac{|\psi _{j,M}(\sqrt{H})f(z)|}{(1+d(x,z)/t)^\lambda }d\mu (z)\\&\quad +\sum _{0\le j\le j_0} 2^{-2(j_0-j)}\int _X \frac{1}{V(y,t)}\Big (1+\frac{d(y,z)}{t}\Big )^{-N}\frac{|\psi _j(\sqrt{H})f(z)|}{(1+d(x,z)/t)^\lambda }d\mu (z)\\&\lesssim \sum _{j = j_0 +1}^\infty 2^{-2M(j-j_0)}\sup _{z\in X}\frac{|\psi _{j,M}(\sqrt{H})f(z)|}{(1+d(x,z)/t)^\lambda }d\mu (z)\\&\quad +\sum _{0\le j\le j_0} 2^{-2(j_0-j)}\sup _{z\in X}\frac{|\psi _j(\sqrt{H})f(z)|}{(1+d(x,z)/t)^\lambda }d\mu (z). \end{aligned} \end{aligned}$$

Since $t\sim 2^{-j_0}$, for $x,y\in X$ we further simplify to that

$$\begin{aligned} \frac{|\varphi (t\sqrt{H})f(y)|}{(1+d(x,y)/t)^\lambda }\lesssim & {} \sum _{j=j_0+1}^\infty 2^{-(2M-\lambda )(j-j_0)}\sup _{z\in X}\frac{|\psi _{j,M}(\sqrt{H})f(z)|}{(1+2^jd(x,z) )^\lambda }d\mu (z)\\{} & {} +\sum _{0\le j\le j_0} 2^{-2(j_0-j)}\sup _{z\in X}\frac{|\psi _j(\sqrt{H})f(z)|}{(1+2^jd(x,z))^\lambda }d\mu (z)\\{} & {} \sim \sum _{j=j_0+1}^\infty 2^{-(2M-\lambda )(j-j_0)}(\psi _{j,M})^*_\lambda f(x)\\{} & {} + \sum _{0\le j\le j_0} 2^{-2(j_0-j)}(\psi _j)^*_\lambda f(x), \end{aligned}$$

which implies that for each $x\in X$ and $t\sim 2^{-j_0} \in (0,1)$,

$$\begin{aligned} \varphi (t\sqrt{H})^*_\lambda f(x) \lesssim \sum _{j= j_0+1}^\infty 2^{-(2M-\lambda )(j-j_0)}(\psi _{j,M})^*_\lambda f(x) + \sum _{0\le j\le j_0} 2^{-2(j_0-j)}(\psi _j)^*_\lambda f(x). \end{aligned}$$

This, along with Lemma 3.8, implies that

$$\begin{aligned} \begin{aligned} \int _0^1 \Vert \varphi ^*_\lambda (t\sqrt{H})f\Vert \frac{dt}{t}&=\sum _{j_0\ge 0}\int _{2^{-j_0-1}}^{2^{-j_0}} \Vert \varphi ^*_\lambda (t\sqrt{H})f\Vert \frac{dt}{t}\\&\lesssim \sum _{j_0\ge 0}\sum _{j=j_0+1}^\infty 2^{-(2M-\lambda )(j-j_0)}\Vert (\psi _{j,M})^*_\lambda f\Vert _1 \\&\quad + \sum _{j_0\ge 0} \sum _{0\le j\le j_0} 2^{-2(j_0-j)}\Vert (\psi _j)^*_\lambda f\Vert _1\\&\lesssim \sum _{j\ge 1} \Vert (\psi _{j,M})^*_\lambda f\Vert _1 + \sum _{j\ge 0} \Vert (\psi _j)^*_\lambda f\Vert _1\\&\lesssim \Vert f\Vert _{B^0_{1,1}(X)}, \end{aligned} \end{aligned}$$

provided that $\lambda >2n$.

This completes our proof. $\square$

We now introduce the notion of atoms for the Besov space $B^0_{1,1}(X)$.

Definition 3.10

Let $\epsilon >0$. A function a is said to be an $\epsilon$-atom if there exists a ball B with $r_B\le 1$ such that

• (i) ${\text {supp}}\,a \subset B$;

• (ii) $|a(x)| \le V(B)^{-1}$;

• (iii) $|a(x)-a(y)| \le V(B)^{-1}\left( \dfrac{d(x,y)}{r_B}\right) ^\epsilon$;

• (iv) $\displaystyle \int a(x)d\mu (x)=0$ if $r_B<1$.

Theorem 3.11

(a) Let $f\in L^1(X)$. Then $f\in B^{0}_{1,1}(X)$ if and only if there exist a sequence of $\epsilon$-atoms $\{a_j\}$ for some $\epsilon >0$ and a sequence of numbers $\{\lambda _j\}\in l^1$ such that

$$\begin{aligned} f=\sum _j\lambda _ja_j \ \ \ \text {in } L^1(X), \end{aligned}$$

(28)

and

$$\begin{aligned} \Vert f\Vert _{B^{0}_{1,1}(X)}\sim \sum _j |\lambda _j|. \end{aligned}$$

(29)

(b) In particular, if $f\in B^{0}_{1,1}(X)$ supported in a ball B with $r_B =1$, then there exist a sequence of $\epsilon$-atoms $\{a_j\}$ supported in 3B for some $\epsilon >0$ and a sequence of numbers $\{\lambda _j\}$ such that (28) and (29) hold true.

Proof

(a) Let $\Phi , \Psi$ be as in Lemma 3.6 such that

$$\begin{aligned} \begin{aligned} f&=\Phi (2^{-2}\sqrt{H})\Psi (2^{-2}\sqrt{H})f-\int _0^{1/4} \widetilde{\Phi }(t\sqrt{H}) \Psi (t\sqrt{H})f\frac{dt}{t} \\&\quad - \int _0^{1/4} \widetilde{\Psi }(t\sqrt{H}) \Phi (t\sqrt{H})f\frac{dt}{t}\\&=\Phi (2^{-2}\sqrt{H})\Psi (2^{-2}\sqrt{H})f +\sum _{j=3}^\infty \Big [-\int _{2^{-j}}^{2^{-j+1}} \widetilde{\Phi }(t\sqrt{H}) \Psi (t\sqrt{H})f\frac{dt}{t} \\&\quad - \int _{2^{-j}}^{2^{-j+1}} \widetilde{\Psi }(t\sqrt{H}) \Phi (t\sqrt{H})f\frac{dt}{t}\Big ]\\&=: f_1 +f_2 \end{aligned} \end{aligned}$$

in $L^1(X)$, where $\widetilde{\Phi }(\lambda )=\lambda \Phi '(\lambda )$ and $\widetilde{\Psi }(\lambda )=\lambda \Psi '(\lambda )$.

Moreover, according to Lemma 2.4, we have, for $t>0$ and $x,y\in X$,

$$\begin{aligned} {\text {supp}}F(t\sqrt{H})(\cdot ,\cdot ) \subset \{(x,y)\in X\times X: d(x,y)<t \}, \end{aligned}$$

(30)

and

$$\begin{aligned} |F(t\sqrt{H})(x,y)| \le \frac{C}{V(x,t)}, \end{aligned}$$

(31)

where $F\in \{\Phi , \Psi , \widetilde{\Phi }, \widetilde{\Psi }\}$.

We first decompose $f_1$ as follows:

$$\begin{aligned} f_1=\sum _{Q\in {\mathcal {D}}_2} \Phi (2^{-2}\sqrt{H})\big [\Psi (2^{-2}\sqrt{H})f\cdot 1_Q\big ]. \end{aligned}$$

For each $Q\in {\mathcal {D}}_{2}$ as in Remark 2.3, we set

$$\begin{aligned} s_Q = V(Q) \Big (\sup _{y\in Q}|\Psi (2^{-2}\sqrt{H})f(y)|\Big ) \end{aligned}$$

and

$$\begin{aligned} a_Q=\frac{1}{s_Q} \Phi (2^{-2}\sqrt{H})\big [(\Psi (2^{-2}\sqrt{H})f)\cdot 1_Q\big ]. \end{aligned}$$

(32)

It is clear that

$$\begin{aligned} f_1=\sum _{Q\in {\mathcal {D}}_2} s_Qa_Q. \end{aligned}$$

For the part $f_2$, we write

$$\begin{aligned} f_2= & {} \sum _{j=3}^\infty \sum _{Q\in {\mathcal {D}}_j}\Big [-\int _{2^{-j}}^{2^{-j+1}} \widetilde{\Phi }(t\sqrt{H}) \big (\Psi (t\sqrt{H})f\cdot 1_Q\big )\frac{dt}{t} \\{} & {} - \int _{2^{-j}}^{2^{-j+1}} \widetilde{\Psi }(t\sqrt{H}) \big (\Phi (t\sqrt{H})f\cdot 1_Q\big )\frac{dt}{t}\Big ]. \end{aligned}$$

For each $Q\in {\mathcal {D}}_j$ with $j\ge 3$, we set

$$\begin{aligned} s_Q= - V(Q) \sup _{y\in Q}\Big [\int ^{2^{-j+1}}_{2^{-j}}|\Psi (t\sqrt{H})f(y)|\frac{dt}{t}+\int ^{2^{-j+1}}_{2^{-j}}|\Phi (t\sqrt{H})f(y)|\frac{dt}{t}\Big ], \end{aligned}$$

and

$$\begin{aligned} a_Q= & {} \frac{1}{s_Q} \Big [\int ^{2^{-j+1}}_{2^{-j}} \widetilde{\Phi }(t\sqrt{H})\big (\Psi (t\sqrt{H})f\cdot 1_Q\big )\frac{dt}{t}\nonumber \\ {}{} & {} +\int ^{2^{-j+1}}_{2^{-j}} \widetilde{\Psi }(t\sqrt{H})\big (\Phi (t\sqrt{H})f\cdot 1_Q\big )\frac{dt}{t}\Big ]. \end{aligned}$$

(33)

Then we have

$$\begin{aligned} f_2=\sum _{j\ge 3}\sum _{Q\in {\mathcal {D}}_j} s_Qa_Q. \end{aligned}$$

Therefore,

$$\begin{aligned} f=f_1+f_2=\sum _{j\ge 2}\sum _{Q\in {\mathcal {D}}_j}s_Qa_Q. \end{aligned}$$

We next claim that $a_Q$ is an atom for each $Q \in {\mathcal {D}}_j, j\ge 2$. Indeed, for $j=2$ we have

$$\begin{aligned} \begin{aligned} a_Q(x) =\frac{1}{s_Q} \int _Q \Phi (2^{-2}\sqrt{H})(x,y)\Psi (2^{-2}\sqrt{H})f(y)d\mu (y). \end{aligned} \end{aligned}$$

It follows, by (30) and Remark 2.3, that ${\text {supp}}a_Q \subset 3 B(x_Q, 2^{-2}) \subset B_Q:=B(x_Q,1)$. Moreover, owing to (31),

$$\begin{aligned} \begin{aligned} |a_Q(x)|&\le \frac{1}{s_Q} \int _Q |\Phi (2^{-2}\sqrt{H})(x,y)|\,|\Psi (2^{-2}\sqrt{H})f(y)|d\mu (y)\\&\le \frac{1}{V(Q)} \int _Q |\Phi (2^{-2}\sqrt{H})(x,y)|d\mu (y)\\&\lesssim \frac{1}{V(Q)}\sim \frac{1}{V(B_Q)}. \end{aligned} \end{aligned}$$

On the other hand, by Lemma 2.7,

$$\begin{aligned} \begin{aligned} |a_Q(x)-a_Q(x')|&\le \frac{1}{s_Q} \int _Q |\Phi (2^{-2}\sqrt{H})(x,y)\\&\quad -\Phi (2^{-2}\sqrt{H})(x',y)|\,|\Psi (2^{-2}\sqrt{H})f(y)|d\mu (y)\\&\le \frac{1}{V(Q)} \int _Q |\Phi (2^{-2}\sqrt{H})(x,y)-\Phi (2^{-2}\sqrt{H})(x',y)|d\mu (y)\\&\lesssim \frac{1}{V(B_Q)}\Big (\frac{d(x,x')}{r_{B_Q}}\Big )^\delta , \end{aligned} \end{aligned}$$

whenever $d(x,x')<r_{B_Q}=1$.

Hence, $a_Q$ is a multiple of an $\epsilon$-atom associated to the ball $B_Q$ for each $Q\in {\mathcal {D}}_{j}$ with $j=2$.

Arguing similarly to above, we can verify that for $Q\in {\mathcal {D}}_j, j\ge 3$, $a_Q$ satisfies (i)-(iii) in Definition 3.10 with the corresponding ball defined by $\widetilde{B}_Q = B(x_Q, 2^{-j})$. The condition $\displaystyle \int a_Q(x)d\mu (x)=0$ follows directly from Lemma 2.9 and the fact that $\widetilde{\Phi }$ and $\widetilde{\Psi }$ are even and $\widetilde{\Phi }(0)=\widetilde{\Psi }(0)=0$. Hence, $a_Q$ is a multiple of an $\epsilon$-atom associated to $B_Q$ with $\epsilon =\delta$ for each $Q\in {\mathcal {D}}_j$, $j\ge 3$.

It remains to show that

$$\begin{aligned} \sum _{j\ge 2}^\infty \sum _{Q\in D_j}|s_Q|\lesssim \Vert f\Vert _{\dot{B}^0_{1,1}(X)}. \end{aligned}$$

Indeed, from the definition of $\{s_Q\}$, we have, for $\lambda > 2D$

$$\begin{aligned} \begin{aligned} \sum _{Q\in {\mathcal {D}}_2}|s_Q|&\lesssim \sum _{Q\in {\mathcal {D}}_0} V(Q) \inf _{x\in Q} \Psi _\lambda ^*f(x)\\&\lesssim \sum _{Q\in {\mathcal {D}}_0}\int _Q \Psi _\lambda ^*f(x)d\mu (x)\\&\sim \Vert \Psi _\lambda ^*f\Vert _1. \end{aligned} \end{aligned}$$

It follows, by using Lemma 3.8, that

$$\begin{aligned} \begin{aligned} \sum _{Q\in {\mathcal {D}}_0}|s_Q|\lesssim \Vert f\Vert _{\dot{B}^0_{1,1}(X)}. \end{aligned} \end{aligned}$$

We now show that

$$\begin{aligned} \begin{aligned} \sum _{j\ge 3}\sum _{Q\in {\mathcal {D}}_j}|s_Q|\lesssim \Vert f\Vert _{\dot{B}^0_{1,1}(X)}. \end{aligned} \end{aligned}$$

Indeed, for $Q\in {\mathcal {D}}_j$ with $j\ge 3$,

$$\begin{aligned} \begin{aligned} s_Q&\le V(Q) \Big [\int ^{2^{-j+1}}_{2^{-j}}\inf _{x\in Q}\Psi (t\sqrt{H})^*_\lambda f(x)|\frac{dt}{t}+\int ^{2^{-j+1}}_{2^{-j}} \inf _{x\in Q}\Phi (t\sqrt{H})^*_\lambda f(x)|\frac{dt}{t}\Big ]\\&\le \Big [\int ^{2^{-j+1}}_{2^{-j}}\Vert \Psi (t\sqrt{H})^*_\lambda f\Vert _{L^1(Q)}\frac{dt}{t}+\int ^{2^{-j+1}}_{2^{-j}} \Vert \Phi (t\sqrt{H})^*_\lambda f\Vert _{L^1(Q)}\frac{dt}{t}\Big ]. \end{aligned} \end{aligned}$$

This, together with Lemma 3.9, implies that

$$\begin{aligned} \begin{aligned} \sum _{j\ge 1}\sum _{Q\in {\mathcal {D}}_j}|s_Q|&\lesssim \int _0^1 \Vert \Psi (t\sqrt{H})^*_\lambda f\Vert _1 f{dt}{t} +\int _0^1 \Vert \Phi (t\sqrt{H})^*_\lambda f\Vert _1 f{dt}{t}\\&\lesssim \Vert f\Vert _{\dot{B}^0_{1,1}(X)}. \end{aligned} \end{aligned}$$

For the reverse direction, it suffices to prove that there exists $C>0$ such that

$$\begin{aligned} \Vert e^{- H}a\Vert _1+ \int _0^{1}\Vert t H e^{-tH}a\Vert _1\frac{dt}{t}\le C \end{aligned}$$

for every $\epsilon$-atom a.

Assume that a is an $\epsilon$-atom associated to a ball B. Since $\Vert a\Vert _1\le 1$, we have

$$\begin{aligned} \Vert e^{-H}a\Vert _1\lesssim \Vert a\Vert _1\lesssim 1. \end{aligned}$$

It remains to prove that

$$\begin{aligned} \int _0^{1}\Vert tH e^{-tH}a\Vert _1\frac{dt}{t}\lesssim 1. \end{aligned}$$

To do this, we write

$$\begin{aligned} \begin{aligned} \Vert a\Vert _{B^{0}_{1,1}(X)}&=\int _0^{4r_B^2}\Vert tH e^{-tH}a\Vert _{L^1(4 B)}\frac{dt}{t}+\int _0^{4r_B^2}\Vert tH e^{-tH}a\Vert _{L^1(X\backslash 4 B)}\frac{dt}{t}\\&~~+\int _{\min \{(4r_B^2),1\}}^1 \Vert tH e^{-tH}a\Vert _1\frac{dt}{t}\\&:=E_1+E_2+E_3. \end{aligned} \end{aligned}$$

For the second term $E_2$, using the Gaussian upper bound of ${\widetilde{q}}_t(x,y)$,

$$\begin{aligned} \begin{aligned} \Vert tH e^{-tH}a\Vert _{L^1(X\backslash 4 B)}&\lesssim \exp \Big (-c\frac{d( B,X\backslash 4 B)^2}{t}\Big )\Vert a\Vert _1\\&\lesssim \frac{\sqrt{t}}{r_B}, \end{aligned} \end{aligned}$$

which implies $E_2\le C$.

To estimate the term $E_1$, using the fact that

$$\begin{aligned} \int _X {\widetilde{q}}_{t}(x,y) d\mu (y)=0, \end{aligned}$$

we obtain

$$\begin{aligned} \begin{aligned} \Vert tH e^{-tH}a\Vert _{L^1(4B)}&\le \int _{4 B}\Big |\int _{B} {\widetilde{q}}_{t}(x,y)(a(y)-a(x))d\mu (y)\Big | d\mu (x). \end{aligned} \end{aligned}$$

By the smoothness condition of the atom a and the Gaussian upper bound of ${\widetilde{q}}_t(x,y)$, we have

$$\begin{aligned} \begin{aligned} \Big |\int _{B} {\widetilde{q}}_t(x,y)(a(y)-a(x))d\mu (y)\Big |&\lesssim \frac{1}{V(B)}\int _{B} \frac{1}{V(x,t)}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )\Big (\frac{d(x,y)}{r_B}\Big )^{\epsilon }d\mu (y)\\&\lesssim \frac{1}{V(B)}\Big (\frac{\sqrt{t}}{r_B}\Big )^{\epsilon }, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned} \int _{4 B}\Big |\int _{B} {\widetilde{q}}_{t}(x,y)(a(y)-a(x))d\mu (y)\Big | d\mu (x)\lesssim \Big (\frac{\sqrt{t}}{r_B}\Big )^{\epsilon }. \end{aligned} \end{aligned}$$

It follows that $E_1 \lesssim 1$.

It remains to estimate $E_3$. Note that if $r_B=1$, then $E_3=0$. Hence, we need only to consider the case $r_B<1$. Due to the cancellation property of the atom a, we have

$$\begin{aligned} \begin{aligned} \Vert tH e^{-tH }a\Vert _1&= \int _{X\backslash 4B} \Big |\int _{B} \big ({\widetilde{q}}_{t }(x,y)-{\widetilde{q}}_{t}(x,x_B)\big ) a(y) d\mu (y)\Big | d\mu (x)\\&\lesssim \int _{X\backslash 4B}\Big |\int _{B} \Big (\frac{d(y,x_B)}{\sqrt{t}}\Big )^{\delta }\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )|a(y)|d\mu (y)d\mu (x)\\&\lesssim \Big (\frac{r_B}{\sqrt{t}}\Big )^{\delta } \Vert a\Vert _1\int _X \frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big ) d\mu (x)\\&\lesssim \Big (\frac{r_B}{\sqrt{t}}\Big )^{\delta }. \end{aligned} \end{aligned}$$

It follows that $E_3 \lesssim 1.$

This completes our proof of (a).

(b) Assume that ${\text {supp}}f\subset B$ with $r_B=1$. Recall that in (a) we have proved that

$$\begin{aligned} f = \sum _{j=2}^\infty \sum _{Q\in {\mathcal {D}}_j} s_Q a_Q, \end{aligned}$$

where $\{s_Q\}$ is a sequence of numbers satisfying (29) and $\{a_Q\}$ is a sequence of $\epsilon$-atoms defined by (32) and (33). From (30), (32) and (33), we have

$$\begin{aligned} f = \sum _{j=0}^\infty \sum _{Q\in {\mathcal {D}}_j: Q\cap \tfrac{3}{2}B \ne \emptyset } s_Qa_Q \end{aligned}$$

and

$$\begin{aligned} {\text {supp}}a_Q\subset 3B \ \ \ \text {whenever } Q\cap \tfrac{3}{2}B \ne \emptyset . \end{aligned}$$

This completes the proof of (b). $\square$

We now introduce a new variant of the inhomogeneous Besov spaces. For $\ell >0$, the Besov space $B^{0,\ell }_{1,1}(X)$ is defined as the set of functions $f\in L^1(X)$ such that

$$\begin{aligned} \Vert f\Vert _{B^{0,\ell }_{1,1}(X)}:=\Vert e^{-\ell ^2H}f\Vert _1+ \int _0^{\ell ^2}\Vert tH e^{-tH}f\Vert _1\frac{dt}{t}. \end{aligned}$$

When $\ell =1$, we simply write $B^{0}_{1,1}(X)$.

Definition 3.12

Let $\epsilon >0$ and $\ell >0$. A function a is said to be an $(\epsilon ,\ell )$-atom if there exists a ball B such that

• (i) ${\text {supp}}\,a \subset B$;

• (ii) $|a(x)| \le V(B)^{-1}$;

• (iii) $|a(x)-a(y)| \le V(B)^{-1}\left( \dfrac{d(x,y)}{r_B}\right) ^\epsilon$;

• (iv) $\displaystyle \int a(x)d\mu (x)=0$ if $r_B<\ell$.

Using the approach in the proof of Theorem 3.11 and the scaling argument, we are also able to prove the following theorem.

Theorem 3.13

Let $\ell >0$ and $f\in L^1(X)$. Then $f\in B^{0,\ell }_{1,1}(X)$ if and only if there exist a sequence of $(\epsilon ,\ell )$-atoms $\{a_j\}$ for some $\epsilon >0$ and a sequence of numbers $\{\lambda _j\}$ such that

$$\begin{aligned} f=\sum _j\lambda _ja_j, \end{aligned}$$

(34)

and

$$\begin{aligned} \Vert f\Vert _{B^{0,\ell }_{1,1}(X)}\sim \sum _j |\lambda _j|. \end{aligned}$$

(35)

In particular, if $f\in B^{0,\ell }_{1,1}(X)$ supported in a ball B with $r_B =\ell$, then there exist a sequence of $(\epsilon ,\ell )$-atoms $\{a_j\}_j$ supported in 3B for some $\epsilon >0$ and a sequence of numbers $\{\lambda _j\}$ such that (34) and (35) hold true.

4 Proofs of Main Results

4.1 Proof of Theorem 1.4

We state the following results in which the proofs of Lemma 4.1 and Proposition 4.2 below are similar to those of Lemmas 3.4, 3.2, 3.3 and Corollary 3.5.

Lemma 4.1

Let $\psi$ be an even function in ${\mathcal {S}}({\mathbb {R}})$ such that ${\text {supp}}\psi \subset \{\lambda : 1/2\le |\lambda |\le 2\}$, and

$$\begin{aligned} \sum _{j \in {\mathbb {Z}}} \psi _j(\lambda )=1, \ \ \ \lambda >0, \end{aligned}$$

where $\psi _j(\lambda )=\psi (2^{-j}\lambda ), \ j \in {\mathbb {Z}}$.

Then we have

$$\begin{aligned} \sum _{j \in {\mathbb {Z}}} \psi _j(\sqrt{L})f=f \ \ \text {in } L^1(X) \end{aligned}$$

for $f\in \dot{B}^{0,L}_{1,1}(X)$.

Proposition 4.2

The following properties hold true for the homogeneous Besov space $\dot{B}^{0,L}_{1,1}(X)$.

• (i) The homogeneous Besov space $\dot{B}^{0,L}_{1,1}(X)$ is complete.

• (ii) The inclusion $\dot{B}^{0,L}_{1,1}(X)\hookrightarrow L^1(X)$ is continuous.

• (iii) For each $p\in [1,\infty )$, the space $L^p(X)$ is dense in $\dot{B}^{0,L}_{1,1}(X)$.

Proposition 4.3

Let $\varphi$ be as in Lemma 2.4 and let $\Phi$ be the Fourier transforms of $\varphi$. For each $m\in {\mathbb {N}}$,

$$\begin{aligned} \begin{aligned} f= c\int _0^{\infty } (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f \frac{dt}{t} \ \ \text {in } L^1(X) \end{aligned} \end{aligned}$$

(36)

for $f\in B^{0,L}_{1,1}(X)$, where $\displaystyle c=\Big [\int _0^{\infty } z^{2\,m}e^{-z^2}\Phi (z) \frac{dz}{z}\Big ]^{-1}$.

Proof

Similarly to the proof of Lemma 3.4, it suffices to prove the proposition for $f\in L^2(X)\cap \dot{B}^{0,L}_{1,1}(X)$. By spectral theory,

$$\begin{aligned} \begin{aligned} f= c\int _0^{\infty } (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f \frac{dt}{t} \end{aligned} \end{aligned}$$

(37)

in $L^2(X)$.

On the other hand, from Lemma 2.7,

$$\begin{aligned} \begin{aligned} \Big \Vert \int _0^{\infty } (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f \frac{dt}{t}\Big \Vert _1&\lesssim \int _0^{\infty } \Vert (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f\Vert _1 \frac{dt}{t}\\&\lesssim \int _0^{\infty } \Vert t^2L e^{-t^2L} f\Vert _1 \frac{dt}{t}\\&\lesssim \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}. \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} \int _0^{\infty } (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f \frac{dt}{t} = g \ \ \ \text {in } L^1(X) \end{aligned}$$

for some $g\in L^1(X)$.

This, in combination with (37), implies that $f=g$ for a.e.. Therefore,

$$\begin{aligned} f= c\int _0^{\infty } (t^2L)^me^{-t^2L}\Phi (t\sqrt{L})f \frac{dt}{t} \ \ \ \text {in } L^1(X) \end{aligned}$$

for $f\in L^2(X)\cap \dot{B}^{0,L}_{1,1}(X)$.

This completes our proof. $\square$

Proof of Theorem 1.4:

The proof of the atomic decomposition for functions $f\in \dot{B}^{0,L}_{1,1}(X)$ is similar to that of Theorem 4.2 in [5] and the proof of Theorem 3.11. Hence, we leave it to the interested reader.

For the reverse direction, it suffices to show that there exists $C>0$ such that

$$\begin{aligned} \int _0^{\infty }\Vert tL e^{-tL}a\Vert _1\frac{dt}{t}\le C \end{aligned}$$

for every (L, M)-atom a.

Suppose that a is an (L, M)-atom associated with a ball B. Then we have

$$\begin{aligned} \int _0^{\infty }\Vert tL e^{-tL}a\Vert _1\frac{dt}{t}= \int _0^{r_B^2}\Vert tL e^{-tL}a\Vert _1\frac{dt}{t}+\int _{r_B^2}^\infty \Vert tL e^{-tL}a\Vert _1\frac{dt}{t}. \end{aligned}$$

For the first term, we have

$$\begin{aligned} \begin{aligned} \int _0^{r_B^2} \Vert tL e^{-tL}a\Vert _1\frac{dt}{t}&=\int _0^{r_B^2} t\Vert e^{-tL}La\Vert _1\frac{dt}{t}\\&\lesssim \int _0^{r_B^2} t\Vert a\Vert _1\frac{dt}{t}\\&\lesssim \int _0^{r_B^2} t r_B^{-2}\frac{dt}{t}\\&\lesssim 1. \end{aligned} \end{aligned}$$

For the second term, using $a=L^Mb$,

$$\begin{aligned} \begin{aligned} \int _{r_B^2}^\infty \Vert tL e^{-tL}a\Vert _1\frac{dt}{t}&=\int _{r_B^2}^\infty \Vert (tL)^{M+1} e^{-tL}b\Vert _1\frac{dt}{t^{M+1}}\\&\lesssim \int _{r_B^2}^\infty \Vert b\Vert _1\frac{dt}{t^{M+1}}\\&\lesssim \int _{r_B^2}^\infty r_B^{2M}\frac{dt}{t^{M+1}}\\&\lesssim 1. \end{aligned} \end{aligned}$$

This completes our proof. $\square$

4.2 Proof of Theorem 1.6

We refer the reader to Sect. 2.1 for the index set ${\mathcal {I}}$, the family functions $\{\psi _\alpha \}_{\alpha }$ and the family of balls $\{B_\alpha \}_{\alpha }$ which will be used in this section.

Lemma 4.4

For each $\alpha \in {\mathcal {I}}$ and $f\in L^1(X)$ we have

$$\begin{aligned} \int _X \int _0^{\rho (x_\alpha )^2} \Big |\big [tH e^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x) \lesssim \Vert f\psi _\alpha \Vert _1. \end{aligned}$$

(38)

Proof

By (L2), we have

$$\begin{aligned} \begin{aligned} \int _X \int _0^{\rho (x_\alpha )^2}&\Big |\big [tH e^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&\lesssim \int _X \int _0^{\rho (x_\alpha )^2}\int _{B_\alpha } \frac{1}{V(x,\sqrt{t})}\Big (\frac{\sqrt{t}}{\sqrt{t}+\rho (y)}\Big )^{\delta }\\&\quad \exp \Big ( -c\frac{d(x,y)^2}{t}\Big )|(f\psi _\alpha )(y)|d\mu (y)\frac{dt}{t}d\mu (x). \end{aligned} \end{aligned}$$

Since $\rho (y)\sim \rho (x_\alpha )$ for all $y\in B_\alpha$, we have

$$\begin{aligned} \begin{aligned} \int _X \int _0^{\rho (x_\alpha )^2}&\Big |\big [tH e^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&\lesssim \int _{B_\alpha } \int _0^{\rho (x_\alpha )^2}\int _X\frac{1}{V(x,\sqrt{t})}\exp \Big (\\&\quad -c\frac{d(x,y)^2}{t}\Big ) \Big (\frac{\sqrt{t}}{\rho (x_\alpha )}\Big )^{\delta }|(f\psi _\alpha )(y)|d\mu (x)\frac{dt}{t}d\mu (y). \end{aligned} \end{aligned}$$

Using the fact that

$$\begin{aligned} \int _X\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )d\mu (x)\lesssim 1, \end{aligned}$$

(39)

we obtain that

$$\begin{aligned} \begin{aligned}&\int _X \int _0^{\rho (x_\alpha )^2} \Big |\big [tH e^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&\lesssim \int _{B_\alpha } \int _0^{\rho (x_\alpha )^2} \Big (\frac{\sqrt{t}}{\rho (x_\alpha )}\Big )^{\delta }|(f\psi _\alpha )(y)|\frac{dt}{t} d\mu (y)\\&\lesssim \int _{B_\alpha }|(f\psi _\alpha )(y)|d\mu (y)\\&\lesssim \Vert f\psi _\alpha \Vert _1. \end{aligned} \end{aligned}$$

This completes our proof. $\square$

Lemma 4.5

For each $f\in L^1(X)$ we have

$$\begin{aligned} \sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2} \Big |tL e^{-tL}f(x)\psi _\alpha (x) -tL e^{-tL}(f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x) \lesssim \Vert f\Vert _1.\nonumber \\ \end{aligned}$$

(40)

Proof

Denote

$$\begin{aligned} {\mathcal {I}}_{1,\alpha } =\{\beta \in {\mathcal {I}}: B_\alpha ^*\cap B_\beta \ne \emptyset \} \ \ \ \text {and} \ \ \ {\mathcal {I}}_{2,\alpha } =\{\beta \in {\mathcal {I}}: B_\alpha ^*\cap B_\beta = \emptyset \}, \end{aligned}$$

where $B^*_\alpha = 4B_\alpha , \alpha \in {\mathcal {I}}$.

Observe that

$$\begin{aligned} \begin{aligned}&tL e^{-tL}f(x)\psi _\alpha (x) -tL e^{-tL}(f\psi _\alpha )(x)\\ {}&=\sum _{\beta \in {\mathcal {I}}}\int _{B_\beta }q_t(x,y)\big (\psi _\alpha (x)-\psi _\alpha (y)\big )(f\psi _\beta )(y)d\mu (y). \end{aligned} \end{aligned}$$

Then we write

$$\begin{aligned} \begin{aligned} \sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2}&\Big |tL e^{-tL}f(x)\psi _\alpha (x) -tL e^{-tL}(f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&=\sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2} \Big |\sum _{\beta \in {\mathcal {I}}}\int _{B_\beta }q_t(x,y)\big (\psi _\alpha (x)\\&\quad -\psi _\alpha (y)\big )(f\psi _\beta )(y)d\mu (y) \Big |\frac{dt}{t}d\mu (x)\\&\le \sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2} \Big |\sum _{\beta \in {\mathcal {I}}_{1,\alpha }}\int _{B_\beta }q_t(x,y)\big (\psi _\alpha (x)\\&\quad -\psi _\alpha (y)\big )(f\psi _\beta )(y)d\mu (y) \Big |\frac{dt}{t}d\mu (x)\\&\ \ \ \ + \sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2} \Big |\sum _{\beta \in {\mathcal {I}}_{2,\alpha }}\int _{B_\beta }q_t(x,y)\big (\psi _\alpha (x)\\&\quad -\psi _\alpha (y)\big )(f\psi _\beta )(y)d\mu (y) \Big |\frac{dt}{t}d\mu (x)\\&=:E_1 + E_2. \end{aligned} \end{aligned}$$

We estimate $E_1$ first. Owing to Lemma 2.1 and the upper bound of $q_t(x,y)$, we have

$$\begin{aligned} \begin{aligned} \sum _{\beta \in {\mathcal {I}}_{1,\alpha }}\int _{B_\beta }&\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )\frac{d(x,y)}{\rho (x_\alpha )}|(f\psi _\beta )(y)|d\mu (y)\\&\lesssim \sum _{\beta \in {\mathcal {I}}_{1,\alpha }}\int _{B_\beta }\frac{1}{V(x,\sqrt{t})}\exp \Big (-c'\frac{d(x,y)^2}{t}\Big )\frac{\sqrt{t}}{\rho (x_\alpha )}|(f\psi _\beta )(y)|d\mu (y). \end{aligned} \end{aligned}$$

This implies that

$$\begin{aligned} \begin{aligned} E_1&\lesssim \sum _{\alpha \in {\mathcal {I}}}\sum _{\beta \in {\mathcal {I}}_{1,\alpha }}\int _X \int _0^{\rho (x_\alpha )^2}\int _{B_\beta }\frac{1}{V(x,\sqrt{t})}\exp \\&\quad \Big (-c'\frac{d(x,y)^2}{t}\Big )\frac{\sqrt{t}}{\rho (x_\alpha )}|(f\psi _\beta )(y)|d\mu (y)\frac{dt}{t}d\mu (x)\\&\lesssim \sum _{\beta \in {\mathcal {I}}} \int _{B_\beta }\sum _{\alpha \in {\mathcal {J}}_{1,\beta }}\int _0^{\rho (x_\alpha )^2}\int _X \frac{1}{V(x,\sqrt{t})}\exp \\&\quad \Big (-c'\frac{d(x,y)^2}{t}\Big )d\mu (x)\frac{\sqrt{t}}{\rho (x_\alpha )}|(f\psi _\beta )(y)|\frac{dt}{t}d\mu (y), \end{aligned} \end{aligned}$$

where

$$\begin{aligned} {\mathcal {J}}_{1,\beta }= \{\alpha \in {\mathcal {I}}: B_\alpha ^*\cap B_\beta \ne \emptyset \}. \end{aligned}$$

Since $\sharp {\mathcal {J}}_{1,\beta }$ is uniformly bounded in $\beta \in {\mathcal {I}}$, using (39) we obtain

$$\begin{aligned} \begin{aligned} \sum _{\alpha \in {\mathcal {I}}}\int _X \int _0^{\rho (x_\alpha )^2}&\Big |tL e^{-tL}f(x)\psi _\alpha (x) -tL e^{-tL}(f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&\lesssim \sum _{\beta \in {\mathcal {I}}} \int _{B_\beta }\sum _{\alpha \in {\mathcal {J}}_{1,\beta }}\int _0^{\rho (x_\alpha )^2}\frac{\sqrt{t}}{\rho (x_\alpha )}|(f\psi _\beta )(y)|\frac{dt}{t}d\mu (y)\\&\lesssim \sum _{\beta \in {\mathcal {I}}}\Vert f\psi _\beta \Vert _1 \sim \Vert f\Vert _1. \end{aligned} \end{aligned}$$

If $\beta \in {\mathcal {I}}_{2,\alpha }$, then $\psi _\alpha (y)=0$ for all $y\in B_\beta$. Therefore,

$$\begin{aligned} E_2 =\sum _{\alpha \in {\mathcal {I}}}\int _{B_\alpha } \int _0^{\rho (x_\alpha )^2} \Big |\sum _{\beta \in {\mathcal {I}}_{2,\alpha }}\int _{B_\beta }q_t(x,y) \psi _\alpha (x)(f\psi _\beta )(y)d\mu (y) \Big |\frac{dt}{t}d\mu (x). \end{aligned}$$

By the upper bound of $q_t(x,y)$ and the fact that $d(x,y)> \rho (x_\alpha )$ whenever $x\in B_\alpha , y\in B_\beta$ with $\beta \in {\mathcal {I}}_{2,\alpha }$, we further simplify to that

$$\begin{aligned} E_2\lesssim & {} \sum _{\alpha \in {\mathcal {I}}}\int _{B_\alpha } \int _0^{\rho (x_\alpha )^2} \sum _{\beta \in {\mathcal {I}}_{2,\alpha }}\int _{B_\beta }\frac{1}{V(x,\sqrt{t})}\exp \\{} & {} \quad \Big (-c\frac{d(x,y)^2}{t}\Big ) \psi _\alpha (x)|(f\psi _\beta )(y)|d\mu (y) \frac{dt}{t}d\mu (x)\\\lesssim & {} \sum _{\alpha \in {\mathcal {I}}}\int _{B_\alpha } \int _0^{\rho (x_\alpha )^2} \sum _{\beta \in {\mathcal {I}}_{2,\alpha }}\int _{B_\beta }\frac{1}{V(x,d(x,y))}\exp \\{} & {} \quad \Big (-c'\frac{d(x,y)^2}{t}\Big ) \frac{\sqrt{t} }{d(x,y)} |(f\psi _\beta )(y)|d\mu (y) \frac{dt}{t}d\mu (x)\\\lesssim & {} \sum _{\alpha \in {\mathcal {I}}}\int _{B_\alpha } \int _0^{\rho (x_\alpha )^2} \sum _{\beta \in {\mathcal {I}}_{2,\alpha }}\int _{B_\beta }\frac{1}{V(x,d(x,y))}\exp \\{} & {} \quad \Big (-c'\frac{d(x,y)^2}{t}\Big ) \frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (y) \frac{dt}{t}d\mu (x)\\\lesssim & {} \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \int _0^{\rho (x_\alpha )^2} \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x,d(x,y))}\exp \\{} & {} \quad \Big (-c'\frac{d(x,y)^2}{t}\Big ) \frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y), \end{aligned}$$

where

$$\begin{aligned} {\mathcal {J}}_{2,\beta }= \{\alpha \in {\mathcal {I}}: B_\alpha ^*\cap B_\beta = \emptyset \}. \end{aligned}$$

Note that $d(x,y)\sim d(x, x_\beta )\sim d(x_\alpha , x_\beta )$ whenever $x\in B_\alpha , y\in B_\beta$ with $\alpha \in {\mathcal {J}}_{2,\beta }$. Hence,

$$\begin{aligned} \begin{aligned} E_2&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \int _0^{\rho (x_\alpha )^2} \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\exp \\&\quad \Big (-c'\frac{d(x_\alpha ,x_\beta )^2}{t}\Big ) \frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y)\\&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \int _0^{\rho (x_\alpha )^2} \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\exp \\&\quad \Big (-c'\frac{d(x_\alpha ,x_\beta )^2}{\rho (x_\alpha )^2}\Big ) \frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y). \end{aligned} \end{aligned}$$

On the other hand, invoking (5) we have

$$\begin{aligned} \begin{aligned} \exp \Big (-c'\frac{d(x_\alpha ,x_\beta )^2}{\rho (x_\alpha )^2}\Big )&\lesssim \frac{\rho (x_\alpha )}{d(x_\alpha ,x_\beta )}\\&\lesssim \frac{\rho (x_\beta )}{d(x_\alpha ,x_\beta )}\Big (1+\frac{d(x_\alpha ,x_\beta )}{\rho (x_\beta )}\Big )^{\frac{k_0}{k_0+1}}\\&\lesssim \Big (\frac{\rho (x_\beta )}{d(x_\alpha ,x_\beta )}\Big )^{\frac{1}{k_0+1}}. \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} E_2&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \int _0^{\rho (x_\alpha )^2} \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\\&\quad \Big (\frac{\rho (x_\beta )}{d(x_\alpha ,x_\beta )}\Big )^{\frac{1}{k_0+1}}\frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y)\\&\sim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \int _0^{\rho (x_\alpha )^2} \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\\&\quad \Big (\frac{\rho (x_\beta )}{d(x,x_\beta )}\Big )^{\frac{1}{k_0+1}}\frac{\sqrt{t} }{\rho (x_\alpha )} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y)\\&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta } \sum _{\alpha \in {\mathcal {J}}_{2,\beta }}\int _{B_\alpha }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\Big (\frac{\rho (x_\beta )}{d(x,x_\beta )}\Big )^{\frac{1}{k_0+1}} |(f\psi _\beta )(y)|d\mu (x) \frac{dt}{t}d\mu (y). \end{aligned} \end{aligned}$$

Since $\{B_\beta \}_{\beta \in {\mathcal {I}}}$ is a finite overlapping family and $\cup _{\alpha \in {\mathcal {J}}_{2,\beta }} B_\alpha \subset X\backslash B^*_\beta$, we also obtain that

$$\begin{aligned} \begin{aligned} E_2&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta }|(f\psi _\beta )(y)| \int _{X\backslash B^*_\beta }\frac{1}{V(x_\beta ,d(x,x_\beta ))}\Big (\frac{\rho (x_\beta )}{d(x,x_\beta )}\Big )^{\frac{1}{k_0+1}} d\mu (x)d\mu (y)\\&\lesssim \sum _{\beta \in {\mathcal {I}}}\int _{B_\beta }|(f\psi _\beta )(y)| d\mu (y)\\&\lesssim \Vert f\Vert _1. \end{aligned} \end{aligned}$$

This completes our proof. $\square$

We are ready to give the proof of Theorem 1.6.

Proof of Theorem 1.6:

We first prove that each function $f\in \dot{B}_{1,1}^{0,L}(X)$ admits an atomic decomposition as in the statement of the theorem.

Indeed, we first observe that from Theorem 3.13,

$$\begin{aligned} \begin{aligned} \int _0^{\rho (x_\alpha )^2} \Vert tH e^{-tH}(f\psi _\alpha ) \Vert _1\frac{dt}{t}\lesssim&\ \ \int _X \int _0^{\rho (x_\alpha )^2} \Big |\big [tH e^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&+ \int _X \int _0^{\rho (x_\alpha )^2} \Big |tL e^{-tL}f(x)\psi _\alpha (x)\\&-tL e^{-tL}(f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\&+ \int _X \int _0^{\infty } \Big |tL e^{-tL}f(x)\psi _\alpha (x)\Big |\frac{dt}{t}d\mu (x).\\ \end{aligned} \end{aligned}$$

By Lemmas 4.4 and 4.5, we have $f\psi _\alpha \in B^{0,\ell _\alpha }_{1,1}(X)$ with $\ell _\alpha =\epsilon _0\rho (x_\alpha )/3$, where $\epsilon _0$ is the constant in Lemma 2.1. Therefore, we can write

$$\begin{aligned} f\psi _\alpha = \sum _{j}\lambda _{j,\alpha }a_{j,\alpha }, \end{aligned}$$

where $a_{j,\alpha }$ is an $(\epsilon ,\ell _\alpha )$-atom associated to a ball $B_{j,\alpha }\subset 3B_\alpha$ for each j, and $\{\lambda _{j,\alpha }\}_j$ is a sequence of numbers satisfying

$$\begin{aligned} \sum _{j}|\lambda _{j,\alpha }|\le \int _0^{\rho (x_\alpha )^2} \Vert tH e^{-tH}(f\psi _\alpha ) \Vert _1\frac{dt}{t}. \end{aligned}$$

Note that $3B_\alpha = B(x_\alpha , \epsilon _0 \rho (x_\alpha ))$, by (5),

$$\begin{aligned} \rho (x_{B_{j,\alpha }})\ge C_\rho ^{-1} \rho (x_\alpha )(1 +\epsilon _0)^{-\frac{k_0}{k_0+1}}, \end{aligned}$$

which implies that

$$\begin{aligned} 3\ell _\alpha = \epsilon _0\rho (x_\alpha )< C_\rho \epsilon _0 (1 +\epsilon _0)^{\frac{k_0}{k_0+1}}\rho (x_{B_{j,\alpha }}). \end{aligned}$$

From (iii) in Lemma 2.1, $C_\rho \epsilon _0 (1 +\epsilon _0)^{\frac{k_0}{k_0+1}}<1$. Hence,

$$\begin{aligned} \ell _\alpha \le \rho (x_{B_{j,\alpha }}). \end{aligned}$$

Consequently, each $a_{j,\alpha }$ is also an $(\epsilon , \rho (\cdot ))$ atom associated to the ball $B_{j,\alpha }$.

Therefore, by Lemmas 4.4, 4.5 and (ii) in Proposition 4.2,

$$\begin{aligned} f= \sum _{\alpha }\sum _{j} \lambda _{j,\alpha }a_{j,\alpha } \end{aligned}$$

such that

$$\begin{aligned} \sum _{\alpha }\sum _{j} |\lambda _{j,\alpha }|\lesssim & {} \sum _{\alpha }\int _X \int _0^{\rho (x_\alpha )^2} \Big |\big [tHe^{-tH} -tL e^{-tL}\big ](f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\{} & {} + \sum _{\alpha }\int _X \int _0^{\rho (x_\alpha )^2} \Big |tL e^{-tL}f(x)\psi _\alpha (x) -tL e^{-tL}(f\psi _\alpha )(x)\Big |\frac{dt}{t}d\mu (x)\\{} & {} + \sum _{\alpha }\int _X \int _0^{\infty } \Big |tL e^{-tL}f(x)\psi _\alpha (x)\Big |\frac{dt}{t}d\mu (x)\\\lesssim & {} \sum _{\alpha } \Vert f\psi _\alpha \Vert _1 + \Vert f\Vert _1 + \int _X \int _0^{\infty } \Big | tL e^{-tL}f(x)\Big |\frac{dt}{t}d\mu (x)\\\lesssim & {} \Vert f\Vert _{1} + \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}\\\lesssim & {} \Vert f\Vert _{\dot{B}^{0,L}_{1,1}(X)}. \end{aligned}$$

This completes the proof of the first direction.

For the reverse direction, it suffices to prove that there exists $C>0$ such that

$$\begin{aligned} \Vert a\Vert _{\dot{B}^{0,L}_{1,1}(X)}\le C \end{aligned}$$

for every $(\epsilon , \rho (\cdot ))$ atom with some $\epsilon >0$.

To do this, suppose that a is an $(\epsilon , \rho (\cdot ))$ atom associated with a ball B. Then we write

$$\begin{aligned} \begin{aligned} \Vert a\Vert _{\dot{B}^{0,L}_{1,1}(X)}&=\int _0^{4r_B^2}\Vert tLe^{-tL}a\Vert _{L^1(3B)} \frac{dt}{t}+\int _0^{4r_B^2}\Vert tLe^{-tL}a\Vert _{L^1(X\backslash 3B)} \frac{dt}{t}\\&\quad +\int _{4r_B^2}^\infty \Vert tLe^{-tL}a\Vert _1\frac{dt}{t}:=A_1+A_2+A_3. \end{aligned} \end{aligned}$$

For the second term $A_2$, using the Gaussian upper bound of $q_t(x,y)$,

$$\begin{aligned} \begin{aligned} \Vert tLe^{-tL}a\Vert _{L^1(X\backslash 3B)}&\lesssim \exp \Big (-c\frac{d(B,X\backslash 3B)^2}{t}\Big )\Vert a\Vert _1\\&\lesssim \Big (\frac{\sqrt{t}}{r_B}\Big )^{\delta }, \end{aligned} \end{aligned}$$

which implies $A_2\le C$.

To estimate the term $A_1$, observe that

$$\begin{aligned} \begin{aligned} \Vert tLe^{-tL}a\Vert _{L^1(3B)}&\le \int _{3B}\Big |\int _{X} q_{t}(x,y)(a(y)-a(x))d\mu (y)\Big | d\mu (x)\\&~~~~+ \int _{3B}\Big |\int _{X} q_{t}(x,y)a(x)d\mu (y)\Big | d\mu (x). \end{aligned} \end{aligned}$$

By the smoothness condition of the atom a, we have

$$\begin{aligned} \begin{aligned} \Big |\int _{B} q_t(x,y)(a(y)-a(x))d\mu (y)\Big |&\lesssim \frac{1}{V(B)}\int _{X} \frac{1}{V(x,t)}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )\Big (\frac{d(x,y)}{r_B}\Big )^{\epsilon }d\mu (y)\\&\lesssim \frac{1}{V(B)}\Big (\frac{\sqrt{t}}{r_B}\Big )^{\epsilon }, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned} \int _{3B}\Big |\int _{X} q_{t}(x,y)(a(y)-a(x))d\mu (y)\Big | d\mu (x)\lesssim \Big (\frac{\sqrt{t}}{r_B}\Big )^{\epsilon }. \end{aligned} \end{aligned}$$

(41)

Invoking the condition (II) to give

$$\begin{aligned} \Big |\int _{X} q_t(x,y)a(x)d\mu (y)\Big |\lesssim |a(x)| \Big (\frac{\sqrt{t}}{\rho (x)}\Big )^{\delta }\Big (1+\frac{\sqrt{t}}{\rho (x)}\Big )^{-K}. \end{aligned}$$

This, along with (41), implies that

$$\begin{aligned} \begin{aligned} A_1&\lesssim \int _0^{4r_B^2} \Big (\frac{\sqrt{t}}{r_B}\Big )^{\epsilon } \frac{dt}{t}+\int _0^{4r_B^2}\int _X |a(x)| \Big (\frac{\sqrt{t}}{\rho (x)}\Big )^{\delta }\Big (1+\frac{\sqrt{t}}{\rho (x)}\Big )^{-K} d\mu (x)\frac{dt}{t}\\&\lesssim 1 + \int _X |a(x)| \int _0^\infty \Big (\frac{\sqrt{t}}{\rho (x)}\Big )^{\delta }\Big (1+\frac{\sqrt{t}}{\rho (x)}\Big )^{-K} \frac{dt}{t}d\mu (x)\\&\lesssim 1 + \int _X |a(x)|d\mu (x)\\&\lesssim 1. \end{aligned} \end{aligned}$$

It remains to estimate $A_3$. To do this, we consider two cases.

Case 1: $0<r_B \le \rho (x_B)$

Due to the cancellation property of the atom a, we have

$$\begin{aligned} \begin{aligned} \Vert tLe^{-tL}a\Vert _1&= \int _X \Big |\int _{3B} \big (q_{t }(x,y)-q_{t}(x,x_B)\big ) a(y) d\mu (y)\Big | d\mu (x)\\&\lesssim \int _X\Big |\int _{3Q} \Big (\frac{d(y,x_B)}{\sqrt{t}}\Big )^{\delta }\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )|a(y)|d\mu (y)d\mu (x)\\&\lesssim \Big (\frac{r_B}{\sqrt{t}}\Big )^{\delta } \sup _{y\in 3Q} \int _X \frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big ) d\mu (x)\\&\lesssim \Big (\frac{r_B}{\sqrt{t}}\Big )^{\delta }. \end{aligned} \end{aligned}$$

It follows that $A_3 \lesssim 1.$

Case 2: $r_B> \rho (x_B)$

Observe that by (5), for $z\in 3B$,

$$\begin{aligned} \begin{aligned} \rho (z)&\lesssim \rho (x_B)\Big (1+\frac{d(z,x_B)}{\rho (x_B)})^{\frac{k_0}{k_0+1}}\\&\lesssim \rho (x_B)\frac{r_B}{\rho (x_B)} = r_B. \end{aligned} \end{aligned}$$

This, together with (L1), yields that

$$\begin{aligned} \Vert tLe^{-tL}a\Vert _1= & {} \int _X\Big |\int _{3B} q_t(x,y)a(y)d\mu (y)\Big | dx\\\lesssim & {} \int _X\int _{3B} \Big (\frac{\rho (y)}{\sqrt{t}}\Big )^{\delta }\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )|a(y)|d\mu (y)d\mu (x)\\{} & {} \int _X\int _{3B} \Big (\frac{r_B}{\sqrt{t}}\Big )^{\delta }\frac{1}{V(x,\sqrt{t})}\exp \Big (-c\frac{d(x,y)^2}{t}\Big )|a(y)|d\mu (y)d\mu (x)\\{} & {} \lesssim \Big (\frac{r_B}{t}\Big )^{\delta }\Vert a\Vert _1\\{} & {} \lesssim \Big (\frac{r_B}{t}\Big )^{\delta }. \end{aligned}$$

It follows that $A_3 \lesssim 1.$

This completes our proof. $\square$

5 Application to Boundedness of Riesz Transforms Associated to Schrödinger Operators on ${\mathbb {R}}^n$

In this section, we show the boundedness of the Riesz transforms associated to Schrödinger operators $L=-\Delta + V$ on ${\mathbb {R}}^n$ on the new Besov space $\dot{B}^{0,L}_{1,1}({\mathbb {R}}^n)$. It is worth noticing that although we restrict ourselves to consider the Schrödinger operators on ${\mathbb {R}}^n$, our approach works well in more general setting including settings listed in Remark 1.1.

Let $L=-\Delta + V$ be a Schrödinger operator on ${\mathbb {R}}^n, n\ge 3$ with $V\in RH_{n/2}$. Our main result in this section is the following theorem.

Theorem 5.1

The Riesz transform $\nabla L^{-1/2}$ is bounded from $\dot{B}^{0,L}_{1,1}({\mathbb {R}}^n)$ to $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)$.

We would like to remark that in the classical case, the Riesz transform $\nabla (-\Delta )^{-1/2}$ is bounded on the classical Besov spaces $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)$. See for example [6, Proposition 2.4]. In the setting of Theorem  5.1, we have a better estimates for the Riesz transform $\nabla L^{-1/2}$ since by Theorem 3.13, $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)\hookrightarrow \dot{B}^{0,L}_{1,1}({\mathbb {R}}^n)$. Therefore, as a consequence of Theorems 3.13 and 5.1, we have:

Corollary 5.2

The Riesz transform $\nabla L^{-1/2}$ is bounded on $\dot{B}^{0}_{1,1}({\mathbb {R}}^n)$.

In order to prove Theorem 5.1 we need the following technical lemma.

Lemma 5.3

Let a be an (L, M) atom associated with a ball B with $M\ge 1$. Then for $\alpha \in (0,1)$, we have

$$\begin{aligned} \Vert L^\alpha a\Vert _p\lesssim r_B^{-2\alpha }|B|^{1/p-1} \end{aligned}$$

for every $p\in [1,\infty ]$.

Proof

We have

$$\begin{aligned} \begin{aligned} L^\alpha a&= c\int _0^\infty s^{1-\alpha } Le^{-sL} a \frac{ds}{s}\\&=\int _0^{r_B^2}s^{1-\alpha } e^{-sL} La \frac{ds}{s} +\int _{r_B^2}^\infty s^{1-\alpha } Le^{-sL} a \frac{ds}{s}, \end{aligned} \end{aligned}$$

which implies

$$\begin{aligned} \begin{aligned} \Vert L^\alpha a\Vert _p&\lesssim \int _0^{r_B^2}s^{1-\alpha } \Vert e^{-sL}\Vert _{p\rightarrow p} \Vert La\Vert _p \frac{ds}{s} +\int _{r_B^2}^\infty s^{-\alpha } \Vert sLe^{-sL}\Vert _{p\rightarrow p} \Vert a\Vert _p \frac{ds}{s}\\&\lesssim \int _0^{r_B^2}s^{1-\alpha } \Vert La\Vert _p \frac{ds}{s}+\int _{r_B^2}^\infty s^{-\alpha } \Vert sLe^{-sL}\Vert _{p\rightarrow p} \Vert a\Vert _p \frac{ds}{s}\\&\lesssim \int _0^{r_B^2}s^{1-\alpha } r_B^{-2}|B|^{1/p-1} \frac{ds}{s}+\int _{r_B^2}^\infty s^{-\alpha }|B|^{1/p-1} \frac{ds}{s}\\&\lesssim r_B^{-2\alpha }|B|^{1/p-1}. \end{aligned} \end{aligned}$$

$\square$

Proof of Theorem 5.1:

Let a be an (L, M) atom associated to a ball B. It suffices to prove that

$$\begin{aligned} \Vert \nabla L^{-1/2}a\Vert _{\dot{B}^{0}_{1,1}({\mathbb {R}}^n)}:=\int _0^{\infty }\Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _{1}\frac{dt}{t}\lesssim 1. \end{aligned}$$

To do this, we write

$$\begin{aligned} \begin{aligned}&=\int _0^{4r_B^2}\Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _{L^1(4 B)}\frac{dt}{t}+\int _0^{4r_B^2}\Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _{L^1({\mathbb {R}}^n\backslash 4 B)}\frac{dt}{t}\\&~~+\int _{ 4r_B^2}^\infty \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _1\frac{dt}{t}\\&:=E_1+E_2+E_3. \end{aligned} \end{aligned}$$

Using the $L^r$-boundedness of $\nabla ^2 L^{-1}$ (see [2]), we have

$$\begin{aligned} \begin{aligned} \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _{L^1(4 B)}&\le \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _{L^r(4 B)} |B|^{1/r'}\\&\lesssim \Vert t\nabla e^{t\Delta }\nabla ^2 L^{-1/2}a\Vert _{L^r(4 B)} |B|^{1/r'}\\&\lesssim \sqrt{t}\Vert t\nabla e^{t\Delta }\Vert _{r\rightarrow r}\Vert \nabla ^2 L^{-1}\Vert _{r\rightarrow r} \Vert L^{1/2}a\Vert _r |B|^{1/r'}\\&\lesssim \frac{\sqrt{t}}{r_B}. \end{aligned} \end{aligned}$$

It follows that $E_1\lesssim 1$.

For the term $E_3$ we have, for $a=Lb$,

$$\begin{aligned} \begin{aligned} \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _1&= \Vert t\nabla (-\Delta ) e^{t\Delta } L^{1/2}b\Vert _1\\&\le \Vert t\nabla (-\Delta ) e^{t\Delta }\Vert _{1\rightarrow 1} \Vert L^{1/2}b\Vert _1\\&\lesssim \frac{r_B}{\sqrt{t}}, \end{aligned} \end{aligned}$$

which implies that $E_3\lesssim 1$.

It remains to estimate $E_2$. To do this, we use the following formula

$$\begin{aligned} L^{-1/2}=c\int _0^\infty s^{3/2}Le^{-sL}\frac{ds}{s} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a&= c\int _0^\infty ts^{3/2}(-\Delta ) e^{t\Delta }\nabla Le^{-sL}a\frac{ds}{s}\\&=c\int _0^t ts^{3/2}(-\Delta ) e^{t\Delta }\nabla e^{-sL}(La)\frac{ds}{s}\\ {}&+c\int _t^\infty ts^{3/2} \nabla e^{t\Delta } \nabla ^2 Le^{-sL}a\frac{ds}{s}. \end{aligned} \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _1&\lesssim \int _0^t ts^{3/2}\Vert (-\Delta ) e^{t\Delta }\nabla e^{-sL}(La)\Vert _1\frac{ds}{s}\\&\quad + \int _t^\infty ts^{3/2} \Vert \nabla e^{t\Delta } \nabla ^2 Le^{-sL}a\Vert _1\frac{ds}{s}. \end{aligned} \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \begin{aligned} \Vert (-\Delta ) e^{t\Delta }\nabla e^{-sL}(La)\Vert _1&\le \Vert (-\Delta ) e^{t\Delta }\Vert _{1\rightarrow 1}\Vert \nabla e^{-sL}\Vert _{1\rightarrow 1}\Vert La\Vert _1\\&\lesssim \frac{1}{\sqrt{s} tr_B^2}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert \nabla e^{t\Delta }\nabla ^2 Le^{-sL}a\Vert _1&\le \Vert \nabla e^{t\Delta }\Vert _{1\rightarrow 1}\Vert \nabla ^2 L^{1-\alpha } e^{-sL}\Vert _{1\rightarrow 1}\Vert L^\alpha a\Vert _1\\&\lesssim \frac{1}{\sqrt{t} s^{2-\alpha } r_B^{2\alpha }}, \ \ \alpha \in (0,1/2). \end{aligned} \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert t(-\Delta ) e^{t\Delta }\nabla L^{-1/2}a\Vert _1\lesssim \int _0^t \frac{s}{r_B^2}\frac{ds}{s} + \int _t^\infty \frac{\sqrt{t}}{s^{1/2-\alpha }}\frac{ds}{s}\sim \frac{t}{r_B^2} + \frac{t^{\alpha }}{r_B^{2\alpha }}, \end{aligned}$$

which implies that

$$\begin{aligned} E_2\lesssim \int _0^{4r_B^2}\Big (\frac{t}{r_B^2} + \frac{t^{\alpha }}{r_B^{2\alpha }}\Big ) \frac{dt}{t} \lesssim 1. \end{aligned}$$

It was proved in [12, 22] that there exists $\beta >0$ such that

$$\begin{aligned} \int _{{\mathbb {R}}^n} |\sqrt{t}\nabla p_t(x,y)|e^{\beta \frac{|x-y|^2}{t}}dx+\int _{{\mathbb {R}}^n} |t\nabla ^2 p_t(x,y)|e^{\beta \frac{|x-y|^2}{t}}dx \le 1. \end{aligned}$$

This completes our proof. $\square$