Fractional powers of the Schrödinger operator on weigthed Lipschitz spaces

Abstract

In the setting of the semigroup generated by the Schrödinger operator $L= -\Delta +V$ with the potential V satisfying an appropriate reverse Hölder condition, we consider some non-local fractional differentiation operators. We study their behaviour on suitable weighted smoothness spaces. Actually, we obtain such continuity results for positive powers of L as well as for the mixed operators $L^{\alpha /2}V^{\sigma /2}$ and $L^{-\alpha /2}V^{\sigma /2}$ with $\sigma >\alpha$, together with their adjoints.

1 Introduction

We will work in the frame of the harmonic analysis related to the Schrödinger differential operator in ${\mathbb {R}}^d$ with $d>2$ as given in [11] by Shen,

$$\begin{aligned} L u = -\Delta u + V\, u\,, \end{aligned}$$

where the potential V is a non-negative locally integrable function belonging to $RH_q$ for some $q>d/2$. We remind that the last property means that there exists a constant C such that

$$\begin{aligned} \left( \frac{1}{|B|} \int _B V^q \right) ^{1/q} \le C \frac{1}{|B|} \int _B V, \end{aligned}$$

(1)

holds for any ball $B \subset {\mathbb {R}}^d$. When the left hand side is replaced by $\sup _B V$, we say that $V \in RH_\infty$. Our work relies strongly on Shen’s estimates which are valid only for $d>2$. His main tool is a comparison between the fundamental solutions of L and $-\Delta$. As it is known, the fundamental solution of the Laplacian has a different form when $d=2$.

During the last two decades many authors have been working in this context, studying the behaviour of associate integral operators such as negative powers of L, the maximal semigroup operator, Riesz Transforms, the Littlewood-Paley function, among others. Results have been obtained for boundedness on weighted $L^p$ spaces as well as on appropriate regularity spaces and their weighted versions.

In order to describe the latter spaces we recall an important feature of this environment. Our assumption on V allows to define a critical radius function $\rho : {\mathbb {R}}^d\longmapsto {\mathbb {R}}^+$ as

$$\begin{aligned} \rho (x)=\sup \left\{ r: \frac{r^2}{|B(x,r))|}\int _{B(x,r)}V \,\, \le 1\right\} . \end{aligned}$$

Under the present assumptions on the potential it is possible to show that $0<\rho (x)<\infty$ for all x and that the following inequalities hold

$$\begin{aligned} c_\rho ^{-1} \rho (x)\left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N_0} \ \le \ \rho (y) \ \le \ c_\rho \,\rho (x)\left( 1+\frac{|x-y|}{\rho (x)}\right) ^{\frac{N_0}{N_0+1}}, \end{aligned}$$

(2)

for some constants $c_\rho$ and $N_0$, independent of x and y. For a proof of these facts see [11].

It turns out that this function $\rho$ plays a special role in defining the suitable weighted regularity spaces we are interested in. Let us point out that the first appearance of the space $BMO_L$, the appropriate substitute of the John-Nirenberg space BMO, can be found in [6]. The main difference with the classical case is that, besides the condition on the oscillations, averages over balls B(x, r) with $r>\rho (x)$ must be also uniformly bounded. This reveals the fact that $-L$ is a perturbation of the Laplacian. Later on, appropriated variants of integral Lipschitz spaces were introduced as well as weighted versions. We remind their precise definition.

Given a weight w and some $\beta$, $0 \le \beta <1$, we say that a locally integrable function f belongs to $BMO^\beta _\rho (w)$ if there is a constant C such that

$$\begin{aligned} \frac{1}{w(B)} \int _B |f-f_B| \le C |B|^{\beta /d}, \end{aligned}$$

(3)

for any ball B, where $f_B$ stands for the average with respect to Lebesgue measure, and

$$\begin{aligned} \frac{1}{w(B)}\int _B |f| \le C |B|^{\beta /d} \end{aligned}$$

(4)

for any ball $B=B(x,r)$ with $r\ge \rho (x)$. Clearly condition (4) implies (3) and moreover, under a mild restriction on the weight, it is enough to know that (4) is satisfied on critical balls, that is, balls of the type $B(x,\rho (x))$. See [1] for the details.

As in the case of Laplacian, these weighted integral Lipschitz spaces can be identified with functions satisfying some point-wise smoothness under certain restriction on the weight. To be precise, we introduce doubling classes of weights adapted to this context. For $\mu \ge 1$ let us denote $D_{\mu }^{\rho }=\bigcup _{\theta \ge 0}D_\mu ^{\rho ,\theta }$, where $D_\mu ^{\rho ,\theta }$ is the class of weights w such that for some constant C,

$$\begin{aligned} w(B(x,R))\le C w(B(x,r))\left( \frac{R}{r}\right) ^{d\mu }\left( 1+\frac{R}{\rho (x)}\right) ^{\theta }, \end{aligned}$$

(5)

for any $x\in {\mathbb {R}}^d$ and $0<r\le R$.

It turns out that when $w \in D_\mu ^\rho$ for some $\mu \ge 1$ and $0<\beta <1$, $BMO_\rho ^\beta (w)$ has a point-wise description and, in fact, it coincides with $\Lambda _\rho ^\beta (w)$, defined as the class of functions f satisfying

$$\begin{aligned} |f(x)|\le C W_{\beta }(x, \rho (x)) \end{aligned}$$

(6)

and

$$\begin{aligned} |f(x)-f(z)|\le C[W_{\beta }(x, |x-z|)+W_{\beta }(z,|x-z|)] \end{aligned}$$

(7)

for all $x,\,z \in {\mathbb {R}}^d$ such that $|x-z|\le \rho (x)$, where

$$\begin{aligned} W_{\beta }(x,r)=\int _{B(x,r)} \frac{w(u)}{|u-x|^{d-\beta }}du. \end{aligned}$$

(8)

For a proof of this equivalence we refer to Proposition 4 in [4]. When we deal with the unweighted case $w=1$ we will simply denote $\Lambda _\rho ^\beta (1)=\Lambda _\rho ^\beta$.

The aim of this work is the study of the behaviour of non-local fractional differentiation operators acting on the above mentioned weighted regularity spaces $\Lambda _\rho ^\beta (w)$ with $0<\beta <1$. Being differentiation operators we expect them to reduce smoothness, and such is the case for positive fractional powers of L. In fact, in Sect. 3 we are able to prove that $L^{\alpha /2}$ maps $\Lambda _\rho ^\beta (w)$ into $\Lambda _\rho ^{\beta -\alpha }(w)$ under suitable assumptions on w. For the case of $w=1$, this result was obtained in [9]. However they achieve that by using a different description of $\Lambda _\rho ^\beta$ based on extensions of functions to the positive half space ${\mathbb {R}}^d \times (0,\infty )$ by means of the generated semigroup. Here we give a direct proof, obtaining the needed pointwise estimates, and our approach also works for a wide class of weighted spaces.

In Sect. 4 we also deal with some mixed fractional differentiation operators that combine powers of L, positives or negatives, with multiplication by positive powers of the potential V. In order to prove continuity of these operators we analyse general pointwise multiplication operators by a function $\phi$, which are bounded from $\Lambda _\rho ^\beta (w)$ into $\Lambda _\rho ^{\eta }(w)$ with $0<\eta \le \beta \le 1$. We find sufficient conditions on the function $\phi$ that for $w=1$ turn to be also necessary. As a consequence we derive boundedness results for fractional differentiation operators as $L^{\alpha /2} V^{\sigma /2}$ and $L^{-\alpha /2} V^{\sigma /2}$ with $\sigma > \alpha$, under stronger assumptions on the potential.

2 Preliminaries

All along this work we deal with weights in the doubling classes $D^\rho _\mu$ defined above. Let us notice that any weight in the class $A^\rho _p$, appearing in [5], belongs to $D^\rho _p$. We recall that For a given $p> 1$, $\displaystyle A_p^{\rho }=\bigcup _{\theta \ge 0}A_p^{\rho ,\theta }$, where $A_p^{\rho ,\theta }$ is defined as those weights w such that

$$\begin{aligned} \left( \int _B w\right) ^{1/p} \left( \int _B w^{-\tfrac{1}{p-1}}\right) ^{1/p'} \le C |B| \left( 1+ \frac{r}{\rho (x)}\right) ^{\theta }, \end{aligned}$$

(9)

for every ball $B=B(x,r)$, where |B| stands for the Lebesgue measure of B. Similarly, when $p=1$, we denote $\displaystyle A_1^{\rho }=\bigcup _{\theta \ge 0}A_1^{\rho ,\theta }$, where $A_1^{\rho ,\theta }$ is the class of weights w such that

$$\begin{aligned} \frac{1}{|B|}\int _B w \le C\left( 1+\frac{r}{\rho (x)}\right) ^{\theta } \inf _{B} w, \end{aligned}$$

(10)

for every ball $B=B(x,r)$.

These classes contain the classical Muckenhoupt families of weights but as it was shown in [5] are strictly larger. For instance if $\rho \equiv 1$, $w(x)=1+|x|^\gamma$ with $\gamma >d(p-1)$ belongs to $A_p^{\rho }$ but it is not in $A_p$. Moreover, this weight belongs to $D^\rho _\mu$ for any $\mu \ge 1$ but it is not in $D_\mu$ when $\mu <1+\gamma /d$.

We will devote this section to develop some results that will be useful in what follows. We start giving some properties for the function $W_\beta$. The first one states that $W_\beta (x,\cdot )$ inherits some kind of doubling property from w.

Lemma 1

Let $w\in D_\mu ^{\rho ,\theta }$, $x\in {\mathbb {R}}^d$ and $r>0$. Then, there exist a constant $C>0$ such that for every $\tau \ge 1$,

$$\begin{aligned} W_{\beta }(x, \tau r) \le C \tau ^{d\mu -d + \beta } W_{\beta }(x, r) \left( 1 + \frac{\tau r}{\rho (x)} \right) ^{\theta } \end{aligned}$$

Proof

Let $\tau >1$ and $j_0\in {\mathbb {Z}}$ such that $2^{j_0}\le \tau <2^{j_0+1}$. Then,

$$\begin{aligned} \begin{aligned} W_\beta (x,\tau r)&= \int _{B(x,r)}\frac{w(z)}{|x-z|^{d-\beta }}dz + \sum _{j=0}^{j_0} \int _{2^jr\le |x-z|<2^{j+1}r}\frac{w(z)}{|x-z|^{d-\beta }}dz\\&\le W_\beta (x,r) + \sum _{j=0}^{j_0} (2^jr)^{\beta -d} w(B(x,2^jr))\\&\le W_\beta (x,r) + C\sum _{j=0}^{j_0} (2^jr)^{\beta -d} w(B(x,r)) 2^{jd\mu } \left( 1+ \frac{2^j r}{\rho (x)}\right) ^\theta \\&\le W_\beta (x,r) + C\frac{w(B(x,r))}{r^{d-\beta }} \left( 1+\frac{2^{j_0} r}{\rho (x)}\right) ^\theta \sum _{j=0}^{j_0}(2^j)^{d\mu -d+\beta }\\&\le W_\beta (x,r) + C W_\beta (x,r) \left( 1+ \frac{\tau r}{\rho (x)}\right) ^\theta \tau ^{d\mu -d+\beta }. \end{aligned} \end{aligned}$$

$\square$

Next we give some estimates concerning the integral of $W_\beta$ over a ball. First notice that from the definition of $W_\beta$, the inequality $W_\beta (x,R)\ge w(B(x,R))R^{\beta -d}$ holds while the opposite is not always true. The next lemma provides a weaker version of that.

Lemma 2

Let $w\in D_\mu ^\rho$ for some $\mu \ge 1$ and $\beta >0$. Then, for any $x\in {\mathbb {R}}^d$ and $R>0$,

$$\begin{aligned} \int _{B(x,R)} W_{\beta }(y, R) dy \le C R^{\beta } w(B(x,R)). \end{aligned}$$

Proof

Let $x\in {\mathbb {R}}^d$ and $R>0$. Applying the definition of $W_\beta$ together with Lemma 1,

$$\begin{aligned} \begin{aligned} \int _{B(x,R)} W_{\beta }(y, R) dy&= \int _{B(x,R)} \int _{B(y,R)} \frac{w(z)}{|y-z|^{d-\beta }} dz dy \\&\le \int _{B(x,2R)} w(z) \int _{B(x,R)} \frac{dy}{|y-z|^{d-\beta }}dz \\&\le \int _{B(x,2R)} w(z) \int _{B(z,3R)} \frac{dy}{|y-z|^{d-\beta }}dz \\&\le C w (B(x,R) ) R^\beta . \end{aligned} \end{aligned}$$

$\square$

Lemma 3

Let $w\in D_\mu ^\rho$, $x\in {\mathbb {R}}^d$, $R>0$ and $\beta >0$. If $0<r\le \rho (x)$,

$$\begin{aligned} \frac{1}{|B(x,r)|}\int _{B(x,r)} W_{\beta }(y, \rho (y)) dy \le C W_{\beta }(x, \rho (x)). \end{aligned}$$

Proof

Let $0<r\le \rho (x)$ and $y\in B(x,r)$. By (2), taking $c=2c_\rho$ we have $B(y,\rho (y))\subset B(x,c\rho (x))$. Therefore, we may write

$$\begin{aligned} \begin{aligned} \int _{B(x,r)}W_\beta (y,\rho (y))dy&= \int _{B(x,r)}\int _{B(y,\rho (y))} \frac{w(z)}{|y-z|^{d-\beta }}dzdy \\&\le \int _{B(x,r)}\int _{B(x,c\rho (x))} \frac{w(z)}{|y-z|^{d-\beta }}dzdy \\&= \int _{B(x,c\rho (x))} w(z) \int _{B(x,r)} \frac{dy}{|y-z|^{d-\beta }}dz. \end{aligned} \end{aligned}$$

(11)

Now, we split the integration domain $B(x,c\rho (x))=B(x,2r)\cup B(x,c\rho (x))\setminus B(x,2r)$ obtaining two terms. For the first one, since w is doubling

$$\begin{aligned} \begin{aligned} \int _{B(x,2r)} w(z) \int _{B(x,r)} \frac{dy}{|y-z|^{d-\beta }}dz&\le C r^\beta w(B(x,2r))\\&\le C r^d W_\beta (x,r) \\&\le Cr^d W_\beta (x,\rho (x)). \end{aligned} \end{aligned}$$

For the remaining term, we may use that $|y-z|\ge |x-z|/2$ when $y\in B(y,r)$ and $z\notin B(x,2r)$ together with Lemma 1 to obtain

$$\begin{aligned} \begin{aligned} \int _{B(x,c\rho (x))\setminus B(x,2r)} w(z) \int _{B(x,r)} \frac{dy}{|y-z|^{d-\beta }}dz&\le C r^d \int _{B(x,c\rho (x))}\frac{ w(z)}{|x-z|^{d-\beta }}dz \\&\le C r^dW_\beta (x,\rho (x)). \end{aligned} \end{aligned}$$

$\square$

We finish this section with some results concerning the diffusion semigroup $\{T_t\}_{t>0}$ associated to $-L$, where

$$\begin{aligned} T_t f(x) = e^{-tL} f(x) = \int _{{\mathbb {R}}^d} k_t(x,y) f(y) dy. \end{aligned}$$

For this kernel $k_t$, the Feynman-Kac formula assures that

$$\begin{aligned} 0\le k_t(x,y) \le h_t(x-y)=(4\pi t)^{-d/2} e^{-\frac{|x-y|^2}{4t}}, \end{aligned}$$

where $h_t$ is the kernel associated to the classical heat diffusion semigroup. In fact, under our assumptions on V, it is shown in [11] that for each $N>0$ there exists a constant $C_N$ such that

$$\begin{aligned} k_t(x,y)\le \frac{C_N}{t^{d/2}} e^{-\frac{|x-y|^2}{5t}} \left( 1 + \frac{\sqrt{t}}{\rho (x)}+ \frac{\sqrt{t}}{\rho (y)}\right) ^{-N}, \end{aligned}$$

(12)

for x, $y\in {\mathbb {R}}^d$ and $t>0$.

It is known that the classical heat kernel satisfies the following smoothness estimate: there exist positive constants c and C such that

$$\begin{aligned} |h_t(x-y)-h_t(z- y)|\le C \frac{|x-z|}{\sqrt{t}} t^{-d/2}e^{-\frac{c|x-y|^2}{t}}, \end{aligned}$$

(13)

for any $y\in {\mathbb {R}}^d$ and $|x-z|\le \sqrt{t}$.

A similar estimate can be obtained for the kernel $k_t$. We state it in the following Lemma. For a proof of this result we refer to Proposition 4.11 in [8].

Lemma 4

For any $0<\delta <\delta _0=\min \{1,2-d/q\}$ and $N>0$, there exist constants c and $C_N$ such that

$$\begin{aligned} |k_t(x,y)-k_t(z,y)|\le C_N \left( \frac{|x-z|}{\sqrt{t}}\right) ^{\delta } t^{-d/2}e^{-\frac{c|x-y|^2}{t}}\left( 1+\frac{\sqrt{t}}{\rho (x)}+\frac{\sqrt{t}}{\rho (y)}\right) ^{-N},\nonumber \\ \end{aligned}$$

(14)

for any $y\in {\mathbb {R}}^d$ and $|x-z|\le \sqrt{t}$.

Now we present the following lemmas that give size and smoothness estimates for the difference between $k_t$ and $h_t$. They will be essential in what follows.

Lemma 5

Let $V\in RH_q$ with $q>d/2$. There exists a constant $C>0$ such that

$$\begin{aligned} |k_t(x,y)-h_t(x-y)|\le Ct^{-d/2}e^{-\frac{C|x-y|^2}{t}} \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{2-d/q}, \end{aligned}$$

(15)

for any x, $y\in {\mathbb {R}}^d$ and $t>0$.

Lemma 6

Let $V\in RH_q$ with $q>d/2$. For any $0<\delta <\delta _0=\min \{1,2-d/q\}$ there exists a constant $C>0$ such that for every x, y and $z\in {\mathbb {R}}^d$ such that $4|x-z|\le |x-y|$ and $|x-z|\le \rho (x)$,

$$\begin{aligned} |k_t(x,y)-h_t(x-y)-[k_t(z,y)-h_t(z-y)]|\le Ct^{-d/2}e^{-\frac{C|x-y|^2}{t}} \left( \frac{|x-z|}{\rho (x)}\right) ^{\delta }.\nonumber \\ \end{aligned}$$

(16)

For a proof of these estimates we refer to Proposition 2.16 and Proposition 2.17 in [7].

3 Positive fractional powers of L

In this section we study the behaviour of the operator $L^{\alpha /2}$ , with $0<\alpha <2$, acting on functions belonging to the appropriate weighted smoothness spaces we just introduced. Let us remind that this operator can be written in terms of the semigroup kernel as

$$\begin{aligned} \begin{aligned} L^{\alpha /2} f(x)&= \int _{0}^{\infty } \left( e^{-tL}-I\right) f(x) \frac{dt}{t^{1+\alpha /2}} . \end{aligned} \end{aligned}$$

The precise result is stated in the following theorem.

Theorem 1

Let $V\in RH_q$ for some $q>d/2$ and $0<\alpha<\beta <\delta _0=\min \{1,2-d/q\}$. Then $L^{\alpha /2}$ is bounded from $\Lambda _{\rho }^{\beta }(w)$ into $\Lambda _{\rho }^{\beta -\alpha }(w)$ as long as $w\in D_{\mu }^{\rho }$ with $1\le \mu < 1 + \frac{\delta _0 - \beta }{d}$.

Before proving this result, we present a few technical lemmas that summarize some of the properties we will need. Observe that estimates like (14), (15) and (16) involve exponentials functions with perhaps different exponents. For that reason the next results are worked out for a generic function of that type.

Lemma 7

Let $\gamma ,\theta \ge 0$ and ${\tilde{h}}_t$ a function of the form

$$\begin{aligned} {\tilde{h}}_t(x)= t^{-d/2}e^{-\frac{C|x|^2}{t}}, \end{aligned}$$

(17)

for some constant $C>0$. Then there exists a constant $C'$ such that

$$\begin{aligned} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) \left( \frac{|x-y|}{\sqrt{t}}\right) ^{\gamma } \left( 1+ \frac{|x-y|}{\rho (x)}\right) ^{\theta } dy\le C' \left( 1+ \frac{\sqrt{t}}{\rho (x)}\right) ^{\theta }, \end{aligned}$$

for $x\in {\mathbb {R}}^d$ and $t>0$.

Proof

To prove this inequality we are going to split the integration domain on $B(x,\rho (x))$ and its complement. For the integral over the ball we obtain

$$\begin{aligned} \begin{aligned}&\int _{B(x,\rho (x))} {\tilde{h}}_t(x-y) \left( \frac{|x-y|}{\sqrt{t}}\right) ^{\gamma } \left( 1+ \frac{|x-y|}{\rho (x)}\right) ^{\theta } dy\\&\quad \le C t^{-d/2}\int _{B(x,\rho (x))} e^{-\frac{C|x-y|^2}{t}} \left( \frac{|x-y|}{\sqrt{t}}\right) ^{\gamma }dy \le C, \end{aligned} \end{aligned}$$

by performing a simple change of variables.

For the remaining term we have

$$\begin{aligned} \begin{aligned}&\int _{B(x,\rho (x))^c} {\tilde{h}}_t(x-y) \left( \frac{|x-y|}{\sqrt{t}}\right) ^{\gamma } \left( 1+ \frac{|x-y|}{\rho (x)}\right) ^{\theta } dy\\&\quad \le C t^{-d/2}\left( \frac{\sqrt{t}}{\rho (x)}\right) ^\theta \int _{B(x,\rho (x))^c} e^{-\frac{C|x-y|^2}{t}} \left( \frac{|x-y|}{\sqrt{t}}\right) ^{\gamma +\theta }dy\\&\quad \le C \left( 1+ \frac{\sqrt{t}}{\rho (x)}\right) ^{\theta }. \end{aligned} \end{aligned}$$

$\square$

Lemma 8

Let ${\tilde{h}}_t$ be a function as in (17), $w\in D_\mu ^\rho$ for some $\mu \ge 1$ and $0<\beta \le 1$. Then, for any $x\in {\mathbb {R}}^d$, $t>0$ and $0\le r< R\le \infty$, we have

$$\begin{aligned} \int _{r<|x-y|\le R} {\tilde{h}}_t(x-y) W_{\beta }(y,|x-y|) dy \le C \int _{r< |x-y|\le 4R} \tilde{h_t}((x-y)/4) W_{\beta }(x,|x-y|) dy \end{aligned}$$

Proof

Suppose first that $r>0$ and $R<\infty$ and let $j_0$ and $k_0\in {\mathbb {Z}}$ such that

$$\begin{aligned} 2^{j_0}\sqrt{t}< r \le 2^{j_0+1}\sqrt{t}, \\ 2^{k_0}\sqrt{t} < R \le 2^{k_0+1}\sqrt{t}. \end{aligned}$$

Applying Lemma 2,

$$\begin{aligned} \begin{aligned}&\int _{r\le |x-y|\le R} {\tilde{h}}_t(x-y) W_{\beta }(y,|x-y|) dy\\&\quad \le \sum _{j=j_0}^{k_0} \int _{ 2^j\sqrt{t}<|x-y|\le 2^{j+1}\sqrt{t}} {{\tilde{h}}_t}(x-y) W_{\beta }(y,|x-y|) dy \\&\quad \le \sum _{j=j_0}^{k_0} t^{-d/2}{\tilde{h}}_1(2^j)\int _{ 2^j\sqrt{t}<|x-y|\le 2^{j+1}\sqrt{t}} W_{\beta }(y,2^{j+1}\sqrt{t}) dy \\&\quad \le C \sum _{j=j_0}^{k_0} t^{-d/2}{\tilde{h}}_1(2^j) (2^{j+1}\sqrt{t})^d W_{\beta }(x,2^{j+1}\sqrt{t}) \\&\quad \le C \sum _{j=j_0}^{k_0} t^{-d/2}\int _{ 2^{j+1}\sqrt{t}<|x-y|\le 2^{j+2}\sqrt{t}} {\tilde{h}}_t(|x-y|/4) W_{\beta }(x,|x-y|)dy \\&\quad \le C \int _{ r<|x-y|\le 4R} {\tilde{h}}_t(|x-y|/4) W_{\beta }(x,|x-y|)dy . \end{aligned} \end{aligned}$$

If $r=0$ or $R=\infty$ the proof is analogous performing the sum from $j=-\infty$ or until $j=\infty$ respectively.

$\square$

Lemma 9

Let ${\tilde{h}}$ be a function as in (17), $w\in D_\mu ^{\rho ,\theta }$ and $0<\beta \le 1$. Then, for any $x\in {\mathbb {R}}^d$, we have that

$$\begin{aligned} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) W_{\beta } (y,\rho (y)) dy\le C W_{\beta } (x,\rho (x)) \left( 1 + \frac{\sqrt{t}}{\rho (x)}\right) ^{d\mu - d + \beta + \theta }. \end{aligned}$$

Proof

First we are going to split ${\mathbb {R}}^d$ and integrate over $B(x,\rho (x))$ and its complement. For the first integral we consider $k_0\in {\mathbb {Z}}$ such that $2^{k_0}\sqrt{t}\le \rho (x)<2^{k_0+1}\sqrt{t}$. Then we may write,

$$\begin{aligned} \begin{aligned} \int _{B(x,\rho (x))}{\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy&\le \sum _{k=-\infty }^{k_0}\int _{ 2^{k}\sqrt{t}<|x-y|\le 2^{k+1}\sqrt{t}} {\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy \\&\le \sum _{k=-\infty }^{k_0} \frac{{\tilde{h}}_1(2^k)}{t^{d/2}}\int _{B(x,2^{k+1}\sqrt{t})} W_\beta (y,\rho (y))dy . \end{aligned} \end{aligned}$$

Therefore, applying Lemma 3 we get

$$\begin{aligned} \begin{aligned}&\int _{B(x,\rho (x))} {\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy\\&\quad \le C W_\beta (x,\rho (x)) \sum _{k=-\infty }^{k_0} \frac{{\tilde{h}}_1(2^k)}{t^{d/2}} (2^k\sqrt{t})^d \\&\quad \le C W_\beta (x,\rho (x)) \sum _{k=-\infty }^{\infty } \int _{2^{k}\sqrt{t}<|x-y|\le 2^{k+1}\sqrt{t}} {\tilde{h}}_t(x-y)dy \\&\quad \le C W_\beta (x,\rho (x)). \end{aligned} \end{aligned}$$

To deal with the integral over the complement we use that $W_\beta (y,\cdot )$ is an increasing function and apply Lemma 8 to obtain

$$\begin{aligned} \begin{aligned}&\int _{|x-y|\ge \rho (x)} {\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy\\&\quad \le \int _{|x-y|\ge \rho (x)} {\tilde{h}}_t(x-y)W_\beta (y,|x-y|)dy \\&\quad \le C \int _{|x-y|\ge c\rho (x)} {\tilde{h}}_t((x-y)/4)W_\beta (x,|x-y|)dy. \end{aligned} \end{aligned}$$

(18)

Now, we apply the doubling property of $W_\beta$ stated on Lemma 1 together with Lemma 7 in the following way,

$$\begin{aligned} \begin{aligned}&\int _{|x-y|\ge \rho (x)} {\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy \\&\quad \le C\frac{W_\beta (x,\rho (x))}{\rho (x)^{d\mu -d+\beta +\theta }} \int {\tilde{h}}_t((x-y)/4) |x-y|^{d\mu -d+\beta +\theta }dy \\&\quad \le C W_\beta (x,\rho (x)) \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{d\mu -d+\beta +\theta }. \end{aligned} \end{aligned}$$

(19)

$\square$

Now we can give the proof of the main result of this section.

Proof of Theorem 1

Let $w\in D_\mu ^{\rho }$ and $f\in \Lambda _{\rho }^{\beta }(w)$ for $0<\beta <\delta _0$. We can assume, without loss of generality, that $\Vert f\Vert _{\Lambda _{\rho }^{\beta }(w)}=1$. First, we are going to check condition (6). Let $x\in {\mathbb {R}}^d$, we may write

$$\begin{aligned} \begin{aligned} L^{\alpha /2} f(x)&= \int _{0}^{\infty } \left( e^{-tL}-I\right) f(x) \frac{dt}{t^{1+\alpha /2}} \pm \int _{0}^{\rho ^2(x)} \left( e^{t\Delta }-I\right) f(x) \frac{dt}{t^{1+\alpha /2}} \\ {}&=I + II + II, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} I= & {} \int _0^{\rho ^2(x)} (e^{-tL}-e^{t\Delta })f(x)\frac{dt}{t^{1+\alpha /2}}, \\ II= & {} \int _0^{\rho (x)^2} (e^{t\Delta } - I )f(x)\frac{dt}{t^{1+\alpha /2}},\\ III= & {} \int _{\rho (x)^2}^{\infty } (e^{-tL} - I) f(x)\frac{dt}{t^{1+\alpha /2}}. \end{aligned}$$

To deal with I, we may apply Lemma 5 together with the size estimate for f to obtain

$$\begin{aligned} \begin{aligned} |I|&\le \int _0^{\rho ^2(x)} \int _{{\mathbb {R}}^d} |k_t(x,y) - h_t(x-y)| |f(y)|dy \frac{dt}{t^{1+\alpha /2}}\\&\le C \int _0^{\rho ^2(x)} \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta } \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) W_{\beta }(y,\rho (y)) dy. \frac{dt}{t^{1+\alpha /2}} \end{aligned} \end{aligned}$$

Now, by Lemma 9,

$$\begin{aligned} \begin{aligned} |I|&\le C W_{\beta }(x,\rho (x)) \rho (x)^{-\delta } \int _0^{\rho ^2(x)} t^{\frac{\delta -\alpha }{2}} \frac{dt}{t}\\&\le C W_{\beta }(x,\rho (x)) \rho (x)^{-\alpha } \le C W_{\beta -\alpha }(x,\rho (x)) . \end{aligned} \end{aligned}$$

To bound II, we use first that $\int h_t(x-y)dy=1$ and then the smoothness estimate for f, obtaining

$$\begin{aligned} \begin{aligned} |II|&\le \int _0^{\rho ^2(x)} \int _{{\mathbb {R}}^d} h_t(x-y) |f(y)-f(x)| dy \frac{dt}{t^{1+\alpha /2}}\\ {}&\le \int _0^{\rho ^2(x)} \left( \int _{{\mathbb {R}}^d} h_t(x-y) W_{\beta }(x,|x-y|) dy\right. \\&\quad +\left. \int _{{\mathbb {R}}^d} h_t(x-y) W_{\beta }(y,|x-y|) dy\right) \frac{dt}{t^{1+\alpha /2}} . \end{aligned} \end{aligned}$$

We will bound the first term only, since the second one can be estimated following the same lines once we have applied Lemma (8). We divide the first term as follows,

$$\begin{aligned} \begin{aligned}&\int _0^{\rho ^2(x)} \int _{{\mathbb {R}}^d} h_t(x-y) W_{\beta }(x,|x-y|) dy \frac{dt}{t^{1+\alpha /2}}\\&\quad \le \int _0^{\rho ^2(x)} \int _{|x-y|< \sqrt{t}} h_t(x-y) W_{\beta }(x,|x-y|) dy \frac{dt}{t^{1+\alpha /2}}\\&\qquad + \int _0^{\rho ^2(x)} \int _{|x-y|\ge \sqrt{t}} h_t(x-y) W_{\beta }(x,|x-y|) dy \frac{dt}{t^{1+\alpha /2}} = II_1 + II_2. \end{aligned} \end{aligned}$$

For $II_1$ we have

$$\begin{aligned} \begin{aligned} II_1&\le \int _0^{\rho ^2(x)} W_{\beta }(x,\sqrt{t}) \int _{|x-y|< \sqrt{t}} h_t(x-y) dy \frac{dt}{t^{1+\alpha /2}} \\&\le \int _0^{\rho ^2(x)} W_{\beta }(x,\sqrt{t}) \frac{dt}{t^{1+\alpha /2}}\\&\le \int _0^{\rho ^2(x)} \int _{|x-u|<\sqrt{t}} \frac{w(u)}{|x-u|^{d-\beta }}du \frac{dt}{t^{1+\alpha /2}}\\&\le \int _{B(x,\rho (x))} \frac{w(u)}{|x-u|^{d-\beta }} \int _{t\ge |x-u|^2}\frac{dt}{t^{1+\alpha /2}} du\\&\le C \int _{B(x,\rho (x))} \frac{w(u)}{|x-u|^{d-(\beta -\alpha )}}du \\&= C W_{\beta -\alpha }(x,\rho (x)). \end{aligned} \end{aligned}$$

For $II_2$ we apply Lemmas 1 and 7 since the integration is performed on $\sqrt{t}\le \rho (x)$.

$$\begin{aligned} \begin{aligned} II_2&\le C \int _0^{\rho ^2(x)} \int _{|x-y|\ge \sqrt{t}} h_t(x-y) W_{\beta }(x,|x-y|) dy \frac{dt}{t^{1+\alpha /2}} \\&\le C \int _0^{\rho ^2(x)} W_{\beta }(x,\sqrt{t}) \int _{|x-y|\ge \sqrt{t}} h_t(x-y) \left( \frac{|x-y|}{\sqrt{t}}\right) ^{d\mu -d+\beta } \\&\quad \times \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{\theta }dy \frac{dt}{t^{1+\alpha /2}}\\&\le C \int _0^{\rho ^2(x)} W_{\beta }(x,\sqrt{t}) \frac{dt}{t^{1+\alpha /2}}\\&\le C W_{\beta -\alpha }(x,\rho (x)), \end{aligned} \end{aligned}$$

where for the last inequality we follow the same steps as for $II_1$.

Now we turn our attention to III. Since $\int k_t(x,y)dy$ is not necessarily one, we split in the two following terms

$$\begin{aligned} |III|\le \int _{\rho ^2(x)}^{\infty } \int _{{\mathbb {R}}^d} |k_t(x,y)||f(y)|dy \frac{dt}{t^{1+\alpha /2}} + \int _{\rho ^2(x)}^{\infty } |f(x)| \frac{dt}{t^{1+\alpha /2}} = III_1 + III_2. \end{aligned}$$

To bound $III_1$ we use the size estimates for f and $k_t$ given in (12). Also we apply Lemma 9 to obtain

$$\begin{aligned} \begin{aligned} III_1&\le C_N \int _{\rho ^2(x)}^{\infty } \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{-N} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) W_{\beta }(y,\rho (y))dy \frac{dt}{t^{1+\alpha /2}}\\&\le C_N \int _{\rho ^2(x)}^{\infty } \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{-N} W_{\beta }(x,\rho (x)) \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{d\mu -d+\beta +\theta } \frac{dt}{t^{1+\alpha /2}}\\&\le C_N \frac{W_{\beta }(x,\rho (x))}{\rho (x)^{d\mu -d+\beta +\theta -N}} \int _{\rho ^2(x)}^{\infty } t^\frac{d\mu -d+\beta +\theta -\alpha -N}{2} \frac{dt}{t}\\&\le C_N \frac{W_{\beta }(x,\rho (x))}{\rho (x)^{d\mu -d+\beta +\theta -N}} \int _{\rho ^2(x)}^{\infty } t^\frac{d\mu -d+\beta +\theta -\alpha -N}{2} \frac{dt}{t}\\&\le C \rho ^{-\alpha }(x) W_{\beta }(x,\rho (x)) \\&\le C W_{\beta -\alpha }(x,\rho (x)), \end{aligned} \end{aligned}$$

choosing N large enough.

Finally, for $III_2$ we simply use the size estimate for f arriving to

$$\begin{aligned} \begin{aligned} III_2&\le W_{\beta }(x,\rho (x)) \int _{\rho ^2(x)}^{\infty } t^{-\alpha /2}\frac{dt}{t}\\&\le C \rho ^{-\alpha }(x) W_{\beta }(x,\rho (x))\\&\le C W_{\beta -\alpha }(x,\rho (x)). \end{aligned} \end{aligned}$$

Now we check condition (7). For $x,\,z\in {\mathbb {R}}^d$ with $|x-z|\le \rho (x)$ we want to show that

$$\begin{aligned} |L^{\alpha /2}f(x)-L^{\alpha /2}f(z)|\le C\, [W_{\beta }(x, |x-z|)+W_{\beta }(z,|x-z|)]. \end{aligned}$$

(20)

Let $x,\,z\in {\mathbb {R}}^d$ such that $|x-z|\le \rho (x)$, we write

$$\begin{aligned} \begin{aligned} L^{\alpha /2}f(x)-L^{\alpha /2}f(z)&= \int _0^{\infty } (e^{-tL}-I)f(x) \frac{dt}{t^{1+\alpha /2}} - \int _0^{\infty } (e^{-tL}-I)f(z) \frac{dt}{t^{1+\alpha /2}}\\&\quad \pm \left[ \int _0^{\rho (x)^2} (e^{t\Delta }-I)f(x) \frac{dt}{t^{1+\alpha /2}} - \int _0^{\rho (x)^2} (e^{t\Delta }-I)f(z) \frac{dt}{t^{1+\alpha /2}} \right] \\&= IV + V + VI, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} IV= & {} \int _0^{\rho (x)^2} (e^{-tL}-e^{t\Delta })f(x)\frac{dt}{t^{1+\alpha /2}} - \int _0^{\rho (x)^2} (e^{-tL}-e^{t\Delta })f(z)\frac{dt}{t^{1+\alpha /2}}\\= & {} \int _0^{\rho (x)^2} \int _{{\mathbb {R}}^d} \left[ k_t(x,y)-h_t(x-y) - \left( k_t(z,y) - h_t(z-y)\right) \right] f(y) dy\frac{dt}{t^{1+\alpha /2}}, \\ V= & {} \int _0^{\rho (x)^2} (e^{t\Delta } - I )f(x)\frac{dt}{t^{1+\alpha /2}} - \int _0^{\rho (x)^2} (e^{t\Delta } - I )f(z)\frac{dt}{t^{1+\alpha /2}},\\ VI= & {} \int _{\rho (x)^2}^{\infty } (e^{-tL} - I )f(x)\frac{dt}{t^{1+\alpha /2}} - \int _{\rho (x)^2}^{\infty } (e^{-tL} - I )f(z)\frac{dt}{t^{1+\alpha /2}}. \end{aligned}$$

First, to estimate IV we decompose the integral in the following way,

$$\begin{aligned} \begin{aligned} IV&\le \int _0^{|x-z|^2} \int _{{\mathbb {R}}^d} \left[ \left| k_t(x,y)-h_t(x,y)\right| + \left| k_t(z,y) - h_t(z,y)\right| \right] |f(y)| \frac{dt}{t^{1+\alpha /2}}\\&\quad + \int _{|x-z|^2}^{\rho (x)^2} \int _{{\mathbb {R}}^d} \left| k_t(x,y)-h_t(x,y) - \left( k_t(z,y) - h_t(z,y)\right) \right| |f(y)| \frac{dt}{t^{1+\alpha /2}} \\&= IV_1 + IV_2. \end{aligned} \end{aligned}$$

Now we decompose $IV_1$ in a sum of two terms. Both of them can be treated in the same way, so we do only the first one. Given $\beta$ and $\mu$ satisfying the assumptions of the theorem, we can fix $\delta >0$ such that $\beta<\delta <\delta _0$ and $\mu \le 1+\frac{\delta -\beta }{d}$. Using again the size estimate for f, applying Lemma 5 for $\delta$ as above, Lemma 9 and then Lemma 1, we obtain

$$\begin{aligned} \begin{aligned} IV_1&\le C \int _0^{|x-z|^2} \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta } \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) W_\beta (y,\rho (y)) dy\frac{dt}{t^{1+\alpha /2}}\\&\le C W_\beta (x,\rho (x)) \rho (x)^{-\delta } \int _0^{|x-z|^2} t^{\frac{\delta -\alpha }{2}-1}{dt}\\&\le C |x-z|^{-\alpha } W_\beta (x,|x-z|) \left( \frac{|x-z|}{\rho (x)}\right) ^{\delta -d\mu +d-\beta } \\&\le C W_{\beta -\alpha }(x,|x-z|), \end{aligned} \end{aligned}$$

since $\alpha <\delta$ and $\mu \le 1+ \frac{\delta -\beta }{d}$.

To handle $IV_2$, we break the integral on ${\mathbb {R}}^d$ into $B(x,4|x-z|)$ and its complement, giving rise to $IV_{21}$ and $IV_{22}$. For the first we proceed as for $IV_1$ producing two similar terms. For each one we apply Lemma  5 together with Lemma 3 to get

$$\begin{aligned} \begin{aligned} IV_{21}&\le \int _{|x-z|^2}^{\rho (x)^2} \left( \frac{\sqrt{t}}{\rho (x)}\right) ^{\delta } \int _{|x-y|\le 4 |x-z|} t^{-d/2} e^{-\frac{|x-y|^2}{t}} W_{\beta } (y,\rho (y)) dy \frac{dt}{t^{1+\alpha /2}} \\&\le \rho (x)^{-\delta } \int _{|x-z|^2}^{\rho (x)^2} \int _{|x-y|\le 4 |x-z|} W_{\beta } (y,\rho (y)) dy \frac{dt}{t^{1+\frac{\alpha -\delta +d}{2}}} \\&\le \rho (x)^{-\delta } |x-z|^{d} W_{\beta } (x,\rho (x)) \int _{|x-z|^2}^{\infty } \frac{dt}{t^{1+\frac{\alpha -\delta +d}{2}}}. \end{aligned} \end{aligned}$$

Now, applying Lemma 1,

$$\begin{aligned} IV_{21}\le |x-z|^{-\alpha } W_{\beta } (x,|x-z|) \left( \frac{|x-z|}{\rho (x)}\right) ^{d-\mu d + \delta - \beta } \le W_{\beta - \alpha } (x,|x-z|) \end{aligned}$$

since $\mu \le 1 + \frac{\delta -\beta }{d}$.

To estimate $IV_{22}$ we use Lemma 6 with $\delta$ as above, together with Lemma 9 and Lemma 1 to obtain

$$\begin{aligned} \begin{aligned} IV_{22}&\le \left( \frac{|x-z|}{\rho (x)}\right) ^\delta \int _{|x-z|^2}^{\rho (x)^2} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y)W_\beta (y,\rho (y))dy \frac{dt}{t^{1+\alpha /2}}\\&\le \left( \frac{|x-z|}{\rho (x)}\right) ^\delta W_\beta (x,\rho (x)) \int _{|x-z|^2}^{\rho (x)^2} \frac{dt}{t^{1+\alpha /2}}\\&\le C |x-z|^{-\alpha } W_\beta (x,|x-z|) \left( \frac{|x-z|}{\rho (x)}\right) ^{\delta -d\mu +d-\beta } \\&\le C W_{\beta -\alpha }(x,|x-z|), \end{aligned} \end{aligned}$$

if $\mu \le 1+ \frac{\delta -\beta }{d}$.

Next, we turn our attention to V. First we write it in the following way

$$\begin{aligned} \begin{aligned} |V|&\le \left| \int _0^{|x-z|^2} \int _{{\mathbb {R}}^d} h_t(x-y)[f(y)-f(x)]dy \right. \\&\quad \left. - \left[ \int _{{\mathbb {R}}^d} h_t(z-y)[f(y)-f(z)]dy\right] \frac{dt}{t^{1+\alpha /2}}\right| \\&\quad + \left| \int _{|x-z|^2}^{\rho (x)^2} \int _{{\mathbb {R}}^d} h_t(x-y)[f(y)-f(x)]dy\right. \\&\quad \left. - \left[ \int _{{\mathbb {R}}^d} h_t(z-y)[f(y)-f(z)]dy\right] \frac{dt}{t^{1+\alpha /2}}\right| \\&= V_1 + V_2. \end{aligned} \end{aligned}$$

In $V_1$ we may bound the difference by the sum of two analogous terms $V_{11}$ and $V_{12}$. To deal with $V_{11}$ we split the inner integral in $B(x,\sqrt{t})$ and its complement, obtaining the terms $V_{111}$ and $V_{112}$ respectively. In $V_{111}$, after applying the smoothness estimate for f, we obtain

$$\begin{aligned} \begin{aligned} V_{111}\le C \int _0^{|x-z|^2}\int _{|x-y|\le \sqrt{t}} h_t(x-y)[W_\beta (x,|x-y|)+W_\beta (y,|x-y|)]dy \frac{dt}{t^{1+\alpha /2}}. \end{aligned} \end{aligned}$$

Now we split the inner sum in two terms. For the first term

$$\begin{aligned} \begin{aligned}&\int _0^{|x-z|^2}\int _{|x-y|\le \sqrt{t}} h_t(x-y)W_\beta (x,|x-y|)dy \frac{dt}{t^{1+\alpha /2}} \\&\quad \le \int _0^{|x-z|^2}W_\beta (x,\sqrt{t})\int _{{\mathbb {R}}^d} h_t(x-y)dy \frac{dt}{t^{1+\alpha /2}}\\&\quad \le C \int _0^{|x-z|^2}W_\beta (x,\sqrt{t}) \frac{dt}{t^{1+\alpha /2}} \\&\quad \le C \int _0^{|x-z|^2} \int _{B(x,\sqrt{t})} \frac{w(u)}{|u-x|^{d-\beta }}du \frac{dt}{t^{1+\alpha /2}} \\&\quad \le C \int _{B(x,|x-z|) }\frac{w(u)}{|u-x|^{d-\beta }} \int _{\sqrt{t}>|x-u|} \frac{dt}{t^{1+\alpha /2}} du \\&\quad \le C \int _{B(x,|x-z|) }\frac{w(u)}{|u-x|^{d-(\beta -\alpha )}} du \\&\quad \le C W_{\beta -\alpha }(x,|x-z|). \end{aligned} \end{aligned}$$

The second term can be treated in the same way once we have applied Lemma 8.

To handle $V_{112}$ we use the smoothness estimate for f, obtaining

$$\begin{aligned} \begin{aligned} V_{112}\le C \int _0^{|x-z|^2}\int _{|x-y|> \sqrt{t}} h_t(x-y)[W_\beta (x,|x-y|)+W_\beta (y,|x-y|)]dy \frac{dt}{t^{1+\alpha /2}}. \end{aligned} \end{aligned}$$

Again, we split the inner sum in two terms. For the first term we apply Lemma 1 together with Lemma 7 since $\sqrt{t}<|x-z|<\rho (x)$. In this way

$$\begin{aligned} \begin{aligned} \int _0^{|x-z|^2}&\int _{|x-y|>\sqrt{t}} h_t(x-y)W_\beta (x,|x-y|)dy \frac{dt}{t^{1+\alpha /2}} \\&\le \int _0^{|x-z|^2}W_\beta (x,\sqrt{t})\int _{{\mathbb {R}}^d} h_t(x-y) \left( \frac{|x-y|}{\sqrt{t}}\right) ^{d\mu -d+\beta } \\&\quad \left( 1+\frac{|x-y|}{\rho (x)}\right) ^\theta dy \frac{dt}{t^{1+\alpha /2}}\\&\le C \int _0^{|x-z|^2}W_\beta (x,\sqrt{t}) \frac{dt}{t^{1+\alpha /2}} \\&\le C W_{\beta -\alpha }(x,|x-z|), \end{aligned} \end{aligned}$$

where for the last line we use the same argument as for $V_{111}$. Once more, after applying Lemma 8, the second term can be treated in the same way.

Now we turn our attention to $V_2$. We add and subtract $h_{t}(z-y)[f(y)-f(x)]$ obtaining $V_2\le V_{21}+V_{22}$, where

$$\begin{aligned} V_{21}= \int _{|x-z|^2}^{\rho ^2(x)} \int _{{\mathbb {R}}^d} h_t(z-y)|f(x)-f(z)| dy \frac{dt}{t^{1+\alpha /2}} \end{aligned}$$

and

$$\begin{aligned} V_{22}= \int _{|x-z|^2}^{\rho ^2(x)} \int _{{\mathbb {R}}^d} |h_t(x-y)-h_t(z-y)||f(y)-f(x)|dy \frac{dt}{t^{1+\alpha /2}}. \end{aligned}$$

In $V_{21}$ we simply apply the smoothness estimate for f arriving to

$$\begin{aligned} \begin{aligned} V_{21}&\le C[W_\beta (x,|x-z|)+W_\beta (z,|x-z|)]\int _{|x-z|^2}^{\rho ^2(x)} \int _{{\mathbb {R}}^d} h_t(z-y)dy \frac{dt}{t^{1+\alpha /2}} \\&\le [W_\beta (x,|x-z|)+W_\beta (z,|x-z|)] |x-z|^{-\alpha } \\&\le W_{\beta -\alpha }(x,|x-z|)+W_{\beta -\alpha }(z,|x-z|). \end{aligned} \end{aligned}$$

To handle $V_{22}$, we use the smoothness of the heat kernel given in (13) together with the smoothness estimate for f to obtain,

$$\begin{aligned} \begin{aligned} V_{22}&= \int _{|x-z|^2}^{\rho ^2(x)} \int _{{\mathbb {R}}^d} |h_t(x-y)-h_t(z-y)||f(y)-f(x)|dy \frac{dt}{t^{1+\alpha /2}} \\&\le C \int _{|x-z|^2}^{\rho ^2(x)} \frac{|x-z|}{\sqrt{t}} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y) W_\beta (x,|x-y|)dy\frac{dt}{t^{1+\alpha /2}} \\&\quad + C\int _{|x-z|^2}^{\rho ^2(x)} \frac{|x-z|}{\sqrt{t}} \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y)W_\beta (y,|x-y|)dy \frac{dt}{t^{1+\alpha /2}}\\&= V_{221} + V_{222}. \end{aligned} \end{aligned}$$

Once more we estimate only $V_{221}$ since $V_{222}$ can be handled following the same lines, after an application of Lemma 9. First, we decompose the inner integral of $V_{221}$ in $B(x,|x-z|)$ and its complement obtaining two terms. For the first one we simply use that $W_\beta$ is increasing on the second variable as follows

$$\begin{aligned} \begin{aligned}&\int _{|x-z|^2}^{\rho ^2(x)} \left( \frac{|x-z|}{\sqrt{t}}\right) \int _{|x-y|< |x-z|} {\tilde{h}}_t(x-y) W_\beta (x,|x-y|)dy\frac{dt}{t^{1+\alpha /2}}\\&\quad \le C\int _{|x-z|^2}^{\rho ^2(x)} \left( \frac{|x-z|}{\sqrt{t}}\right) W_\beta (x,|x-z|) \int _{|x-y|< |x-z|} {\tilde{h}}_t(x-y) dy \frac{dt}{t^{1+\alpha /2}}\\&\quad \le C W_\beta (x,|x-z|) |x-z| \int _{|x-z|^2}^{\infty } \frac{dt}{t^{1+\alpha /2+1/2}}\\&\quad \le C W_\beta (x,|x-z|) |x-z|^{-\alpha }\le C W_{\beta -\alpha }(x,|x-z|). \end{aligned} \end{aligned}$$

For the second term we can apply Lemma 1 together with Lemma 7 and, since $\sqrt{t}<\rho (x)$, we get

$$\begin{aligned} \begin{aligned}&\int _{|x-z|^2}^{\rho ^2(x)} \frac{|x-z|}{\sqrt{t}} \int _{|x-y|\ge |x-z|} {\tilde{h}}_t(x-y) W_\beta (x,|x-y|)dy\frac{dt}{t^{1+\alpha /2}}\\&\quad \le C\int _{|x-z|^2}^{\rho ^2(x)} \left( \frac{|x-z|}{\sqrt{t}}\right) \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y)W_\beta (x,|x-z|) \left( \frac{|x-y|}{|x-z|}\right) ^{d\mu -d+\beta } \\&\qquad \times \left( 1+\frac{|x-y|}{\rho (x)}\right) ^\theta dy \frac{dt}{t^{1+\alpha /2}}\\&\quad \le C W_\beta (x,|x-z|) |x-z|^{1-d\mu +d-\beta } \int _{|x-z|^2}^{\infty } \sqrt{t}^{d\mu -d+\beta -1} \frac{dt}{t^{1+\alpha /2}}\\&\quad \le C W_\beta (x,|x-z|) |x-z|^{-\alpha }\le C W_{\beta -\alpha }(x,|x-z|). \end{aligned} \end{aligned}$$

Finally, we need to bound VI. First we write

$$\begin{aligned} \begin{aligned} VI&= \int _{\rho (x)^2}^{\infty } \left[ \int _{{\mathbb {R}}^d} k_t(x,y) f(y) dy - f(x) -\left( \int _{{\mathbb {R}}^d} k_t(z,y) f(y) dy - f(z) \right) \right] \frac{dt}{t^{1+\alpha /2}} \\&\le \int _{\rho (x)^2}^{\infty } \left( \int _{{\mathbb {R}}^d} |k_t(x,y) - k_t(z,y)| |f(y)| dy + |f(x) - f(z)|\right) \frac{dt}{t^{1+\alpha /2}} \\&= VI_1 + VI_2, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} VI_1= \int _{\rho (x)^2}^{\infty } \int _{{\mathbb {R}}^d} |k_t(x,y) - k_t(z,y)| |f(y)| dy \frac{dt}{t^{1+\alpha /2}} \end{aligned}$$

and

$$\begin{aligned} VI_2= |f(x) - f(z)|\int _{\rho (x)^2}^{\infty }\frac{dt}{t^{1+\alpha /2}}. \end{aligned}$$

To take care of $VI_1$, we apply Lemma 4 with $\delta$ as above, since $|x-z|<\rho (x)<\sqrt{t}$. That, together with the size estimate for f, Lemmas 1 and 9 give us

$$\begin{aligned} \begin{aligned} VI_{1}&= \int _{\rho (x)^2}^{\infty }\int _{{\mathbb {R}}^d} |k_t(x,y)-k_t(z,y)| |f(y)|dy \frac{dt}{t^{1+\alpha /2}}\\&\le C_N \int _{\rho (x)^2}^{\infty } \left( \frac{|x-z|}{\sqrt{t}}\right) ^{\delta }\left( \frac{\rho (x)}{\sqrt{t}}\right) ^N \int _{{\mathbb {R}}^d} {\tilde{h}}_t(x-y)W_{\beta }(y,\rho (y)) dy \frac{dt}{t^{1+\alpha /2}}\\&\le C_N W_{\beta }(x,\rho (x))|x-z|^{\delta }\int _{\rho (x)^2}^{\infty } \left( \frac{\rho (x)}{\sqrt{t}}\right) ^{N-d\mu +d+-\beta -\theta } \frac{dt}{t^{1+\alpha /2+\delta /2}}\\&\le C W_{\beta }(x,\rho (x))\left( \frac{|x-z|}{\rho (x)}\right) ^{\delta }\rho (x)^{-\alpha }\\&\le C W_{\beta }(x,|x-z|)\left( \frac{|x-z|}{\rho (x)}\right) ^{\delta -d\mu +d-\beta }\rho (x)^{-\alpha }\\&\le C W_{\beta -\alpha }(x,|x-z|)\left( \frac{|x-z|}{\rho (x)}\right) ^{\delta -d\mu +d-\beta +\alpha }\\&\le C W_{\beta -\alpha }(x,|x-z|), \end{aligned} \end{aligned}$$

for N large enough and as long as $\mu \le 1+\frac{\delta -(\beta -\alpha )}{d}$.

It only remains to bound $VI_2$. To do that, we just apply the smoothness estimate for f to obtain

$$\begin{aligned} \begin{aligned} VI_2&\le \left( W_{\beta }(x,|x-z|) + W_{\beta }(z,|x-z|) \right) \int _{\rho (x)^2}^{\infty }\frac{dt}{t^{1+\alpha /2}} \\&\le \rho (x)^{-\alpha } \left( W_{\beta }(x,|x-z|) + W_{\beta }(z,|x-z|) \right) \\&\le |x-z|^{-\alpha } \left( W_{\beta }(x,|x-z|) + W_{\beta }(z,|x-z|) \right) \\ {}&\le \left( W_{\beta - \alpha }(x,|x-z|) + W_{\beta - \alpha }(z,|x-z|) \right) , \end{aligned} \end{aligned}$$

since $|x-z|\le \rho (x)$.

$\square$

4 Other fractional differentiation operators

We will consider in this section other fractional differentiation operators related to L. These will be defined as the composition of powers of L (positive or negative) together with a multiplication by a power of the potential V. We have already results concerning the behaviour on $\Lambda ^{\beta }_\rho (w)$ of powers of L so we need a tool to handle point-wise multiplication operators. That is the first aim of this section.

For a measurable function $\phi :{\mathbb {R}}^d\longmapsto {\mathbb {R}}$ we consider a multiplication operator $T_\phi f (x) = \phi (x) f(x)$. The next theorem gives sufficient conditions on the function $\phi$ to guarantee the boundedness of $T_\phi$ from $\Lambda _\rho ^\beta (w)$ to $\Lambda _\rho ^\sigma (w)$ when $0<\sigma \le \beta$.

Theorem 2

Let $w\in D_\mu ^{\rho }$ for some $\mu \ge 1$ and $\eta$, $\beta$ such that $0<\eta \le \beta \le 1$. Suppose there exists a constant $C>0$ such that the function $\phi$ satisfies:

$$\begin{aligned} |\phi (x)|\le C \rho ^{\eta -\beta }(x)\text {, for all }x\in {\mathbb {R}}^d \end{aligned}$$

(21)

and

$$\begin{aligned} |\phi (x)-\phi (z)|\le C \rho ^{\eta -\beta }(x) \left( \frac{|x-z|}{\rho (x)}\right) ^{\eta +d(\mu -1)} , \text { if } |x-z|\le \rho (x). \end{aligned}$$

(22)

Then, the operator $T_\phi$ is bounded from $\Lambda _\rho ^\beta (w)$ into $\Lambda _\rho ^\eta (w)$.

Proof

Let $w\in D_\mu ^{\rho }$ and $f\in \Lambda _\rho ^\beta (w)$. First we are going to check condition (6). For $x\in {\mathbb {R}}^d$ we have that

$$\begin{aligned} \begin{aligned} |T_\phi (f)(x)|&= |\phi (x)||f(x)|\\&\le C \rho ^{\eta -\beta }(x)W_{\beta }(x,\rho (x)) \\&\le C W_{\eta }(x,\rho (x)), \end{aligned} \end{aligned}$$

where we have used (21) for $\phi$ and that $f\in \Lambda _\rho ^\beta (w)$.

Now, to check (7) for $T_\phi f$, we apply both conditions asked to $\phi$. Let x, $z\in {\mathbb {R}}^d$ with $|x-z|\le \rho (x)$.

$$\begin{aligned} \begin{aligned} |T_\phi (f)(x)-T_\phi (f)(z)|&= |\phi (x)f(x)-\phi (z)f(z)|\\&\le |\phi (x)||f(x)-f(z)| + |\phi (x)-\phi (z)||f(z)|. \end{aligned} \end{aligned}$$

(23)

For the first term of (23), we use again assumption (21) on $\phi$ to obtain,

$$\begin{aligned} \begin{aligned} |\phi (x)||f(x)-f(z)|&\le C\rho ^{\eta -\beta }(x) [W_\beta (x,|x-z|)+W_\beta (z,|x-z|) ]\\&\le C [W_\eta (x,|x-z|)+W_\eta (z,|x-z|)] \end{aligned} \end{aligned}$$

as above, since $f\in \Lambda _\rho ^\beta (w)$.

As for the second term of (23), we apply now (22) in the following way,

$$\begin{aligned} \begin{aligned} |\phi (x)-\phi (z)||f(z)|&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\eta +d(\mu -1)} \rho ^{\eta -\beta }(x) W_\beta (z,\rho (z))\\&\le C \left( \frac{|x-y|}{\rho (z)}\right) ^{\eta +d(\mu -1)} \rho ^{\eta -\beta }(z) W_\beta (z,\rho (z))\\&\le C \left( \frac{|x-y|}{\rho (z)}\right) ^{\eta +d(\mu -1)} W_\eta (z,\rho (z))\\&\le C W_\eta (z,|x-z|). \end{aligned} \end{aligned}$$

Here, we have used that $\rho (x)\simeq \rho (z)$ and the duplication property for $W_\eta$ given in Lemma 1. $\square$

Remark 1

In Theorem 2 the conditions on the function $\phi$ depend only on w through its doubling exponent $\mu$. In particular, note that if $w\in A_1^\rho$ (hence in $D_1^\rho$), condition (22) on $\phi$ becomes

$$\begin{aligned} |\phi (x)-\phi (z)|\le C \left( \frac{|x-z|}{\rho (x)}\right) ^{\eta } \rho ^{\eta -\beta }(x) \;\;\text { if } |x-z|\le \rho (x). \end{aligned}$$

(24)

Remark 2

It is worth noting that the point-wise conditions for $\phi$ given in Theorem 2 are equivalent to the following integral conditions:

$$\begin{aligned} \frac{1}{|B(x,\rho (x))|}\int _{B(x,\rho (x))}|\phi (y)|dy\le C \rho ^{\eta -\beta }(x)\text {, for all }x\in {\mathbb {R}}^d \end{aligned}$$

(25)

and

$$\begin{aligned}&\frac{1}{|B(x,s)|}\int _{B(x,s)}|\phi (y)-\phi _B|dy\nonumber \\&\quad \le C \left( \frac{s}{\rho (x)}\right) ^{\eta +d(\mu -1)} \rho ^{\eta -\beta }(x)\text {, whenever }0<s\le \rho (x). \end{aligned}$$

(26)

It is immediate that the point-wise conditions imply the integral ones. On the other hand, if $\phi$ satisfies the integral conditions above we can use Lebesgue Differentiation Theorem to obtain the point-wise ones. Given x, $y\in {\mathbb {R}}^d$ with $|x-y|\le \rho (x)$ we write

$$\begin{aligned}&|\phi (x)-\phi (y)|\le |\phi (x)-\phi _{B(x,|x-y|)}|+|\phi (y)-\phi _{B(y,|x-y|)}|\\&\quad +\,|\phi _{B(x,|x-y|)}-\phi _{B(y,|x-y|)}| \end{aligned}$$

For the the first term, defining $B_i=B(x,2^{-i}|x-y|)$,

$$\begin{aligned} \begin{aligned} |\phi (x)-\phi _{B(x,|x-y|)}|&\le \lim _{m\rightarrow \infty }\left( |\phi (x)-\phi _{B_m}| +\sum _{i=0}^{m-1} |\phi _{B_{i+1}}-\phi _{B_i}|\right) \\&\le C \sum _{i=0}^{\infty } \frac{1}{|B_i|}\int _{B_i}|\phi (z)-\phi _{B_i}|dz \\&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\eta +d(\mu -1)} \rho ^{\eta -\beta }(x) \sum _{i=0}^{\infty } 2^{-i(\eta +d(\mu -1))} \\&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\eta +d(\mu -1)} \rho ^{\eta -\beta }(x). \end{aligned} \end{aligned}$$

The second and third terms of the sum can be bounded in a similar way, obtaining (22).

Finally, to obtain (21) we set $B=B(x,\rho (x)/2)$

$$\begin{aligned} \begin{aligned} |\phi (x)|&\le |\phi (x)-\phi _B+\phi _B| \\&\le \frac{1}{|B|}\int _B |\phi (x)-\phi (z)|dz + \frac{1}{|B|}\int _B |\phi (z)|dz \\&\le C \rho (x)^{\eta -\beta }, \end{aligned} \end{aligned}$$

applying condition (25) and the estimate we just obtained.

If we consider the case $w=1$ in Theorem 2, we are able to prove that the conditions imposed on $\phi$ are also necessary for the operator $T_\phi$ to be bounded from $\Lambda _\rho ^\beta$ into $\Lambda _\rho ^\eta$, as we show in the next proposition.

Proposition 1

Let $\phi$ a real function such that the operator $T_\phi$ is bounded from $\Lambda _\rho ^\beta$ into $\Lambda _\rho ^\eta$, for some $0<\eta \le \beta$. Then the function $\phi$ satisfies:

$$\begin{aligned}&|\phi (x)|\le C \rho ^{\eta -\beta }(x), \text { for all } x\in {\mathbb {R}}^d\end{aligned}$$

(27)

$$\begin{aligned}&|\phi (x)-\phi (z)|\le C \left( \frac{|x-z|}{\rho (x)}\right) ^{\eta } \rho ^{\eta -\beta }(x)\text {, if } |x-z|\le \rho (x). \end{aligned}$$

(28)

Proof

Consider first, for $x\in {\mathbb {R}}^d$ and $s\le \rho (x)$, the function

$$\begin{aligned} f_{x,s}(y)=\chi _{[0,s]}(|x-y|)(\rho (x)^\beta -s^\beta ) + \chi _{[s,\rho (x)]}(|x-y|)(\rho (x)^\beta -|x-y|^\beta ). \end{aligned}$$

As it was shown in Lemma 2.5 of [10], this function belongs to $\Lambda _{\rho }^\beta$ and its norm does not depend neither on x nor on s.

To check (27) we take $s=\rho (x)/2$. Since $T_\phi$ is bounded from $\Lambda _\rho ^\beta$ into $\Lambda _\rho ^\eta$,

$$\begin{aligned} |\phi (x)f_{x,s}(x)|= |\phi (x)| (\rho (x)^\beta -s^{\beta })\le C \rho (x)^{\eta } \end{aligned}$$

Then,

$$\begin{aligned} |\phi (x)|\le C \rho (x)^{\eta -\beta }. \end{aligned}$$

Now, to verify inequality (28), we take x and $z\in {\mathbb {R}}^d$ such that $|x-z|\le \rho (x)$. If $|x-z|\le \rho (x)/2$, choose $s=|x-z|$. Then

$$\begin{aligned} |\phi (x)f_{x,s}(x)-\phi (z)f_{x,s}(z)|= |\phi (x)-\phi (z)|(\rho (x)^{\beta }-s^\beta )\le C |x-z|^{\eta }. \end{aligned}$$

Therefore, since $\rho (x)^\beta -s^\beta \simeq \rho (x)^\beta$,

$$\begin{aligned} |\phi (x)-\phi (z)|\le C |x-z|^\eta \rho (x)^{-\beta } =C \left( \frac{|x-z|}{\rho (x)}\right) ^\eta \rho (x)^{\eta -\beta }. \end{aligned}$$

On the other hand, if $\rho (x)/2<|x-z|\le \rho (x)$, then

$$\begin{aligned} |\phi (x)-\phi (y)|\le \rho (x)^{\eta -\beta } + \rho (z)^{\eta -\beta } \le C \left( \frac{|x-z|}{\rho (x)}\right) ^\eta \rho (x)^{\eta -\beta }. \end{aligned}$$

$\square$

Remark 3

In view of Remarks 1, 2 and Proposition 1, we are able to extend Proposition 3.2 in [10] in two directions. On one side, for $\eta =\beta$, we prove that the same conditions on $\phi$ required in [10] imply not only the boundedness on $\Lambda ^\beta _\rho (w)$ for $w=1$, but also for any $w\in A_1^\rho$. On the other side, for $w=1$, we obtain equivalence between some conditions on $\phi$ and the boundedness of the multiplication operator $T_\phi$ from $\Lambda ^\beta _\rho$ into $\Lambda ^\eta _\rho$ with $\eta \le \beta$, extending the already known result for the case $\eta =\beta$.

With this tool at hand, we will continue this section by stating and proving boundedness results for some other fractional differentiation operators in this context. First we take a look at compositions of two fractional differentiation operators, namely, given two positives numbers $\sigma$ and $\alpha$ such that $\alpha +\sigma <1$ we consider $V^{\sigma /2}L^{\alpha /2}$ and its adjoint, $L^{\alpha /2} V^{\sigma /2}$. In this case we have the following result.

Theorem 3

Let us assume that $V\in RH_{d}$ and that we are given $\sigma >0$ and $\alpha >0$ with $\alpha +\sigma <1$.

• (i) If there exists some $\varepsilon _0>\alpha$ such that for each $0<\varepsilon <\varepsilon _0$, V satisfies

  $$\begin{aligned} |V^{\sigma /2}(x) - V^{\sigma /2} (y)|\le C \rho (x)^{-\sigma } \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon } \text { for } |x-y| \le \rho (x), \end{aligned}$$

  (29)

  then $L^{\alpha /2} V^{\sigma /2}$ maps continuously $\Lambda _{\rho }^\beta (w)$ into $\Lambda _{\rho }^{\beta -(\alpha +\sigma )}(w)$, for any $\alpha +\sigma<\beta <\min \{\varepsilon _0+\sigma ,1\}$ and $w\in D_\mu ^\rho$ with $1\le \mu < 1+ \frac{\min \{\varepsilon _0+\sigma ,1\}-\beta }{d}$.

• (ii) If there exists some $\varepsilon _0>0$ such that for each $0<\varepsilon <\varepsilon _0$, V satisfies (29), then $V^{\sigma /2} L^{\alpha /2}$ maps continuously $\Lambda _{\rho }^\beta (w)$ into $\Lambda _{\rho }^{\beta -(\alpha +\sigma )}(w)$, for any $\alpha +\sigma<\beta <\min \{\varepsilon _0+\alpha +\sigma ,1\}$ and $w\in D_\mu ^\rho$ with $1\le \mu < 1+ \frac{\min \{\varepsilon _0+\alpha +\sigma ,1\}-\beta }{d}$.

Proof

We are going to prove (ii) in the first place. Applying Theorem 1 with $q=d$ we obtain that $L^{\alpha /2}$ maps continuously $\Lambda _{\rho }^{\beta }(w)$ into $\Lambda _{\rho }^{\beta -\alpha }(w)$, provided $0<\alpha<\beta <1$ and $1\le \mu < 1+ \frac{1-\beta }{d}$.

To conclude (ii), we make the following observation. For $x\in {\mathbb {R}}^d$ and $y\in B=B(x,\rho (x))$, condition (29) implies

$$\begin{aligned} V(x)^{\sigma /2} \le V(y)^{\sigma /2} + |V(x)^{\sigma /2}-V(y)^{\sigma /2}| \le V(y))^{\sigma /2} + \frac{C}{\rho ^{\sigma }(x)}. \end{aligned}$$

Therefore, averaging over $B=B(x,\rho (x))$ in the y-variable

$$\begin{aligned} V(x)^{\sigma /2} = \frac{1}{|B|} \int _{B} V(x)^{\sigma /2}dy \le \frac{1}{|B|} \int _{B} V^{\sigma /2} + \frac{C}{\rho ^{\sigma }(x)} \le \frac{C'}{\rho ^{\sigma }(x)}, \end{aligned}$$

where we have used Hölder’s inequality and the definition of $\rho (x)$.

So, (29) together with the above inequality give the sufficient conditions to get that multiplication by $V^{\sigma /2}$ is bounded from $\Lambda _{\rho }^{\beta -\alpha }(w)$ into $\Lambda _{\rho }^{\beta -\alpha -\sigma }(w)$ as long as $\alpha +\sigma<\beta <\varepsilon _0+\alpha +\sigma$ and $1\le \mu < 1+ \frac{\varepsilon _0+\alpha +\sigma -\beta }{d}$. Therefore, under our assumptions on $\beta$ and $\mu$ both operators are bounded and the conclusion follows.

Item (i) can be proved in an analogous way, noting that condition (29) allows us to apply Theorem 2 obtaining that multiplication by $V^{\sigma /2}$ is bounded from $\Lambda _{\rho }^{\beta }(w)$ into $\Lambda _{\rho }^{\beta -\sigma }(w)$. Applying again Theorem 1 we obtain the desired result. $\square$

Next we consider the case of compositions of negative powers of L with multiplication by positive powers of the potential, that is, operators of the form $L^{-\alpha /2}V^{\sigma /2}$, with $\alpha <\sigma$. In order to do that we look at the operator as the composition of $L^{-\alpha /2}V^{\alpha /2}$ with multiplication by $V^{(\sigma -\alpha )/2}$. The behaviour on $\Lambda _\rho ^\beta (w)$ of the first of these operators was analysed in [2]. However, the hypotheses considered there were different from those needed here. In fact, since we must assume some regularity on $V^{(\sigma -\alpha )/2}$ to get boundedness of the multiplication operator, we have a fortiori a size estimate for V, namely $V(x)\le C \rho (x)^{-2}$. For that reason we state and prove the following result. Before, let us recall that $A_\infty ^\rho = \bigcup _{p\ge 1} A_p^\rho$.

Proposition 2

Let us assume that $V\in RH_{d/2}$ and that there exists a constant C such that $V(x)\le C\rho (x)^{-2}$. Let $0<\alpha \le 2$, the operator $L^{-\alpha /2}V^{\alpha /2}$ is bounded on $\Lambda ^{\beta }_{\rho }(w)$ as long as $0\le \beta <\min \{\alpha ,1\}$ and $w \in A_\infty ^\rho \cap D_\mu ^\rho$ with $1\le \mu < 1+ \frac{\min \{\alpha ,1\}-\beta }{d}$.

Proof

We are going to apply the general criterium given in [2], Corollary 4. Let ${\mathcal {J}}_{\alpha /2}$ be the kernel of the fractional integral $L^{-\alpha /2}$. By Lemma 4 in [3], we know that for each $N>0$, there exists a constant $C_N$ such that

$$\begin{aligned} {\mathcal {J}}_{\alpha /2}(x,y)\le \frac{C_N}{|x-y|^{d-\alpha }} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N} \end{aligned}$$

and

$$\begin{aligned} |{\mathcal {J}}_{\alpha /2}(x,y)-{\mathcal {J}}_{\alpha /2}(x+h,y)| \le C \frac{|h|}{|x-y|^ {d-\alpha +1}} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N}, \end{aligned}$$

if $|h|\le |x-y|/2$.

Now, we consider $K_{\alpha /2}={\mathcal {J}}_{\alpha /2}(x,y)V^{\alpha /2}(y)$ the kernel of $L^{-\alpha /2}V^{\alpha /2}$. First, we make the following observation. For $f\in L^1_{loc }$, by (2), the hypothesis on V and the size estimate for ${\mathcal {J}}_{\alpha /2}$ given above, we get

$$\begin{aligned} \begin{aligned} L^{-\alpha /2}V^{\alpha /2} f(x)&\le \int _{{\mathbb {R}}^d} |{\mathcal {J}}_{\alpha /2}(x,y)|V^{\alpha /2}(y)|f(y)|dy \\&\le C_N \int _{{\mathbb {R}}^d} \frac{1}{|x-y|^{d-\alpha }} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N} \rho ^{-\alpha }(y)|f(y)|dy\\&\le C_N \rho ^{-\alpha }(x)\int _{{\mathbb {R}}^d} \frac{1}{|x-y|^{d-\alpha }} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N+\alpha N_0} |f(y)|dy\\&\le C_N \rho ^{-\alpha }(x) \sum _{j=-\infty }^{j=\infty } \frac{(1+2^j)^{-N+\alpha N_0}}{(2^j\rho (x))^{d-\alpha }}\int _{|x-y|\le 2^j\rho (x)} |f(y)|dy\\&\le C_N Mf(x) \sum _{j=-\infty }^{j=\infty }(1+2^j)^{-N+\alpha N_0}2^{j\alpha }\\&\le C Mf(x), \end{aligned} \end{aligned}$$

where M is the Hardy–Littlewood maximal function and N is chosen large enough. This point-wise estimate guarantee that $L^{-\alpha /2}V^{\alpha /2}$ is bounded from $L^1$ into $L^{1,\infty }$.

Now, we are going to derive size and smoothness estimates for $K_{\alpha /2}$ from the ones given for ${\mathcal {J}}_{\alpha /2}$. For each $N>0$ there exists $C_N$ such that

$$\begin{aligned} ~ \begin{aligned} |V^{\alpha /2}(y){\mathcal {J}}_{\alpha /2}(x,y)|&\le \frac{C_N}{\rho ^{\alpha }(y)|x-y|^{d-\alpha }} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N}\\&\le \frac{C_N}{|x-y|^{d}} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N+\alpha + \alpha N_0}. \end{aligned} \end{aligned}$$

(30)

In a similar way, taking $|h|\le |x-y|/2$ ,

$$\begin{aligned} \begin{aligned} |V^{\alpha /2}(y){\mathcal {J}}_{\alpha /2}(x,y)&-V^{\alpha /2}(y){\mathcal {J}}_{\alpha /2}(x+h,y)| \\&\le V^{\alpha /2}(y)|{\mathcal {J}}_{\alpha /2}(x,y)-{\mathcal {J}}_{\alpha /2}(x+h,y)| \\&\le \frac{C_N|h|}{\rho ^{\alpha }(y)|x-y|^{d-\alpha +1}} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N} \\&\le C_N \frac{|h|}{|x-y|^{d+1}} \left( 1+\frac{|x-y|}{\rho (x)}\right) ^{-N+\alpha +\alpha N_0}. \end{aligned} \end{aligned}$$

(31)

Finally, we are going to show a T1 condition for $L^{-\alpha /2}V^{\alpha /2}$. Consider $x_0\in {\mathbb {R}}^d$ and $r>0$ such that $r\le \rho (x)/2$. If x, $z\in B(x_0,r)$

$$\begin{aligned}&|L^{-\alpha /2}V^{\alpha /2} 1(x)-L^{-\alpha /2}V^{\alpha /2}1(z)|\\&\quad =\left| \int _{{\mathbb {R}}^d} {\mathcal {J}}_{\alpha /2}(x,y)V^{\alpha /2}(y)dy-\int _{{\mathbb {R}}^d} {\mathcal {J}}_{\alpha /2}(z,y)V^{\alpha /2}(y)dy \right| \\&\quad \le \left| \int _{B_\rho } {\mathcal {J}}_{\alpha /2}(x,y)V^{\alpha /2}(y)dy-\int _{B_\rho } {\mathcal {J}}_{\alpha /2}(z,y)V^{\alpha /2}(y)dy \right| \\&\qquad + \int _{B_\rho ^c} \left| V^{\alpha /2}(y){\mathcal {J}}_{\alpha /2}(x,y)-V^{\alpha /2}(y){\mathcal {J}}_{\alpha /2}(z,y) \right| dy \\&\quad = I + II, \end{aligned}$$

where $B_\rho =B(x_0,\rho (x_0))$.

To bound II, we may simply apply the smoothness estimate obtained for the kernel to get

$$\begin{aligned} \begin{aligned} II&\le \int _{B_\rho ^c} \frac{|x-z|}{|x-y|^{d+1}}dy\le C \frac{r}{\rho (x_0)}, \end{aligned} \end{aligned}$$

since $\rho (x)\simeq \rho (x_0)$.

The point-wise estimates for ${\mathcal {J}}_{\alpha /2}$ allow us to apply Proposition 3 in [2] to obtain that $L^{-\alpha /2}$ is bounded from $L^p$ into $\Lambda _\rho ^{\alpha -d/p}$, for any $p\in (d/\alpha ,d/(\alpha -1)^+)$. From this property we obtain,

$$\begin{aligned} \begin{aligned} I&= |L^{-\alpha /2}(V^{\alpha /2} \chi _{B_\rho })(x)-L^{-\alpha /2}(V^{\alpha /2}\chi _{B_\rho })(z)|\\&\le C \Vert V^{\alpha /2} \chi _{B_\rho }\Vert _p |x-z|^{\alpha -d/p} \\&\le C \left( \int _{B_\rho } V^{p\alpha /2}(y)dy\right) ^{1/p} r^{\alpha -d/p} \\&\le C \left( \frac{r}{\rho (x_0)}\right) ^{\alpha -d/p}. \end{aligned} \end{aligned}$$

Now, since we can take any $p\in (d/\alpha ,d/(\alpha -1)^+)$, we obtain that

$$\begin{aligned} |L^{-\alpha /2}V^{\alpha /2} 1(x)-L^{-\alpha /2}V^{\alpha /2}1(z)| \le C \left( \frac{r}{\rho (x_0)}\right) ^{\varepsilon }, \end{aligned}$$

(32)

for any $\varepsilon <\min \{\alpha ,1\}$.

To conclude, the weak type (1, 1) together with estimates (30), (31) and (32), allow us to apply Corollary 4 of [2] to obtain the stated result.

$\square$

Remark 4

Notice that we get the same conclusions as in Theorem 6 in [2] under the assumption $V\in RH_q$ for all $q\ge 1$. However, as it was pointed out in [3], there are potentials that satisfy the hypothesis of Proposition 2 above but they are not in $RH_q$ for all $q\ge 1$.

Now, as a consequence of Theorem 2 and the previous result, we can obtain the following boundedness properties for differentiation operators of the form $L^{-\alpha /2} V^{\sigma /2}$ and $V^{\sigma /2}L^{-\alpha /2}$, where $\sigma >\alpha$. In what follows, for a better understanding of the statements, we change the parameters $\alpha$ and $\sigma$ by $\alpha$ and $\gamma$ with $\gamma =\sigma -\alpha$.

Theorem 4

Let us assume that $V\in RH_{d/2}$, $0<\alpha \le 2$ and $0<\gamma <1$. If there exists some $\varepsilon _0>0$ such that for each $0<\varepsilon <\varepsilon _0$, V satisfies

$$\begin{aligned} |V^{\gamma /2}(x) - V^{\gamma /2} (y)|\le C \rho (x)^{-\gamma } \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon }, \end{aligned}$$

(33)

then $L^{-\alpha /2} V^{\alpha /2+\gamma /2}$ maps continuously $\Lambda _{\rho }^\beta (w)$ into $\Lambda _{\rho }^{\beta -\gamma }(w)$, for any $\gamma<\beta <\min \{1,\alpha +\gamma ,\varepsilon _0+\gamma \}$ and $w\in w\in A_\infty ^\rho \cap D_\mu ^\rho$ with $1\le \mu < 1+ \frac{\min \{1,\alpha +\gamma ,\varepsilon _0+\gamma \}-\beta }{d}$.

Proof

Let $w\in A_{\infty }^{\rho }\cap D_\mu ^\rho$ and $f\in \Lambda _\rho ^\beta (w)$. First, observe that condition (33) implies that $V^{\gamma /2}(x)\le C \rho ^{-\gamma }(x)$, as it was done in the proof of Theorem 3.

This observation together with condition (33), let us apply Theorem 2 obtaining that the multiplication by $V^{\gamma /2}$ is an operator bounded from $\Lambda _\rho ^{\beta }(w)$ into $\Lambda _\rho ^{\beta -\gamma }(w)$, provided $\gamma<\beta <\min \{1,\gamma +\varepsilon _0\}$ and $1\le \mu < 1+ \frac{\min \{1,\gamma +\varepsilon _0\}-\beta }{d}$.

On the other hand, we apply Proposition 2 to assure that $L^{-\alpha /2}V^{\alpha /2}$ is bounded on $\Lambda _\rho ^{\beta -\gamma }(w)$ as long as $\gamma<\beta <\min \{1,\gamma +\alpha \}$ and $1\le \mu < 1+ \frac{\min \{1,\gamma +\alpha \}-\beta }{d}$.

Since our assumptions on $\beta$ and $\mu$ allow us to conclude the boundedness of both operators, the proof is complete.

$\square$

Theorem 5

Let $V\in RH_{d/2}$ and $0<\alpha <2$, $0<\gamma <1$. Assume further that for some $\varepsilon >0$, V satisfies

$$\begin{aligned} |V^{\alpha /2}(x) - V^{\alpha /2} (y)|\le C \rho (x)^{-\alpha } \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon }, \end{aligned}$$

for $|x-y| \le \rho (x)$. Then, for $\gamma< \beta <\beta _0=\min \{1,\varepsilon ,\gamma +\varepsilon \gamma /\alpha \}$, $V^{\alpha /2+\gamma /2}L^{-\alpha /2}$ maps continuously $\Lambda _{\rho }^\beta (w)$ into $\Lambda _{\rho }^{\beta -\gamma }(w)$ as long as $w \in A_\infty ^\rho \cap D_\mu ^\rho$ with $1\le \mu < 1+\frac{\beta _0-\beta }{d}$.

Proof

Let $w\in A_{\infty }^{\rho }\cap D_\mu ^\rho$ and $f\in \Lambda _\rho ^\beta (w)$. The assumptions on V and w allow us to apply Theorem 1 in [3], to show that for $0<\beta <\min \{1,\varepsilon \}$ and $1\le \mu <1+\frac{\min \{1,\varepsilon \}-\beta }{d}$,

$$\begin{aligned} \Vert V^{\alpha /2}L^{-\alpha /2}f\Vert _{\Lambda _\rho ^{\beta }(w)}\le C\Vert f\Vert _{\Lambda _\rho ^{\beta }(w)}. \end{aligned}$$

Now we will show that the same hypothesis on V allows us to establish the boundedness of the multiplier operator $Tg=V^{\gamma /2}g$ from $\Lambda _{\rho }^\beta (w)$ into $\Lambda _{\rho }^{\beta -\gamma }(w)$, via Theorem 2. It would be sufficient to find all possible $0<\beta <1$ and $\mu \ge 1$ such that, for $|x-y|\le \rho (x)$,

$$\begin{aligned} |V^{\gamma /2}(x) - V^{\gamma /2} (y)|\le C \rho (x)^{-\gamma } \left( \frac{|x-y|}{\rho (x)}\right) ^{\beta -\gamma +d(\mu -1)}, \end{aligned}$$

(34)

since, as in the previous results, (34) implies the size condition on V.

We will do that considering two cases: $\alpha >\gamma$ and $\alpha \le \gamma$. Let x, $y\in {\mathbb {R}}^d$ such that $|x-y|\le \rho (x)$. If $\alpha >\gamma$, we may apply the hypothesis on V to obtain

$$\begin{aligned} \begin{aligned} |V^{\gamma /2}(x) - V^{\gamma /2} (y)|&\le C |V^{\alpha /2}(x) - V^{\alpha /2} (y)|^{\gamma /\alpha }\\&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon \gamma /\alpha } \rho (x)^{-\gamma }, \end{aligned} \end{aligned}$$

so (34) holds provided $\gamma<\beta <\min \{1,\gamma +\varepsilon \gamma /\alpha \}$ and $1\le \mu < 1+\frac{\min \{1,\gamma +\varepsilon \gamma /\alpha \}-\beta }{d}$.

Next, if $\alpha \le \gamma$, we can use the Mean Value Theorem together with the hypothesis on V to obtain, for a point $\xi$ lying between x and y,

$$\begin{aligned} \begin{aligned} |V^{\gamma /2}(x) - V^{\gamma /2} (y)|&\le C |V^{\alpha /2}(x) - V^{\alpha /2} (y)| V(\xi )^{\alpha (\gamma /\alpha -1)/2}\\&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon } \rho (x)^{-\alpha } \rho (x)^{-\alpha (\gamma /\alpha -1)} \\&\le C \left( \frac{|x-y|}{\rho (x)}\right) ^{\varepsilon }\rho (x)^{-\gamma }, \end{aligned} \end{aligned}$$

which implies (34) for $\gamma<\beta <\min \{1,\gamma +\varepsilon \}$ and $1\le \mu < 1+\frac{\min \{1,\gamma +\varepsilon \}-\beta }{d}$.

Therefore, the three conditions on $\beta$ and $\mu$ are satisfied for $\gamma<\beta <\beta _0=\min \{1,\varepsilon ,\gamma +\varepsilon \gamma /\alpha \}$ and $1\le \mu < 1+\frac{\beta _0-\beta }{d}$. Altogether, we obtain that

$$\begin{aligned} \Vert V^{\gamma /2}(L^{-\alpha /2}V^{\alpha /2}f)\Vert _{\Lambda _\rho ^{\beta -\gamma }(w)}\le C\Vert L^{-\alpha /2}V^{\alpha /2}f\Vert _{\Lambda _\rho ^{\beta }(w)}\le C\Vert f\Vert _{\Lambda _\rho ^{\beta }(w)}. \end{aligned}$$

$\square$