Riesz Transforms Characterizations of Hardy Spaces $H^1$ for the Rational Dunkl Setting and Multidimensional Bessel Operators

Abstract

We characterize the Hardy space $H^1$ in the rational Dunkl setting associated with the reflection group $\mathbb {Z}_2^n$ by means of special Riesz transforms. As a corollary we obtain Riesz transforms characterization of $H^1$ for product of Bessel operators in $(0,\infty )^n$.

Appendix

It is well known that

$$\begin{aligned} \int _{{\mathbb {R}}} h^{\{j\}}_{ t}(x,y)\, d\mu _j(y)=1, \quad d\mu _j(y)= |y|^{2k_j}\, dy. \end{aligned}$$

(7.2)

It was proved in [2] that $h^{\{j\}}_t (x,y)$ has the following global behavior :

$$\begin{aligned} h^{\{j\}}_{ t}(x,y)\,\asymp \, {\left\{ \begin{array}{ll} t^{-k_j-\frac{1}{2}}\, e^{-\frac{x^2+ y^2}{4 t}} &{}\quad \text {if }| x y|\le t,\\ t^{-\frac{1}{2}}\,(x y)^{-k_j}\, e^{-\frac{(x-y)^2}{4 t}} &{}\quad \text {if }x y\ge t,\\ t^{\frac{1}{2}}\,(- x y)^{-k_j-1}\, e^{-\frac{(x+y)^2}{4 t}} &{}\quad \text {if }-x y\ge t.\\ \end{array}\right. } \end{aligned}$$

(7.3)

From (7.3) we easily conclude that

$$\begin{aligned}&0<{\mathbf {h}}_{t}({\mathbf {x}},{\mathbf {y}})\le \frac{C}{{\varvec{\mu }}(B({\mathbf {x}},\sqrt{t}))} \text {for all } \ {\mathbf {x}}, {\mathbf {y}} \in {\mathbb {R}}^n \text {and } t>0; \end{aligned}$$

(7.4)

$$\begin{aligned}&{\mathbf {h}}_{t}({\mathbf {x}},{\mathbf {y}})\le \frac{C}{{\varvec{\mu }}(B({\mathbf {x}},\sqrt{t}))} e^{-c|{\mathbf {x}}|^2\slash t} \text { for } |{\mathbf {x}}|>2n|{\mathbf {y}}|. \end{aligned}$$

(7.5)

We shall need the following inequalities for volumes of the Euclidean balls (see [2])

$$\begin{aligned} \big (\tfrac{R}{r}\big )^{n}\lesssim \tfrac{\varvec{\mu }(\mathbf {B}(\mathbf {x}, R))}{\varvec{\mu }(\mathbf {B}(\mathbf {x}, r))} \lesssim \big (\tfrac{R}{r}\big )^{\mathbf {N}}, \qquad \forall \;\mathbf {x}\in {\mathbb {R}}^n,\,\forall \;R\ge r>0. \end{aligned}$$

(7.6)

The subordination formula (2.1) combined with (7.1) and (7.2) implies

$$\begin{aligned} \int _{{\mathbb {R}}^n} P_t({\mathbf {x}},{\mathbf {y}})\, d{\varvec{\mu }} ({\mathbf {x}})=\int _{{\mathbb {R}}^n} P_t({\mathbf {x}},{\mathbf {y}})\, d{\varvec{\mu }} ({\mathbf {y}})=1. \end{aligned}$$

(7.7)

Lemma 7.1

There is a constant $C>0$ such that

$$\begin{aligned} 0< P_t({\mathbf {x}},{\mathbf {y}})\le \frac{C}{{\varvec{\mu }} (B({\mathbf {x}}, t))}. \end{aligned}$$

(7.8)

Moreover, for every $0<\delta <\frac{1}{\mathbf N}$ there is a constant $C_\delta$ such that

$$\begin{aligned} P_t({\mathbf {x}},{\mathbf {y}})\le \frac{C_\delta }{{\varvec{\mu }} (B({\mathbf {x}}, t))}\left( 1+\frac{ {\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}},t))}\right) ^{-1-\delta } \text { for } |{\mathbf {x}}|>2n|{\mathbf {y}}|. \end{aligned}$$

(7.9)

Proof

To see (7.8) we use (2.1) together with (7.4) and (7.6) and obtain

$$\begin{aligned}\begin{aligned} P_t({\mathbf {x}},{\mathbf {y}})&\lesssim \int _0^{\frac{1}{4}} \frac{1}{{\varvec{\mu }} (B({\mathbf {x}},t))} \frac{{\varvec{\mu }} (B({\mathbf {x}},t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))}\frac{du}{\sqrt{u}}+ \int _{\frac{1}{4}}^\infty \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}},t))} \frac{{\varvec{\mu }} (B({\mathbf {x}},t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))}\frac{du}{\sqrt{u}}\\&\lesssim \frac{1}{\mu (B({\mathbf {x}},t))} \left( \int _0^{\frac{1}{4}} u^{n\slash 2}\frac{d}{\sqrt{u}}+\int _{\frac{1}{4}}^\infty e^{-u} u^{\mathbf N\slash 2}\frac{du}{\sqrt{u}}\right) \lesssim \frac{1}{\mu (B({\mathbf {x}},t))}. \end{aligned}\end{aligned}$$

The proof of the lower bound of $P_t({\mathbf {x}},{\mathbf {y}})$ is obvious.

In order to prove (7.9) it suffices to consider $t\le |{\mathbf {x}}|\slash 2$. By (7.6), for every $\delta >0$ and $c>0$, we have

$$\begin{aligned} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}},\sqrt{s}))} \right) ^{1+\delta }\le C_\delta \left( 1+\frac{|{\mathbf {x}}|}{\sqrt{s}}\right) ^{(1+\delta ) \mathbf N} \le C_{\delta , c} e^{c|{\mathbf {x}}|^2\slash s}, \text {for } \ s>0. \end{aligned}$$

(7.10)

Utilizing (7.5) together with (7.10) and proceeding similarly to the proof of (7.8) we have

$$\begin{aligned}&P_t({\mathbf {x}},{\mathbf {y}}) \lesssim \int _0^\infty \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))}\right) ^{-1-\delta }\frac{du}{\sqrt{u}}\\&\quad = \int _0^{\infty } \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}},t))} \frac{{\varvec{\mu }} (B({\mathbf {x}},t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))}\\&\qquad \times \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \frac{{\varvec{\mu }} (B({\mathbf {x}}, t))}{{\varvec{\mu }} (B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))}\right) ^{-1-\delta } \frac{du}{\sqrt{u}}\\&\quad \lesssim \int _0^{\frac{1}{4}} \frac{1}{{\varvec{\mu }} (B({\mathbf {x}},t))} \frac{{\varvec{\mu }} (B({\mathbf {x}},t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, t))}\right) ^{-1-\delta }\\&\quad \quad \times \left( \frac{{\varvec{\mu }} (B({\mathbf {x}}, t))}{{\varvec{\mu }} (B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))}\right) ^{-1-\delta } \frac{du}{\sqrt{u}}\\&\qquad + \int _{\frac{1}{4}}^{\infty } \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}},t))} \frac{{\varvec{\mu }} (B({\mathbf {x}},t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, t))}\right) ^{-1-\delta } \frac{du}{\sqrt{u}}.\\ \end{aligned}$$

Fix $0<\delta <\mathbf {N}^{-1}$. Applying (7.6) we obtain

$$\begin{aligned} \begin{aligned} P_t({\mathbf {x}},{\mathbf {y}})&\lesssim \int _0^{\frac{1}{4}} \frac{1}{{\varvec{\mu }} (B({\mathbf {x}},t))} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, t))}\right) ^{-1-\delta } u^{-\delta \mathbf N\slash 2} \frac{du}{\sqrt{u}}\\&\qquad + \int _{\frac{1}{4}}^{\infty } \frac{e^{-u}u^{\mathbf N\slash 2}}{{\varvec{\mu }} (B({\mathbf {x}},t))} \left( 1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }} (B({\mathbf {x}}, t))}\right) ^{-1-\delta } \frac{du}{\sqrt{u}},\\ \end{aligned} \end{aligned}$$

which proves (7.9). $\square$

Proof of Theorem 2.1

The proof, which is in its spirit similar to that of the heat kernel characterization of $H^1_\mathrm{{atom}}$ (see [2]), is based on the following result due to Uchiyama [26].

Theorem 7.2

Assume that a set X is equipped with

• a quasi-distance $\widetilde{d}$ i.e., a distance except that the triangular inequality is replaced by the weaker condition

  $$\begin{aligned} \widetilde{d}(x,y)\le A\,\{\widetilde{d}(x,z)+\widetilde{d}(z,y)\}, \qquad \forall \;x,y,z\in X; \end{aligned}$$

• a measure $\mu$ whose values on quasi-balls satisfy

  $$\begin{aligned} \frac{r}{A}\le \mu (\widetilde{B}(x,r))\le r, \qquad \forall \;x\in X,\,\forall \;r>0\,; \end{aligned}$$

• a continuous kernel $K_r(x,y)\ge 0$ such that, for every $r>0$ and $x,y,y^{\prime }\in X$,

  • $K_r(x,x)\ge \frac{1}{Ar}$,

  • $K_r(x,y)\le r^{-1}\bigl ( 1+\frac{\widetilde{d}(x, y)}{r} \bigr )^{-1-\delta }$ ,

  • $\bigl | K_r(x,y)-K_r(x,y^{\prime })\bigr |\le r^{-1}\bigl (1+\frac{\widetilde{d}(x, y)}{r}\bigr )^{-1-2\delta } \bigl (\frac{\widetilde{d}(y, y^{\prime })}{r}\bigr )^{\delta }$ when $\widetilde{d}(y,y^{\prime }) \le \frac{r+\widetilde{d}(x, y)}{4 A}$ .

Here $A\ge 1$ and $\delta >0$. Then the following definitions of the Hardy space $H^1(X)$ and their corresponding norms are equivalent :

• Maximal definition : $H^1_{\mathrm{max}, K_r}(X)$ consists of all functions $f\in L^1(X,d\mu )$ such that

  $$\begin{aligned} K_*f(x)=\sup \nolimits _{\,r>0}\, \Bigl |{\displaystyle \int _X}K_r(x,y)\,f(y)\,d\mu (y)\Bigr | \end{aligned}$$

  belongs to $L^1(X,d\mu )$ and the norm $\Vert f\Vert _{H^1_{\mathrm{max}, K_r}(X)} = \Vert K_*f\Vert _{L^1(X,d\mu )}$.

• Atomic definition : An atom for $H^1_\mathrm{atom}(X,\widetilde{d})$ is a measurable function $a:X\rightarrow {\mathbb {C}}$  such that: a is supported in a quasi-ball $\widetilde{B}$, $\Vert a\Vert _{L^\infty }\lesssim \mu (\widetilde{B})^{-1}$ and $\displaystyle \int _Xa \,d\mu =0$ (see [5, 15, 26]). Then $H^1_\mathrm{atom}(X, \widetilde{d})$ consists of all functions $f\in L^1(X,d\mu )$ which can be written as $f=\sum _{\ell }\lambda _{\ell } a_{\ell }$, where the $a_{\ell }$’s are atoms and $\sum _{\ell }|\lambda _{\ell }|<+\infty$, and the norm $\Vert f\Vert _{H^1_\mathrm{atom}(X,\widetilde{d})}= \inf \sum _{\ell }|\lambda _{\ell }|$, where the infimum is taken over all such representations.

For $X={\mathbb {R}}^n$, equipped with the Euclidean distance $d(\mathbf {x},\mathbf {y})=|\mathbf {x}-\mathbf {y}|$ and the measure ${\varvec{\mu }}$ (see (1.5)), set

$$\begin{aligned} \widetilde{d}(\mathbf {x},\mathbf {y})=\inf \varvec{\mu }(B), \qquad \forall \;\mathbf {x},\mathbf {y}\in {\mathbb {R}}^n, \end{aligned}$$

where the infimum is taken over all closed balls B containing $\mathbf {x}$ and $\mathbf {y}$. Let $t=t(\mathbf {x},r)$ be defined by $\varvec{\mu }(B(\mathbf {x},\sqrt{t}))=r$. Then

$$\begin{aligned} {\varvec{\mu }}(\widetilde{B}({\mathbf {x}},r))\sim r \end{aligned}$$

and there exists a constant $c>0$ such that

$$\begin{aligned} B(\mathbf {x},\sqrt{ t\,}) \subset \widetilde{B}(\mathbf {x},r) \subset B(\mathbf {x},c\sqrt{ t\,}), \end{aligned}$$

(7.11)

where $\widetilde{B}(\mathbf {x},r)=\{ {\mathbf {y}}\in {\mathbb {R}}^n: \widetilde{d}({\mathbf {x}},{\mathbf {y}})<r\}$ (see, e.g., [2]).

Let us remark that thanks to (7.11) and (7.6) the atomic spaces $H^1_\mathrm{atom}(X,\widetilde{d})$ and $H^1_\mathrm{atom}$ (defined in Section 1) do coincide and $\Vert f\Vert _{H^1_\mathrm{atom}(X,\widetilde{d})} \sim \Vert f\Vert _{H^1_\mathrm{atom}}$.

It was proved in [2] that the kernel ${\mathbf {h}}_t$ can be written in the form

$$\begin{aligned} {\mathbf {h}}_{ t}(\mathbf {x},{\mathbf {y}}) ={\mathbf {H}}_{ t}(\mathbf {x},{\mathbf {y}}) +{\mathbf {S}}_{t}(\mathbf {x},{\mathbf {y}}), \end{aligned}$$

where ${\mathbf {H}}_{ t}(\mathbf {x},{\mathbf {y}})$ and ${\mathbf { S}}_{t}(\mathbf {x},{\mathbf {y}})$ are nonnegative continuous functions such that there are $C_1,\, C_2,\, C_4,\, \delta >0$ such that

$$\begin{aligned}&{\mathbf {H}}_t(\mathbf {x},\mathbf {x})\ge \tfrac{C_1}{{\varvec{\mu }} (B({\mathbf {x}},\sqrt{t}))}; \end{aligned}$$

(7.12)

$$\begin{aligned}&{\mathbf {H}}_t(\mathbf {x},\mathbf {y})\le \tfrac{C_2}{ {\varvec{\mu }} (B({\mathbf {x}},\sqrt{t}))} \,\bigl (1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{ {\varvec{\mu }} (B({\mathbf {x}},\sqrt{t}))} \bigr )^{-1-\delta }; \end{aligned}$$

(7.13)

$$\begin{aligned}&\bigl | {\mathbf {H}}_t (\mathbf {x},\mathbf {y})-{\mathbf {H}}_t(\mathbf {x},\mathbf {y}^{\prime })\bigr | \le \tfrac{C_4}{{\varvec{\mu }} ( B ({\mathbf {x}},\sqrt{t}))} \bigl ( 1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{{\varvec{\mu }} (B({\mathbf {x}},\sqrt{t}))} \bigr )^{-1-2\delta } \bigl (\tfrac{\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })}{{\varvec{\mu }} ( B({\mathbf {x}},\sqrt{t}))} \bigr )^{\frac{1}{\mathbf {N}}}\nonumber \\ \end{aligned}$$

(7.14)

for $\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })\le C_3\max \,\{ {\varvec{\mu }} (B({\mathbf {x}},\sqrt{t})),\widetilde{d}(\mathbf {x},\mathbf {y})\}$, (the kernel ${\mathbf {S}}_t$ is denoted in [2] by $\mathbf {P}_t$). Moreover, the maximal function

$$\begin{aligned} {\mathbf {S}}_*f(x)=\sup _{t>0} \Big |\int {\mathbf {S}}_t({\mathbf {x}},{\mathbf {y}})f({\mathbf {y}})d{\varvec{\mu }}({\mathbf {y}}) \Big | \end{aligned}$$

is a bounded operator on $L^1({\mathbb {R}}^n,\, d{\varvec{\mu }} )$.

Using subordination formula (2.1) we write

$$\begin{aligned} P_t({\mathbf {x}},{\mathbf {y}}) = U_t({\mathbf {x}},{\mathbf {y}}) + W_t({\mathbf {x}},{\mathbf {y}}), \end{aligned}$$

(7.15)

where

$$\begin{aligned} U_t({\mathbf {x}},{\mathbf {y}})=c_1\int _0^\infty e^{-u}{\mathbf {H}}_{t^2\slash 4u}({\mathbf {x}},{\mathbf {y}})\frac{du}{\sqrt{u}}, W_t({\mathbf {x}},{\mathbf {y}}) = c_1\int _0^\infty e^{-u}{\mathbf {S}}_{t^2\slash 4u}({\mathbf {x}},{\mathbf {y}})\frac{du}{\sqrt{u}}. \end{aligned}$$

Clearly, the maximal operator

$$\begin{aligned} W_*f({\mathbf {x}})=\sup _{t>0} \Big |\int W_t({\mathbf {x}},{\mathbf {y}})f({\mathbf {y}}) d{\varvec{\mu }}({\mathbf {y}})\Big | \end{aligned}$$

is bounded on $L^1({\mathbb {R}}^n,\, d{\varvec{\mu }} )$, that is,

$$\begin{aligned} \Vert W_*f\Vert _{L^1({\mathbb {R}}^n,\, d{\varvec{\mu }} )} \le C \Vert f\Vert _{L^1({\mathbb {R}}^n,\, d{\varvec{\mu }} )}. \end{aligned}$$

(7.16)

$\square$

Our task is to prove the following lemma.

Lemma 7.3

There are constants $C_1,\, C_2,\, C_4,\, \delta ' >0$ such that

$$\begin{aligned}&U_t(\mathbf {x},\mathbf {x})\ge \tfrac{C_1}{{\varvec{\mu }} (B ({\mathbf {x}},t))}; \end{aligned}$$

(7.17)

$$\begin{aligned}&U_t(\mathbf {x},\mathbf {y})\le \tfrac{C_2}{ {\varvec{\mu }} (B({\mathbf {x}},t))} \,\bigl (1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{ {\varvec{\mu }} (B({\mathbf {x}},t))} \bigr )^{-1-\delta '}; \end{aligned}$$

(7.18)

$$\begin{aligned}&\bigl | U_t (\mathbf {x},\mathbf {y})-U_t(\mathbf {x},\mathbf {y}^{\prime })\bigr | \le \tfrac{C_4}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \bigl ( 1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{{\varvec{\mu }} (B({\mathbf {x}},t))} \bigr )^{-1-2\delta '} \bigl (\tfrac{\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })}{{\varvec{\mu }} ( B({\mathbf {x}},t))} \bigr )^{\delta '} \end{aligned}$$

(7.19)

for $\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })\le C_3\max \,\{ {\varvec{\mu }} (B({\mathbf {x}},t )),\widetilde{d}(\mathbf {x},\mathbf {y})\}$ .

Proof

Take $0<\delta <\mathbf {N}^{-1}$. By (7.12) and the subordination formula we have

$$\begin{aligned} \begin{aligned} U_t({\mathbf {x}},{\mathbf {x}})&\gtrsim \int _1^\infty \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))} \frac{du}{\sqrt{u}} \gtrsim \int _1^\infty \frac{e^{-u}}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))} \frac{du}{\sqrt{u}}\\&\gtrsim \frac{1}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \int _1^\infty \frac{e^{-u}}{\sqrt{u}}\, du \gtrsim \frac{1}{{\varvec{\mu }}(B({\mathbf {x}}, t))}, \end{aligned} \end{aligned}$$

which proves (7.17).

The proof of (7.18) is similar to that of (7.9). Indeed, by (7.13) we have

$$\begin{aligned} \begin{aligned} U_t({\mathbf {x}},{\mathbf {y}})&\le \int _0^\infty \frac{e^{-u}}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }}(B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))}\\&\qquad \left( 1+ \frac{\tilde{d}({\mathbf {x}},{\mathbf {y}})}{{{\varvec{\mu }}(B({\mathbf {x}}, t))}}\frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }}(B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))}\right) ^{-1-\delta }\frac{du}{\sqrt{u}}\\&\le \int _0^{\frac{1}{4}} \frac{e^{-u}}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }}(B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))} \left( 1+ \frac{\tilde{d}({\mathbf {x}},{\mathbf {y}})}{{{\varvec{\mu }}(B({\mathbf {x}}, t))}}\right) ^{-1-\delta }\\&\qquad \left( \frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }}(B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))} \right) ^{-1-\delta } \frac{du}{\sqrt{u}}\\&\qquad + \int _{\frac{1}{4}}^\infty \frac{e^{-u}}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \frac{{\varvec{\mu }}(B({\mathbf {x}}, t))}{{\varvec{\mu }}(B({\mathbf {x}}, \frac{t}{2\sqrt{u}}))} \left( 1+ \frac{\tilde{d}({\mathbf {x}},{\mathbf {y}})}{{{\varvec{\mu }}(B({\mathbf {x}}, t))}}\right) ^{-1-\delta }\frac{du}{\sqrt{u}}.\\ \end{aligned} \end{aligned}$$

Now using (7.6) we obtain (7.18).

Now we turn to the proof of (7.19). First we show that there is a constant $C_4>0$ such that

$$\begin{aligned} \bigl | U_t (\mathbf {x},\mathbf {y})-U_t(\mathbf {x},\mathbf {y}^{\prime })\bigr | \le \tfrac{C_4}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \bigl (\tfrac{\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })}{{\varvec{\mu }} ( B({\mathbf {x}},t))} \bigr )^{\frac{1}{\mathbf {N}}} \text { for every } {\mathbf {x}},{\mathbf {y}}, {\mathbf {y}}'\in {\mathbb {R}}^n. \end{aligned}$$

(7.20)

Since $U_t({\mathbf {x}},{\mathbf {y}})\le C{\varvec{\mu }} (B({\mathbf {x}},t))^{-1}$ (see (7.18)), it suffices to prove (7.20) for $\tilde{d}({\mathbf {y}},{\mathbf {y}}')\le {\varvec{\mu }} (B({\mathbf {x}},t))$. Let $u_0\ge 1\slash 4$ be such that ${\varvec{\mu }}(B({\mathbf {x}},\frac{t}{2\sqrt{u_0}}))=\tilde{d}({\mathbf {y}},{\mathbf {y}}')$. Then, using (7.14) and (7.6), we have

$$\begin{aligned}&\displaystyle \int _0^{u_0} e^{-u}|{\mathbf {H}}_{\frac{t^2}{4u}}({\mathbf {x}},{\mathbf {y}})-{\mathbf {H}}_{\frac{t^2}{ 4u}}({\mathbf {x}},{\mathbf {y}}')|\frac{du}{\sqrt{u}}\\&\quad \displaystyle \lesssim \int _0^{u_0} \frac{e^{-u}}{{\varvec{\mu }} (B(\mathbf x,\frac{t}{2\sqrt{u}}))} \left( \frac{\tilde{d}({\mathbf {y}},\mathbf y')}{{\varvec{\mu }}(B({\mathbf {x}},\frac{t}{2\sqrt{u}}))}\right) ^{\frac{1}{ \mathbf N} } \frac{du}{\sqrt{u}}\\&\quad \displaystyle = \int _0^{u_0} \frac{e^{-u}}{{\varvec{\mu }} (B({\mathbf {x}},t))} \left( \frac{\tilde{d}({\mathbf {y}},\mathbf y')}{{\varvec{\mu }}(B({\mathbf {x}},t))}\right) ^{\frac{1}{ \mathbf N} } \left( \frac{\mu (B({\mathbf {x}},t))}{\mu (B(\mathbf {x},\frac{t}{2\sqrt{u}}))}\right) ^{1+\frac{1}{\mathbf N}} \frac{du}{\sqrt{u}}\\&\quad \displaystyle \lesssim \frac{1}{{\varvec{\mu }} (B({\mathbf {x}},t))} \left( \frac{\tilde{d}({\mathbf {y}},\mathbf y')}{{\varvec{\mu }}(B({\mathbf {x}},t))}\right) ^{\frac{1}{ \mathbf N} }\\&\quad \quad \displaystyle \left( \int _0^{1\slash 4} e^{-u}u^{n(1+\frac{1}{\mathbf N})\slash 2}\frac{du}{\sqrt{u}}+\int _{1\slash 4}^{u_0}e^{-u}u^{\mathbf N(1+\mathbf N^{-1})\slash 2} \frac{du}{\sqrt{u}}\right) \\&\quad \displaystyle \lesssim \frac{1}{{\varvec{\mu }} (B({\mathbf {x}},t))} \left( \frac{\tilde{d}({\mathbf {y}},\mathbf y')}{{\varvec{\mu }}(B({\mathbf {x}},t))}\right) ^{\frac{1}{ \mathbf N} }. \end{aligned}$$

Similarly, by (7.13), we get

$$\begin{aligned}\begin{aligned} \int _{u_0}^\infty e^{-u}|{\mathbf {H}}_{\frac{t^2}{4u}}({\mathbf {x}},{\mathbf {y}})-{\mathbf {H}}_{\frac{t^2}{ 4u}}({\mathbf {x}},{\mathbf {y}}')|\frac{du}{\sqrt{u}}&\lesssim \int _{u_0}^\infty \frac{e^{-u}}{{\varvec{\mu }}(B({\mathbf {x}},t))}\left( \frac{\mu (B({\mathbf {x}},t))}{ \mu (B({\mathbf {x}},\frac{t}{2\sqrt{u}}))}\right) \frac{du}{\sqrt{u}} \\&\lesssim \frac{1}{{\varvec{\mu }}(B({\mathbf {x}},t))} \int _{u_0}^\infty e^{-u}u^{\mathbf N\slash 2}\frac{du}{\sqrt{u}}\\&\lesssim \frac{1}{{\varvec{\mu }}(B({\mathbf {x}},t))}u_0^{-\mathbf N\slash 2}. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \frac{\tilde{d}({\mathbf {y}},{\mathbf {y}}')}{ {\varvec{\mu }} ( B({\mathbf {x}},t))}= \frac{{\varvec{\mu }} (B({\mathbf {x}},\frac{t}{2\sqrt{u_0}}))}{{\varvec{\mu }} (B({\mathbf {x}},t))}\gtrsim u_0^{-\mathbf N\slash 2}, \end{aligned}$$

(see (7.6)), we obtain (7.20).

We are now in a position to continue the proof of (7.19).

If $\tilde{d}({\mathbf {x}},{\mathbf {y}})\le {\varvec{\mu }} (B({\mathbf {x}},t))$ then (7.19) follows from (7.20).

If $\tilde{d}({\mathbf {x}},{\mathbf {y}})>{\varvec{\mu }} (B({\mathbf {x}},t))$ and $\tilde{d}({\mathbf {y}},{\mathbf {y}}') <\tilde{d}({\mathbf {x}},{\mathbf {y}})\slash (2A)$, then $\tilde{d}({\mathbf {x}},{\mathbf {y}})\le 2A\tilde{d} ({\mathbf {x}},{\mathbf {y}}')$. Hence, from (7.18) we conclude that

$$\begin{aligned} \bigl | U_t (\mathbf {x},\mathbf {y})-U_t(\mathbf {x},\mathbf {y}^{\prime })\bigr | \le \tfrac{C_2'}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \bigl ( 1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{{\varvec{\mu }} (B({\mathbf {x}},t))} \bigr )^{-1-\delta } . \end{aligned}$$

(7.21)

Consequently, we deduce (7.19) (with perhaps small $\delta '>0$) from (7.20) and (7.21).

It remains to consider the case when $\tilde{d}({\mathbf {x}},{\mathbf {y}}) > {\varvec{\mu }} (B({\mathbf {x}},t))$ and $\tilde{d}({\mathbf {y}},{\mathbf {y}}') \ge \tilde{d}({\mathbf {x}},{\mathbf {y}})\slash (2A)$. Recall that $\tilde{d}({\mathbf {y}},{\mathbf {y}}')\le {\varvec{\mu }} (B({\mathbf {x}},t))$. Thus $\tilde{d}({\mathbf {x}},{\mathbf {y}})\sim {\varvec{\mu }} (B({\mathbf {x}},t))$. So, finally, using (7.20) we have

$$\begin{aligned}&\bigl | U_t (\mathbf {x},\mathbf {y})-U_t(\mathbf {x},\mathbf {y}^{\prime })\bigr | \le \tfrac{C_4}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \bigl (\tfrac{\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })}{{\varvec{\mu }} ( B({\mathbf {x}},t))} \bigr )^{\frac{1}{\mathbf {N}}}\\&\quad \le \tfrac{C_4}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \bigl (\tfrac{\widetilde{d}(\mathbf {y},\mathbf {y}^{\prime })}{{\varvec{\mu }} ( B({\mathbf {x}},t))} \bigr )^{\frac{1}{\mathbf {N}}}\bigl ( 1+\tfrac{\widetilde{d}(\mathbf {x},\mathbf {y})}{{\varvec{\mu }} (B({\mathbf {x}},t))} \bigr )^{-1-\delta }. \end{aligned}$$

This completes the proof of Lemma 7.3. $\square$

Set $K_r({\mathbf {x}},{\mathbf {y}})=U_t({\mathbf {x}},{\mathbf {y}})$, where $r={\varvec{\mu }} (B({\mathbf {x}},t))$. Now Theorem 2.1 follows from (7.15), boundedness of the maximal function $W_*$ on $L^1({\mathbb {R}}^n, {\varvec{\mu }})$, and the Uchiyama theorem (see Theorem 7.2) combined with Lemma 7.3.

Now we turn to prove (2.4). Recall that $P_t({\mathbf {x}},{\mathbf {y}})>0$. So, by (7.7), the operator $P_*$ is bounded on $L^\infty ({\mathbb {R}}^n, d{\varvec{\mu }})$. Thanks to (7.18) and (7.16), it is of weak-type (1,1). Finally, from the Marcinkiewicz interpolation theorem we conclude that $P_*$ is bounded on $L^p({\mathbb {R}}^n,\, d{\varvec{\mu }})$ for $1<p<\infty$. $\square$

Proof of Proposition 2.2

Fix $\varepsilon >0$. There is $R>0$ such that $|g({\mathbf {x}})|<\varepsilon$ for $|{\mathbf {x}}|>R$. Write

$$\begin{aligned} g=g\chi _{B(0,R)}+g\chi _{B(0,R)^c}=: g_1+ g_2. \end{aligned}$$

From (7.7) we get $|P_tg_2({\mathbf {x}})|<\varepsilon$ for every $t>0$ and ${\mathbf {x}}\in {\mathbb {R}}^n$. Now using (7.8) we obtain

$$\begin{aligned} | P_tg_1({\mathbf {x}})| \le \frac{C}{{\varvec{\mu }} (B({\mathbf {x}}, t))} \Vert g_1\Vert _{L^1({\mathbb {R}}^n, \, d{\varvec{\mu }})}\rightarrow 0 \text { as }\ t\rightarrow \infty . \end{aligned}$$

On the other hand, if t remains in a bounded interval and $|{\mathbf {x}}|>2nR$, applying (7.9) we have

$$\begin{aligned} |P_tg_1({\mathbf {x}})| \le \frac{C}{{\varvec{\mu }}(B({\mathbf {x}}, t))} \Big (1+\frac{{\varvec{\mu }} (B({\mathbf {x}}, |{\mathbf {x}}|))}{{\varvec{\mu }}(B({\mathbf {x}}, t))}\Big )^{-1-\delta } \Vert g_1\Vert _{L^1({\mathbb {R}}^n, \, d{\varvec{\mu }})}\rightarrow 0 \text { as } |{\mathbf {x}}|\rightarrow \infty . \end{aligned}$$

The proof of the first part of Proposition 2.2 is complete.

In order to prove the second part of the proposition we fix $\varepsilon >0$. We claim that

$$\begin{aligned} \lim _{|{\mathbf {x}}|\rightarrow \infty } P_\varepsilon f({\mathbf {x}}) =0. \end{aligned}$$

To prove the claim let $\varepsilon ' >0$. Take $R>0$ large enough such that $\int _{|{\mathbf {y}}|>R} |f({\mathbf {y}})|\, d{\varvec{\mu }} ({\mathbf {y}})\le \varepsilon ' {\varvec{\mu }}(B(0, \varepsilon ))$. Write $f=f\chi _{B(0,R)}+f\chi _{B(0,R)^c}=:f_1+f_2$. Then, by (7.7) and (7.8) we have $|P_\varepsilon f_2|\le \varepsilon '$. On the other hand from the first part of the proposition we conclude that $\lim _{|{\mathbf {x}}|\rightarrow \infty } P_\varepsilon f_1({\mathbf {x}})=0$, which gives the claim. Now (2.5) follows from the first part of Proposition 2.2, since $P_{t+\varepsilon } f=P_t(P_\varepsilon f)$. $\square$