The proof of Theorem 1.10 requires some weighted estimates in Dunkl analysis, which are well-known in the Euclidean setting corresponding to \(\mathbf{k}=\mathbf{0}\). Let us first prove a weak analog of the Euclidean estimate
$$\begin{aligned} \Vert (1\!+\!|\varvec{\xi }|)^\sigma \widehat{f}(\varvec{\xi })\Vert _{L^1(d\varvec{\xi })}\lesssim \Vert f\Vert _{W_{2}^{\sigma +n/2+\epsilon }}\,. \end{aligned}$$
Lemma 5.1
For every \(\ell \!\in \!\mathbb {N}\) and \(r\!>\!0\), there is a constant \(C=C_{\ell ,r}\!>\!0\) such that
$$\begin{aligned} \sup \nolimits _{\,\varvec{\xi }\in \mathbb {R}^n}(1\!+\!|\varvec{\xi }|)^\ell \,|{\mathcal {F}}f(\varvec{\xi })| \le C\,\Vert f\Vert _{C^\ell } \end{aligned}$$
for every \(f\!\in \!C^\ell (\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Proof
By using the Riemann–Lebesgue lemma for the Fourier transform (1.9), we get
$$\begin{aligned} \sup \nolimits _{\,\varvec{\xi }\in \mathbb {R}^n}(1\!+\!|\varvec{\xi }|)^\ell \,|{\mathcal {F}}f(\varvec{\xi })|&\lesssim \,\sup \nolimits _{\,\varvec{\xi }\in \mathbb {R}^n} \left( 1+\sum \nolimits _{j=1}^{\,n}|\xi _j|^\ell \right) \, |{\mathcal {F}}f(\varvec{\xi })|\\&\le \,\Vert f\Vert _{L^1(\mathrm{d}\varvec{\mu })}+\sum \nolimits _{j=1}^{\,n} \Vert D_j^{\ell }f\Vert _{L^1(\mathrm{d}\varvec{\mu })}\,. \end{aligned}$$
The last expression is bounded by \(\Vert f\Vert _{C^\ell }\) as, by induction on \(\ell \), \(\mathrm{supp }( D_j^{\ell }f)\!\subset \!B(0,r)\) and \(\Vert D_j^{\ell }f\Vert _{L^\infty }\! \lesssim \Vert f\Vert _{C^\ell }.\)
\(\square \)
Corollary 5.2
For every \(\ell \!\in \!\mathbb {N}\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\ell ,r,\epsilon }\!>\!0\) such that
$$\begin{aligned} \Vert (1\!+\!|\varvec{\xi }|)^{\ell -\mathbf{N}/2-\epsilon } {\mathcal {F}}f(\varvec{\xi })\Vert _{L^2(\mathrm{d}\varvec{\mu }(\varvec{\xi }))}\le C\,\Vert f\Vert _{W_{2}^{\ell +n/2+\epsilon }}\,, \end{aligned}$$
for every \(f\!\in \!W_{2}^{\ell +n/2+\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\), where \(\mathbf{N}\) denotes the homogeneous dimension (1.6).
Proof
This result is deduced from Lemma 5.1, by using on the left-hand side the finiteness of the integral
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\varvec{\xi })\, (1\!+\!|\varvec{\xi }|)^{-\mathbf{N}-2\epsilon } \end{aligned}$$
and on the right-hand side the Euclidean Sobolev embedding theorem. \(\square \)
Proposition 5.3
For every \(\sigma \!>\!0\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\sigma ,r,\epsilon }\!>\!0\) such that
$$\begin{aligned} \Vert (1\!+\!|\varvec{\xi }|)^\sigma {\mathcal {F}}f(\varvec{\xi })\Vert _{L^2(\mathrm{d}\varvec{\mu }(\varvec{\xi }))}\le C\,\Vert f\Vert _{W_{2}^{\sigma +\epsilon }} \end{aligned}$$
for every \(f\!\in \!W_{2}^{\sigma +\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Proof
Let \(\chi \!\in \!C_c^\infty (\mathbb {R}^n)\). Following an argument due to Mauceri and Meda [13], this result is obtained by interpolation between the \(L^2\) estimate
$$\begin{aligned} \Vert {\mathcal {F}}(\chi f)\Vert _{L^2(\mathrm{d}\varvec{\mu })}=\mathrm{const. }\Vert \chi f\Vert _{L^2(\mathrm{d}\varvec{\mu })} \lesssim \Vert f\Vert _{L^2(\mathrm{d}\mathbf{x})}, \end{aligned}$$
which is deduced from Plancherel’s formula, and the following estimate for \(\ell \!\in \!\mathbb {N}\) large, which is deduced from Corollary 5.2:
$$\begin{aligned} \Vert (1\!+\!|\varvec{\xi }| )^{\ell -\mathbf{N}/2-\epsilon ^{\prime }}{\mathcal {F}}(\chi f)(\varvec{\xi })\Vert _{L^2(\mathrm{d}\varvec{\mu }(\varvec{\xi }))} \lesssim \Vert \chi f\Vert _{ W_{2}^{\ell +n/2+\epsilon ^{\prime }}} \lesssim \Vert f\Vert _{ W_{2}^{\ell +n/2+\epsilon ^{\prime }}}. \end{aligned}$$
\(\square \)
By using the Cauchy–Schwarz inequality, we deduce finally the following result.
Corollary 5.4
For every \(\sigma \!>\!0\), \(r\!>\!0\), and \(\epsilon \!>\!0\), there is a constant \(C=C_{\sigma ,r,\epsilon }\!>\!0\) such that
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\varvec{\xi })\, (1\!+\!|\varvec{\xi }|)^\sigma \,|{\mathcal {F}}f(\varvec{\xi })| \le C\,\Vert f\Vert _{ W_{2}^{\sigma +\mathbf{N}/2+\epsilon }} \end{aligned}$$
for every \(f\!\in \! W_{2}^{\sigma +\mathbf{N}/2+\epsilon }(\mathbb {R}^n)\) with \(\mathrm{supp }f\!\subset \mathbf{B}(0,r)\).
Let us next prove analogs in the Dunkl setting of the Euclidean estimates
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\mathbf{x}\, (1\!+\!|\mathbf{x}|)^\delta \,|f*g(\mathbf{x})|\, \le \int _{\mathbb {R}^n}\!\mathrm{d}\mathbf{z}\, (1\!+\!|\mathbf{z}|)^\delta \,|f(\mathbf{z})|\, \int _{\mathbb {R}^n}\!\mathrm{d}\mathbf{y}\, (1\!+\!|\mathbf{y}|)^\delta \,|g(\mathbf{y})| \end{aligned}$$
and
$$\begin{aligned} \int _{\mathbb {R}^n\backslash \mathbf{B}(\mathbf{y},r)}\mathrm{d}\mathbf{x}\, |f(\mathbf{x}\!-\!\mathbf{y})|\,\lesssim \,r^{-\delta }\; \Vert (1\!+\!|\mathbf{x}|)^\delta f(\mathbf{x})\Vert _{L^1(\mathrm{d}\mathbf{x})}\,. \end{aligned}$$
Recall that Dunkl translations are defined via the Fourier transform (1.9) by
$$\begin{aligned} (\tau _{\mathbf{y}}f)(\mathbf{x}) =\mathbf{c}_{\mathbf{k}}^{-1}\! \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\varvec{\xi })\,{\mathcal {F}}f(\mathbf{\varvec{\xi }})\, \mathbf{E}(\mathbf{x},i\varvec{\xi })\, \mathbf{E}(\mathbf{y},i\varvec{\xi }) \end{aligned}$$
(see [17, 19, 23, 24]) and have an explicit integral representation
$$\begin{aligned} (\tau _{\mathbf{y}}f)(\mathbf{x})=\int _{\mathbb {R}^n} \mathrm{d}\varvec{\nu }_{\mathbf{x},\mathbf{y}}(\mathbf{z})\, f(\mathbf{z}) \end{aligned}$$
in dimension \(1\) (see [2, 16, 23]) and hence in the product case. Specifically,
$$\begin{aligned} \mathrm{d}\varvec{\nu }_{\mathbf{x},\mathbf{y}}(\mathbf{z}) =d\nu _{x_1,y_1}^{(1)}(z_1)\cdots \, \mathrm{d}\nu _{x_n,y_n}^{(n)}(z_n)\,, \end{aligned}$$
where
$$\begin{aligned} \mathrm{d}\nu _{x_j,y_j}^{(j)}(z_j)\, =\,{\left\{ \begin{array}{ll} \,\nu _j(x_j,y_j,z_j)\,|z_j|^{2k_j}\mathrm{d}z_j &{}\text {if }\, x_j,y_j\!\in \mathbb {R}^*,\\ \,\mathrm{d}\delta _{y_j}(z_j) &{}\text {if }\, x_j\!=0,\\ \,\mathrm{d}\delta _{x_j}(z_j) &{}\text {if} \, y_j\!=0, \end{array}\right. } \end{aligned}$$
and
$$\begin{aligned}\begin{aligned} \nu _j(x_j,y_j,z_j)\,&=\frac{\Gamma (k_j+\frac{1}{2})}{\sqrt{\pi }\,2^{2k_j}\Gamma (k_j)}\, \frac{(x_j+y_j+z_j)(-x_j+y_j+z_j)(x_j-y_j+z_j)}{x_jy_jz_j}\\&\quad \times \tfrac{\left\{ \left( |x_j|+|y_j|+|z_j|\right) \left( -|x_j|+|y_j|+|z_j|\right) \left( |x_j|-|y_j|+|z_j|\right) \left( |x_j|+|y_j|-|z_j|\right) \right\} ^{k_j-1}}{|x_jy_jz_j|^{2k_j-1}}\\&\quad \times \mathrm{1}\mathrm{I}_{\,\left[ \left| |x_j|-|y_j|\right| , |x_j|+|y_j|\right] }(|z_j|) \,. \end{aligned}\end{aligned}$$
Thus \(\varvec{\nu }_{\mathbf{x},\mathbf{y}}\) is a signed measure which is supported in the product
$$\begin{aligned} \varvec{{\mathcal {I}}}_{\mathbf{x},\mathbf{y}} ={\mathcal {I}}_{x_1,y_1} \times \cdots \times \, {\mathcal {I}}_{x_n,y_n} \end{aligned}$$
of the one-dimensional sets
$$\begin{aligned} {\mathcal {I}}_{x_j,y_j}&=\left\{ z_j\!\in \!\mathbb {R}\,| \left| |x_j|-|y_j|\right| \le |z_j|\le |x_j|+|y_j|\right\} \\&=\left[ -|x_j|\!-\!|y_j|,-\left| |x_j|\!-\!|y_j|\right| \right] \cup \left[ \left| |x_j|\!-\!|y_j|\right| ,|x_j|\!+\!|y_j|\right] \end{aligned}$$
and which is generically given by
$$\begin{aligned} \mathrm{d}\varvec{\nu }_{\mathbf{x},\mathbf{y}}(\mathbf{z}) =\underbrace{\nu _1(x_1,y_1,z_1)\cdots \,\nu _n(x_n,y_n,z_n)}_{ \varvec{\nu }(\mathbf{x},\mathbf{y},\mathbf{z})}\, \mathrm{d}\varvec{\mu }(\mathbf{z})\,. \end{aligned}$$
Moreover, it is known (see [2, 16, 23]) that
$$\begin{aligned} \sup \nolimits _{\,\mathbf{x},\mathbf{y}\in \mathbb {R}^n}|\varvec{\nu }_{\mathbf{x},\mathbf{y}}|(\mathbb {R}^n) <+\infty \,. \end{aligned}$$
Lemma 5.5
For every \(\delta \ge 0\), \(L^1((1\!+\!|\mathbf{x}|)^\delta \mathrm{d}\varvec{\mu }(\mathbf{x}))\) is an algebra with respect to the Dunkl convolution product
$$\begin{aligned} f*g(\mathbf{x}) =\!\int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{y})\, (\tau _{-\mathbf{y}}f)(\mathbf{x})\,g(\mathbf{y}). \end{aligned}$$
Proof
By using the symmetries
$$\begin{aligned} \varvec{\nu }(\mathbf{x},-\mathbf{y},\mathbf{z}) =\varvec{\nu }(-\mathbf{z},-\mathbf{y},-\mathbf{x}) =\varvec{\nu }(\mathbf{z},\mathbf{y},\mathbf{x})\,, \end{aligned}$$
we have
$$\begin{aligned} f*g(\mathbf{x}) =\!\int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{z})\,f(\mathbf{z}) \int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{y})\,g(\mathbf{y})\, \varvec{\nu }(\mathbf{z},\mathbf{y},\mathbf{x})\,. \end{aligned}$$
We conclude by estimating
$$\begin{aligned} \int _{\varvec{{\mathcal {I}}}_{\mathbf{z},\mathbf{y}}} \mathrm{d}\varvec{\mu }(\mathbf{x})\,\left( 1\!+\!\left| \mathbf{x}\right| \right) ^\delta \, \left| \varvec{\nu }(\mathbf{z},\mathbf{y},\mathbf{x})\right| \lesssim \left( 1\!+\!\left| \mathbf{z}\right| \right) ^\delta \left( 1\!+\!\left| \mathbf{y}\right| \right) ^\delta . \end{aligned}$$
\(\square \)
Lemma 5.6
For every \(\delta \!>\!0\), there is a constant \(C\!>\!0\) such that, for every \(\mathbf{y}\in \mathbb {R}^n\) and \(r\!>\!0\),
$$\begin{aligned} \int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{y},r)}\mathrm{d}\varvec{\mu }(\mathbf{x}) \left| (\tau _{-\mathbf{y}}f)(\mathbf{x})\right| \le C\,r^{-\delta }\,\left\| f\right\| _{L^1((1+|\mathbf{x}|)^\delta \mathrm{d}\varvec{\mu }(\mathbf{x}))}\!, \end{aligned}$$
where
$$\begin{aligned} \mathcal {O}(\mathbf{y},r) =\big \{\,\mathbf{x}\!\in \!\mathbb {R}^n\,|\;\big ||x_j|-|y_j|\big |\le r, \forall \;1\le j\le n\,\big \} \end{aligned}$$
is the orbit of the cube \(\mathbf Q(\mathbf{y},r)=\prod _{j=1}^nB(y_j,r)\) under the group generated by the reflections (1.1).
Proof
As \(\mathbb {R}^n\!\backslash \!\mathcal {O}(\mathbf{y},r)\) is contained in the union of the sets
$$\begin{aligned} A_j=\big \{\,\mathbf{x}\!\in \!\mathbb {R}^n\,|\;\big ||x_j|-|y_j|\big |>r\,\big \} \qquad (j=1,\ldots ,n), \end{aligned}$$
we have
$$\begin{aligned} \int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{y},r)}\mathrm{d}\varvec{\mu }(\mathbf{x}) \left| (\tau _{-\mathbf{y}}f)(\mathbf{x})\right| \le \,\sum \nolimits _{j=1}^{\,n} \int _{A_j}\mathrm{d}\varvec{\mu }(\mathbf{x}) \int _{\varvec{{\mathcal {I}}}_{\mathbf{x},\mathbf{y}}} \mathrm{d}\varvec{\mu }(\mathbf{z}) \,\left| \varvec{\nu }(\mathbf{x},-\mathbf{y},\mathbf{z})\right| \,\left| f(\mathbf{z})\right| . \end{aligned}$$
As
$$\begin{aligned} |\mathbf{z}|\ge |z_j|\ge \big ||x_j|-|y_j|\big |>r \end{aligned}$$
when \(\mathbf{x}\!\in \!A_j\) and \(\mathbf{z}\!\in \!{\mathcal {I}}_{\mathbf{x},\mathbf{y}}\), the latter expression is bounded above by
$$\begin{aligned} r^{-\delta } \int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{z})\, \left| \mathbf{z}\right| ^\delta \,|f(\mathbf{z})| \int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{x}) \,|\varvec{\nu }(\mathbf{x},-\mathbf{y},\mathbf{z})|\,. \end{aligned}$$
We conclude by using the uniform estimate
$$\begin{aligned} \int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{x}) \,|\varvec{\nu }(\mathbf{x},-\mathbf{y},\mathbf{z})| =\int _{\mathbb {R}^n}\mathrm{d}\varvec{\mu }(\mathbf{x}) \,|\varvec{\nu }(\mathbf{z},\mathbf{y},\mathbf{x})| \le C. \end{aligned}$$
\(\square \)
Let us turn to the proof of Theorem 1.10, which consists in estimating
$$\begin{aligned} \Vert \mathbf{h}_{*}({\mathcal {T}}_{m}a)\Vert _{L^1(d\varvec{\mu })} \lesssim M\end{aligned}$$
(5.7)
for every atom \(a\) in the Hardy space \(H^1\!=H^1_\mathrm{atom}\). By rescaling, it suffices to prove (5.7) for any atom \(a\) associated with a unit ball \(\mathbf{B}(\mathbf{z},1)\). As \(\mathbf{h}_{*}\) and \({\mathcal {T}}_{m}\) are bounded on \(L^2(\mathbb {R}^n,d\varvec{\mu })\), we have
$$\begin{aligned} \left\| \mathbf{h}_{*}({\mathcal {T}}_{m}a)\right\| _{L^1(\mathcal {O}(\mathbf{z},2),\mathrm{d}\varvec{\mu })} \lesssim M. \end{aligned}$$
Thus it remains to show that
$$\begin{aligned} \Vert \mathbf{h}_{*}({\mathcal {T}}_{m}a)\Vert _{L^1(\mathbb {R}^n\backslash \mathcal {O}(\mathbf{z},2),\mathrm{d}\varvec{\mu })}\lesssim M. \end{aligned}$$
(5.8)
For this purpose, let us introduce a dyadic partition of unity on the Dunkl transform side. More precisely, given a smooth radial function \(\psi \) on \(\mathbb {R}^n\) such that
$$\begin{aligned} \mathrm{supp }\psi \subset \{\,\varvec{\xi }\!\in \!\mathbb {R}^n\,|\,\frac{1}{2}\le |\varvec{\xi }|\le 2\,\} \quad \text {and}\quad \sum \nolimits _{\ell \in \mathbb {Z}}\psi (2^{-\ell }\varvec{\xi })^2=1, \quad \forall \;\varvec{\xi }\!\in \!\mathbb {R}^n\!\backslash \!\{0\}\,, \end{aligned}$$
let us split up
$$\begin{aligned} e^{-t|\varvec{\xi }|^2}m(\varvec{\xi }) =\sum \nolimits _{\ell \in \mathbb {Z}} \psi (2^{-\ell }\varvec{\xi })\,e^{-t|\varvec{\xi }|^2}\psi (2^{-\ell }\varvec{\xi })\,m(\varvec{\xi })\,. \end{aligned}$$
Set
$$\begin{aligned} m_{t,\ell }(\varvec{\xi }) =\overbrace{ \psi (\varvec{\xi })\,e^{-t|2^{\ell }\varvec{\xi }|^2} }^{\psi _{t,\ell }(\varvec{\xi })} \overbrace{ \psi (\varvec{\xi })\,m(2^{\ell }\varvec{\xi }) }^{m_\ell (\varvec{\xi })},\\ f_{t,\ell }={\mathcal {F}}^{-1}(m_{t,\ell }) =\underbrace{{\mathcal {F}}^{-1}(\psi _{t,\ell })}_{g_{t,\ell }}*\underbrace{{\mathcal {F}}^{-1}(m_\ell )}_{w_\ell }. \end{aligned}$$
Then \(e^{-t|\varvec{\xi }|^2}m(\varvec{\xi })={\displaystyle \sum \nolimits _{\ell \in \mathbb {Z}}}m_{t,\ell }(2^{-\ell }\varvec{\xi })\). Consider the convolution kernel
$$\begin{aligned} F_{t,\ell }(\mathbf{x},\mathbf{y})= \tau _{-\mathbf{y}}\,{\mathcal {F}}^{-1}\left\{ m_{t,\ell }(2^{-\ell }.)\right\} (\mathbf x) =2^{\mathbf{N}\ell }(\tau _{-2^{\ell }\mathbf{y}}f_{t,\ell })(2^{\ell }\mathbf{x})\end{aligned}$$
of the multiplier operator \(\mathcal T_{m_{t,\ell } (2^{-\ell }. )}\).
Lemma 5.9
-
(a)
On the one hand, for every \(0\!\le \!\delta \!<\!\epsilon \), we have
$$\begin{aligned} \int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{z},2)} \mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})|\, \lesssim \,M\,2^{-\delta \ell }, \quad \forall \;\ell \!\in \!\mathbb {Z}, \,\forall \,\mathbf{z}\in \!\mathbb {R}^n, \;\forall \;\mathbf{y}\!\in \!\mathcal {O}(\mathbf{z},1). \end{aligned}$$
-
(b)
On the other hand,
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\,\sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})-F_{t,\ell }(\mathbf{x},\mathbf{y}^{\prime })|\, \lesssim \,M\,2^{\ell }|\mathbf{y}\!-\!\mathbf{y}^{\prime }|\, , \quad \forall \;\ell \!\in \!\mathbb {Z},\; \forall \;\mathbf{y},\mathbf{y}^{\prime }\in \!\mathbb {R}^n. \end{aligned}$$
Proof
On the one hand, as
$$\begin{aligned} \left| \partial _{\varvec{\xi }}^{\varvec{\alpha }} \left( \psi (\varvec{\xi })e^{-t|\varvec{\xi }|^2}\right) \right| \le C_{\varvec{\alpha }}, \qquad \forall \;t\!>\!0,\;\forall \,\varvec{\xi }\!\in \!\mathbb {R}^n, \end{aligned}$$
Lemma 5.1 yields the estimate
$$\begin{aligned} |g_{t,\ell }(\mathbf{x})|\le C_d\,(1\!+\!|\mathbf{x}|)^{-d}, \qquad \forall \,\mathbf{x}\!\in \!\mathbb {R}^n, \end{aligned}$$
which holds for any \(d\!\in \!\mathbb {N}\) and which is uniform in \(t\!>\!0\) and \(\ell \!\in \!\mathbb {Z}\). On the other hand, Corollary 5.4 yields the estimate
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\,(1\!+\!|\mathbf{x}|)^\delta \, |w_\ell (\mathbf{x})|\lesssim M, \end{aligned}$$
which holds uniformly in \(\ell \!\in \!\mathbb {Z}\). By resuming the proof of Lemma 5.5, we deduce that
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\,(1\!+\!|\mathbf{x}|)^\delta \sup _{t>0}|f_{t,\ell }(\mathbf{x})|\lesssim M. \end{aligned}$$
(5.10)
We reach our first conclusion by rescaling and by using Lemma 5.6:
$$\begin{aligned}&\int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{z},2)} \mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})|\, \le \int _{\mathbb {R}^n\backslash \mathcal {O}(\mathbf{y},1)} \mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})|\\&=\int _{\mathbb {R}^n\backslash \mathcal {O}(2^{\ell }\mathbf{y},2^{\ell })} \mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|(\tau _{-2^{\ell }\mathbf{y}}f_{t,\ell })(\mathbf{x})|\, \lesssim \,M\,2^{-\delta \ell }. \end{aligned}$$
Let us turn to the proof of (b). This time we factorize
$$\begin{aligned} m_{t,\ell }(\varvec{\xi }) =\underbrace{\overbrace{ \psi (\varvec{\xi })\, e^{|\varvec{\xi }|^2}e^{-t|2^{\ell }\varvec{\xi }|^2} }^{\widetilde{\psi }_{t,\ell }(\varvec{\xi })} \overbrace{ \psi (\varvec{\xi })\, m(2^{\ell }\varvec{\xi }) }^{m_\ell (\varvec{\xi })} }_{\widetilde{m}_{t,\ell }(\varvec{\xi })}\, e^{-|\varvec{\xi }|^2},\\ \end{aligned}$$
and accordingly
$$\begin{aligned} f_{t,\ell } ={\mathcal {F}}^{-1}(m_{t,\ell })=\,\underbrace{ {\mathcal {F}}^{-1}(\widetilde{m}_{t,\ell }) }_{\widetilde{f}_{t,\ell }}\, *\,\underbrace{ {\mathcal {F}}^{-1}(e^{-|\varvec{\xi }|^2}) }_{h}\,. \end{aligned}$$
On the one hand, by resuming the proof of (5.10), we get
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\, \sup _{t>0}|\widetilde{f}_{t,\ell }(\mathbf{x})|\lesssim M. \end{aligned}$$
On the other hand, \(\mathbf{h}(\mathbf{x},\mathbf{y})=(\tau _{-\mathbf{y}}h)(\mathbf{x})\) is the heat kernel at time \(t=1\), which satisfies
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu } (\mathbf{x})\, |\mathbf{h}(\mathbf{x},\mathbf{y})-\mathbf{h}(\mathbf{x},\mathbf{y}^{\prime })| \lesssim |\mathbf{y}-\mathbf{y}^{\prime }|, \qquad \forall \;\mathbf{y},\mathbf{y}^{\prime }\in \!\mathbb {R}^n, \end{aligned}$$
according to the next lemma. After rescaling, we reach our second conclusion:
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu }(\mathbf{x})\,\sup _{t>0}|F_{t,\ell }(\mathbf{x},\mathbf{y})-F_{t,\ell }(\mathbf{x},\mathbf{y}^{\prime })|\lesssim M\,2^{\ell }\,|\mathbf{y}-\mathbf{y}^{\prime }|\,. \end{aligned}$$
\(\square \)
Lemma 5.11
The following gradient estimate holds for the heat kernel:
$$\begin{aligned} \int _{\mathbb {R}^n}\!\mathrm{d}\varvec{\mu } (\mathbf{x})\, |\nabla _{\mathbf{y}}\mathbf{h}_{t}(\mathbf{x},\mathbf{y})|\, \lesssim \,t^{-\frac{1}{2}}, \qquad \forall \;t\!>\!0,\;\forall \;\mathbf{y}\!\in \!\mathbb {R}^n. \end{aligned}$$
Proof
We can reduce to the one-dimensional case and moreover to \(t=\!1\) by rescaling. It follows from our gradient estimates for the heat kernel in dimension \(1\) (see Proposition 2.3) that
$$\begin{aligned} \left| \frac{\partial }{\partial y}h_1(x,y)\right| \lesssim \frac{1}{1+\,|xy|^k}\,e^{-\frac{1}{8}(|x|-|y|)^2}. \end{aligned}$$
-
Case 1: Assume that \(|y|\le 2\). Then \(|\partial _{y}h_1(x,y)|\lesssim e^{-x^2/16}\), hence
$$\begin{aligned} \int _{-\infty }^{+\infty }\mathrm{d}x\,|x|^{2k}\, \left| \frac{\partial }{\partial y}h_1(x,y)\right| \,\lesssim \,1\,. \end{aligned}$$
-
Case 2: Assume that \(|y|\ge 2\). Then \(|x|/|y|\le 1+\frac{1}{2}\left| |x|-|y|\right| \), hence
$$\begin{aligned}&|x|^{2k}\,\left| \frac{\partial }{\partial y}h_1(x,y)\right| \lesssim \left( \frac{|x|}{|y|}\right) ^{k}\, e^{-\frac{1}{8}(|x|-|y|)^2}\\&\qquad \lesssim \left( 1+\left| |x|-|y|\right| \right) ^{k}\, e^{-\frac{1}{8}(|x|-|y|)^2} \lesssim e^{-\frac{1}{16}(|x|-|y|)^2} \end{aligned}$$
and
$$\begin{aligned} \int _{-\infty }^{+\infty }\mathrm{d}x\,|x|^{2k}\, \left| \frac{\partial }{\partial y}h_1(x,y)\right| \, \lesssim \int _{\,0}^{+\infty }\mathrm{d}x\,e^{-\frac{1}{16}(x-|y|)^2} \lesssim \int _{-\infty }^{+\infty }\mathrm{d}z\,e^{-\frac{1}{16}z^2} \lesssim \,1\,. \end{aligned}$$
Conclusion of proof of Theorem 1.10
Let us split up and estimate
$$\begin{aligned} |\mathbf{h}_{*}({\mathcal {T}}_{m}a)(\mathbf{x})|&\le \sum \nolimits _{\ell \ge 0}|\mathbf{h}_{*} ({\mathcal {T}}_{\psi (2^{-\ell }.)^2m}a)(\mathbf{x})| +\sum \nolimits _{\ell <0}|\mathbf{h}_{*} ({\mathcal {T}}_{\psi (2^{-\ell }.)^2m}a)(\mathbf{x})|\\&=\sum \nolimits _{\ell \ge 0}\sup \nolimits _{t>0}\, \left| \,\int _{\mathbf{B}(\mathbf{z},1)} \mathrm{d}\varvec{\mu }(\mathbf{y})\, F_{t,\ell }(\mathbf{x},\mathbf{y})\,a(\mathbf{y})\,\right| \\&+\sum \nolimits _{\ell <0}\sup \nolimits _{t>0}\, \left| \,\int _{\mathbf{B}(\mathbf{z},1)} \mathrm{d}\varvec{\mu }(\mathbf{y})\, \left\{ F_{t,\ell }(\mathbf{x},\mathbf{y})\!-\!F_{t,\ell }(\mathbf{x},\mathbf{z})\right\} \, a(\mathbf{y})\,\right| \\&\le \sum \nolimits _{\ell \ge 0} \int _{\mathbf{B}(\mathbf{z},1)} \mathrm{d}\varvec{\mu }(\mathbf{y})\;|a(\mathbf{y})|\, \sup \nolimits _{t>0}\,\left| F_{t,\ell }(\mathbf{x},\mathbf{y})\right| \\&+\sum \nolimits _{\ell <0} \int _{\mathbf{B}(\mathbf{z},1)} \mathrm{d}\varvec{\mu }(\mathbf{y})\;|a(\mathbf{y})|\, \sup \nolimits _{t>0}\, \left| F_{t,\ell }(\mathbf{x},\mathbf{y})-F_{t,\ell }(\mathbf{x},\mathbf{z})\right| \,. \end{aligned}$$
Then (5.8) follows from Lemma 5.9. \(\square \)
Example 5.1
The Riesz transforms \({\mathcal {R}}_j=D_j(-\mathbf L)^{-1\slash 2}\) in the Dunkl setting correspond to the multipliers \(\xi _j/|\varvec{\xi }|\), up to a constant. Hence, by Theorem 1.10, they are bounded operators on the Hardy space \(H^1\).