Proof. Each \(\alpha^\perp\) is an affine linear hyperplane through the origin and has Lebesgue measure zero. The set \(\mathcal{W}\) is a finite union of such hyperplanes, hence is also Lebesgue null. Since \(d\omega=w\,dx\) and the weight is locally integrable for \(\kappa\ge0\), every Lebesgue null set is \(\omega\)-null. Thus \(\omega(\mathcal{W})=0\). The possible overlaps of distinct closed chambers are contained in \(\mathcal{W}\), so the chamber decomposition is disjoint after removing an \(\omega\)-null set. ◻

Proof. We prove first that the representative of an orbit which lies in the closed chamber is closest to every point of the same chamber. It is enough to show that, for every \(\sigma\in G\), \[\|x-y\|\le \|\sigma x-y\|, \qquad x,y\in\mathcal{C}.\] We use the usual successive-reflection argument, and we record the elementary distance computation because it is the point at which the chamber geometry enters. Suppose that \(z\) and \(y\in\mathcal{C}\) lie on opposite sides of the wall \(\alpha^\perp\), so that \(\langle z,\alpha\rangle\le0, \; \langle y,\alpha\rangle\ge0.\) Since \[\sigma_\alpha z =z-2\frac{\langle z,\alpha\rangle}{\|\alpha\|^2}\alpha,\] a direct computation gives \[\begin{align} \|\sigma_\alpha z-y\|^2-\|z-y\|^2 &=-2\left\langle 2\frac{\langle z,\alpha\rangle}{\|\alpha\|^2}\alpha, z-y \right\rangle + 4\frac{\langle z,\alpha\rangle^2}{\|\alpha\|^2} =4\frac{\langle z,\alpha\rangle\langle y,\alpha\rangle}{\|\alpha\|^2} \le0. \end{align}\] Thus reflecting \(z\) across a wall toward the chamber does not increase its distance to \(y\). Starting with \(z=\sigma x\), choose a finite sequence of simple reflections which moves \(z\) back into the closed chamber. This is possible because the Weyl chambers are fundamental domains for the action of \(G\). At each step the distance to \(y\) does not increase. The final point is the unique chamber representative of the orbit of \(x\), namely \(x\), up to the boundary where the assertion is unchanged by continuity. Therefore \(\|x-y\|\le \|\sigma x-y\|, \;\forall \sigma\in G.\) Taking the minimum over \(\sigma\) gives \(d(x,y)=\|x-y\|\).

For the reflected identity, use the invariance of the orbit distance in both variables: \[\begin{align} d(\sigma_\rho x,\sigma_\tau y) &=d(x,\sigma_\rho^{-1}\sigma_\tau y). \end{align}\] Since both \(x\) and \(y\) lie in \(\mathcal{C}\), the first part of the proof applied to the orbit of \(y\) gives \[d(x,\sigma_\rho^{-1}\sigma_\tau y)=d(x,y)=\|x-y\|.\] This proves the lemma. ◻

Proof. The inequality \(\omega(B_{\mathcal{C}}(x,r))\le \omega(B(x,r))\) follows from the inclusion \(B_{\mathcal{C}}(x,r)\subset B(x,r)\). We prove the reverse inequality. Since chamber boundaries are \(\omega\)-null, we may decompose \(B(x,r)\) into its intersections with the chambers \(\sigma\mathcal{C}\), up to a null set. For each \(\sigma\in G\), the invariance of \(d\omega\) gives \[\omega(B(x,r)\cap\sigma\mathcal{C}) = \omega(B(\sigma^{-1}x,r)\cap\mathcal{C}).\] We claim that \[B(\sigma^{-1}x,r)\cap\mathcal{C} \subset B(x,r)\cap\mathcal{C}.\] Indeed, if \(y\in B(\sigma^{-1}x,r)\cap\mathcal{C}\), then \(\|\sigma^{-1}x-y\|<r.\) By Lemma 7, the chamber representative \(x\) is no farther from \(y\) than any other representative of its orbit; therefore \(\|x-y\|\le \|\sigma^{-1}x-y\|<r.\) This proves the claimed inclusion. Summing over the finitely many chambers, we obtain \[\begin{align} \omega(B(x,r)) &=\sum_{\sigma\in G}\omega(B(x,r)\cap\sigma\mathcal{C}) \le |G|\,\omega(B(x,r)\cap\mathcal{C}) =|G|\,\omega(B_{\mathcal{C}}(x,r)). \end{align}\] Thus \(\omega(B(x,r))\approx\omega(B_{\mathcal{C}}(x,r))\).

Now let \(x,y\in\mathcal{C}\). By reflection invariance of the measure, \[\omega(B(\sigma_\rho x,r))=\omega(B(x,r)), \qquad \omega(B(\sigma_\tau y,r))=\omega(B(y,r)).\] Using the comparison just proved for \(x\) and \(y\), we get \[\begin{align} V(\sigma_\rho x,\sigma_\tau y,r) &=\max\{\omega(B(x,r)),\omega(B(y,r))\} \approx \max\{\omega(B_{\mathcal{C}}(x,r)),\omega(B_{\mathcal{C}}(y,r))\} =V_{\mathcal{C}}(x,y,r), \end{align}\] with constants independent of \(\rho,\tau\). Finally, \[\omega(B_{\mathcal{C}}(x,2r)) \approx \omega(B(x,2r)) \lesssim \omega(B(x,r)) \approx \omega(B_{\mathcal{C}}(x,r)),\] so \((\mathcal{C},\|\cdot\|,d\omega|_\mathcal{C})\) is a space of homogeneous type. ◻

Proof. For any \(\sigma\in G\), the two points \(x\) and \(\sigma x\) lie on the same orbit, hence \(d(x,\sigma x)=0.\) The orbit-Lipschitz condition gives \(|b(x)-b(\sigma x)|\le \|b\|_{\operatorname{Lip}_d}d(x,\sigma x)=0,\) so \(b(\sigma x)=b(x)\). The Euclidean Lipschitz estimate is obtained by using the identity element in the orbit-distance minimum: \[d(x,y)=\min_{\tau\in G}\|x-\tau y\|\le\|x-y\|.\] Therefore \(|b(x)-b(y)|\le \|b\|_{\operatorname{Lip}_d}d(x,y) \le \|b\|_{\operatorname{Lip}_d}\|x-y\|.\) The proof is complete. ◻

Proof. Take \(b_\varepsilon=b*\varphi_\varepsilon\) with a radial mollifier. Radiality preserves \(G\)-invariance, and the Euclidean Lipschitz bound gives \(\|\nabla b_\varepsilon\|_\infty\le\|b\|_{\operatorname{Lip}_d}\). Since \(b_\varepsilon\) is \(G\)-invariant, \[|b_\varepsilon(x)-b_\varepsilon(y)|=|b_\varepsilon(x)-b_\varepsilon(\sigma y)| \le\|b\|_{\operatorname{Lip}_d}\|x-\sigma y\|.\] Taking the infimum over \(\sigma\in G\) gives the \(\operatorname{Lip}_d\) bound. ◻

Proof. Using the definition of \(T_j\), \[\begin{align} T_j(bf)(x) &=\partial_j(bf)(x) + \sum_{\alpha\in R}\frac{\kappa(\alpha)}{2}\alpha_j \frac{b(x)f(x)-b(\sigma_\alpha x)f(\sigma_\alpha x)}{\langle\alpha,x\rangle}. \end{align}\] Since \(b\) is \(G\)-invariant, \(b(\sigma_\alpha x)=b(x)\). Thus the difference part factors as \[b(x) \sum_{\alpha\in R}\frac{\kappa(\alpha)}{2}\alpha_j \frac{f(x)-f(\sigma_\alpha x)}{\langle\alpha,x\rangle}.\] The ordinary derivative contributes \(\partial_j(bf)=(\partial_jb)f+b\partial_jf.\) Combining the derivative term with the factored difference part gives \[T_j(bf)=(\partial_jb)f+bT_jf.\] The summands with \(\kappa(\alpha)=0\) are present only notationally and vanish. ◻

Proof. By Lemma 13, \[|T_{\ell,x}h_t(x,y)| \le \frac{\|x-y\|}{2t}|h_t(x,y)|.\] Applying Lemma 12 gives \[\begin{align} |T_{\ell,x}h_t(x,y)| &\lesssim \frac{\|x-y\|}{t} \frac{1}{V(x,y,\sqrt t)} \left(1+\frac{\|x-y\|}{\sqrt t}\right)^{-2} e^{-cd(x,y)^2/t}\\ &= \frac{1}{\sqrt t} \left(\frac{\|x-y\|}{\sqrt t}\right) \left(1+\frac{\|x-y\|}{\sqrt t}\right)^{-2} \frac{1}{V(x,y,\sqrt t)}e^{-cd(x,y)^2/t}. \end{align}\] The function \(u(1+u)^{-2}\) is bounded on \([0,\infty)\). This proves 16 . Setting \(t=s^2\) and multiplying by \(s\) gives 17 . ◻

Proof. Set \[a_j(x,y,t)=\frac{y_j-x_j}{2t}.\] Lemma 13 gives \(T_{j,x}h_t(x,y)=a_j(x,y,t)h_t(x,y),\) hence \[T_{i,x}T_{j,x}h_t(x,y)=T_i(a_jh_t)(x).\] Expanding the Dunkl operator, we obtain \[\begin{align} T_i(a_jh_t)(x) &=\partial_i(a_jh_t)(x) + \sum_{\alpha\in R}\frac{\kappa(\alpha)}{2}\alpha_i \frac{a_j(x)h_t(x,y)-a_j(\sigma_\alpha x)h_t(\sigma_\alpha x,y)}{\langle\alpha,x\rangle}. \end{align}\] We insert and subtract the term \(a_j(x)h_t(\sigma_\alpha x,y)\) in the numerator. The part involving \(h_t(x,y)-h_t(\sigma_\alpha x,y)\), together with the ordinary derivative term \(a_j(x)\partial_i h_t(x,y)\), reconstructs \(a_j(x)T_{i,x}h_t(x,y)\). Thus \[\begin{align} T_i(a_jh_t)(x) &=(\partial_i a_j)(x)h_t(x,y)+a_j(x)T_{i,x}h_t(x,y)+ \sum_{\alpha\in R}\frac{\kappa(\alpha)}{2}\alpha_i \frac{a_j(x)-a_j(\sigma_\alpha x)}{\langle\alpha,x\rangle} h_t(\sigma_\alpha x,y). \end{align}\] The first two factors are \[\partial_i a_j=-\frac{\delta_{ij}}{2t}, \qquad T_{i,x}h_t(x,y)=\frac{y_i-x_i}{2t}h_t(x,y).\] For the reflected coefficient we use \[\sigma_\alpha x=x-2\frac{\langle x,\alpha\rangle}{\|\alpha\|^2}\alpha.\] Then \[\begin{align} a_j(x,y,t)-a_j(\sigma_\alpha x,y,t) &=\frac{y_j-x_j}{2t}-\frac{y_j-(\sigma_\alpha x)_j}{2t} =\frac{(\sigma_\alpha x)_j-x_j}{2t} =-\frac{\langle x,\alpha\rangle}{t}\frac{\alpha_j}{\|\alpha\|^2}. \end{align}\] Therefore \[\frac{a_j(x,y,t)-a_j(\sigma_\alpha x,y,t)}{\langle\alpha,x\rangle} =-\frac{\alpha_j}{t\|\alpha\|^2}.\] Substituting these identities gives the formula. The summands corresponding to roots with \(\kappa(\alpha)=0\) vanish identically. ◻

Proof. Put \(s=\sqrt t\). We use the formula of Lemma 15. Thus \(A_t^{ij}(x,y)\) is the sum of \[\frac{(y_i-x_i)(y_j-x_j)}{4t^2}h_t(x,y), \qquad -\frac{\delta_{ij}}{2t}h_t(x,y),\] and the finitely many reflected terms \[-\frac{1}{t} \frac{\kappa(\alpha)}{2} \frac{\alpha_i\alpha_j}{\|\alpha\|^2} h_t(\sigma_\alpha x,y), \qquad \alpha\in R.\] We estimate these terms separately. For the first term, this factored heat-kernel bound gives \[\begin{align} \frac{|y_i-x_i|\,|y_j-x_j|}{t^2}|h_t(x,y)| &\lesssim \frac{\|x-y\|^2}{t^2} \frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2} \exp\left(-c\frac{d(x,y)^2}{s^2}\right) \\ &= \frac{1}{t} \left(\frac{\|x-y\|}{s}\right)^2 \left(1+\frac{\|x-y\|}{s}\right)^{-2} \frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right) \\ &\lesssim \frac{1}{t}\frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right). \end{align}\] This is the only place where the Euclidean decay factor is needed. For the scalar term one simply writes \[\frac{1}{t} |h_t(x,y)| \le \frac{C}{t}\frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2} \exp\left(-c\frac{d(x,y)^2}{s^2}\right) \le \frac{C}{t}\frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right).\] For a reflected term with \(\kappa(\alpha)>0\), the reflection \(\sigma_\alpha\) belongs to \(G\), and therefore \[d(\sigma_\alpha x,y)=d(x,y), \qquad V(\sigma_\alpha x,y,s)=V(x,y,s).\] Applying the scalar heat estimate at the point \((\sigma_\alpha x,y)\), and then again discarding its Euclidean decay factor, gives \[\frac{1}{t} |h_t(\sigma_\alpha x,y)| \le \frac{C}{t}\frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right).\] The coefficients of the reflected terms are fixed constants depending only on \(R\) and \(\kappa\), and the root system is finite. Summing the preceding bounds proves the size estimate.

We next prove the regularity in the first variable. Let \(\delta=\|x-x'\|\le s\). The scalar term satisfies \[\begin{align} \frac{1}{t} |h_t(x,y)-h_t(x',y)| &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2} \exp\left(-c\frac{d(x,y)^2}{s^2}\right) \\ &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right). \end{align}\] For each reflected term we use \(\|\sigma_\alpha x-\sigma_\alpha x'\|=\delta,\) together with the invariance \(d(\sigma_\alpha x,y)=d(x,y)\) and \(V(\sigma_\alpha x,y,s)=V(x,y,s).\) The heat-kernel regularity estimate applied to \(h_t(\sigma_\alpha x,y)-h_t(\sigma_\alpha x',y)\) gives \[\begin{align} \frac{1}{t} |h_t(\sigma_\alpha x,y)-h_t(\sigma_\alpha x',y)| &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \left(1+\frac{\|\sigma_\alpha x-y\|}{s}\right)^{-2} \exp\left(-c\frac{d(x,y)^2}{s^2}\right) \\ &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \exp\left(-c\frac{d(x,y)^2}{s^2}\right). \end{align}\] Thus no comparison between \(\|\sigma_\alpha x-y\|\) and \(\|x-y\|\) is used; the Euclidean decay factor is only discarded. It remains only to check the principal term. Set \[P_t(x,y)=\frac{(y_i-x_i)(y_j-x_j)}{4t^2}h_t(x,y).\] Then \[\begin{align} P_t(x,y)-P_t(x',y) &=\frac{(y_i-x_i)(y_j-x_j)-(y_i-x'_i)(y_j-x'_j)}{4t^2}h_t(x,y) \\ &\quad+ \frac{(y_i-x'_i)(y_j-x'_j)}{4t^2} \big(h_t(x,y)-h_t(x',y)\big) \\ &=:P_1+P_2. \end{align}\] Expanding the coefficient gives \[\begin{align} &(y_i-x_i)(y_j-x_j)-(y_i-x'_i)(y_j-x'_j) =(x'_i-x_i)(y_j-x_j)+(x'_j-x_j)(y_i-x'_i). \end{align}\] Since \(\delta\le s\), we have \(|(y_i-x_i)(y_j-x_j)-(y_i-x'_i)(y_j-x'_j)| \le C\delta(\|x-y\|+s).\) Using the size estimate with this extra factor, \[\begin{align} |P_1| &\lesssim \frac{\delta(\|x-y\|+s)}{t^2} \frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2} e^{-cd(x,y)^2/s^2} \lesssim \frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] For the second term, again \(\delta\le s\) gives \(|y_i-x'_i|\,|y_j-x'_j|\le C(\|x-y\|+s)^2.\) The first-variable heat-kernel regularity gives \[|h_t(x,y)-h_t(x',y)| \lesssim \frac{\delta}{s}\frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2}.\] Consequently, \[\begin{align} |P_2| &\lesssim \frac{(\|x-y\|+s)^2}{t^2} \frac{\delta}{s}\frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2} \lesssim \frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] Combining the principal, scalar and reflected terms proves ?? .

The proof in the second variable is the same in structure, but we give the principal estimate to make clear that no new cancellation is used. Let \(\delta=\|y-y'\|\le s\). Then \[\begin{align} P_t(x,y)-P_t(x,y') &=\frac{(y_i-x_i)(y_j-x_j)-(y'_i-x_i)(y'_j-x_j)}{4t^2}h_t(x,y)\\ &\quad+ \frac{(y'_i-x_i)(y'_j-x_j)}{4t^2} \big(h_t(x,y)-h_t(x,y')\big). \end{align}\] For the coefficient in the first term, expand \[\begin{align} &(y_i-x_i)(y_j-x_j)-(y'_i-x_i)(y'_j-x_j)=(y_i-y'_i)(y_j-x_j)+(y'_i-x_i)(y_j-y'_j). \end{align}\] Since \(\delta\le s\), this gives \[|(y_i-x_i)(y_j-x_j)-(y'_i-x_i)(y'_j-x_j)| \le C\delta(\|x-y\|+s).\] Consequently the first part is bounded by \[\begin{align} & C\frac{\delta(\|x-y\|+s)}{t^2} \frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2} \le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}, \end{align}\] because \((u+1)(1+u)^{-2}\le C\). For the second part, the coefficient is bounded by \[|y'_i-x_i|\,|y'_j-x_j|\le C(\|x-y\|+s)^2.\] Using the second-variable heat-kernel regularity of \(h_t\), we obtain \[\begin{align} \frac{(\|x-y\|+s)^2}{t^2}|h_t(x,y)-h_t(x,y')| &\le C\frac{(\|x-y\|+s)^2}{t^2}\frac{\delta}{s} \frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2} \\ &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] The scalar term satisfies the explicit estimate \[\begin{align} \frac{1}{t}|h_t(x,y)-h_t(x,y')| &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2} \le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] For a reflected term one estimates \(h_t(\sigma_\alpha x,y)-h_t(\sigma_\alpha x,y')\) with \(\sigma_\alpha x\) fixed. Since \[d(\sigma_\alpha x,y)=d(x,y),\qquad V(\sigma_\alpha x,y,s)=V(x,y,s),\] the heat-kernel regularity estimate gives \[\begin{align} \frac{1}{t}|h_t(\sigma_\alpha x,y)-h_t(\sigma_\alpha x,y')| &\le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)} \left(1+\frac{\|\sigma_\alpha x-y\|}{s}\right)^{-2}e^{-cd(x,y)^2/s^2} \le C\frac{\delta}{s}\frac{1}{t}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] After multiplication by the fixed reflected coefficients and summation over \(R\), this proves ?? . ◻

Proof. The size estimate follows from the orbit Lipschitz condition and ?? : \[\begin{align} |\theta_s(x,y)| &\le s\|b\|_{\operatorname{Lip}_d}d(x,y) \frac{1}{s^2}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2} =\|b\|_{\operatorname{Lip}_d}\frac{d(x,y)}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\\ &\lesssim \|b\|_{\operatorname{Lip}_d}\frac{1}{V(x,y,s)}e^{-c'd(x,y)^2/s^2}. \end{align}\] For the first-variable regularity, assume \(\|x-x'\|\le s\) and write \[\begin{align} \theta_s(x,y)-\theta_s(x',y) &=s[b(x)-b(x')]A_{s^2}^{ij}(x,y)+s[b(x')-b(y)] \big(A_{s^2}^{ij}(x,y)-A_{s^2}^{ij}(x',y)\big) =:I_1+I_2. \end{align}\] For the first term, the Euclidean Lipschitz bound following from \(b\in\operatorname{Lip}_d\) gives \[|b(x)-b(x')|\le \|b\|_{\operatorname{Lip}_d}\|x-x'\|.\] Using ?? with \(t=s^2\), \[\begin{align} |I_1| &\le s\|b\|_{\operatorname{Lip}_d}\|x-x'\| \frac{1}{s^2}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2} = \|b\|_{\operatorname{Lip}_d}\frac{\|x-x'\|}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] For the second term, the orbit triangle inequality gives \[|b(x')-b(y)|\le \|b\|_{\operatorname{Lip}_d}d(x',y) \le \|b\|_{\operatorname{Lip}_d}\big(d(x,y)+\|x-x'\|\big) \le \|b\|_{\operatorname{Lip}_d}\big(d(x,y)+s\big).\] By ?? , again with \(t=s^2\), \[\begin{align} |I_2| &\lesssim s\|b\|_{\operatorname{Lip}_d}(d(x,y)+s) \frac{\|x-x'\|}{s}\frac{1}{s^2} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\\ &= \|b\|_{\operatorname{Lip}_d}\frac{\|x-x'\|}{s} \left(1+\frac{d(x,y)}{s}\right) \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] The elementary inequality \((1+u)e^{-cu^2}\le C e^{-c'u^2}\), \(u\ge0\), absorbs the factor \(1+d(x,y)/s\) into the Gaussian. Hence \(|I_1|+|I_2|\) satisfies 22 .

The estimate in the second variable is obtained by the same decomposition, but we spell it out to avoid hiding the Lipschitz factor. If \(\|y-y'\|\le s\), then \[\begin{align} \theta_s(x,y)-\theta_s(x,y') &=s[b(y')-b(y)]A_{s^2}^{ij}(x,y)+s[b(x)-b(y')] \big(A_{s^2}^{ij}(x,y)-A_{s^2}^{ij}(x,y')\big). \end{align}\] For the first term, the second heat-size estimate gives \[\begin{align} s|b(y')-b(y)|\,|A_{s^2}^{ij}(x,y)| &\le s\|b\|_{\operatorname{Lip}_d}\|y-y'\| \frac{1}{s^2}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2} = \|b\|_{\operatorname{Lip}_d}\frac{\|y-y'\|}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] For the second term, \[|b(x)-b(y')|\le \|b\|_{\operatorname{Lip}_d}d(x,y') \le \|b\|_{\operatorname{Lip}_d}(d(x,y)+\|y-y'\|) \le \|b\|_{\operatorname{Lip}_d}(d(x,y)+s).\] Using ?? , we obtain \[\begin{align} s|b(x)-b(y')|\, |A_{s^2}^{ij}(x,y)-A_{s^2}^{ij}(x,y')| &\lesssim s\|b\|_{\operatorname{Lip}_d}(d(x,y)+s) \frac{\|y-y'\|}{s}\frac{1}{s^2} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\\ &= \|b\|_{\operatorname{Lip}_d}\frac{\|y-y'\|}{s} \left(1+\frac{d(x,y)}{s}\right) \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}. \end{align}\] Again \((1+u)e^{-cu^2}\le C e^{-c'u^2}\) absorbs the extra factor, and the second-variable estimate follows. ◻

Proof. By the orbit Lipschitz condition, we have \(|b(x)-b(y)|\le\|b\|_{\operatorname{Lip}_d}r.\) Combining this with ?? gives \[\begin{align} |K_b^{ij}(x,y)| &\lesssim \|b\|_{\operatorname{Lip}_d}r \int_0^\infty t^{-3/2}V(x,y,\sqrt t)^{-1}e^{-cr^2/t}\,dt. \end{align}\] The integral is finite and bounded by \(C/(rV(x,y,r))\) by Lemma 17. This proves both absolute convergence and 29 . ◻

Proof. For readability we write \(A_t=A_t^{ij}\). Set \(r'=d(x',y).\) By Lemma 18, \(r'\approx r, \; V(x',y,r')\approx V(x,y,r).\) We decompose the difference according to whether the change falls on the Lipschitz factor or on the heat-kernel factor: \[\begin{align} K_b^{ij}(x,y)-K_b^{ij}(x',y) &=c[b(x)-b(x')] \int_0^\infty A_t(x,y)\frac{dt}{\sqrt t}+c[b(x')-b(y)] \int_0^\infty (A_t(x,y)-A_t(x',y))\frac{dt}{\sqrt t}\\ &=:I+II. \end{align}\] For the first term, \(|b(x)-b(x')|\le \|b\|_{\operatorname{Lip}_d}\delta\), and ?? gives \[\int_0^\infty |A_t(x,y)|\frac{dt}{\sqrt t} \lesssim \int_0^\infty t^{-3/2}V(x,y,\sqrt t)^{-1}e^{-cr^2/t}\,dt.\] By 24 , the last integral is bounded by \(C/(rV(x,y,r))\). Hence \[|I|\lesssim \|b\|_{\operatorname{Lip}_d} \frac{\delta}{r}V(x,y,r)^{-1}.\]

For the second term, the orbit Lipschitz condition gives \[|b(x')-b(y)|\le \|b\|_{\operatorname{Lip}_d}d(x',y) \lesssim \|b\|_{\operatorname{Lip}_d}r.\] It remains to estimate the integral of the difference of \(A_t\). We split at the scale \(t=\delta^2\). On the small interval \(0<t<\delta^2\), no regularity estimate is available, so we use two size estimates: \[\begin{align} &\int_0^{\delta^2}|A_t(x,y)-A_t(x',y)|\frac{dt}{\sqrt t} \le \int_0^{\delta^2}|A_t(x,y)|\frac{dt}{\sqrt t} + \int_0^{\delta^2}|A_t(x',y)|\frac{dt}{\sqrt t}. \end{align}\] The first integral is bounded by \[C\int_0^{\delta^2}t^{-3/2}V(x,y,\sqrt t)^{-1}e^{-cr^2/t}\,dt,\] whereas the second is bounded by \[C\int_0^{\delta^2}t^{-3/2}V(x',y,\sqrt t)^{-1}e^{-c(r')^2/t}\,dt.\] The first short-time integral is controlled by 25 applied to the unperturbed pair \((x,y)\). The second is controlled by Corollary 2, applied to the perturbed pair \((x',y)\). Hence \[\int_0^{\delta^2}|A_t(x,y)-A_t(x',y)|\frac{dt}{\sqrt t} \lesssim \frac{\delta}{r^2V(x,y,r)}.\] On the complementary interval \(t\ge\delta^2\), one has \(\delta\le\sqrt t\). The heat-scale regularity ?? gives \[\begin{align} &\int_{\delta^2}^{\infty}|A_t(x,y)-A_t(x',y)|\frac{dt}{\sqrt t} \lesssim \delta\int_{\delta^2}^{\infty} t^{-2}V(x,y,\sqrt t)^{-1}e^{-cr^2/t}\,dt \lesssim \frac{\delta}{r^2V(x,y,r)}, \end{align}\] again by 25 . Combining the two ranges, \[\int_0^\infty |A_t(x,y)-A_t(x',y)|\frac{dt}{\sqrt t} \lesssim \frac{\delta}{r^2V(x,y,r)}.\] Therefore \[|II|\lesssim \|b\|_{\operatorname{Lip}_d}r\frac{\delta}{r^2V(x,y,r)} = C\|b\|_{\operatorname{Lip}_d}\frac{\delta}{r}V(x,y,r)^{-1}.\] The estimates for \(I\) and \(II\) prove the proposition. ◻

Proof. The argument is the same as the proof of Proposition 8; we record the substitutions to avoid any ambiguity about the normalization.

Put \(r''=d(x,y').\) By Lemma 18, \(r''\approx r, \; V(x,y',r'')\approx V(x,y,r).\) We decompose \[\begin{align} K_b^{ij}(x,y)-K_b^{ij}(x,y') &=c[b(y')-b(y)] \int_0^\infty A_t^{ij}(x,y)\frac{dt}{\sqrt t} +c[b(x)-b(y')] \int_0^\infty \bigl(A_t^{ij}(x,y)-A_t^{ij}(x,y')\bigr)\frac{dt}{\sqrt t} \\ &=:I+II . \end{align}\] The first term is estimated exactly as before: \(|b(y')-b(y)| \le \|b\|_{\operatorname{Lip}_d}\|y-y'\| =\|b\|_{\operatorname{Lip}_d}\delta ,\) and \[\int_0^\infty |A_t^{ij}(x,y)|\frac{dt}{\sqrt t} \lesssim \frac{1}{rV(x,y,r)}.\] Thus \[|I| \lesssim \|b\|_{\operatorname{Lip}_d} \frac{\delta}{r}V(x,y,r)^{-1}.\]

For the second term, note that \(|b(x)-b(y')| \le \|b\|_{\operatorname{Lip}_d}d(x,y') \lesssim \|b\|_{\operatorname{Lip}_d}r .\) It remains to estimate the integrated heat-kernel difference. Split the integral at \(t=\delta^2\). On \(0<t<\delta^2\), use the two size estimates for \(A_t^{ij}(x,y)\) and \(A_t^{ij}(x,y')\). The first is controlled by 25 for the pair \((x,y)\), and the second is controlled by Corollary 2 for the pair \((x,y')\). Hence \[\int_0^{\delta^2} |A_t^{ij}(x,y)-A_t^{ij}(x,y')| \frac{dt}{\sqrt t} \lesssim \frac{\delta}{r^2V(x,y,r)}.\] On \(t\ge\delta^2\), we have \(\delta\le\sqrt t\), so ?? gives \[\int_{\delta^2}^{\infty} |A_t^{ij}(x,y)-A_t^{ij}(x,y')| \frac{dt}{\sqrt t} \lesssim \delta\int_{\delta^2}^{\infty} t^{-2}V(x,y,\sqrt t)^{-1}e^{-cr^2/t}\,dt \lesssim \frac{\delta}{r^2V(x,y,r)}.\] Therefore \[\int_0^\infty |A_t^{ij}(x,y)-A_t^{ij}(x,y')| \frac{dt}{\sqrt t} \lesssim \frac{\delta}{r^2V(x,y,r)}.\] Consequently \[|II| \lesssim \|b\|_{\operatorname{Lip}_d}r \frac{\delta}{r^2V(x,y,r)} = C\|b\|_{\operatorname{Lip}_d} \frac{\delta}{r}V(x,y,r)^{-1}.\] Combining the estimates for \(I\) and \(II\) proves the proposition. ◻

Proof. The statement is obtained by collecting the estimates established in the preceding three results, but we spell out the correspondence between their parameters and the present notation. If \(d(x,y)>0\), set \(r=d(x,y)\). Lemma 19 gives \(|K_b^{ij}(x,y)|\le C\|b\|_{\operatorname{Lip}_d}V(x,y,r)^{-1},\) which is 30 .

For the first-variable estimate, assume \(\|x-x'\|\le d(x,y)/4\). Then Proposition 8 applies with \(r=d(x,y),\;\delta=\|x-x'\|,\) and yields \[|K_b^{ij}(x,y)-K_b^{ij}(x',y)| \le C\|b\|_{\operatorname{Lip}_d}\frac{\|x-x'\|}{d(x,y)}V(x,y,d(x,y))^{-1}.\] This is exactly 31 .

For the second-variable estimate, assume \(\|y-y'\|\le d(x,y)/4\). Applying Proposition 9 with \(r=d(x,y),\;\delta=\|y-y'\|,\) gives \[|K_b^{ij}(x,y)-K_b^{ij}(x,y')| \le C\|b\|_{\operatorname{Lip}_d}\frac{\|y-y'\|}{d(x,y)}V(x,y,d(x,y))^{-1},\] which is 32 . The volume comparisons needed in the two propositions are part of their proofs, through Lemma 18. ◻

Proof. We first prove the measure estimate. The weight \[w(x)=\prod_{\alpha\in R}|\langle x,\alpha\rangle|^{\kappa(\alpha)}\] is an \(A_\infty\)-weight with respect to Lebesgue measure. Indeed, for a fixed root \(\alpha\), an orthogonal change of variables writes \[|\langle x,\alpha\rangle|^{\kappa(\alpha)} = c_\alpha |u|^{\kappa(\alpha)}\] in the normal variable \(u\) to the hyperplane \(\alpha^\perp\), up to a harmless constant. Since \(\kappa(\alpha)\ge0\), the one-dimensional power weight \(|u|^{\kappa(\alpha)}\) belongs to \(A_p(\mathbb{R})\) for every \(p>1+\kappa(\alpha)\), and hence the corresponding hyperplane weight belongs to \(A_\infty(\mathbb{R}^N)\). As the root system is finite, the product over \(\alpha\in R\) still belongs to \(A_\infty\). Therefore there are constants \(C,\varepsilon_0>0\) such that for every Euclidean ball \(B\) and every measurable set \(E\subset B\), \[\frac{\omega(E)}{\omega(B)} \le C\left(\frac{|E|}{|B|}\right)^{\varepsilon_0}.\]

Since \(\mathcal{W}\) is a finite union of hyperplanes, the Euclidean set \(B\cap\{\delta_\mathcal{W}\le\lambda r\}\), where \(B=B(x_0,r)\), is contained in a finite union of slabs of thickness comparable to \(\lambda r\). Hence \[|B\cap\{\delta_\mathcal{W}\le\lambda r\}|\le C\lambda |B|.\] Combining this with the \(A_\infty\) estimate gives \[\omega(B\cap\{\delta_\mathcal{W}\le\lambda r\}) \le C\lambda^{\varepsilon_0}\omega(B),\] which is 33 .

We next prove the Carleson estimate. Decompose \(B\), up to the \(\omega\)-null set \(B\cap\mathcal{W}\), into \[B_\infty=B\cap\{\delta_\mathcal{W}>r\} \quad{\rm and }\quad B_k=B\cap\{2^{-k-1}r<\delta_\mathcal{W}\le2^{-k}r\}, \;k=0,1,2,\ldots.\] If \(x\in B_\infty\), then \(m_s(x)\le s/r\) for \(0<s\le r\), and hence \[\int_0^r m_s(x)^2\frac{ds}{s}\le \int_0^r\left(\frac{s}{r}\right)^2\frac{ds}{s}\le C.\] If \(x\in B_k\), write \(\delta=\delta_\mathcal{W}(x)\). Then \(\delta\approx2^{-k}r\) and \[\begin{align} \int_0^r m_s(x)^2\frac{ds}{s} &\le \int_0^\delta\left(\frac{s}{\delta}\right)^2\frac{ds}{s} + \int_\delta^r\frac{ds}{s} \lesssim k+1. \end{align}\] By the measure estimate, we have \(\omega(B_k)\le \omega(B\cap\{\delta_\mathcal{W}\le2^{-k}r\}) \lesssim 2^{-k\varepsilon_0}\omega(B).\) Thus \[\begin{align} \int_0^r\int_Bm_s(x)^2\,d\omega(x)\frac{ds}{s} &\lesssim \omega(B)+\sum_{k\ge0}(k+1)2^{-k\varepsilon_0}\omega(B) \lesssim\omega(B), \end{align}\] as required. ◻

Proof. Decompose the space into the full-orbit annuli \(E_0=\{y:d(x,y)<2s\},\) and, for \(\ell\ge1\), \(E_\ell=\{y:2^\ell s\le d(x,y)<2^{\ell+1}s\}.\) For \(y\in E_\ell\), with the convention \(2^0s=s\) in the constants, we have \(e^{-cd(x,y)^2/s^2}\le C e^{-c2^{2\ell}}\) and \(E_\ell\subset \mathcal{O}(B(x,2^{\ell+1}s)), \; V(x,y,s)\ge \omega(B(x,s)).\) By Lemma 3 and the upper growth estimate, we obtain that \(\omega(\mathcal{O}(B(x,2^{\ell+1}s))) \lesssim \omega(B(x,2^{\ell+1}s))\lesssim 2^{\ell\mathbf{N}}\omega(B(x,s)).\) It follows that, for \(y\in E_\ell\), \[\frac{1}{V(x,y,s)} \le \frac{1}{\omega(B(x,s))} \lesssim \frac{2^{\ell\mathbf{N}}}{\omega(\mathcal{O}(B(x,2^{\ell+1}s)))}.\] Therefore \[\begin{align} \int_{E_\ell}V(x,y,s)^{-1}e^{-cd(x,y)^2/s^2}F(y)\,d\omega(y) &\lesssim e^{-c2^{2\ell}}2^{\ell\mathbf{N}} \frac{1}{\omega(\mathcal{O}(B(x,2^{\ell+1}s)))} \int_{\mathcal{O}(B(x,2^{\ell+1}s))}F(y)\,d\omega(y) \\ &= e^{-c2^{2\ell}}2^{\ell\mathbf{N}}A_{2^{\ell+1}s}^{\mathcal{O}}F(x). \end{align}\] Since for every fixed \(\mathbf{N}\) one has \(2^{\ell\mathbf{N}}e^{-c2^{2\ell}}\le C e^{-c'2^{2\ell}}, \;\ell=0,1,2,\ldots,\) we may decrease the Gaussian constant and absorb the polynomial factor. The sum over all annuli gives the asserted domination. ◻

Proof. By Fubini, \[\begin{align} \int_{B_0}A_\rho^\mathcal{O}F(x)\,d\omega(x) &=\int_{\mathbb{R}^N}F(y) \int_{B_0}\frac{{\boldsymbol{1}}_{\mathcal{O}(B(x,\rho))}(y)}{\omega(\mathcal{O}(B(x,\rho)))} \,d\omega(x)d\omega(y). \end{align}\] If \({\boldsymbol{1}}_{\mathcal{O}(B(x,\rho))}(y)\ne0\) and \(x\in B_0\), then \(d(x,y)<\rho\). Since \(\|x-x_0\|<r\), the triangle inequality for \(d\) gives \[d(x_0,y)\le \|x_0-x\|+d(x,y)<r+\rho.\] Thus the inner integral can be nonzero only when \(y\in\mathcal{O}(B(x_0,r+\rho))\). Moreover, if \(d(x,y)<\rho\), choose \(\sigma\in G\) such that \(\|x-\sigma y\|<\rho\). By \(G\)-invariance, \(\omega(B(y,\rho))=\omega(B(\sigma y,\rho)).\) The two inclusions \(B(x,\rho)\subset B(\sigma y,2\rho), \; B(\sigma y,\rho)\subset B(x,2\rho)\) and doubling imply \(\omega(B(x,\rho))\approx \omega(B(y,\rho)).\) By Lemma 3, this also gives \(\omega(\mathcal{O}(B(x,\rho)))\approx \omega(\mathcal{O}(B(y,\rho))).\) Hence \[\begin{align} \int_{B_0}\frac{{\boldsymbol{1}}_{\mathcal{O}(B(x,\rho))}(y)}{\omega(\mathcal{O}(B(x,\rho)))} \,d\omega(x) &\lesssim \frac{\omega(B_0\cap\mathcal{O}(B(y,\rho)))}{\omega(\mathcal{O}(B(y,\rho)))} \le C. \end{align}\] Substitution in the Fubini identity gives the desired localization bound. ◻

Proof. Since \[\mathcal{O}(B(x_0,Lr))=\bigcup_{\sigma\in G}B(\sigma x_0,Lr),\] it is enough to estimate the integral over each ball in this finite union. For a fixed \(\sigma\), Lemma 20 applied to the ball \(B(\sigma x_0,Lr)\) gives \[\int_0^{Lr}\int_{B(\sigma x_0,Lr)}m_s(y)^2\,d\omega(y)\frac{ds}{s} \lesssim \omega(B(\sigma x_0,Lr)).\] The left-hand side dominates the same integral with the upper limit \(r\), since \(L\ge1\). By \(G\)-invariance and upper growth, we have \(\omega(B(\sigma x_0,Lr))=\omega(B(x_0,Lr)) \lesssim L^{\mathbf{N}}\omega(B(x_0,r)).\) Summing over the finitely many \(\sigma\in G\) proves the estimate. ◻

Proof. By Lemma 21, \[\mathcal{G}_sm_s(x) \le C\sum_{\ell\ge0}a_\ell A_{2^{\ell+1}s}^{\mathcal{O}}m_s(x), \qquad a_\ell=e^{-c2^{2\ell}}.\] Since \(\sum_\ell a_\ell<\infty\), Cauchy’s inequality gives \[|\mathcal{G}_sm_s(x)|^2 \lesssim \sum_{\ell\ge0}a_\ell \big(A_{2^{\ell+1}s}^{\mathcal{O}}m_s(x)\big)^2.\] The orbit average is normalized by a positive measure, hence Jensen’s inequality yields \[\big(A_{2^{\ell+1}s}^{\mathcal{O}}m_s(x)\big)^2 \le A_{2^{\ell+1}s}^{\mathcal{O}}(m_s^2)(x).\] After decreasing the Gaussian constant in the coefficients, we obtain \[|\mathcal{G}_sm_s(x)|^2 \lesssim \sum_{\ell\ge0}e^{-c2^{2\ell}} A_{2^{\ell+1}s}^{\mathcal{O}}(m_s^2)(x).\] Integrating over \(B_0\) and applying Lemma 22, \[\begin{align} &\int_0^r\int_{B_0}|\mathcal{G}_sm_s(x)|^2\,d\omega(x)\frac{ds}{s} \lesssim \sum_{\ell\ge0}e^{-c2^{2\ell}} \int_0^r\int_{\mathcal{O}(B(x_0,r+2^{\ell+1}s))}m_s(y)^2\,d\omega(y)\frac{ds}{s}. \end{align}\] Since \(0<s\le r\), we have \(r+2^{\ell+1}s\le C2^\ell r.\) Lemma 23 then gives \[\int_0^r\int_{\mathcal{O}(B(x_0,C2^\ell r))}m_s(y)^2\,d\omega(y)\frac{ds}{s} \lesssim 2^{\ell\mathbf{N}}\omega(B_0).\] The resulting series \(\sum_{\ell\ge0}e^{-c2^{2\ell}}2^{\ell\mathbf{N}}\) is finite. This proves the Carleson bound for the Gaussian average of the wall layer. ◻

Proof. We first compare the discontinuous chamber indicator with the smoothed cutoff. By construction, \(\eta_{\tau,s}\) is equal to \(1\) at points of \(\Omega_\tau\) whose distance from every reflection hyperplane is at least \(2s\), and it is equal to \(0\) outside \(\Omega_\tau\) whenever the point is at distance at least \(s\) from the reflection wall which separates it from \(\Omega_\tau\). Thus \(\chi_\tau-\eta_{\tau,s}\) is supported in a fixed multiple of the \(s\)-neighborhood of \(\mathcal{W}\). On this set \(m_s\) is bounded from below by a positive constant, whereas outside this set the difference is zero. This gives \[|\chi_\tau-\eta_{\tau,s}|\lesssim m_s.\]

We now estimate the Dunkl derivative. The ordinary derivative satisfies \[|\partial_\ell\eta_{\tau,s}(x)|\lesssim s^{-1} \mathbf{1}_{\{\operatorname{dist}(x,\mathcal{W})\le Cs\}},\] and hence \(|s\partial_\ell\eta_{\tau,s}(x)|\lesssim m_s(x).\)

For the difference part in \(T_\ell\), write \[T_\ell\eta_{\tau,s}(x) = \partial_\ell\eta_{\tau,s}(x) + \sum_{\alpha\in R} \frac{\kappa(\alpha)}{2}\alpha_\ell \frac{\eta_{\tau,s}(x)-\eta_{\tau,s}(\sigma_\alpha x)}{\langle\alpha,x\rangle}.\] It is useful to keep the following root-wise comparison in mind. Put \[\delta_\alpha(x)=\operatorname{dist}(x,\alpha^\perp),\qquad m_{\alpha,s}(x)=\min\left\{1,\frac{s}{\delta_\alpha(x)}\right\}.\] Since \(\delta_{\mathcal{W}}(x)=\min_\beta\delta_\beta(x)\), one has \(m_{\alpha,s}(x)\le m_s(x)\). Thus after estimating each root contribution by \(m_{\alpha,s}\), the finite sum over roots is still controlled by \(m_s\).

The summands with \(\kappa(\alpha)=0\) vanish. For the remaining summands we use two elementary bounds. Since \(0\le\eta_{\tau,s}\le1\), \(|\eta_{\tau,s}(x)-\eta_{\tau,s}(\sigma_\alpha x)|\le1.\) Since \(\eta_{\tau,s}\) is \(Cs^{-1}\)-Lipschitz, \(|\eta_{\tau,s}(x)-\eta_{\tau,s}(\sigma_\alpha x)| \lesssim s^{-1}\|x-\sigma_\alpha x\|.\) Combining the two estimates, we have \[|\eta_{\tau,s}(x)-\eta_{\tau,s}(\sigma_\alpha x)| \lesssim \min\left\{1,\frac{\|x-\sigma_\alpha x\|}{s}\right\}.\] Note also that \[\|x-\sigma_\alpha x\| =2\frac{|\langle x,\alpha\rangle|}{\|\alpha\|}.\] Therefore, away from the wall \(\langle x,\alpha\rangle=0\), \[\begin{align} s\frac{|\eta_{\tau,s}(x)-\eta_{\tau,s}(\sigma_\alpha x)|}{|\langle\alpha,x\rangle|} &\lesssim \min\left\{\frac{s}{|\langle\alpha,x\rangle|},1\right\}. \end{align}\] Equivalently, the last expression is bounded by a constant multiple of \(m_{\alpha,s}(x)\), and hence by a constant multiple of \(m_s(x)\). The hyperplanes themselves are \(\omega\)-null, so the estimate holds for \(\omega\)-almost every \(x\). Summing over the finite root system proves the lemma. ◻

Proof. The proof is local in the testing box. We fix \(x\in B_0\) and \(0<s\le r_0\). Since \(\psi_R\) is radial, it is \(G\)-invariant, and hence \(T_j\psi_R=\partial_j\psi_R.\) Moreover, we have \(|\partial_j\psi_R|\lesssim R^{-1}, \; \operatorname{supp}\partial_j\psi_R\subset A_R:=\{R\le |y|\le2R\}.\) Since \(\psi_R\) is \(G\)-invariant, the Dunkl product rule gives \[T_j(\eta_{\tau,s}\psi_R) =\psi_R T_j\eta_{\tau,s}+\eta_{\tau,s}\partial_j\psi_R.\] Thus the only new term created by compact truncation is the ordinary derivative of the radial cutoff; there is no additional reflection-difference term coming from \(\psi_R\). Choose \(R\ge4(1+|x_0|+r_0)\). If \(y\notin B(0,R)\), then for every \(\sigma\in G\), \(\|x-\sigma y\|\ge |y|-|x| \ge R-(|x_0|+r_0) \ge cR.\) Thus \[d(x,y)\ge cR\] whenever \(x\in B_0\) and \(y\notin B(0,R)\). We shall use the following tail bound: for every fixed \(M\ge0\), \[\label{eq:tail-bound-compact-cutoff} \int_{\{d(x,y)\ge cR\}} \left(1+\frac{d(x,y)}{s}\right)^M \frac{1}{V(x,y,s)}e^{-c d(x,y)^2/s^2}\,d\omega(y) \le C_Me^{-c_MR^2/s^2}.\tag{37}\] Indeed, decompose \(\{d(x,y)\ge cR\}\) into orbit annuli \(2^kR\le d(x,y)<2^{k+1}R, \;k=0,1,2,\ldots.\) On such an annulus the Gaussian gives \(e^{-c4^kR^2/s^2}\), while upper growth and Lemma 3 give at most a polynomial factor in \(2^kR/s\). The polynomial factor is absorbed by the Gaussian, and the sum in \(k\) gives 37 .

We now list the error terms. First, replacing \(\eta_{\tau,s}\) by \(\eta_{\tau,s}^R\) in \(\Theta_s\eta_{\tau,s}\) leaves the tail \(\eta_{\tau,s}(1-\psi_R)\). By the scale kernel estimate, \[\begin{align} |\Theta_s(\eta_{\tau,s}(1-\psi_R))(x)| &\lesssim \|b\|_{\operatorname{Lip}_d} \int_{\{|y|>R\}} \frac{1}{V(x,y,s)}e^{-c d(x,y)^2/s^2}\,d\omega(y) \lesssim \|b\|_{\operatorname{Lip}_d}e^{-cR^2/s^2}. \end{align}\] There is also the corresponding tail in the first-order commutator term, namely \[s[M_b,T_iH_{s^2}]\big((1-\psi_R)T_j\eta_{\tau,s}\big).\] Using \(|sT_j\eta_{\tau,s}|\lesssim m_s\le1\), the first-derivative heat estimate, and the orbit-Lipschitz bound for \(b\), we obtain \[\begin{align} \left|s[M_b,T_iH_{s^2}]\big((1-\psi_R)T_j\eta_{\tau,s}\big)(x)\right| &\lesssim \|b\|_{\operatorname{Lip}_d} \int_{\{|y|>R\}} \frac{d(x,y)}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\,d\omega(y) \lesssim \|b\|_{\operatorname{Lip}_d}e^{-cR^2/s^2}, \end{align}\] where the last step follows from 37 with \(M=1\). Second, in the term \(-sT_iH_{s^2}((\partial_jb)\eta_{\tau,s})\), the corresponding tail is controlled by the first-derivative heat estimate: \[\begin{align} |sT_iH_{s^2}((\partial_jb)\eta_{\tau,s}(1-\psi_R))(x)| &\lesssim \|b\|_{\operatorname{Lip}_d} \int_{\{|y|>R\}} \frac{1}{V(x,y,s)}e^{-c d(x,y)^2/s^2}\,d\omega(y) \lesssim \|b\|_{\operatorname{Lip}_d}e^{-cR^2/s^2}. \end{align}\] Third, differentiating the cutoff produces the term \[s[M_b,T_iH_{s^2}](\eta_{\tau,s}\partial_j\psi_R).\] Using \(|\partial_j\psi_R|\lesssim R^{-1}\), the orbit-Lipschitz bound for \(b\), and Lemma 14, we get \[\begin{align} |s[M_b,T_iH_{s^2}](\eta_{\tau,s}\partial_j\psi_R)(x)| &\le s\int_{A_R}|b(x)-b(y)|\,|T_{i,x}h_{s^2}(x,y)|\,|\partial_j\psi_R(y)|\,d\omega(y)\\ &\lesssim \|b\|_{\operatorname{Lip}_d} \int_{A_R} \frac{d(x,y)}{R} \frac{1}{V(x,y,s)}e^{-c d(x,y)^2/s^2}\,d\omega(y). \end{align}\] Since \(0<s\le r_0\) and \(R\ge4(1+|x_0|+r_0)\), we have \(s/R\le1\). Hence \[\frac{d(x,y)}{R}\le \frac{d(x,y)}{s} \le 1+\frac{d(x,y)}{s}.\] The last integral is therefore bounded by \(\|b\|_{\operatorname{Lip}_d}e^{-cR^2/s^2}\), after applying 37 with \(M=1\).

Thus every error term \(\mathcal{E}_{R,s}\) which appears in the local testing argument satisfies \[|\mathcal{E}_{R,s}(x)| \lesssim \|b\|_{\operatorname{Lip}_d}e^{-cR^2/s^2}, \qquad x\in B_0,\quad 0<s\le r_0.\] Consequently, \[\begin{align} \int_0^{r_0}\int_{B_0}|\mathcal{E}_{R,s}(x)|^2\,d\omega(x)\frac{ds}{s} &\lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_0) \int_0^{r_0}e^{-cR^2/s^2}\frac{ds}{s}. \end{align}\] The last integral tends to zero as \(R\to\infty\), for example after the change of variables \(u=R^2/s^2\). This proves the lemma. ◻

Proof. By Plancherel for the Dunkl transform, \[\begin{align} \int_0^\infty\|sT_\ell H_{s^2}f\|_2^2\frac{ds}{s} &= \int_{\mathbb{R}^N} \left(\int_0^\infty s^2\xi_\ell^2e^{-2s^2\|\xi\|^2}\frac{ds}{s}\right) |\mathcal{F}_\kappa f(\xi)|^2\,d\omega(\xi). \end{align}\] For \(\xi\ne0\), \[\int_0^\infty s\xi_\ell^2e^{-2s^2\|\xi\|^2}\,ds =\frac{\xi_\ell^2}{4\|\xi\|^2} \le \frac{1}{4}.\] At \(\xi=0\) the inner integrand is zero. Hence the preceding display is bounded by \(C\|f\|_{L^2(\mathbb{R}^N,d\omega)}^2\). ◻

Proof. Let \[g_1=g\mathbf{1}_{\mathcal{O}(B(x_0,4r))}, \qquad g_2=g\mathbf{1}_{\mathbb{R}^N\setminus\mathcal{O}(B(x_0,4r))}.\] For the local part, the vertical square-function estimate gives \[\begin{align} \int_0^r\int_{B_0}|sT_\ell H_{s^2}g_1(x)|^2\,d\omega(x)\frac{ds}{s} &\le \int_0^\infty\|sT_\ell H_{s^2}g_1\|_2^2\frac{ds}{s} \lesssim \|g_1\|_2^2 \le \|g\|_\infty^2\omega(\mathcal{O}(B(x_0,4r))) \lesssim \|g\|_\infty^2\omega(B_0), \end{align}\] where the last step uses Lemma 3 and doubling.

For the far part, take \(x\in B_0\), \(0<s\le r\), and \(y\notin\mathcal{O}(B(x_0,4r))\). Then \[d(x,y)\ge d(x_0,y)-\|x-x_0\|\ge4r-r=3r.\] Using Lemma 14, \[\begin{align} |sT_\ell H_{s^2}g_2(x)| &\le \|g\|_\infty \int_{d(x,y)\ge3r} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\,d\omega(y). \end{align}\] We estimate the last integral explicitly. Let \(E_k=\{y:2^kr\le d(x,y)<2^{k+1}r\}, \;k=1,2,\ldots.\) For \(y\in E_k\), \[e^{-cd(x,y)^2/s^2}\le e^{-c4^kr^2/s^2}, \qquad V(x,y,s)\ge \omega(B(x,s)).\] Moreover, \(E_k\subset \mathcal{O}(B(x,2^{k+1}r)),\) and hence, using Lemma 3 and upper growth, \[\begin{align} \omega(E_k) &\le \omega(\mathcal{O}(B(x,2^{k+1}r))) \lesssim \left(\frac{2^kr}{s}\right)^{\mathbf{N}}\omega(B(x,s)). \end{align}\] Therefore \[\begin{align} &\int_{E_k}\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\,d\omega(y) \lesssim e^{-c4^kr^2/s^2}\left(\frac{2^kr}{s}\right)^{\mathbf{N}}. \end{align}\] Since \(0<s\le r\), the polynomial factor is absorbed by the Gaussian after summing in \(k\), and we obtain \[|sT_\ell H_{s^2}g_2(x)| \lesssim \|g\|_\infty e^{-cr^2/s^2}.\] Thus \[\int_0^r |sT_\ell H_{s^2}g_2(x)|^2\frac{ds}{s} \lesssim \|g\|_\infty^2\int_0^r e^{-cr^2/s^2}\frac{ds}{s} \lesssim \|g\|_\infty^2.\] Integrating in \(x\in B_0\) completes the proof. ◻

Proof. We use the Dunkl transform. On the Schwartz class, and hence by closure on \(L^2(\mathbb{R}^N,d\omega)\), \[\mathcal{F}_\kappa(T_k f)(\xi)=i\xi_k\mathcal{F}_\kappa f(\xi), \qquad \mathcal{F}_\kappa(H_{s^2}f)(\xi)=e^{-s^2\|\xi\|^2}\mathcal{F}_\kappa f(\xi).\] Consequently \(L_s=T_iT_jH_{s^2}\) is the multiplier \[-\xi_i\xi_j e^{-s^2\|\xi\|^2}.\] This multiplier is real-valued and bounded. Therefore \(\langle L_s f,g\rangle_{L^2(\mathbb{R}^N,d\omega)} = \langle f,L_s g\rangle_{L^2(\mathbb{R}^N,d\omega)}\) for Schwartz functions, and the identity extends to all of \(L^2(\mathbb{R}^N,d\omega)\) by boundedness. ◻

Proof. We state the identity at the kernel level, because the symbol \(b\) is not assumed to be bounded on \(\mathbb{R}^N\). Thus one should not regard \(M_b\) as a bounded multiplication operator on \(L^2(\mathbb{R}^N,d\omega)\). By Lemma 29, the operator \(L_s=T_iT_jH_{s^2}\) is self-adjoint on \(L^2(\mathbb{R}^N,d\omega)\). Hence its kernel satisfies \[A_{s^2}^{ij}(x,y)=\overline{A_{s^2}^{ij}(y,x)}\] in the usual off-diagonal associated-kernel sense. Therefore the adjoint kernel of \(\Theta_{s,b}\) is \[\begin{align} \overline{s(b(y)-b(x))A_{s^2}^{ij}(y,x)} &=s\big(\overline{b}(y)-\overline{b}(x)\big)A_{s^2}^{ij}(x,y) =-s\big(\overline{b}(x)-\overline{b}(y)\big)A_{s^2}^{ij}(x,y), \end{align}\] which is exactly the kernel of \(-\Theta_{s,\overline{b}}\). The Gaussian bound for the scale kernel justifies the identity first for bounded compactly supported test functions on \(\mathbb{R}^N\) and then for the chamber indicators used in the Carleson testing argument. ◻

Proof. For compactly supported smooth \(\varphi\), both \(\varphi\) and \(b\varphi\) are admissible test functions for the finite heat operator. By the definition of the kernel \(A_{s^2}^{ij}\), \[\Theta_{s,b}\varphi =s\big(M_bL_s\varphi-L_s(b\varphi)\big), \qquad L_s=T_iT_jH_{s^2}.\] We next justify the commutation used below on this class of test functions. Since the Dunkl operators commute with one another and \(\Delta=\sum_kT_k^2\), one has \[T_j\Delta^m\varphi=\Delta^mT_j\varphi, \qquad m=0,1,2,\ldots.\] Applying this identity to the heat semigroup, first for the heat-polynomial approximants and then passing to the heat operator on test functions, gives \[T_jH_{s^2}\varphi=H_{s^2}T_j\varphi.\] Equivalently, the same equality follows from the Dunkl-transform multiplier representation of \(H_{s^2}\). Therefore \[T_iT_jH_{s^2}\varphi=T_iH_{s^2}(T_j\varphi),\] that is, \(L_s\varphi=T_iH_{s^2}(T_j\varphi).\)

Since \(b\) is \(G\)-invariant, the product rule gives \(T_j(b\varphi)=(\partial_jb)\varphi+bT_j\varphi.\) Consequently \[\begin{align} L_s(b\varphi) &=T_iH_{s^2}T_j(b\varphi) =T_iH_{s^2}\big((\partial_jb)\varphi\big) +T_iH_{s^2}\big(bT_j\varphi\big), \end{align}\] while \(M_bL_s\varphi=bT_iH_{s^2}(T_j\varphi).\)

Subtracting these two identities gives 38 . The point of the compact support assumption is only to justify the algebra before any limiting argument is taken; the noncompact chamber cutoffs are obtained from this identity by Lemma 26. ◻

Proof. We prove the estimate on a fixed ball \(B_0=B(x_0,r),\) with constants independent of \(B_0\) and of the chamber. The argument has two separate roles. The first is to replace the discontinuous chamber indicator by a heat-scale cutoff; the second is to use the algebraic commutator identity only for smooth, compactly supported objects and then pass to the limiting ones.

Set \(\beta_{\tau,s}:=\chi_\tau-\eta_{\tau,s}.\) Then \[\Theta_s\chi_\tau=\Theta_s\eta_{\tau,s}+\Theta_s\beta_{\tau,s}.\] We first estimate the boundary part using the scale kernel bound. From the integral representation of \(\Theta_s\), \[\Theta_s\beta_{\tau,s}(x) = \int_{\mathbb{R}^N}\theta_s(x,y)\beta_{\tau,s}(y)\,d\omega(y),\] whenever the right-hand side is interpreted by the same heat-scale kernel estimate. Since \(|\beta_{\tau,s}|\lesssim m_s\), Corollary 1 gives \[\begin{align} |\Theta_s\beta_{\tau,s}(x)| &\le \int |\theta_s(x,y)|\,|\beta_{\tau,s}(y)|\,d\omega(y) \lesssim \|b\|_{\operatorname{Lip}_d} \int\frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}m_s(y)\,d\omega(y) =C\|b\|_{\operatorname{Lip}_d}\mathcal{G}_sm_s(x). \end{align}\] After integration over \(B_0\times(0,r)\), Lemma 24 gives \[\int_0^r\int_{B_0}|\Theta_s\beta_{\tau,s}(x)|^2\,d\omega(x)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_0).\]

We next treat \(\Theta_s\eta_{\tau,s}\). At first assume that \(b\) is smooth and \(G\)-invariant, with \(\|\nabla b\|_\infty\le \|b\|_{\operatorname{Lip}_d}\). Since \(\eta_{\tau,s}\) is not compactly supported, one should first replace it by \(\eta_{\tau,s}^R=\eta_{\tau,s}\psi_R\), apply the following computation to the compactly supported function \(\eta_{\tau,s}^R\), and then let \(R\to\infty\). Lemma 26 shows that the error tends to zero in the local Carleson norm on every testing box. We therefore write the identity for \(\eta_{\tau,s}\) itself.

Applying Lemma 31 first to \(\eta_{\tau,s}^R\), and then letting \(R\to\infty\) by Lemma 26, gives \[\begin{align} \Theta_s\eta_{\tau,s} &=s[M_b,T_iH_{s^2}](T_j\eta_{\tau,s}) -sT_iH_{s^2}\big((\partial_jb)\eta_{\tau,s}\big) =:A_s+B_s. \end{align}\] This is the only algebraic identity needed in the testing argument, and it has now been obtained without applying the product rule directly to a noncompact cutoff.

For \(A_s\), the cutoff derivative estimate gives \(|sT_j\eta_{\tau,s}(y)|\lesssim m_s(y)\). Hence, by the first derivative heat estimate and the orbit Lipschitz bound for \(b\), \[\begin{align} |A_s(x)| &\le s\int |b(x)-b(y)|\,|T_{i,x}h_{s^2}(x,y)|\, |T_j\eta_{\tau,s}(y)|\,d\omega(y)\\ &\lesssim \|b\|_{\operatorname{Lip}_d} \int \frac{d(x,y)}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}m_s(y)\,d\omega(y)\\ &\lesssim \|b\|_{\operatorname{Lip}_d}\mathcal{G}_sm_s(x). \end{align}\] Thus \(A_s\) satisfies the same Carleson estimate as the boundary term.

For \(B_s\), write again \(\eta_{\tau,s}=\chi_\tau-\beta_{\tau,s}\). Then \[\begin{align} B_s &=-sT_iH_{s^2}\big((\partial_jb)\chi_\tau\big) +sT_iH_{s^2}\big((\partial_jb)\beta_{\tau,s}\big) =:B_s^{\rm main}+B_s^{\rm bd}. \end{align}\] Since \((\partial_jb)\chi_\tau\in L^\infty\) and \(\|\partial_jb\|_\infty\le\|b\|_{\operatorname{Lip}_d}\), Lemma 28 gives \[\int_0^r\int_{B_0}|B_s^{\rm main}(x)|^2\,d\omega(x)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_0).\] The boundary part satisfies \[\begin{align} |B_s^{\rm bd}(x)| &\le s\int |T_{i,x}h_{s^2}(x,y)|\,|\partial_jb(y)|\, |\beta_{\tau,s}(y)|\,d\omega(y) \lesssim \|b\|_{\operatorname{Lip}_d}\mathcal{G}_sm_s(x), \end{align}\] and is controlled by Lemma 24. This proves the component testing estimate for smooth \(b\).

It remains to remove the smoothness assumption. Let \(b_\varepsilon\) be the smooth \(G\)-invariant approximants from Lemma 10, and let \(\Theta_{s,\varepsilon}\) be the corresponding scale operators. The preceding smooth argument gives, with a constant independent of \(\varepsilon\), \[\label{eq:smooth-approx-uniform-testing} \int_0^r\int_{B_0}|\Theta_{s,\varepsilon}\chi_\tau(x)|^2 \,d\omega(x)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_0).\tag{39}\] We now pass to the limit in this local Carleson estimate. This passage uses only the scale kernels; it does not require treating \(M_b\) or \(M_{b_\varepsilon}\) as bounded multiplication operators.

Fix \(s>0\) and fix \(x\) outside the exceptional null set on which the heat kernels are not represented by their pointwise formulae. Then \[\label{eq:smooth-approx-pointwise-difference} \begin{align} &\Theta_{s,\varepsilon}\chi_\tau(x)-\Theta_s\chi_\tau(x) =\int_{\mathbb{R}^N}s\Big[(b_\varepsilon(x)-b_\varepsilon(y))-(b(x)-b(y))\Big] A_{s^2}^{ij}(x,y)\chi_\tau(y)\,d\omega(y). \end{align}\tag{40}\] We prove that the right-hand side tends to zero. Choose \(M>2|x|+1\), and split the integral into the compact part \(B(0,M)\) and the tail \(\mathbb{R}^N\setminus B(0,M)\).

On \(B(0,M)\), local uniform convergence gives \[\label{eq:smooth-approx-local-uniform} \sup_{y\in B(0,M)} \left|(b_\varepsilon(x)-b_\varepsilon(y))-(b(x)-b(y))\right| \longrightarrow0.\tag{41}\] For this fixed \(s\), the function \(y\mapsto sA_{s^2}^{ij}(x,y)\) is locally integrable. Indeed, Proposition 7 gives \[\label{eq:smooth-approx-local-integrable-majorant} s|A_{s^2}^{ij}(x,y)| \lesssim \frac{1}{s}\frac{1}{V(x,y,s)} e^{-cd(x,y)^2/s^2},\tag{42}\] and the right-hand side has finite integral over bounded sets by the same orbit-annular estimate used for Lemma 16. Hence 41 and 42 imply that the compact part of 40 tends to zero as \(\varepsilon\to0\).

For the tail, use the uniform orbit-Lipschitz bounds for both \(b\) and \(b_\varepsilon\): \[\begin{align} &\left|(b_\varepsilon(x)-b_\varepsilon(y))-(b(x)-b(y))\right| \le |b_\varepsilon(x)-b_\varepsilon(y)|+|b(x)-b(y)| \le 2\|b\|_{\operatorname{Lip}_d}d(x,y). \end{align}\] Together with the second heat estimate, this gives the uniform majorant \[\label{eq:smooth-approx-tail-majorant} \begin{align} s\left|(b_\varepsilon(x)-b_\varepsilon(y))-(b(x)-b(y))\right| |A_{s^2}^{ij}(x,y)| &\lesssim \|b\|_{\operatorname{Lip}_d} \frac{d(x,y)}{s} \frac{1}{V(x,y,s)}e^{-cd(x,y)^2/s^2}\\ &\lesssim \|b\|_{\operatorname{Lip}_d} \frac{1}{V(x,y,s)}e^{-c'd(x,y)^2/s^2}. \end{align}\tag{43}\] The last bound is independent of \(\varepsilon\). Since every element of \(G\) is an isometry, \[d(x,y)=\min_{\sigma\in G}\|x-\sigma y\| \ge |y|-|x|,\] so for \(y\notin B(0,M)\) and \(M>2|x|+1\) one has \(d(x,y)\ge M/2\). The orbit-annular estimate gives \[\label{eq:smooth-approx-tail-vanishing} \int_{\mathbb{R}^N\setminus B(0,M)} \frac{1}{V(x,y,s)}e^{-c'd(x,y)^2/s^2}\,d\omega(y) \longrightarrow0, \qquad M\to\infty.\tag{44}\] Thus the tail in 40 is small uniformly in \(\varepsilon\) once \(M\) is large. Combining the compact part and the tail proves \[\label{eq:smooth-approx-pointwise-convergence} \Theta_{s,\varepsilon}\chi_\tau(x)\to\Theta_s\chi_\tau(x)\tag{45}\] for almost every \((x,s)\).

Now apply Fatou’s lemma on the fixed testing box \(B_0\times(0,r)\). From 39 and 45 , \[\begin{align} \int_0^r\int_{B_0}|\Theta_s\chi_\tau(x)|^2\,d\omega(x)\frac{ds}{s} \le \liminf_{\varepsilon\to0} \int_0^r\int_{B_0}|\Theta_{s,\varepsilon}\chi_\tau(x)|^2 \,d\omega(x)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_0). \end{align}\] Thus the component testing estimate holds for every \(b\in\operatorname{Lip}_d\).

Finally, we handle the adjoint. By Lemma 30, \[\Theta_{s,b}^\ast\chi_\tau=-\Theta_{s,\overline{b}}\chi_\tau.\] The estimate just proved for \(\Theta_s\chi_\tau\) was established for an arbitrary complex-valued symbol in \(\operatorname{Lip}_d\). Applying it to \(\overline{b}\) therefore gives \[\int_0^r\int_{B_0}|\Theta_{s,b}^*\chi_\tau(x)|^2\,d\omega(x)\frac{ds}{s} \lesssim \|\overline{b}\|_{\operatorname{Lip}_d}^2\omega(B_0) = \|b\|_{\operatorname{Lip}_d}^2\omega(B_0).\] This proves the adjoint testing estimate. The passage uses only the scale-kernel adjoint identity and the Lipschitz norm of \(\overline{b}\); it does not require \(M_b\) to be a bounded multiplication operator on \(L^2\). ◻

Proof. We first take \(1\le p<\infty\). By definition of \(U\), \[\begin{align} \|Uf\|_{L^p(\mathcal{C};\ell^p_M)}^p &= \int_{\mathcal{C}} \sum_{\rho=1}^M |f(\sigma_\rho x)|^p \,d\omega(x) = \sum_{\rho=1}^M \int_{\mathcal{C}}|f(\sigma_\rho x)|^p\,d\omega(x). \end{align}\] Since the map \(x\mapsto\sigma_\rho x\) is a measure-preserving bijection from \(\mathcal{C}\) onto \(\sigma_\rho\mathcal{C}\), it follows that \[\int_{\mathcal{C}}|f(\sigma_\rho x)|^p\,d\omega(x) = \int_{\sigma_\rho\mathcal{C}}|f(x)|^p\,d\omega(x).\] Therefore \[\begin{align} \|Uf\|_{L^p(\mathcal{C};\ell^p_M)}^p &= \sum_{\rho=1}^M \int_{\sigma_\rho\mathcal{C}}|f(x)|^p\,d\omega(x) = \int_{\mathbb{R}^N}|f(x)|^p\,d\omega(x), \end{align}\] because the reflected chambers cover \(\mathbb{R}^N\) disjointly up to an \(\omega\)-null set. Hence \[\|Uf\|_{L^p(\mathcal{C};\ell^p_M)} = \|f\|_{L^p(\mathbb{R}^N,d\omega)}.\]

For \(p=\infty\), \[\begin{align} \|Uf\|_{L^\infty(\mathcal{C};\ell^\infty_M)} &= \operatorname*{ess\,sup}_{x\in\mathcal{C}} \max_{1\le\rho\le M}|f(\sigma_\rho x)| = \operatorname*{ess\,sup}_{x\in\mathbb{R}^N}|f(x)| = \|f\|_{L^\infty(\mathbb{R}^N,d\omega)}. \end{align}\] The formula for \(U^{-1}\) above shows that \(U\) is onto. This proves that \(U\) is an isometric isomorphism. For \(p=2\), the same computation also gives preservation of the inner product, so \(U\) is unitary. ◻

Proof. Fix \(x,y\in\mathcal{C}\), and write \(z=\sigma_\rho x, \; w=\sigma_\tau y .\) These are full-space points, whereas \(x\) and \(y\) remain chamber points. By the chamber representative lemma, \[d(z,w)=d(\sigma_\rho x,\sigma_\tau y)=\|x-y\|.\] Moreover, chamber volume comparability gives \[V(z,w,s) = V(\sigma_\rho x,\sigma_\tau y,s) \approx V_{\mathcal{C}}(x,y,s),\] with constants independent of \(\rho\) and \(\tau\). Applying Corollary 1 to the full-space points \(z\) and \(w\) gives \[\begin{align} |\mathbb{\theta}_s^{\rho\tau}(x,y)| &= |\theta_s(\sigma_\rho x,\sigma_\tau y)| = |\theta_s(z,w)| \le C\|b\|_{\operatorname{Lip}_d} \frac{1}{V(z,w,s)} \exp\left(-c\frac{d(z,w)^2}{s^2}\right), \end{align}\] which is exactly 47 after the two preceding comparisons.

For regularity in the first chamber variable, assume \(x,x',y\in\mathcal{C}\) and \(\|x-x'\|\le s\). Put \(z'=\sigma_\rho x'.\) Since reflections are Euclidean isometries, \[\|z-z'\| = \|\sigma_\rho x-\sigma_\rho x'\| = \|x-x'\| \le s.\] By Corollary 1, \[\begin{align} |\mathbb{\theta}_s^{\rho\tau}(x,y) -\mathbb{\theta}_s^{\rho\tau}(x',y)| & = |\theta_s(z,w)-\theta_s(z',w)| \le C\|b\|_{\operatorname{Lip}_d} \frac{\|z-z'\|}{s} \frac{1}{V(z,w,s)} \exp\left(-c\frac{d(z,w)^2}{s^2}\right). \end{align}\] Using \[\|z-z'\|=\|x-x'\|, \qquad d(z,w)=\|x-y\|, \qquad V(z,w,s)\approx V_{\mathcal{C}}(x,y,s),\] we obtain 48 .

For the second chamber variable, assume \(x,y,y'\in\mathcal{C}\) and \(\|y-y'\|\le s\). Put \(w'=\sigma_\tau y'.\) Then \[\|w-w'\| = \|\sigma_\tau y-\sigma_\tau y'\| = \|y-y'\| \le s.\] The second-variable regularity estimate in Corollary 1 gives \[\begin{align} |\mathbb{\theta}_s^{\rho\tau}(x,y) -\mathbb{\theta}_s^{\rho\tau}(x,y')| & = |\theta_s(z,w)-\theta_s(z,w')| \le C\|b\|_{\operatorname{Lip}_d} \frac{\|w-w'\|}{s} \frac{1}{V(z,w,s)} \exp\left(-c\frac{d(z,w)^2}{s^2}\right). \end{align}\] Again \[\|w-w'\|=\|y-y'\|, \qquad d(z,w)=\|x-y\|, \qquad V(z,w,s)\approx V_{\mathcal{C}}(x,y,s),\] and hence 49 follows. The constants do not depend on \(\rho\) or \(\tau\). ◻

Proof. Let \[\chi_\tau=\mathbf{1}_{\Omega_\tau}=\mathbf{1}_{\sigma_\tau\mathcal{C}}.\] For \(x\) in the interior of \(\mathcal{C}\), the point \(\sigma_\rho x\) lies in \(\Omega_\tau\) exactly when \(\rho=\tau\). Since the walls are \(\omega\)-null, this gives \[U\chi_\tau=e_\tau\] in every \(L^p(\mathcal{C},d\omega;\ell^p_{|G|})\). Hence \[\boldsymbol{\Theta}_s e_\tau=U\Theta_s\chi_\tau.\]

Let \(B_{\mathcal{C}}(x_0,r)=B(x_0,r)\cap\mathcal{C}\) be a chamber ball. Then \[\begin{align} &\int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\boldsymbol{\Theta}_s e_\tau(x)|_{\ell^2}^2\,d\omega(x)\frac{ds}{s} = \sum_{\rho=1}^{|G|} \int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\Theta_s\chi_\tau(\sigma_\rho x)|^2\,d\omega(x)\frac{ds}{s}. \end{align}\] For the \(\rho\)-th term, we use that \(\sigma_\rho:B_{\mathcal{C}}(x_0,r)\to\sigma_\rho B_{\mathcal{C}}(x_0,r)\) is a measure-preserving bijection. Therefore \[\begin{align} \int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\Theta_s\chi_\tau(\sigma_\rho x)|^2\,d\omega(x)\frac{ds}{s} = \int_0^r\int_{\sigma_\rho B_{\mathcal{C}}(x_0,r)} |\Theta_s\chi_\tau(z)|^2\,d\omega(z)\frac{ds}{s}. \end{align}\] Moreover, \[\sigma_\rho B_{\mathcal{C}}(x_0,r) \subset B(\sigma_\rho x_0,r),\] because \(\sigma_\rho\) is a Euclidean isometry. Hence Proposition 16 gives \[\begin{align} \int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\boldsymbol{\Theta}_s e_\tau(x)|_{\ell^2}^2\,d\omega(x)\frac{ds}{s} & \le \sum_{\rho=1}^{|G|} \int_0^r\int_{B(\sigma_\rho x_0,r)} |\Theta_s\chi_\tau(z)|^2\,d\omega(z)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2 \sum_{\rho=1}^{|G|}\omega(B(\sigma_\rho x_0,r)). \end{align}\] By invariance of \(d\omega\), \(\omega(B(\sigma_\rho x_0,r))=\omega(B(x_0,r)),\) and by chamber volume comparability, \(\omega(B(x_0,r))\approx \omega(B_{\mathcal{C}}(x_0,r)).\) Thus \[\int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\boldsymbol{\Theta}_s e_\tau(x)|_{\ell^2}^2\,d\omega(x)\frac{ds}{s} \lesssim \|b\|_{\operatorname{Lip}_d}^2\omega(B_{\mathcal{C}}(x_0,r)),\] which is the desired Carleson estimate on \(\mathcal{C}\).

For the adjoint, use that \(U\) is unitary on \(L^2\). Hence \[\boldsymbol{\Theta}_s^*=U\Theta_s^*U^{-1}, \qquad \boldsymbol{\Theta}_s^*e_\rho=U\Theta_s^*\chi_\rho.\] Repeating the preceding argument and using the adjoint part of Proposition 16 proves the adjoint coordinate testing estimate. ◻

Proof. For each pair of indices \(a,b\), define the scalar truncated operator \[S_{\varepsilon,R}^{ab}f := \int_\varepsilon^R\Theta_s^{ab}f\,\frac{ds}{s}.\] The scalar kernel assumptions in Theorem 19 hold for every \(\Theta_s^{ab}\), with the same constant \(\mathfrak K\). We also need the scalar testing conditions for each entry. These are contained in the coordinate testing hypotheses. Indeed, \[\Theta_s^{ab}1=(\boldsymbol{\Theta}_s e_b)_a.\] Similarly, since the adjoint matrix has entries \(((\Theta_s^{ab})^*)_{b,a}\), \[(\Theta_s^{ab})^*1=(\boldsymbol{\Theta}_s^*e_a)_b.\] Thus the Carleson norms of \(\Theta_s^{ab}1\) and \((\Theta_s^{ab})^*1\) are bounded by the coordinate testing constants, uniformly in \(a,b\). Notice that this implication is one-way and entrywise: no vector-valued testing theorem is being used here.

All constants in the preceding hypotheses are independent of the truncation endpoints \(\varepsilon\) and \(R\); those endpoints only specify the interval of scale integration. The quoted scalar continuous \(T1\) theorem gives, for each \(a,b\), a weak operator limit \[S^{ab}= \operatorname*{w\!-\!lim}_{\varepsilon\to0,\,R\to\infty} S_{\varepsilon,R}^{ab}\] on \(L^2(X,\mu)\), and \[\|S^{ab}\|_{2\to2} \le C(\mathfrak K+\mathfrak C+\mathfrak C^*).\] Now define the matrix operator \[(\mathbb{S} F)_a:=\sum_{b=1}^M S^{ab}F_b.\] Since \(M\) is finite, \[\begin{align} \|\mathbb{S} F\|_{L^2(X;\ell_M^2)} \le C_M \sum_{a,b=1}^M \|S^{ab}F_b\|_{L^2(X)} \le C_M(\mathfrak K+\mathfrak C+\mathfrak C^*) \sum_{b=1}^M \|F_b\|_{L^2(X)} \le C_M(\mathfrak K+\mathfrak C+\mathfrak C^*) \|F\|_{L^2(X;\ell_M^2)}. \end{align}\] This proves the norm bound.

It remains only to check weak convergence of the matrix truncations. Let \(F,G\in L^2(X;\ell_M^2)\). Expanding the vector inner product and using the finite number of coordinates, we obtain \[\begin{align} \langle \mathbb{S}_{\varepsilon,R}F,G\rangle &= \int_X\sum_{a=1}^M \left(\sum_{b=1}^M S_{\varepsilon,R}^{ab}F_b(x)\right) \overline{G_a(x)}\,d\mu(x) = \sum_{a,b=1}^M \langle S_{\varepsilon,R}^{ab}F_b,G_a\rangle. \end{align}\] For each pair \((a,b)\), the scalar weak convergence gives \(\langle S_{\varepsilon,R}^{ab}F_b,G_a\rangle \longrightarrow \langle S^{ab}F_b,G_a\rangle.\)

Since the double sum contains only \(M^2\) terms, the limit may be taken term by term. Hence \[\begin{align} \lim_{\varepsilon\to0,\;R\to\infty} \langle \mathbb{S}_{\varepsilon,R}F,G\rangle &= \sum_{a,b=1}^M\langle S^{ab}F_b,G_a\rangle =\langle \mathbb{S} F,G\rangle. \end{align}\] This is precisely weak convergence of \(\mathbb{S}_{\varepsilon,R}\) to \(\mathbb{S}\) on \(L^2(X;\ell_M^2)\). ◻

Proof. We give the verification in some detail, since this is the point where the full-space problem is converted into a finite-dimensional continuous \(T1\) problem. The underlying scalar space is \[(X,\rho,\mu)=(\mathcal{C},\|\cdot\|,d\omega|_{\mathcal{C}}).\] By Lemma 8, \(\mu(B_{\mathcal{C}}(x,2r)) =\omega(B_{\mathcal{C}}(x,2r)) \lesssim \omega(B_{\mathcal{C}}(x,r)),\) so \((\mathcal{C},\|\cdot\|,d\omega|_{\mathcal{C}})\) is a space of homogeneous type. We take \(M=|G|,\) and regard \(\boldsymbol{\Theta}_s\) as an \(M\times M\) matrix of scalar operators. For its \((\rho,\tau)\)-entry, the kernel is \(\mathbb{\theta}_s^{\rho\tau}(x,y) =\theta_s(\sigma_\rho x,\sigma_\tau y).\)

We first check the scalar kernel assumptions. By the lifted kernel estimate, \[|\mathbb{\theta}_s^{\rho\tau}(x,y)| \le C\|b\|_{\operatorname{Lip}_d} \frac{1}{V_{\mathcal{C}}(x,y,s)} \exp\left(-c\frac{\|x-y\|^2}{s^2}\right).\] Choose once and for all an exponent \(\Gamma\) larger than the doubling order required in Theorem 19. Since \(e^{-cu^2}\le C_\Gamma(1+u)^{-\Gamma}, \;u\ge0,\) we obtain, with \(u=\|x-y\|/s\), \[|\mathbb{\theta}_s^{\rho\tau}(x,y)| \le C_\Gamma\|b\|_{\operatorname{Lip}_d} \frac{1}{V_{\mathcal{C}}(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-\Gamma}.\] Thus the size constant in Theorem 19 is bounded by \(C\|b\|_{\operatorname{Lip}_d}\), uniformly in \(\rho,\tau\). If \(\|x-x'\|\le s\), the lifted regularity estimate gives \[\begin{align} |\mathbb{\theta}_s^{\rho\tau}(x,y) -\mathbb{\theta}_s^{\rho\tau}(x',y)| &\le C\|b\|_{\operatorname{Lip}_d} \frac{\|x-x'\|}{s} \frac{1}{V_{\mathcal{C}}(x,y,s)} \exp\left(-c\frac{\|x-y\|^2}{s^2}\right) \\ &\le C_{\Gamma}\|b\|_{\operatorname{Lip}_d} \left(\frac{\|x-x'\|}{s}\right) \frac{1}{V_{\mathcal{C}}(x,y,s)} \left(1+\frac{\|x-y\|}{s}\right)^{-\Gamma}. \end{align}\] This is the \(x\)-regularity condition in the scalar theorem with \(\alpha=1\). The proof in the \(y\)-variable is the same, using the second-variable regularity estimate for \(\mathbb{\theta}_s^{\rho\tau}\). Hence all scalar entries have the required kernel constants bounded by \(C\|b\|_{\operatorname{Lip}_d}\).

We next check the scalar testing assumptions entry by entry. Let \(e_\tau\) denote the \(\tau\)-th coordinate vector in \(\mathbb{C}^{|G|}\). Since \(U\chi_\tau=e_\tau\) modulo null chamber walls, \[\boldsymbol{\Theta}_s e_\tau=U\Theta_s\chi_\tau.\] The coordinate testing estimate already proved gives, for every chamber ball \(B_{\mathcal{C}}(x_0,r)\), \[\int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\boldsymbol{\Theta}_s e_\tau(x)|_{\ell^2}^2 \,d\omega(x)\frac{ds}{s} \le C\|b\|_{\operatorname{Lip}_d}^2\omega(B_{\mathcal{C}}(x_0,r)).\] For a fixed entry \((\rho,\tau)\), \[\boldsymbol{\Theta}_s^{\rho\tau}1 =(\boldsymbol{\Theta}_s e_\tau)_\rho.\] Since the square of one coordinate is bounded by the square of the \(\ell^2\)-norm of the vector, the preceding estimate implies \[\int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |\boldsymbol{\Theta}_s^{\rho\tau}1(x)|^2 \,d\omega(x)\frac{ds}{s} \le C\|b\|_{\operatorname{Lip}_d}^2\omega(B_{\mathcal{C}}(x_0,r)).\] Thus the scalar \(T1\) Carleson norm of every entry is bounded by \(C\|b\|_{\operatorname{Lip}_d}\). For the adjoint entry, we use the coordinate testing estimate for \(\boldsymbol{\Theta}_s^*\) rather than any new kernel calculation. Since \[(\boldsymbol{\Theta}_s^{\rho\tau})^*1 =(\boldsymbol{\Theta}_s^*e_\rho)_\tau,\] we have pointwise \[|(\boldsymbol{\Theta}_s^{\rho\tau})^*1(x)|^2 \le |\boldsymbol{\Theta}_s^*e_\rho(x)|_{\ell^2}^2.\] Integrating this inequality over the testing box and applying the adjoint coordinate testing estimate gives \[\int_0^r\int_{B_{\mathcal{C}}(x_0,r)} |(\boldsymbol{\Theta}_s^{\rho\tau})^*1(x)|^2 \,d\omega(x)\frac{ds}{s} \le C\|b\|_{\operatorname{Lip}_d}^2\omega(B_{\mathcal{C}}(x_0,r)).\] Consequently the constants \(\mathfrak K\), \(\mathfrak C\), and \(\mathfrak C^*\) in the finite-dimensional continuous \(T1\) criterion are all bounded by \(C\|b\|_{\operatorname{Lip}_d}\), uniformly in the chamber indices. None of these bounds depends on the finite scale endpoints \(a\) and \(B\).

Corollary 3 therefore applies to the family \(\boldsymbol{\Theta}_s\). It gives weak convergence of \[\mathbb{S}_{a,B}F=\int_a^B\boldsymbol{\Theta}_sF\,\frac{ds}{s}\] on \(L^2(\mathcal{C},d\omega;\ell^2_{|G|})\) as \(a\downarrow0\) and \(B\uparrow\infty\). The norm estimate is \[\|\mathbb{S} F\|_{L^2(\mathcal{C};\ell^2_{|G|})} \le C\|b\|_{\operatorname{Lip}_d} \|F\|_{L^2(\mathcal{C};\ell^2_{|G|})},\] where the constant may depend on the fixed finite group \(G\), but not on \(b\), the truncation parameters, or \(F\). ◻

Proof. For finite truncations the heat parameter and scale parameter are related by \(t=s^2\). Since \(dt/\sqrt t=2\,ds/s\), and the constant \(2\) is absorbed in the normalization constant, we have \[C_{\varepsilon,R}^{ij,b} =c\int_{\sqrt\varepsilon}^{\sqrt R}\Theta_s\,\frac{ds}{s}.\] Let \(f,g\in L^2(\mathbb{R}^N,d\omega)\) and put \(F=Uf\), \(G=Ug\). Because \(U\) is unitary on \(L^2\), \[\begin{align} \langle C_{\varepsilon,R}^{ij,b}f,g\rangle_{L^2(\mathbb{R}^N,d\omega)} &=c\left\langle \int_{\sqrt\varepsilon}^{\sqrt R}\boldsymbol{\Theta}_sF\,\frac{ds}{s},G \right\rangle_{L^2(\mathcal{C};\ell^2_{|G|})}. \end{align}\] Set \(a=\sqrt\varepsilon, B=\sqrt R.\) As \(\varepsilon\downarrow0\) and \(R\uparrow\infty\), we have \(a\downarrow0\) and \(B\uparrow\infty\). Proposition 22 therefore gives the limit \[\left\langle \int_a^B\boldsymbol{\Theta}_sF\,\frac{ds}{s},G \right\rangle \longrightarrow \langle \mathbb{S} F,G\rangle.\] Consequently the full-space bilinear forms \(\langle C_{\varepsilon,R}^{ij,b}f,g\rangle\) converge to \(c\langle \mathbb{S} Uf,Ug\rangle.\)

Absorbing the fixed scalar \(c\) into the definition of the limiting operator, we define \[[M_b,T_i\mathcal{R}_j]f:=U^{-1}\mathbb{S} Uf.\] Then \[\begin{align} |\langle [M_b,T_i\mathcal{R}_j]f,g\rangle_{L^2(\mathbb{R}^N,d\omega)}| &\le \|\mathbb{S} Uf\|_{L^2(\mathcal{C},d\omega;\ell^2_{|G|})} \|Ug\|_{L^2(\mathcal{C},d\omega;\ell^2_{|G|})} \\ &\le C\|b\|_{\operatorname{Lip}_d} \|Uf\|_{L^2(\mathcal{C},d\omega;\ell^2_{|G|})} \|Ug\|_{L^2(\mathcal{C},d\omega;\ell^2_{|G|})} \\ &= C\|b\|_{\operatorname{Lip}_d} \|f\|_{L^2(\mathbb{R}^N,d\omega)} \|g\|_{L^2(\mathbb{R}^N,d\omega)}. \end{align}\] Taking the supremum over \(g\) with \(\|g\|_{L^2(\mathbb{R}^N,d\omega)}=1\) gives the asserted \(L^2(\mathbb{R}^N,d\omega)\)-operator norm. The preceding convergence of bilinear forms is exactly weak convergence of the regularized operators. ◻

Proof. Let \(\delta_0= \operatorname{dist}_d(\operatorname{supp}g,\operatorname{supp}f)>0.\) For \(0<\varepsilon<R<\infty\), the truncated bilinear form is \[\label{eq:full-space-truncated-pairing-detailed} \begin{align} \langle C_{\varepsilon,R}^{ij,b}f,g\rangle &= c\iint_{\mathbb{R}^N\times\mathbb{R}^N} (b(x)-b(y)) \left(\int_\varepsilon^R A_t^{ij}(x,y)\frac{dt}{\sqrt t}\right) f(y)\overline{g(x)}\,d\omega(y)d\omega(x). \end{align}\tag{53}\] Indeed, the supports of \(f\) and \(g\) are compact, and the finite interval \([\varepsilon,R]\) removes both heat-parameter endpoints. If the supports are contained in a fixed ball \(B(0,M)\), then for \(t\in[\varepsilon,R]\) the estimate in Proposition 7 gives \[|A_t^{ij}(x,y)| \le C_{\varepsilon,R,M}\] for \(x,y\in B(0,M)\). Since \(d(x,y)\le \|x-y\|\), the assumption \(b\in\operatorname{Lip}_d\) implies that \(b\) is Euclidean Lipschitz. In particular, \(b\) is bounded on \(B(0,M)\), and hence \[|b(x)-b(y)|\,|A_t^{ij}(x,y)|\,|f(y)|\,|g(x)|\] is integrable on \(\operatorname{supp}g\times\operatorname{supp}f\times[\varepsilon,R]\). This justifies Fubini and gives 53 .

On \(\operatorname{supp}g\times\operatorname{supp}f\), one has \(d(x,y)\ge\delta_0\). Lemma 19 gives \[|b(x)-b(y)|\int_0^\infty |A_t^{ij}(x,y)|\frac{dt}{\sqrt t} \le C\|b\|_{\operatorname{Lip}_d}\frac{1}{V(x,y,d(x,y))}.\] We check that this right-hand side is integrable over the product of the supports. Since the supports are compact, there is \(M>0\) such that both are contained in \(B(0,M)\). If \(x\in \operatorname{supp}g\), then \[V(x,y,d(x,y))\ge \omega(B(x,\delta_0)).\] Using the volume formula, \[\omega(B(x,\delta_0)) \approx \delta_0^N \prod_{\alpha\in R} \big(|\langle x,\alpha\rangle|+\delta_0\big)^{\kappa(\alpha)}.\] The last expression is bounded below by a positive constant depending on \(M\), \(\delta_0\), and the Dunkl data. Hence \[\frac{|f(y)||g(x)|}{V(x,y,d(x,y))}\] is integrable on the product of the supports. Dominated convergence in 53 yields \[\lim_{\varepsilon\to0,\;R\to\infty} \langle C_{\varepsilon,R}^{ij,b}f,g\rangle = \iint K_b^{ij}(x,y)f(y)\overline{g(x)}\,d\omega(y)d\omega(x).\] On the other hand, Theorem 24 says that the same truncations converge weakly on \(L^2(\mathbb{R}^N,d\omega)\) to \(T_b^{ij}\). Since \(f,g\in L^2(\mathbb{R}^N,d\omega)\), the left-hand side also tends to \(\langle T_b^{ij}f,g\rangle_{L^2(\mathbb{R}^N,d\omega)}\). This proves ?? . ◻

Proof. Set \(f=U^{-1}F, \; g=U^{-1}G.\) The supports of \(f\) and \(g\) are compact, because they are finite unions of reflections of compact coordinate supports in \(\mathcal{C}\).

We first check the separation in the full orbit metric. Let \(z\in\operatorname{supp}g\) and \(w\in\operatorname{supp}f\). Up to the \(\omega\)-null chamber walls, there exist indices \(\rho,\tau\) and chamber points \(x\in\operatorname{supp}G_\rho, \; y\in\operatorname{supp}F_\tau\) such that \[z=\sigma_\rho x, \qquad w=\sigma_\tau y.\] By the chamber identity, \[d(z,w) = d(\sigma_\rho x,\sigma_\tau y) = \|x-y\| \ge \delta_{\mathcal{C}}(F,G).\] Consequently \(\operatorname{supp}g\) and \(\operatorname{supp}f\) are separated with respect to the full orbit metric. Hence Proposition 27 applies and gives \[\langle T_b^{ij}f,g\rangle_{L^2(\mathbb{R}^N,d\omega)} = \iint_{\mathbb{R}^N\times\mathbb{R}^N} K_b^{ij}(z,w)f(w)\overline{g(z)}\,d\omega(w)d\omega(z).\]

We now decompose both full-space variables into reflected chambers. Since the chamber walls are \(\omega\)-null, \(\mathbb{R}^N\) is the disjoint union, modulo null sets, of the chambers \(\sigma_\rho\mathcal{C}\), \(1\le\rho\le |G|\). Therefore \[\begin{align} &\iint_{\mathbb{R}^N\times\mathbb{R}^N} K_b^{ij}(z,w)f(w)\overline{g(z)}\,d\omega(w)d\omega(z) = \sum_{\rho,\tau=1}^{|G|} \iint_{\sigma_\rho\mathcal{C}\times\sigma_\tau\mathcal{C}} K_b^{ij}(z,w)f(w)\overline{g(z)}\,d\omega(w)d\omega(z). \end{align}\] For the \((\rho,\tau)\)-term, we use the measure-preserving bijections \[\sigma_\rho:\mathcal{C}\to\sigma_\rho\mathcal{C}, \qquad \sigma_\tau:\mathcal{C}\to\sigma_\tau\mathcal{C}.\] Thus the \((\rho,\tau)\)-term becomes \[\begin{align} \iint_{\mathcal{C}\times\mathcal{C}} K_b^{ij}(\sigma_\rho x,\sigma_\tau y) f(\sigma_\tau y)\overline{g(\sigma_\rho x)} \,d\omega(y)d\omega(x). \end{align}\] By the definition of \(U\), \[f(\sigma_\tau y)=F_\tau(y), \qquad g(\sigma_\rho x)=G_\rho(x), \qquad x,y\in\mathcal{C}.\] Therefore \[\begin{align} \langle T_b^{ij}f,g\rangle_{L^2(\mathbb{R}^N,d\omega)} &= \sum_{\rho,\tau=1}^{|G|} \iint_{\mathcal{C}\times\mathcal{C}} K_b^{ij}(\sigma_\rho x,\sigma_\tau y) F_\tau(y)\overline{G_\rho(x)} \,d\omega(y)d\omega(x) \\ &= \sum_{\rho,\tau=1}^{|G|} \iint_{\mathcal{C}\times\mathcal{C}} \mathbb{K}_b^{ij,\rho\tau}(x,y) F_\tau(y)\overline{G_\rho(x)} \,d\omega(y)d\omega(x), \end{align}\] where \(\mathbb{K}_b^{ij,\rho\tau}(x,y) = K_b^{ij}(\sigma_\rho x,\sigma_\tau y).\) Finally, since \(\mathbb{T}_b^{ij}=UT_b^{ij}U^{-1},\) we have \[\langle \mathbb{T}_b^{ij}F,G\rangle_ {L^2(\mathcal{C},d\omega;\ell^2_{|G|})} = \langle T_b^{ij}f,g\rangle_{L^2(\mathbb{R}^N,d\omega)}.\] Combining the preceding identities gives 56 . ◻

Proof. Fix \(x,y\in\mathcal{C}\), \(x\ne y\), and set \(z=\sigma_\rho x, \; w=\sigma_\tau y, \; r=\|x-y\|.\) Here \(x,y\) are chamber points, whereas \(z,w\) are the corresponding full-space reflected points. The chamber identity gives \[d(z,w) = d(\sigma_\rho x,\sigma_\tau y) = \|x-y\| = r.\] The full-space estimate 30 yields \[|K_b^{ij}(z,w)| \le C\|b\|_{\operatorname{Lip}_d}\frac{1}{V(z,w,r)}.\] By chamber volume comparability, \(V(z,w,r) = V(\sigma_\rho x,\sigma_\tau y,r) \approx V_{\mathcal{C}}(x,y,r),\) with constants independent of \(\rho\) and \(\tau\). Since \[\mathbb{K}_b^{ij,\rho\tau}(x,y) = K_b^{ij}(\sigma_\rho x,\sigma_\tau y) = K_b^{ij}(z,w),\] this proves the size estimate 57 .

We next prove the first-variable regularity. Suppose \(x,x',y\in\mathcal{C}\) and \(\|x-x'\|\le r/4\). Put \(z'=\sigma_\rho x'.\) Since \(\sigma_\rho\) is a Euclidean isometry, \[\|z-z'\| = \|\sigma_\rho x-\sigma_\rho x'\| = \|x-x'\| \le \frac{r}{4} = \frac{1}{4} d(z,w).\] Applying the full-space regularity estimate 31 , we get \[\begin{align} |\mathbb{K}_b^{ij,\rho\tau}(x,y) -\mathbb{K}_b^{ij,\rho\tau}(x',y)| = |K_b^{ij}(z,w)-K_b^{ij}(z',w)| \le C\|b\|_{\operatorname{Lip}_d} \frac{\|z-z'\|}{d(z,w)} \frac{1}{V(z,w,d(z,w))}. \end{align}\] Using \[\|z-z'\|=\|x-x'\|, \qquad d(z,w)=\|x-y\|, \qquad V(z,w,d(z,w))\approx V_{\mathcal{C}}(x,y,\|x-y\|),\] we obtain 58 .

Finally, suppose \(x,y,y'\in\mathcal{C}\) and \(\|y-y'\|\le r/4\). Put \(w'=\sigma_\tau y'.\) Then \[\|w-w'\| = \|\sigma_\tau y-\sigma_\tau y'\| = \|y-y'\| \le \frac{r}{4} = \frac{1}{4} d(z,w).\] Applying the full-space second-variable regularity estimate 32 , we obtain \[\begin{align} |\mathbb{K}_b^{ij,\rho\tau}(x,y) -\mathbb{K}_b^{ij,\rho\tau}(x,y')| = |K_b^{ij}(z,w)-K_b^{ij}(z,w')| \le C\|b\|_{\operatorname{Lip}_d} \frac{\|w-w'\|}{d(z,w)} \frac{1}{V(z,w,d(z,w))}. \end{align}\] Using again \[\|w-w'\|=\|y-y'\|, \qquad d(z,w)=\|x-y\|, \qquad V(z,w,d(z,w))\approx V_{\mathcal{C}}(x,y,\|x-y\|),\] we obtain 59 . Thus the second-variable chamber regularity is normalized at the original chamber pair \((x,y)\), exactly as in the scalar Calderón–Zygmund convention. ◻

Proof. Apply Lemma 35 to the vector-valued functions \(F=J_\tau u, \; G=J_\rho v.\) The separation of \(\operatorname{supp}u\) and \(\operatorname{supp}v\) is exactly the separation required in that lemma, because all other coordinates of \(F\) and \(G\) vanish. In the finite sum 56 , only the pair of indices \((\rho,\tau)\) contributes, and the left-hand side is \[\langle \mathbb{T}_b^{ij}J_\tau u,J_\rho v\rangle_ {L^2(\mathcal{C},d\omega;\ell^2_{|G|})} = \langle P_\rho\mathbb{T}_b^{ij}J_\tau u,v\rangle_ {L^2(\mathcal{C},d\omega)} = \langle T^{\rho\tau}u,v\rangle_{L^2(\mathcal{C},d\omega)}.\] This gives the desired scalar associated-kernel identity. ◻

Proof. Let \(J_b\) denote the injection of a scalar function into the \(b\)-th coordinate of \(\ell_M^2\), and let \(P_a\) denote projection onto the \(a\)-th coordinate. Then, on \(L^2\), \[T^{ab}=P_a\mathbb{T} J_b.\] Since both \(P_a\) and \(J_b\) have norm one on \(L^2\), we have \(\|T^{ab}\|_{2\to2}\le A.\) By assumption, the kernel associated with \(T^{ab}\) has standard Calderón–Zygmund constants bounded by \(B\). The scalar Calderón–Zygmund theorem gives a bounded \(L^p\)-extension, denoted by \(\widetilde{T}^{ab}_p\), such that \[\label{eq:scalar-entry-Lp-bound} \|\widetilde{T}^{ab}_pf\|_{L^p(X)} \le C_p(A+B)\|f\|_{L^p(X)},\tag{64}\] with a constant independent of \(a,b\). On \(L^p(X)\cap L^2(X)\), this extension agrees with the original \(L^2\)-operator \(T^{ab}\).

For \(F=(F_1,\ldots,F_M)\in L^p(X;\ell_M^p)\), define \[(\widetilde{\mathbb{T}}_pF)_a :=\sum_{b=1}^M \widetilde{T}^{ab}_pF_b, \qquad a=1,\ldots,M.\] This is a finite sum of scalar \(L^p\)-operators. Pointwise in the fibre, \[\begin{align} |\widetilde{\mathbb{T}}_pF(x)|_{\ell_M^p} &=\left(\sum_{a=1}^M\left|\sum_{b=1}^M \widetilde{T}^{ab}_pF_b(x)\right|^p\right)^{1/p} \le \sum_{a=1}^M\sum_{b=1}^M |\widetilde{T}^{ab}_pF_b(x)|. \end{align}\] Taking the \(L^p(X)\)-norm and using Minkowski’s inequality gives \[\begin{align} \|\widetilde{\mathbb{T}}_pF\|_{L^p(X;\ell_M^p)} &\le \sum_{a,b=1}^M \|\widetilde{T}^{ab}_pF_b\|_{L^p(X)} \le C_p(A+B)\sum_{a,b=1}^M \|F_b\|_{L^p(X)} = C_pM(A+B)\sum_{b=1}^M \|F_b\|_{L^p(X)}. \end{align}\] Since \(M<\infty\), \[\sum_{b=1}^M\|F_b\|_{L^p(X)} \le M^{1-1/p}\left(\sum_{b=1}^M\|F_b\|_{L^p(X)}^p\right)^{1/p} =M^{1-1/p}\|F\|_{L^p(X;\ell_M^p)}.\] This proves the asserted \(L^p\)-bound for \(\widetilde{\mathbb{T}}_p\).

Finally, if \(F\in L^p(X;\ell_M^p)\cap L^2(X;\ell_M^2)\), then each \(F_b\in L^p(X)\cap L^2(X)\), and the scalar extensions agree with the original entries. Hence \[(\widetilde{\mathbb{T}}_pF)_a =\sum_{b=1}^M T^{ab}F_b =(\mathbb{T}F)_a\] for every \(a\). Thus \(\widetilde{\mathbb{T}}_p\) is the bounded \(L^p\)-extension of the original finite-dimensional operator. Uniqueness follows from the density of \(L^p\cap L^2\) in \(L^p\). ◻

Proof. For \(1\le\rho,\tau\le |G|\), let \(P_\rho\) be the projection onto the \(\rho\)-th coordinate and let \(J_\tau\) be the injection into the \(\tau\)-th coordinate. Define \[T^{\rho\tau}:=P_\rho\mathbb{T}_b^{ij}J_\tau.\] The lifted operator is bounded on \(L^2(\mathcal{C};\ell^2_{|G|})\) by Proposition 22; hence \[\|T^{\rho\tau}\|_{L^2(\mathcal{C})\to L^2(\mathcal{C})} \le C\|b\|_{\operatorname{Lip}_d}.\] By Corollary 4, each entry \(T^{\rho\tau}\) is associated, in the separated-support sense, with \(\mathbb{K}_b^{ij,\rho\tau}\). By Lemma 36, these scalar kernels have standard Calderón–Zygmund constants bounded by \(C\|b\|_{\operatorname{Lip}_d}\), uniformly in \(\rho\) and \(\tau\). Thus the hypotheses of Theorem 32 are satisfied with \[A+B\le C\|b\|_{\operatorname{Lip}_d}.\] Applying that finite-dimensional entrywise reduction gives \[\|\mathbb{T}_b^{ij}F\|_{L^p(\mathcal{C};\ell^p_{|G|})} \le C_p\|b\|_{\operatorname{Lip}_d} \|F\|_{L^p(\mathcal{C};\ell^p_{|G|})},\] where the dependence on \(|G|\) is absorbed into the structural constant. This is exactly ?? . ◻

Proof of Theorem 1. We first take \(f\in L^p(\mathbb{R}^N,d\omega)\cap L^2(\mathbb{R}^N,d\omega)\) and set \(F=Uf\). The map \(U\) is an isometry on \(L^p\), so \[\|F\|_{L^p(\mathcal{C},d\omega;\ell^p_{|G|})} = \|f\|_{L^p(\mathbb{R}^N,d\omega)}.\] For bounded compactly supported functions, the finite regularized operators satisfy \[U C_{\varepsilon,R}^{ij,b} U^{-1} = c\int_{\sqrt\varepsilon}^{\sqrt R}\boldsymbol{\Theta}_s\,\frac{ds}{s}.\] This identity passes to the weak \(L^2\)-limits in the following concrete sense. Let \(h,k\in L^2(\mathbb{R}^N,d\omega)\). Approximate them in \(L^2(\mathbb{R}^N,d\omega)\) by bounded compactly supported functions \(h_n,k_n\). The finite-truncation identity holds for \((h_n,k_n)\), and the uniform \(L^2\)-bounds from Theorem 24 and Proposition 22 allow \(n\to\infty\) at fixed truncation parameters and then in the weak limit. Consequently \[U(T_b^{ij}h)=\mathbb{T}_b^{ij}(Uh), \qquad h\in L^2(\mathbb{R}^N,d\omega).\] Applying this to the present \(f\in L^p(\mathbb{R}^N,d\omega)\cap L^2(\mathbb{R}^N,d\omega)\) gives \(U(T_b^{ij}f)=\mathbb{T}_b^{ij}(Uf).\) Therefore Proposition 35 yields \[\begin{align} \|T_b^{ij}f\|_{L^p(\mathbb{R}^N,d\omega)} &= \|\mathbb{T}_b^{ij}(Uf)\|_{L^p(\mathcal{C},d\omega;\ell^p_{|G|})} \le C_p\|b\|_{\operatorname{Lip}_d}\|Uf\|_{L^p(\mathcal{C},d\omega;\ell^p_{|G|})} = C_p\|b\|_{\operatorname{Lip}_d}\|f\|_{L^p(\mathbb{R}^N,d\omega)}. \end{align}\] Finally, \(L^p(\mathbb{R}^N,d\omega)\cap L^2(\mathbb{R}^N,d\omega)\) is dense in \(L^p(\mathbb{R}^N,d\omega)\), as recalled in Remark 36. The preceding estimate therefore extends \(T_b^{ij}\) uniquely to a bounded operator on all of \(L^p(\mathbb{R}^N,d\omega)\). This extension is the \(L^p\)-closure of the same regularized-kernel weak limit. ◻