$\mathcal{H}(\zeta )$, belongs in $S(\mathfrak{a^{\ast }}%
)^{W}\subset L^{1}(\mathfrak{a^{\ast }})$, (see Section 2). So, $$\Vert \mathcal{H}(\zeta )\Vert _{L^{1}(\mathfrak{a^{\ast }})}\leq c(\zeta
)<\infty .$$From (\[tj02\]), (\[Tj,r\]) and (\[alexo7\]) it follows that $$\begin{aligned}
\Vert T_{j,r}^{0}\Vert_{L^{2}\rightarrow L^{2}}&\leq c(\zeta)\Vert
s_{j,r}\Vert _{L^{\infty }(\mathfrak{a^{\ast }})}\leq c(\zeta) \| h_{j,r}(\sqrt{\cdot })e^{-1/r^{2}(\cdot)}\|_{L^{\infty }(\mathfrak{a^{\ast }})} \notag \\&\leq c(\zeta
) \| h_{j,r}(\sqrt{\cdot })\|_{L^{\infty }(\mathfrak{a^{\ast }})}\leq
c(\zeta )2^{-j\operatorname{Re}z}.  \label{Tjr0L2}\end{aligned}$$ Further, by (\[Tjr0L2\]) and the fact that $%
S_{R}^{z,0}=\sum_{j\geq 0}T_{j,r}^{0}$, it follows that $$\Vert S_{R}^{z,0}\Vert _{L^{2}\rightarrow L^{2}} \leq \sum_{j\geq 0}\Vert
T_{j,r}^{0}\Vert _{L^{2}\rightarrow L^{2}} \leq c\sum_{j\geq 0}2^{-j\operatorname{Re}%
z}\leq c<\infty .  \label{t2}$$

*End of the proof of Proposition \[T0X\]*: Since $\kappa
_{R}^{z}=\sum\limits_{j\geq 0}\kappa _{j,r}$, by Lemma \[L1ball\], we have $$\Vert \kappa _{R}^{z,0}\Vert _{L^{1}(X)}=\Vert \zeta \kappa _{R}^{z}\Vert
_{L^{1}(X)}\leq c\Vert \kappa _{R}^{z}\Vert _{L^{1}(B(0,1))}<c.$$This implies that $$\Vert S_{R}^{z,0}\Vert _{L^{\infty }\rightarrow L^{\infty }}\leq c(z).
\label{tapeiro}$$

By interpolation and duality, it follows from (\[tapeiro\]) and (\[t2\]), that for all $p\in \lbrack 1,\infty ]$, $\Vert S_{R}^{z,0}\Vert _{p\rightarrow p}\leq
c(z)$, with $\operatorname{Re}z>n/2$.

The part at infinity
--------------------

For the part at infinity $S_R^{z, \infty}$ of the operator, we proceed as in [@LOMAanna] to obtain estimates of its kernel $\kappa_{R}^{z, \infty}$. Let $l=\operatorname{rank}(X)$.

To begin with, recall that $\kappa _{R}^{z}=\mathcal{H}^{-1}s_{R}^{z}$. Recall also the following result from [@LOMAanna p.650], based on the Abel transform conservation property.

