$$\kappa _{R}^{z}=\zeta \kappa _{R}^{z}+(1-\zeta )\kappa _{R}^{z}:=\kappa
    _{R}^{z,0}+\kappa _{R}^{z,\infty },  \label{decomp}$$ where $\zeta \in C^{\infty }(K\backslash G/K)$ is a cut-off function such that $$\zeta (x)=%
    \begin{cases}
    1, & \text{if }|x|\leq 1/2, \\
    0, & \text{if }|x|\geq 1.%
    \end{cases}
    \label{zita}$$ Denote by ${S}_{R}^{z,0}$ (resp. ${S}_{R}^{z,\infty }$) the convolution operator on $X$ with kernel $\kappa _{R}^{z,0}$ (resp. $\kappa
    _{R}^{z,\infty })$.

The local part
--------------

We shall prove the following proposition.

\[T0X\] Assume that $\operatorname{Re}z >n/2$. Then the operator ${S}_{R}^{z,0}$ is bounded on $L^{p}(X)$, $1\leq p\leq \infty$, and $\|S_R^{z,
0}\|_{p\rightarrow p}\leq c(z)$, for some constant $c(z)>0$.

The proof is lengthy and it will be given in several steps. First, we shall express the kernel $\kappa _{R}^{z}$ in terms of the heat kernel $%
p_{t} $ of $X$. Then, we shall use the heat kernel estimates (\[ostellari\]) to prove that $\kappa _{R}^{z}$ is integrable in the unit ball $B(0,1)$ of $X$. This implies that $S_{R}^{z,0}$ is bounded on $L^{\infty }(X)$. We then prove that $S_{R}^{z,0}$ is bounded on $L^2(X)$, and an interpolation argument between $L^{\infty}(X)$ and $L^2(X)$ allows us to conclude.

To express the kernel $\kappa _{R}^{z}$ in terms of $p_{t}$, we follow [ALEXOLO]{} and we write $$s_{R}^{z}(\lambda )=s_{R}^{z}(\Vert \lambda \Vert ) =\left( 1-\frac{\Vert
\lambda 