$\operatorname{Re}z>0$. Then, $S_*^{z}$ is continuous on $L^2(X)$.

Recall the following decomposition of the kernel $\kappa_{R}^z$ of the operator $S_R^z$: $$\label{decomp1}
\kappa_{R}^{z}=\zeta \kappa_{R}^{z}+(1-\zeta
)\kappa_{R}^z:=\kappa_{R}^{z,0}+\kappa_{R}^{z,\infty },$$where $\zeta \in C^{\infty }(K\backslash G/K)$ is a cut-off function such that $$\zeta (x)= 
\begin{cases}
1, & \text{if }|x|\leq 1/2, \\ 
0, & \text{if }|x|\geq 1.%
\end{cases}
\label{zita1}$$ Denote by ${S}_R^{z,0}$ (resp. ${S}_R^{z,\infty}$) the convolution operators on $X$ with kernel $\kappa _R^{z,0}$ (resp. $\kappa_R^{z,\infty})$. Then, $$\
S_*^zf\leq\sup\limits_{R\geq \|\rho\|^2}|S_R^{z,0}f|+\sup\limits_{R\geq \|\rho\|^2}|S_R^{z, \infty}f|.$$ The following holds true for the part at infinity $S_{\ast}^{z, \infty}$ of the operator $S_{\ast}^{z}$.

\[propinf\] \[TinfXR\] Let $ \operatorname{Re}z \geq n-\frac{1}{2}$. Then, for every $q>2$ and $p\in[1, q^{\prime}]$, $S_*^{z,\infty}$ is continuous from $L^p(X)$ to $L^r(X)$ for every $r\in[qp^{\prime}/(p^{\prime} - q), \infty]$.

The proof relies on the uniform kernel estimates for $\kappa_{R}^{z, \infty}$ implied by Lemmata 9 and 10. It is similar to the proof of Proposition 11, thus omitted.

We shall now prove the following result concerning the local part $S_{\ast}^{z, 0}$ of the Riesz means maximal operator.

\[proplocal\] Let $\operatorname{Re}z\geq n-\frac{1}{2}$. Then, $S_*^{z,0}$ is continuous on $L^p(X)$, for every $p \in (1, \infty)$, and it maps $L^1(X)$ continuously into $L^{1,w}(X)$.

Denote by $e^{t\Delta}, \;t>0$, the heat operator on $X$. Then, $e^{t\Delta}=\ast p_t$, where $p_t$ is 