as in [@AN]. More precisely, by using a smooth, radial partition of unity (and thus invariant by the Weyl group), we decompose the multiplier $m^{\gamma}(\lambda)=(\|\lambda\|^2+\|\rho\|^2)^{i\gamma}$ as follows $$m^{\gamma}(\lambda)=\sum_{k=0}^{+\infty}m^{\gamma}_k(2^{-k}\lambda),$$ where $\operatorname{supp}m^{\gamma}_0 \subset \{\|\lambda\|\leq 2 \}$ and $\operatorname{supp}m^{\gamma}_k \subset \{1/2\leq \|\lambda\|\leq 2 \}$ for $k\geq 1$. Then, for every $p\in (1, +\infty)$, we have $$\label{Lpimag}
    \|(-\Delta)^{i\gamma,0}\|_{p\rightarrow p}\leq c_p \sup_{k\geq 0}\|m^{\gamma}_k\|_{H_2^{\sigma/2}},$$ with $\sigma >n$ and $H_2^{\sigma/2}$ the usual Sobolev space, [@AN Corollary 17, ii]. Note that the same upper bound also holds for the $L^1\rightarrow L^{1, w}$ norm of $(-\Delta)^{i\gamma,0}$, [@AN]. A straightforward computation yields $$\label{mknorm}
    \|m^{\gamma}_k\|_{H_2^{\sigma/2}}\leq c(1+|\gamma|)^{\sigma /2},$$ for $\sigma /2$ an integer, and Lemma \[Anker\] follows from (\[Lpimag\]).

We shall complete the proof for the $L^p$ boundedness of $S_{*}^{z, 0}$, $p\in(1, \infty)$; the $L^1\rightarrow L^{1,w}$ result is similar, thus omitted. Recall that (\[1,0\]) states that $$S_{\ast}^{z,0}f\leq \sup_{R\geq \|\rho\|^2} |M^0(-R^{-1}\Delta)f|+\sup_{R\geq \|\rho\|^2}|f\ast p^0_{R^{-1}}|.$$ Note that since $p_t(x)\geq 0$, for every $x\in X$, we have $p_t^0(x)\leq p_t(x)$. Thus, $$\label{L1w}
    |(f\ast p_t^0)(x)|\leq (|f|\ast p_t)(x).$$ Also, it is known (see for example [@AN2 Corollary 3.2]) that the heat maximal operator $\sup\limits_{t>0}|e^{t\Delta}f|$ is $L^p$-bounded and also $L^1\rightarrow L^{1,w}$ bounded. This implies that the operator $\sup\limits_{R\geq\|\rho\|^2}|\ast p_{R^{-1}}^0|$ is also $L^p$-bounded and $L^1\rightarrow L^{1,w}$ bounded. Thus, from (\[1,0\]), it 