the heat kernel on $X$. Recall that $p_t$ is given as the inverse spherical Fourier transform of $$w_t(\lambda)=e^{-t(\|\lambda\|^2+\|\rho\|^2)}, \; \lambda\in \mathfrak{\ a^{\ast}}.$$Consider the radial multiplier $$\label{Mdef1}
M(R^{-1}\lambda):=s_R^z(\lambda)-w_{R^{-1}}(\lambda), \; R\geq \|\rho\|^2.$$

Denote by $K_R(x)$ the kernel of the operator $M(-R^{-1}\Delta)$ and set $K_R^0(x):=\zeta(x)K_R(x)$. Similarly, set $s_R^{z,0}=\mathcal{H}(\zeta \kappa_{R}^{z})=\mathcal{H}( \kappa_{R}^{z,0})$ and $w_{R^{-1}}^0=\mathcal{H}(\zeta p_{R^{-1}})=\mathcal{H}( p^0_{R^{-1}})$. Then, using (\[Mdef1\]), we have that $$\label{mult1}
\mathcal{H}(\kappa_{R}^0):=M^0(-R^{-1}\cdot)=s_R^{z,0}-w_{R^{-1}}^0,$$ From (\[mult1\]) we have that $$\label{1,0}
S_*^{z,0}f=\sup_{R\geq\|\rho\|^2}|s_R^{z,0}(-\Delta)f|\leq \sup_{R\geq\|\rho\|^2}|M^0(-R^{-1} \Delta)f|+\sup_{R\geq\|\rho\|^2}|f \ast p^0_{R^{-1}} |.$$

Consider the operator $(-\Delta)^{i\gamma}$, $\gamma \in \mathbb{R}$, which in the spherical Fourier transform variables is given by $$\
\mathcal{H}((-\Delta)^{i\gamma}f)=(\|\lambda\|^2+\|\rho\|^2)^{i\gamma}\mathcal{H}(f), \; \lambda\in \mathfrak{\ a^{\ast}}.$$ Denote by $$\kappa^{\gamma}=\mathcal{H}^{-1}((\|\lambda\|^2+|\rho\|^2)^{i\gamma})$$ the kernel of $(-\Delta)^{i\gamma}$. As in [@ALEXOLO; @GIUMA], using the Mellin transform $\mathcal{M}(\gamma)$ of the radial function $M(\lambda)$, one can express the operator $M(-R^{-1}\Delta)$ as follows: $$\label{Mellin1}
M(-R^{-1}\Delta)=\int_{-\infty}^{+\infty}\mathcal{M}(\gamma)R^{-i\gamma}(-\Delta)^{i\gamma}d\gamma,$$ where $$\label{calM}
|\mathcal{M}(\gamma)|\leq c(1+|\gamma|)^{-(\operatorname{Re}z+1)},$$ [@GIUMA]. Using (\[Mellin1\]), the kernel $K_R$ of $M(-R^{-1}\Delta)$ is given by $$K_R=\int_{-\infty}^{+\infty}\mathcal{M}(\gamma)R^{-i\gamma} \kappa^{\gamma}d\gamma,$$ and thus $$\begin{aligned}
    K_R^0(x)&=\zeta(x) K_R(x)=\int_{-\infty}^{+\infty}\mathcal{M}(\gamma)R^{-i\gamma} \zeta(x)\kappa^{\gamma}(x)d\gamma\\
    &=\int_{-\infty}^{+\infty}\mathcal{M}(\gamma)R^{-i\gamma} \kappa^{\gamma,0}(x)d\gamma .
    \end{aligned}$$ It follows that $$\
    M^0(-R^{-1}\Delta)=\int_{-\infty}^{+\infty}\mathcal{M}(\gamma)R^{-i\gamma} (-\Delta)^{i\gamma,0}d\gamma .$$ Hence, $$\label{Mellin}
\sup_{R>\|\rho\|^2} |M^0(-R^{-1}\Delta)f|\leq \int_{-\infty}^{+\infty}|\mathcal{M}(\gamma)||(-\Delta)^{i\gamma,0}f|d\gamma$$

\[Anker\] The operator $(-\Delta)^{i\gamma,0}$ is bounded on $L^p$, $p\in(1, \infty)$, with $$\label{Lpnorm}
    \|(-\Delta)^{i\gamma,0}\|_{L^p\rightarrow L^p}\leq c_p(1+|\gamma|)^{[n/2]+1}.$$ Moreover, the operator $(-\Delta)^{i\gamma,0}$ is also $L^1\rightarrow L^{1,w}$ bounded, with $$\label{L1wnorm}
    \|(-\Delta)^{i\gamma,0}\|_{L^1\rightarrow L^{1,w}}\leq c(1+|\gamma|)^{[n/2]+1}.$$

To prove the lemma, we shall proceed 