$$\begin{aligned}
I_{2}& \leq c2^{p/2}2^{-j\operatorname{Re}z}(r^{-2})^{-n/4}e^{-2^{p-1}/2Dr^{-2}} \\
& \leq c2^{-j\operatorname{Re}z}2^{p/2}r^{n/2}e^{-2^{p}r^{2}/4D} \\
& \leq c2^{-j\operatorname{Re}z}2^{(p+i)n/2}e^{-D_1 2^{p+i}}.\end{aligned}$$Using the elementary estimate $$e^{-D_1x}x^{n/2}\leq c_kx^{-k},\text{ for all }x>1,\;k\in \mathbb{N},$$we obtain $$I_{2}\leq 2^{-j\operatorname{Re}z}2^{-k(p+i)}.  \label{I1}$$Also, from (\[L2heat\]) we have that $$I_{1}\leq c2^{p/2}(r^{-2})^{-n/4}\int_{|t|\geq 2^{p-1}}|\hat{h}_{j,r}(t)|dt.$$Then, applying (\[hhat\]) for $k>n/2$, we obtain $$\begin{aligned}
I_{1}& \leq c_{n}2^{(p+i)n/2}2^{-pk}r^{-k}2^{(k-\operatorname{Re}z)j}  \notag
\label{I2} \\
& \leq c2^{-(p+i)(k-n/2)}2^{-j(\operatorname{Re}z-n/2)}.\end{aligned}$$Finally, using (\[I1\]) and (\[I2\]), (\[L1norm\]) implies that $$\Vert \kappa _{j,r}\Vert _{L^{1}(A_{p})}\leq c2^{-(p+i)(k-n/2)}2^{-j(\operatorname{Re%
}z-n/2)}.  \label{L1Ap}$$*End of proof of Lemma \[L1ball\].* It follows from (\[B1/r\]) and (\[L1Ap\]) that $$\begin{aligned}
\Vert \kappa _{j,r}\Vert _{L^{1}(B(0,1))}& \leq c2^{-j\operatorname{Re}%
z}+c\sum_{p=-i}^{0}2^{-(p+i)(k-n/2)}2^{-j(\operatorname{Re}z-n/2)} \\
& \leq c2^{-j(\operatorname{Re}z-n/2)}.  \notag\end{aligned}$$So, for $\operatorname{Re}z>n/2$, $$\begin{aligned}
\Vert \kappa _{R}^{z}\Vert _{L^{1}(B(0,1))} &\leq c\sum_{j\geq 0}\Vert
\kappa _{j,r}\Vert _{L^{1}(B(0,1))} \\
&\leq c\sum_{j\geq 0}2^{-j(\operatorname{Re}z-n/2)}\leq c.\end{aligned}$$

$S_{R}^{z,0}$ is bounded on $L^{2}(X)$.

Set $$\kappa _{j,r}^{0}=\zeta \kappa _{j,r},\;T_{j,r}^{0}=\ast \kappa _{j}^{0}%
\text{ and }s_{j,r}^{0}=\mathcal{H}(\kappa _{j,r}^{0}),  \label{cutoff}$$where $\zeta $ is the cut-off function given in (\[zita\]).

By Plancherel theorem and using (\[cutoff\]), we get that $$\begin{aligned}
\Vert T_{j,r}^{0}\Vert _{L^{2}\rightarrow L^{2}}& \leq \Vert
s_{j,r}^{0}\Vert _{L^{\infty }(\mathfrak{a^{\ast }})}=\Vert \mathcal{H}%
(\kappa _{j,r}^{0})\Vert _{L^{\infty }(\mathfrak{a^{\ast }})}  \notag \\
& =\Vert \mathcal{H}(\zeta \kappa _{j,r})\Vert _{L^{\infty }(\mathfrak{%
a^{\ast }})}=\Vert \mathcal{H}(\zeta )\ast \mathcal{H}(\kappa _{j,r})\Vert
_{L^{\infty }(\mathfrak{a^{\ast }})}  \label{tj02} \\
& \leq \Vert \mathcal{H}(\zeta )\Vert _{L^{1}(\mathfrak{a^{\ast }})}\Vert
s_{j,r}\Vert _{L^{\infty }(\mathfrak{a^{\ast }})}.  \notag\end{aligned}$$But $\zeta \in S(K\backslash G/K)$. Therefore, its spherical Fourier transform 