\Vert ^{2}+\Vert \rho \Vert ^{2}}{R}\right) _{+}^{z}.$$ \[Srz\] Set $r=\sqrt{R}$, $\xi =\Vert \lambda \Vert $ and consider the function $$h_{r}^{z}(\lambda )=h_{r}^{z}(\xi ):=\left( 1-\left( \frac{\sqrt{\xi ^{2}+\Vert \rho
\Vert ^{2}}}{r} \right)^2\right) _{+}^{z}e^{(\sqrt{\xi ^{2}+\Vert \rho \Vert ^{2}}/{%
r})^{2}}.  \label{hr,z}$$ Then, from (\[Srz\]) and (\[hr,z\]) we have $$s_r^z(\lambda)=h_R^z(\lambda)e^{-(\|\lambda\|^2+\| \rho \|^{2})/r^2},$$ and thus $$s_{R}^{z}(-\Delta )=h_{r}^{z}(\sqrt{-\Delta })e^{-1/{r^{2}}(-\Delta )}.
\label{heatexpr}$$

Next, we recall the construction of the partition of unity of [@ALEXOLO p.213] we shall use for the splitting of the operator $s_{R}^{z}(-\Delta )$. For that we set $\psi (\xi )=e^{-1/\xi ^{2}}$, $\xi \geq 0$, and $\psi _{1}(\xi )=\psi (\xi )\psi (1-\xi )$. Then $\psi
_{1}\in C^{\infty }\left( [0,1]\right) $. Set also $\phi (\xi )=\psi
_{1}(\xi +\frac{5}{4})$, and $$\phi _{j}(\xi )=\phi (2^{j}(\xi -1)),\text{ }j\in \mathbb{N}.$$Then $\phi _{j}(\xi )$ is a $C^{\infty }$ function with support in $%
I_{j}=[1-5/2^{j+2},1-1/2^{j+2}]$. The functions $$\chi _{j}(\xi )=\frac{\phi _{j}(\xi )}{\sum_{i\geq 0}\phi _{i}(\xi )},\text{
}$$form the required partition of unity.

Set $$\chi _{j,r}(\xi )=\chi _{j}\left( (\xi /r)^{2}\right) ,\text{ }$$and$$h_{j,r}(\xi ):=h_{R}^{z}(\xi )\chi _{j,r}(\xi ).$$

Consider the operator $$T_{j,r}:=s_{j,r}(-\Delta )=h_{j,r}(\sqrt{-\Delta })e^{-1/r^{2}(-\Delta )}.
\label{Tj,r}$$Note that by (\[Tj,r\]) and (\[heatexpr\]), $$\sum_{j\in \mathbb{N}}T_{j,r}=\sum_{j\in \mathbb{N}}h_{r,j}(\sqrt{-\Delta })e^{-1/r^{2}(-\Delta
)}=h_{r}(\sqrt{-\Delta })e^{-1/r^{2}(-\Delta
)}=s_{R}^{z}(-\Delta ).  \label{sum}$$Denote by $\kappa _{j,r}$ the kernel of the operator $T_{j,r}$. Then, ([Tj,r]{}) implies that $$\kappa _{j,r}(x)=T_{j,r}\delta _{0}\left( x\right) =h_{j,r}(\sqrt{-\Delta }%
