$and given $h\in {\mathbb{N}}$ we denote $$\rho _{h}=\frac{(a+b)m_{0}+q+2d/p_{\ast }}{2h}.  \label{H5'}$$Notice that this is equal to the constant $\rho _{h}$ defined in (\[reg5\]) corresponding to $k=(a+b)m_{0}$ and $q$ and to $2d$ (instead of $d).$

**Step 1: a Lindeberg-type method to decompose $P_t-P^n_t$.** We fix (once for all) $t\in(0,1]$ and we write$$P_{t}f-P_{t}^{n}f=\int_{0}^{t}\partial
_{s}(P_{t-s}^{n}P_{s})fds=\int_{0}^{t}P^n
_{t-s}(L-L_{n})P_{s}fds=\int_{0}^{t}P^n_{t-s}\Delta _{n}P_{s}fds$$We iterate this formula $m_{0}$ times (with $m_{0}$ chosen in (\[H4\])) and we obtain$$P_{t}f(x)-P_{t}^{n}f(x)=\sum_{m=1}^{m_{0}-1}I_{n}^{m}f(x)+R_{n}^{m_{0}}f(x)
\label{R2}$$with (we put $t_{0}=t)$$$\begin{aligned}
I_{n}^{m}f(x)&=\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}...%
\int_{0}^{t_{m-1}}dt_{m}\prod_{i=0}^{m-1}(P_{t_{i}-t_{i+1}}^{n}\Delta
_{n})P_{t_{m}}^{n}f(x),\quad 1\leq m\leq m_0-1, \\
R_{n}^{m_{0}}f(x)&=\int_{0}^{t}dt_{1}\int_{0}^{t_{1}}dt_{2}...%
\int_{0}^{t_{m_{0}-1}}dt_{m_{0}}\prod_{i=0}^{m_{0}-1}(P_{t_{i}-t_{i+1}}^{n}%
\Delta _{n})P_{t_{m_{0}}}f(x).\end{aligned}$$

In order to analyze $I_{n}^{m}f$ we use Lemma \[Reg\] for the semigroup $%
S_{t}=P_{t}^{n}$ and for the operators $U_{i}=\Delta _{n}=L-L_{n}$ (the same for each $i$), with $\delta _{i}=t_{i}-t_{i+1}$, $i=0,\ldots ,m$ (with $%
t_{m+1}=0$). So the hypotheses (\[h1\]) and (\[h1’\]) in Assumption \[H1H\*1\] coincide with the requests (\[TR3\]) and (\[TR3’\]) in Assumption \[A1A\*1\]. And we have $%
C_{q,\kappa ,\infty }(U)=C_{q,\kappa ,p}(U)=C\varepsilon _{n}.$ Moreover the hypotheses (\[h2\]) and (\[h2’\]) in Assumption \[H2H\*2\] coincide with the hypotheses (\[TR2\]) and (\[TR2’\]) in Assumption \[A2A\*2\]. And we have $%
C_{q,\kappa ,\infty }(P^{n})=C_{q,\kappa ,p}(P^{n})=\Lambda _{n}$. Hence, $$C_{q,\kappa ,\infty ,p}(\Delta _{n},P^{n})=C\,\varepsilon _{n}\times \Lambda
_{n},  \label{app1}$$Finally, the hypothesis (\[h3\]) in Assumption \[HH3\] coincides with (\[TR5\]) in Assumption \[A3\]. So, we can apply Lemma \[Reg\]: by using 