$t_{j+1}$ configuration $(C)$ becomes $$\dots  0\ \underbrace{\underbrace{0\ 0}\ \underbrace{0 \ 0}\ \dots\  \underbrace{0 \ 0}}_{2l\mbox{ elements}}\ ? \dots  \tag{$\mathrm{C}'$}$$ while configuration $(\mathrm{D})$ becomes $$\dots  0\ \underbrace{\underbrace{0\ 0}\ \underbrace{0 \ 0}\ \dots\  \underbrace{0 \ 0}}_{2l-2\mbox{ elements}}\ 0 \ 1\ 0\ ? \dots  \tag{$\mathrm{D}'$}$$ In both cases $(\mathrm{C})\to(\mathrm{C}')$ and $(\mathrm{D})\to(\mathrm{D}')$ the number of $1$s among the $v_i$ does not increase, and in the first case it goes down by $1$. However, it is easy to see that whether $M$ is odd or even (in the latter case due to Corollary \[cor\_even\]) there will be at least one chunk of type $(\mathrm{C})$, and hence with positive probability $v(t)$ reaches state ${\bf 0}$ in at most $M^2$ steps (since $t_{j+1}-t_j\le M$). The observation that after $t_{j_0}$ the Markov chain $v(t)$ lives on a finite state space $\{0,1,2\}^{M}$ finishes the proof.

The first part easily follows from Proposition \[prop\_a\_conv\] and the definition of potentials $v$. Indeed, for $j\ge j_0$ and all $i$ we have $v_i(t_j)\in\{0,1\}$, while for $t\in (t_j,t_{j+1})$ we have $v_i(t)\in\{0,1,2\}$. On the other hand, omitting $(t)$, we can write $v_{i+1}-v_{i}=u_{i+1}-u_i=\xi_{i+2}-\xi_{i}$, $i\in\M$, yielding (\[eq\_at\_most2\]).

Next, iterating this argument, we obtain $|\xi_{i+2l}-\xi_i|\le
2l$. Because of the periodic boundary condition, in the case when $M$ is 