$$\pi_{N,k}(\xi) \le \pi_{N,k}(\xi^y)\le {\lambda}\pi_{N,k}(\xi).$$ The reader can check indeed that $\pi_{N,k}(\xi^y)= {\lambda}\pi_{N,k}(\xi)$ if $\xi\in A_y \setminus A_{y+1}$ and that $\xi^y=\xi$ if $\xi\in A_y \cap A_{y+1}$. The desired inequality is then obtained by summing over $\xi\in A_y$.

By iterating we obtain $${\lambda}^{x-N}\pi_{N,k}(A_1) \le  \pi_{N,k}(A_{x}) \le  {\lambda}^{x-1} \pi_{N,k}(A_1).$$ By monotony of $\pi_{N,k}(A_y)$ in $y$ and the fact that there are $k$ particles $$N\pi_{N,k}(A_1) \le \sum_{y=1}^N  \pi_{N,k}(A_y)=k \le N  \pi_{N,k}(A_N),$$ and thus can be deduced.

We pass to the proof of . We can perform the same reasoning as above but limiting ourselves to configurations with no particles in some set $I\subset {\llbracket}1, N{\rrbracket}$. Setting $B_I:=\{ \forall y\in I, \xi(y)=0 \}$ we obtain similarly to (exchanging directly the content of $x$ and $y$ instead of nearest neighbors) that for every $x,y \in  {\llbracket}1, N{\rrbracket}\setminus I$ with $x< y$ $$\pi_{N,k}(A_x  \cap B_I )  \le \pi_{N,k}(A_y \cap B_I ) \le  {\lambda}^{y-x} \pi_{N,k}(A_x  \cap B_I ).$$ This allows to deduce that $$\pi_{N,k}(\xi(x)=1 \ | \ \forall y\in I, \xi(y)=0) \ge \frac{k}{N-|I|}{\lambda}^{x-N},$$ and yields by induction $$\pi_{N,k}(\forall x\in I, \  \xi(x)=0)\le \left(1-{\lambda}^{-N} \frac{k}{N}\right)^{|I|} \le \exp\left(-|I| {\lambda}^{-N} \frac{k}{N}\right).$$ Then noticing that $\{Q_1(\xi)\ge 2m\}$ implies that an interval of the type ${\llbracket}mi+1, m(i+1){\rrbracket}$ is empty, a union bound yields 