Moreover as $x^n_i-x^n_j$ is a $C^2$ function, by Taylor’s formula $$\begin{aligned}
\label{on}
x^n_i(s)-x^n_j(s)=s\cdot\left(v^n_i(0)-v^n_j(0)\right)+o_n(s)=s\left(v^n_i(s)-v^n_j(s)-r_n(s)\right)+o_n(s),\end{aligned}$$ where $$\begin{aligned}
o_n(s):=\int_0^s (\dot{v}^n_i-\dot{v}^n_j)(s-\theta)d\theta,\ \ \ \ \ \ |o_n(s)|\leq 2|s|^2Mn.\end{aligned}$$ Thus [$$\begin{aligned}
|(v_i^n(s)-v^n_j(s))(x^n_i(s)-x^n_j(s))|=|s(v_i^n(s)-v^n_j(s))^2-s(v_i^n(s)-v^n_j(s))r_n(s)+ (v_i^n(s)-v^n_j(s))o_n(s)|\geq\nonumber\\
\geq s(v_i^n(s)-v^n_j(s))^2-s|(v_i^n(s)-v^n_j(s))r_n(s)|-|(v_i^n(s)-v^n_j(s))o_n(s)|\geq \frac{s}{2}(v_i^n(s)-v^n_j(s))^2\label{licznik}\end{aligned}$$ ]{} assuming that $s\in(0,t_n]$, where $t_n$ is the supremum of all times $s_n$, such that for all $s\in(t,s_n]$, we have $$\begin{aligned}
\label{war1}
|r_n(s)|\leq \frac{1}{4}|v_i^n(s)-v^n_j(s)|,\ \ \ \ \ \ |o_n(s)|\leq \frac{s}{4}|v_i^n(s)-v^n_j(s)|.\end{aligned}$$ To check that $t_n>0$, we notice that for $$\begin{aligned}
\label{tn}
s_n:=\frac{|v^n_i(0)-v^n_j(0)|}{10M(n)n}\end{aligned}$$ and $s\in [0,s_n]$, we have $$\begin{aligned}
\label{war2}
|r_n(s)|\leq \frac{1}{5}|v^n_i(0)-v^n_j(0)|,\ \ \ \ \ \ |o_n(s)|\leq \frac{s}{5}|v^n_i(0)-v^n_j(0)|,\end{aligned}$$ which together with (\[rn\]) implies that $$\begin{aligned}
\frac{4}{5}|v^n_i(0)-v^n_j(0)|\leq |v^n_i(s)-v^n_j(s)|,\end{aligned}$$ condition (\[war1\]) is satisfied. Therefore by taking $s_n$ given by (\[tn\]) we get (\[licznik\]). Now by (\[rn\]) and (\[on\]) on $(0,s_n]$ we also have $$\begin{aligned}
\label{war3}
|x^n_i(s)-x^n_j(s)|\leq 2s|v^n_i(s)-v^n_j(s)|, \end{aligned}$$ which together with (\[licznik\]) proves that there exists $s_n>0$ such that on $(0,s_n]$ the assertion holds. Now we define $t_n$ as the supremum of all such times $s_n$. This finishes the proof.

Next we can proceed with the proof of Lemma \[equi\].

The proof will follow by 2 steps. In step 1 we prove that for each $n$ there exists an interval $[0, t_n]$ on which $v^n$ is Hölder continuous with a constant 