which belongs to ${\mathcal P}(B_R^Y)$ whenever ${\boldsymbol{y}}\in (B_R^Y)^N$. Consider the map ${\boldsymbol{b}}^N\colon Y^N\to\overline Y^N$ whose components are defined through $$\label{T134}
b_i^N({\boldsymbol{y}})\coloneqq b_{\Lambda^N}(y^i).$$ Then the Cauchy problem associated with can be written as $$\label{T135}
\begin{cases}
\dot {\boldsymbol{y}}_t={\boldsymbol{b}}^N({\boldsymbol{y}}_t), \\
{\boldsymbol{y}}_0=\bar {\boldsymbol{y}}.
\end{cases}$$ In order to apply Corollary \[cor:Brezis\] to the system above, we first notice that assumption (ii) is automatically satisfied since the system is autonomous. To check the other assumptions, we fix a ball $B_R^{Y^N}$ and notice that $B_R^{Y^N}\subset (B_{R}^Y)^N$.

Applying with $\Psi=\Lambda^N$ to each component $y^i$ of ${\boldsymbol{y}}$, we get that assumption (iii) of Corollary \[cor:Brezis\] is satisfied.

We now show that assumption (i’) holds. Fix ${\boldsymbol{y}}_1,{\boldsymbol{y}}_2\in B_R^{Y^N}\subset (B_R^Y)^N$, and let $\Lambda_1^N,\Lambda_2^N$ be the associated empirical measures. Notice that $$\label{T138}
{\mathcal W}_1(\Lambda_1^N,\Lambda_2^N) {\leqslant}\frac1N\sum_{i=1}^N \lVert y_1^i - y_2^i \rVert_{\overline Y} = \lVert {\boldsymbol{y}}_1 - {\boldsymbol{y}}_2 \rVert_{\overline Y^N}.$$ With this, by the triangle inequality, , and , we estimate $$\label{T137}
\begin{split}
\lVert {\boldsymbol{b}}^N({\boldsymbol{y}}_1)-{\boldsymbol{b}}^N({\boldsymbol{y}}_2)\rVert_{\overline Y^N} = & \frac1N \sum_{i=1}^N \lVert b_{\Lambda_1^N}(y_1^i) - b_{\Lambda_2^N}(y_2^i) \rVert_{\overline Y} \\
{\leqslant}& L_R {\mathcal W}_1(\Lambda_1^N,\Lambda_2^N) + \frac{L_R}N\sum_{i=1}^N \lVert y_1^i - y_2^i \rVert_{\overline Y} {\leqslant}2L_R \lVert {\boldsymbol{y}}_1 - {\boldsymbol{y}}_2 \rVert_{\overline Y^N},
\end{split}$$ which yields . To see that also , and therefore assumption (i”), holds, we apply and obtain, 