Dobrushin coupling estimate and actual particle-to-Vlasov limit

Fix a coupling ${\pi }_0$ of ${\mu }_0$ and ${\nu }_0$. Write the two characteristics as $(X_s,V_s)$ and $(Y_s,W_s)$, and push ${\pi }_0$ forward by their product flow to obtain ${\pi }_s$. Define $D_x(s)=\int d_{{\mathbb{T}}^d}(X_s,Y_s)d{\pi }_0$, $D_v(s)=\int \mathrm{\mid }{V}_s-W_s\mathrm{\mid }d{\pi }_0$, and $D=D_x+D_v$. The torus distance along absolutely continuous curves has upper derivative at most $\mathrm{\mid }{V}_s-W_s\mathrm{\mid }$. For an outer pair $(z,w)$, write $F_{{\mu }_s}(X_s(z))=\int k(X_s(z)-X_s(z^{\mathrm{\prime }}))d{\pi }_0(z^{\mathrm{\prime }},w^{\mathrm{\prime }})$ and $F_{{\nu }_s}(Y_s(w))=\int k(Y_s(w)-Y_s(w^{\mathrm{\prime }}))d{\pi }_0(z^{\mathrm{\prime }},w^{\mathrm{\prime }})$, using the two marginals of ${\pi }_0$. The torus triangle inequality and the $A$-Lipschitz bound for $k$ give a force difference at most $A(d_{{\mathbb{T}}^d}(X_s(z),Y_s(w))+d_{{\mathbb{T}}^d}(X_s(z^{\mathrm{\prime }}),Y_s(w^{\mathrm{\prime }})))$. Integrating the inner pair $(z^{\mathrm{\prime }},w^{\mathrm{\prime }})$ against ${\pi }_0$ bounds this by $A(d_{{\mathbb{T}}^d}(X_s(z),Y_s(w))+D_x(s))$. On every bounded time slab the velocity difference is bounded by its integrable initial difference plus $2\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }s\mathrm{\mid }$, and the acceleration difference is at most $2\mathrm{\parallel }k{\mathrm{\parallel }}_{\mathrm{\infty }}$. Integrate the pathwise distance inequalities in time and against the outer ${\pi }_0$; Tonelli [F6] justifies the order of integration and gives $D(s)\le D(0)+{\int }_0^s(D_v(r)+2A{D}_x(r))dr\le D(0)+(1+2A){\int }_0^{s}D(r)dr$. Gronwall [F9] gives $D(s)\le e^{(1+2A)s}D(0)$. For negative time, the transformation $R(x,v)=(x,-v)$ sends a solution run backward to a forward solution starting from $R_{\mathrm{\#}}{\mu }_0$ (and likewise for ${\nu }_0$). Explicitly, if the original characteristic starts at $(x,v)$ with law ${\mu }_0$, set ${\tilde{X}}_r=X_{-r}(x,v)$ and ${\tilde{V}}_r=-V_{-r}(x,v)$; the reversed initial state is $(x,-v)$. Then $\dot{\tilde{X}}=\tilde{V}$ and $\dot{\tilde{V}}=F_{{\mu }_{-r}}(\tilde{X})=F_{R_{\mathrm{\#}}{\mu }_{-r}}(\tilde{X})$, because the force depends only on position. Thus $R_{\mathrm{\#}}{\mu }_{-r}$ is a forward solution starting at $R_{\mathrm{\#}}{\mu }_0$. Since $R$ preserves $d_Z$, the positive-time estimate for these reversed solutions proves the estimate with $\mathrm{\mid }s\mathrm{\mid }$.

Let ${\sigma }_s$ be a weak solution in the Statement and set $F_s(x)=\int k(x-y)d{\sigma }_s(y,w)$. Narrow continuity and boundedness of $k$ make $F_s(x)$ continuous in $(s,x)$; every state derivative is bounded by the corresponding bound on $k$. By [F5] the field $(v,F_s(x))$ has a complete state flow on every bounded time interval and bounded state derivatives of that flow there. For fixed terminal $t$ and $\psi \in C_c^{\mathrm{\infty }}(Z)$, compose $\psi$ with the characteristic map from time $s$ to $t$. Bounded acceleration makes the preimage of its compact velocity support remain in a bounded velocity set on the finite slab, so the resulting time-dependent test $\varphi (s,z)$, its first state derivatives, and its time derivative are bounded. It satisfies ${\mathrm{\partial }}_s\varphi +v\cdot {\mathrm{\nabla }}_x\varphi +F_s\cdot {\mathrm{\nabla }}_v\varphi =0$. The weak identity makes $\int \varphi (s,z)d{\sigma }_s(z)$ constant, hence $\int \psi d{\sigma }_t=\int \psi (S_{t,0}(z))d{\sigma }_0(z)$. This is equality on every smooth compactly supported test, so the phase-space measure lemma [F8] gives ${\sigma }_t=(S_{t,0}{)}_{\mathrm{\#}}{\sigma }_0$.

The measure ${\pi }_s$ is a coupling of ${\mu }_s$ and ${\nu }_s$, so [F2] gives ${\mathcal{W}}_1({\mu }_s,{\nu }_s)\le D(s)$. The estimate in step 1.1 holds for every coupling ${\pi }_0$; taking the infimum of its right side gives the first claimed inequality without assuming an optimal coupling exists.

The trajectory defined in the Statement is complete by [F3]. Differentiating its coordinate and time rescaling using [F1] shows that $Z_i(s)=(X_i(s),V_i(s))$ satisfies ${\dot{X}}_i=V_i$ and ${\dot{V}}_i=N^{-1}{\sum }_{j\mathrm{\ne }i}k(X_i-X_j)$. Since $k(0)=0$, this equals $N^{-1}{\sum }_{j}k(X_i-X_j)$, the characteristic force generated by ${\mu }_N(s)=N^{-1}{\sum }_i{\delta }_{Z_i(s)}$. The coordinatewise map ${\mathrm{\Xi }}_N$ commutes with label permutations, so the conjugated flow retains the permutation equivariance in [F3]. The finite sum $F_{{\mu }_N(s)}$ is continuous in time, smooth in position and bounded with all state derivatives; [F5] therefore supplies its characteristic flow for every initial state. ODE uniqueness identifies this flow at each initial atom with the particle trajectory, so its push-forward of ${\mu }_N(0)$ is ${\mu }_N(s)$. The initial empirical law is in ${\mathcal{P}}_1(Z)$, and uniqueness of the global characteristic evolution in [F4] identifies ${\mu }_N(s)$ with that evolution from ${\mu }_N(0)$. Applying step 2.1 to initial laws ${\mu }_N(0)$ and ${\mu }_0$ gives the particle estimate.

By step 1.2 any two weak solutions with the same initial law are the characteristic evolutions of that law. Step 2.1 gives zero ${\mathcal{W}}_1$ between them at each time; alternatively their representation in step 1.2 already identifies both with the same push-forward. This proves uniqueness in the stated weak class.