dominated convergence theorem), there exists some $T > 0$ such that, for all $t \in [0, T]$, the function $\psi_t$ belongs to $C_b^4(\mathbb{R}^n)$. This proves that, for all $0 \le t \le T$, $\bar{P}_t G$ belongs to $\mathcal{P}^4$.

Then, since the process $(\bar{\mu}_t)_t$ is Markov, we have: for all $t \in [T, 2T]$, for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ \bar{P}_t G(m) = \bar{P}_{t-T} (\bar{P}_T G) (m). $$

Whence, since $\bar{P}_T G \in \mathcal{P}^4$ and $t - T \in [0, T]$, the above equality implies that, for all $T \le t \le 2T$, the function $\bar{P}_t G$ also belongs to $\mathcal{P}^4$. Iterating this argument concludes the proof.

The proof of Item (b) follows from the fact that we have the explicit expression of $\bar{P}_t G$ at (26) and the control given at Lemma 3.6 (recalling Proposition 2.31). The proof of Item (c) relies on the same arguments as for Item (b) with the additional use of the Dominated Convergence Theorem. $\square$

### 3.4. Proof of Theorem 1.7

In this section, we prove that the empirical measure $(\mu_t^N)_{t \ge 0}$ of the particle system defined at (8) is a Markov process, with an explicit expression for its generator.

The fact that $\mu^N$ is a Markov process follows from Proposition 2.3.3 of Dawson (1993). It can also be proved as in Theorem 1.5 using Proposition 3.3 and the following computation that are in any case required to prove the remains of the statement of the theorem.

So, as in the proof of Theorem 1.5, let $G \in \mathcal{P}^2$ such that: for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ G(m) = \int_{\mathbb{R}^n} \varphi(x) dm^{\otimes n}(x), $$

with $n \in \mathbb{N}^*$ and $\varphi \in C_b^2(\mathbb{R}^n)$.

Then,

$$ G(\mu_t^N) = \frac{1}{N^n} \sum_{k_1, \dots, k_n = 1}^{N} \varphi(X_t^{N, k_1}, \dots, X_t^{N, k_n}). $$

For a given tuple $\mathbf{k} = (k_1, \dots, k_n) \in [1, N]^n$, let us denote

$$ \mathbf{X}_t^{N, \mathbf{k}} = (X_t^{N, k_1}, \dots, X_t^{N, k_n}), $$

and $X^{N, k_i, c}$ the continuous part of the semimartingale $X^{N, k_i}$.

By Ito's formula, for all $t \ge r \ge 0$,

$$ \begin{aligned} \varphi(\mathbf{X}_t^{N, \mathbf{k}}) &= \varphi(\mathbf{X}_r^{N, \mathbf{k}}) + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) dX_s^{N, k_i, c} \\ & \quad + \frac{1}{2} \sum_{i,j=1}^{n} \int_r^t \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) d\langle X^{N, k_i, c}, X^{N, k_j, c} \rangle_s \\ & \quad + \sum_{r<s \le t} (\varphi(\mathbf{X}_s^{N, \mathbf{k}}) - \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}})) \\ & = \varphi(\mathbf{X}_r^{N, \mathbf{k}}) + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) b(\mu_s^N, X_s^{N, k_i}) ds \end{aligned} $$
(a) the function

$$m \in \mathcal{P}_1(\mathbb{R}) \mapsto \bar{P}_t G(m)$$

belongs to $\mathcal{P}^4$,

(b) there exists $C_t > 0$ independent of $G$ such that, for all $0 \le k \le 4$,

$$\sup_{s \le t} \|\bar{P}_s G\|_k \le C_t \|G\|_k,$$

where $\|\cdot\|_k$ is introduced in Definition 2.14,

(c) for all $m \in \mathcal{P}_1(\mathbb{R})$ and $x, y \in \mathbb{R}$, the following functions are continuous

$$\begin{align*}
t \in \mathbb{R}_+ &\mapsto \bar{P}_t G(m), \\
t \in \mathbb{R}_+ &\mapsto \delta(\bar{P}_t G)(m, x), \\
t \in \mathbb{R}_+ &\mapsto \partial_x \delta(\bar{P}_t G)(m, x), \\
t \in \mathbb{R}_+ &\mapsto \partial_{xx}^2 \delta(\bar{P}_t G)(m, x), \\
t \in \mathbb{R}_+ &\mapsto \partial_{xy}^2 \delta^2(\bar{P}_t G)(m, x, y).
\end{align*}$$

*Proof.* Let $G \in \mathcal{P}^4$ satisfying for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$G(m) = \int_{\mathbb{R}^n} \varphi(x) dm^{\otimes n}(x),$$

with $n \in \mathbb{N}^*$ and $\varphi \in C_b^4(\mathbb{R}^n)$.

Let us prove Item (a). For all $t \ge 0$,

$$G(\bar{\mu}_t) = \mathbb{E} \left[ \varphi(\bar{X}_t^1, \dots, \bar{X}_t^n) \mid W_t \right],$$

where the processes $\bar{X}^k$ ($1 \le k \le n$) are solutions to (30) w.r.t. the same $W$, but independent $B^k$ and $\bar{X}_0^k$ (similarly as in (25)). Since the variables $\bar{X}_0^1, \dots, \bar{X}_0^n$ are i.i.d. $\bar{\mu}_0$-distributed, we have, for any $m \in \mathcal{P}_1(\mathbb{R})$,

$$\bar{P}_t G(m) = \mathbb{E}_m [G(\bar{\mu}_t)] = \int_{\mathbb{R}^n} \mathbb{E} \left[ \varphi \left( \bar{X}_t^{1,(x_1)}, \dots, \bar{X}_t^{n,(x_n)} \right) \right] dm^{\otimes n}(x) = \int_{\mathbb{R}^n} \psi_t(x) dm^{\otimes n}(x), \quad (31)$$

with

$$\psi_t : x \in \mathbb{R}^n \mapsto \mathbb{E} \left[ \varphi \left( \bar{X}_t^{1,(x_1)}, \dots, \bar{X}_t^{n,(x_n)} \right) \right],$$

and $\bar{X}^{k,(x_k)}$ the process $\bar{X}^k$ starting from initial condition $\bar{X}_0^{k,(x_k)} = x_k$ ($1 \le k \le n$). Then, by Lemma 3.6 and by differentiation under the integral sign (using uniform integrability condition, like, for example, in Lemma 6.1 of ?), there exists some $T > 0$ such that, for all $t \in [0, T]$, the function $\psi_t$ belongs to $C_b^4(\mathbb{R}^n)$. This proves that, for all $0 \le t \le T$, $\bar{P}_t G$ belongs to $\mathcal{P}^4$.

Then, since the process $(\bar{\mu}_t)_t$ is Markov, we have: for all $t \in [T, 2T]$, for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$\bar{P}_t G(m) = \bar{P}_{t-T} (\bar{P}_T G)(m).$$

Whence, since $\bar{P}_T G \in \mathcal{P}^4$ and $t - T \in [0, T]$, the above equality implies that, for all $T \le t \le 2T$, the function $\bar{P}_t G$ also belongs to $\mathcal{P}^4$. Iterating this argument concludes the proof.

The proof of Item (b) follows from the fact that we have the explicit expression of $\bar{P}_t G$ at (31) and the control given at Lemma 3.6 (recalling Proposition 2.32). The proof of Item (c) relies on the same arguments as for Item (b) with the additional use of the Dominated Convergence Theorem. $\square$
$$
\begin{align*}
&+ \sum_{i=1}^{n} \int_{r}^{t} \partial_i \varphi(\mathbf{X}_{s}^{N, \mathbf{k}}) \sigma(\mu_{s}^{N}, X_{s}^{N, k_{i}}) dB_{s}^{k_{i}} \\
&+ \sum_{i=1}^{n} \int_{r}^{t} \partial_i \varphi(\mathbf{X}_{s}^{N, \mathbf{k}}) \varsigma(\mu_{s}^{N}, X_{s}^{N, k_{i}}) dW_{s} \\
&+ \frac{1}{2} \sum_{i,j=1}^{n} \partial_{ij}^{2} \varphi(\mathbf{X}_{s}^{N, \mathbf{k}}) \sigma(\mu_{s}^{N}, X_{s}^{N, k_{i}}) \sigma(\mu_{s}^{N}, X_{s}^{N, k_{j}}) \mathbf{1}_{\{k_{i}=k_{j}\}} ds \\
&+ \frac{1}{2} \sum_{i,j=1}^{n} \partial_{ij}^{2} \varphi(\mathbf{X}_{s}^{N, \mathbf{k}}) \varsigma(\mu_{s}^{N}, X_{s}^{N, k_{i}}) \varsigma(\mu_{s}^{N}, X_{s}^{N, k_{j}}) ds \\
&+ \sum_{l=1}^{N} \int_{[r,t] \times \mathbb{R}_{+} \times \mathbb{R}} \mathbf{1}_{\{z \le f(\mu_{s-}^{N}, X_{s-}^{N,l})\}} \\
&\qquad \left[ \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}} + h(\mu_{s-}^{N}, X_{s-}^{N,l}, u) \cdot \mathbf{1}) - \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}}) \right] d\pi^{l}(s, z, u),
\end{align*}
$$

with $\mathbf{1} = (1, 1, \dots, 1) \in \mathbb{R}^n$. Consequently,

$$
\begin{align}
G(\mu_t^N) &= \frac{1}{N^n} \sum_{k_1, \dots, k_n = 1}^N \varphi(\mathbf{X}_t^{N, \mathbf{k}}) \nonumber \\
&= G(\mu_r^N) + M_t^{N,r} + \sum_{i=1}^n \int_r^t \int_{\mathbb{R}^n} \partial_i \varphi(x) b(\mu_s^N, x_i) d(\mu_s^N)^{\otimes n} (x) ds \nonumber \\
&\quad + \frac{1}{N^n} \sum_{\mathbf{k} \in [1, N]^n} \frac{1}{2} \sum_{i,j=1}^n \int_r^t \int_{\mathbb{R}^n} \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i}) \sigma(\mu_s^N, X_s^{N, k_j}) \mathbf{1}_{\{k_i=k_j\}} ds \tag{27}
\end{align}
$$

$$
\begin{align}
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \int_r^t \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i}) \sigma(\mu_s^N, X_s^{N, k_j}) \mathbf{1}_{\{k_i=k_j\}} ds \nonumber \\
& + \frac{1}{2} \sum_{i,j=1}^n \int_r^t \int_{\mathbb{R}^n} \partial_{ij}^2 \varphi(x) \varsigma(\mu_s^N, x_i) \varsigma(\mu_s^N, x_j) d(\mu_s^N)^{\otimes n} (x) ds \nonumber \\
& + \sum_{l=1}^N \int_r^t \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} f(\mu_s^N, X_s^{N,l}) \tag{28}
\end{align}
$$

$$
[\varphi(x + h(\mu_s^N, X_s^{N,l}, u) \cdot \mathbf{1}) - \varphi(x)] d(\mu_s^N)^{\otimes n}(x) d\nu(u) ds,
$$

with $(M_t^{N,r})_{t \ge r}$ some martingale (the fact that $M^{N,r}$ is not only a local martingale holds true thanks to controls like (5) and the assumption that $f, \sigma, \varsigma$ are sublinear).

Then, noticing that, for any $x_1, \dots, x_N \in \mathbb{R}$ and $\lambda \in \mathbb{R}$,

$$
S \left( \frac{1}{N} \sum_{k=1}^{N} \delta_{x_k}, \lambda \right) = \frac{1}{N} \sum_{k=1}^{N} \delta_{x_k + \lambda},
$$

it is possible to rewrite the expression at (28) as
### 3.4. Proof of Theorem 1.7

In this section, we prove that the empirical measure $(\mu_t^N)_{t \ge 0}$ of the particle system defined at (8) is a Markov process, with an explicit expression for its generator.

The fact that $\mu^N$ is a Markov process follows from Proposition 2.3.3 of ?. It can also be proved as in Theorem 1.5 using Proposition 3.3 and the following computation that are in any case required to prove the remains of the statement of the theorem.

So, as in the proof of Theorem 1.5, let $G \in \mathcal{P}^2$ such that: for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$G(m) = \int_{\mathbb{R}^n} \varphi(x) dm^{\otimes n}(x),$$

with $n \in \mathbb{N}^*$ and $\varphi \in C_b^2(\mathbb{R}^n)$.

Then,

$$G(\mu_t^N) = \frac{1}{N^n} \sum_{k_1, \dots, k_n = 1}^{N} \varphi(X_t^{N, k_1}, \dots, X_t^{N, k_n}).$$

For a given tuple $\mathbf{k} = (k_1, \dots, k_n) \in [1, N]^n$, let us denote

$$\mathbf{X}_t^{N, \mathbf{k}} = (X_t^{N, k_1}, \dots, X_t^{N, k_n}),$$

and $X^{N,k_i,c}$ the continuous part of the semimartingale $X^{N,k_i}$.

By Ito's formula, for all $t \ge r \ge 0$,

$$
\begin{align*}
\varphi(\mathbf{X}_t^{N, \mathbf{k}}) &= \varphi(\mathbf{X}_r^{N, \mathbf{k}}) + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) dX_s^{N, k_i, c} \\
&\quad + \frac{1}{2} \sum_{i,j=1}^{n} \int_r^t \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) d\langle X^{N, k_i, c}, X^{N, k_j, c} \rangle_s \\
&\quad + \sum_{r<s \le t} (\varphi(\mathbf{X}_s^{N, \mathbf{k}}) - \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}})) \\
&= \varphi(\mathbf{X}_r^{N, \mathbf{k}}) + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) b(\mu_s^N, X_s^{N, k_i}) ds \\
&\quad + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i}) dB_s^{k_i} \\
&\quad + \sum_{i=1}^{n} \int_r^t \partial_i \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \varsigma(\mu_s^N, X_s^{N, k_i}) dW_s \\
&\quad + \frac{1}{2} \sum_{i,j=1}^{n} \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i}) \sigma(\mu_s^N, X_s^{N, k_j}) \mathbf{1}_{\{k_i=k_j\}} ds \\
&\quad + \frac{1}{2} \sum_{i,j=1}^{n} \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \varsigma(\mu_s^N, X_s^{N, k_i}) \varsigma(\mu_s^N, X_s^{N, k_j}) ds
\end{align*}
$$
$$
\begin{align*}
& N \int_r^t \int_{\mathbb{R}} \int_{\mathbb{R}^n} \int_{\mathbb{R}} f(\mu_s^N, y) [\varphi(x+h(\mu_s^N, y, u) \cdot \mathbf{1}) - \varphi(x)] d\mu_s^N(y) d(\mu_s^N)^{\otimes n}(x) d\nu(u) ds \\
&= N \int_r^t \int_{\mathbb{R}} \int_{\mathbb{R}} f(\mu_s^N, y) [G(\mathcal{S}(\mu_s^N, h(\mu_s^N, y, u))) - G(\mu_s^N)] d\mu_s^N(y) d\nu(u) ds,
\end{align*}
$$

Besides, separating the terms $i = j$ and $i \neq j$ in the double-sum of the expression at (27), this expression can be rewritten as

$$
\begin{align*}
& \frac{1}{2} \int_r^t \sum_{i=1}^n \frac{1}{N^n} \sum_{\mathbf{k} \in [1,N]^n} \partial_{ii}^2 \varphi(\mathbf{X}_s^{N,\mathbf{k}}) \sigma(\mu_s^N, X_s^{N,k_i})^2 ds \\
& \quad + \frac{1}{2} \int_r^t \sum_{\substack{i,j=1 \\ i \neq j}}^n \frac{1}{N} \cdot \frac{1}{N^{n-1}} \sum_{\mathbf{k} \setminus \{j\} \in [1,N]^{n-1}} \partial_{ij}^2 \varphi(\mathbf{X}_s^{N,(\mathbf{k}\setminus\{j\}k_i)}) \sigma(\mu_s^N, X_s^{N,k_i})^2 ds \\
&= \frac{1}{2} \int_r^t \sum_{i=1}^n \int_{\mathbb{R}^n} \partial_{ii}^2 \varphi(x) \sigma(\mu_s^N, x_i)^2 d(\mu_s^N)^{\otimes n}(x) ds \\
& \quad + \frac{1}{2N} \int_r^t \sum_{\substack{i,j=1 \\ i \neq j}}^n \int_{\mathbb{R}^{n-1}} \partial_{ij}^2 \varphi(x \setminus j, x_i) \sigma(\mu_s^N, x_i)^2 d(\mu_s^N)^{\otimes n-1}(x \setminus j) ds \\
&= \frac{1}{2} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 \partial_{xx}^2 \delta G(\mu_s^N, x) d\mu_s^N(x) ds \\
& \quad + \frac{1}{2N} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(\mu_s^N, y_1, y_2))_{|y_1=y_2=x} d\mu_s^N(x) ds,
\end{align*}
$$

where the last equality uses (23) and (24), and is mostly a notation matter.

In particular, doing the same computation trick as the one done at the end of the proof of
Theorem 1.5, it is possible to write the dynamics of $G(\mu_t^N)$ as follows

$$
\begin{align*}
G(\mu_t^N) ={}& G(\mu_r^N) + M_t^{N,r} + \int_r^t \int_{\mathbb{R}} b(\mu_s^N, x) \partial_x \delta G(m, x) d\mu_s^N(x) ds \\
& + \frac{1}{2} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 \partial_{xx}^2 \delta G(\mu_s^N, x) d\mu_s^N(x) ds \\
& + \frac{1}{2N} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(\mu_s^N, y_1, y_2))_{|y_1=y_2=x} d(\mu_s^N(x))^{\otimes 2}(x) ds \\
& + \frac{1}{2} \int_r^t \int_{\mathbb{R}^2} \varsigma(\mu_s^N, x) \varsigma(\mu_s^N, y) \partial_{xy}^2 \delta^2 G(\mu_s^N, x, y) d(\mu_s^N)^{\otimes 2}(x, y) ds \\
& + N \int_r^t \int_{\mathbb{R}} \int_{\mathbb{R}} f(\mu_s^N, y) [G(\mathcal{S}(\mu_s^N, h(\mu_s^N, y, u))) - G(\mu_s^N)] d\mu_s^N(y) d\nu(u) ds,
\end{align*}
$$

with $(M_t^{N,r})_t$ a martingale that depends only on the process $\mu^N$, on the Brownian motions $W, B^k$
$(1 \le k \le N)$ and on the Poisson measures $\pi^l$ $(1 \le l \le N)$ such that it adapted to the union
of the filtrations of these objects. The equality above being true for all $G \in \mathcal{P}^2$, Proposition 3.3
entails that the measure-valued process $(\mu_t^N)_t$ is a Markov process (which was already guaranteed
$$ + \sum_{l=1}^{N} \int_{[r,t] \times \mathbb{R}_{+} \times \mathbb{R}} \mathbf{1}_{\{z \le f(\mu_{s-}^{N}, X_{s-}^{N,l})\}} \left[ \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}} + h(\mu_{s-}^{N}, X_{s-}^{N,l}, u) \cdot \mathbf{1}) - \varphi(\mathbf{X}_{s-}^{N, \mathbf{k}}) \right] d\pi^{l}(s, z, u), $$

with $\mathbf{1} = (1, 1, ..., 1) \in \mathbb{R}^n$. Consequently,

$$ G(\mu_t^N) = \frac{1}{N^n} \sum_{k_1, \dots, k_n = 1}^N \varphi(\mathbf{X}_t^{N, \mathbf{k}}) \\ = G(\mu_r^N) + M_t^{N,r} + \sum_{i=1}^n \int_r^t \int_{\mathbb{R}^n} \partial_i \varphi(x) b(\mu_s^N, x_i) d(\mu_s^N)^{\otimes n} (x) ds \\ + \frac{1}{N^n} \sum_{\mathbf{k} \in [1, N]^n} \frac{1}{2} \sum_{i,j=1}^n \int_r^t \partial_{ij}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i}) \sigma(\mu_s^N, X_s^{N, k_j}) \mathbf{1}_{\{k_i=k_j\}} ds \quad (32) $$

$$ + \frac{1}{2} \sum_{i,j=1}^{n} \int_{r}^{t} \int_{\mathbb{R}^{n}} \partial_{ij}^{2} \varphi(x) \zeta(\mu_{s}^{N}, x_{i}) \zeta(\mu_{s}^{N}, x_{j}) d(\mu_{s}^{N})^{\otimes n}(x) ds \\ + \sum_{l=1}^{N} \int_{r}^{t} \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} f(\mu_{s}^{N}, X_{s}^{N,l}) \quad (33) $$

$$ [\varphi(x + h(\mu_s^N, X_s^{N,l}, u) \cdot \mathbf{1}) - \varphi(x)] d(\mu_s^N)^{\otimes n}(x) d\nu(u) ds, $$

with $(M_t^{N,r})_{t \ge r}$ some martingale (the fact that $M^{N,r}$ is not only a local martingale holds true thanks to controls like (5) and the assumption that $f, \sigma, \zeta$ are sublinear).

Then, noticing that, for any $x_1, \dots, x_N \in \mathbb{R}$ and $\lambda \in \mathbb{R}$,

$$ S \left( \frac{1}{N} \sum_{k=1}^{N} \delta_{x_k}, \lambda \right) = \frac{1}{N} \sum_{k=1}^{N} \delta_{x_k + \lambda}, $$

it is possible to rewrite the expression at (33) as

$$ N \int_{r}^{t} \int_{\mathbb{R}} \int_{\mathbb{R}^{n}} \int_{\mathbb{R}} f(\mu_{s}^{N}, y) [\varphi(x + h(\mu_{s}^{N}, y, u) \cdot \mathbf{1}) - \varphi(x)] d\mu_{s}^{N}(y) d(\mu_{s}^{N})^{\otimes n}(x) d\nu(u) ds \\ = N \int_{r}^{t} \int_{\mathbb{R}} \int_{\mathbb{R}} f(\mu_{s}^{N}, y) [G(S(\mu_{s}^{N}, h(\mu_{s}^{N}, y, u))) - G(\mu_{s}^{N})] d\mu_{s}^{N}(y) d\nu(u) ds, $$

Besides, separating the terms $i = j$ and $i \ne j$ in the double-sum of the expression at (32), this expression can be rewritten as

$$ \frac{1}{2} \int_{r}^{t} \sum_{i=1}^{n} \frac{1}{N^n} \sum_{\mathbf{k} \in [1, N]^n} \partial_{ii}^2 \varphi(\mathbf{X}_s^{N, \mathbf{k}}) \sigma(\mu_s^N, X_s^{N, k_i})^2 ds $$
by Proposition 2.3.3 of Dawson (1993)), but also that the domain of the generator of $\mu^N$ contains the set $\mathcal{P}^2$, and that the expression of the generator on any $G \in \mathcal{P}^2$ is the one given in the statement of Theorem 1.7.

# 4. Proof of the results of Section 1.1

## 4.1. Proof of Theorem 1.4

It is sufficient to prove the result for $n = 1$. Indeed, the general case can be deduced inductively using the following Markovian properties of the processes: if $(\mu_t)_t$ is an E-valued homogeneous Markov process with semigroup $(P_t)_t$, then for $t_1 < t_2$, and $G_1, G_2$ “smooth enough”,

$$ \mathbb{E}[G_1(\mu_{t_1})G_2(\mu_{t_2})] = \mathbb{E}[G_1(\mu_{t_1})P_{t_2-t_1}G_2(\mu_{t_1})]. $$

So we only prove the result for $n = 1$. Let us recall that, for each $N \in \mathbb{N}^*$, the processes $X^{N,k}$ ($1 \le k \le N$) are defined as the (strong) solutions of (3), and $\bar{X}$ as the (strong) solution of (4). And, by definition, for all $t \ge 0$,

$$ \mu_t^N = \frac{1}{N} \sum_{k=1}^{N} \delta_{X_t^{N,k}} \quad \text{and} \quad \bar{\mu}_t = \mathcal{L}(\bar{X}_t | \mathcal{W}_t). $$

Let us denote

$$ \mathcal{E}_N = \left\{ m_w = \frac{1}{N} \sum_{k=1}^{N} \delta_{w_k} : w_1, \dots, w_N \in \mathbb{R} \right\} \subseteq \mathcal{P}_{\infty}(\mathbb{R}). $$

It is classical that for any $t \ge 0$, there exists $C_t > 0$ such that for all $p \in \mathbb{N}^*$, $m \in \mathcal{P}_p(\mathbb{R})$ and $w_1, \dots, w_N \in \mathbb{R}$

$$ \mathbb{E}_m \left[ \sup_{s \le t} |\bar{X}_s|^p \right] \le C_t \left( 1 + \int_{\mathbb{R}} |x|^p dm(x) \right), \qquad (29) $$

$$ \frac{1}{N} \sum_{k=1}^{N} \mathbb{E}_{m_w} \left[ \sup_{s \le t} |X_s^{N,k}|^p \right] \le C_t \left( 1 + \frac{1}{N} \sum_{k=1}^{N} |w_k|^p \right), \qquad (30) $$

where $m_w = N^{-1} \sum_{k=1}^{N} w_k$, $\mathbb{E}_m$ (resp. $\mathbb{E}_{m_w}$) denotes the expectation w.r.t. the probability measure under which the Markov process $\bar{\mu}$ (resp. $\mu^N$) has $m$ (resp. $m_w$) as initial condition.

**Step 1.** The first step of the proof consists in showing that, for any fixed $N \in \mathbb{N}^*$, the $\mathcal{P}_1(\mathbb{R})$-valued Markov processes $\mu^N$ and $\bar{\mu}$ satisfy the following Trotter-Kato formula: for all $G \in \mathcal{P}^4$, $m \in \mathcal{E}_N$ and $t \ge 0$,

$$ (\bar{P}_t - P_t^N) G(m) = \int_0^t P_{t-s}^N (\bar{A} - A^N) \bar{P}_s G(m) ds, $$

with $P^N, \bar{P}$ (resp. $A^N, \bar{A}$) the semigroups (resp. generators) of $\mu^N, \bar{\mu}$, in the sense of Definition 3.1.

To prove this, it is sufficient to check that $\mu^N, \bar{\mu}$ satisfy the four conditions of Proposition 3.4. Condition 1 holds true for $\mathcal{A} = \mathcal{P}^4$ by Proposition 3.7. Besides, Condition 2 is a direct consequence of Theorems 1.5 and 1.7.
$$
\begin{align*}
&+ \frac{1}{2} \int_r^t \sum_{\substack{i,j=1 \\ i \neq j}}^n \frac{1}{N} \cdot \frac{1}{N^{n-1}} \sum_{\mathbf{k} \setminus \{j\} \in [1,N]^n} \partial_{ij}^2 \varphi(\mathbf{X}_s^{N,(\mathbf{k} \setminus \{j\} k_i)}) \sigma(\mu_s^N, X_s^{N,k_i})^2 ds \\
&= \frac{1}{2} \int_r^t \sum_{i=1}^n \int_{\mathbb{R}^n} \partial_{ii}^2 \varphi(x) \sigma(\mu_s^N, x_i)^2 d(\mu_s^N)^{\otimes n}(x) ds \\
&\quad + \frac{1}{2N} \int_r^t \sum_{\substack{i,j=1 \\ i \neq j}}^n \int_{\mathbb{R}^{n-1}} \partial_{ij}^2 \varphi(x \setminus \{j\} x_i) \sigma(\mu_s^N, x_i)^2 d(\mu_s^N)^{\otimes n-1}(x \setminus \{j\}) ds \\
&= \frac{1}{2} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 \partial_{xx}^2 \delta G(\mu_s^N, x) d\mu_s^N(x) ds \\
&\quad + \frac{1}{2N} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(\mu_s^N, y_1, y_2))_{|y_1=y_2=x} d\mu_s^N(x) ds,
\end{align*}
$$

where the last equality uses (28) and (29), and is mostly a notation matter.

In particular, doing the same computation trick as the one done at the end of the proof of
Theorem 1.5, it is possible to write the dynamics of $G(\mu_t^N)$ as follows

$$
\begin{align*}
G(\mu_t^N) ={}& G(\mu_r^N) + M_t^{N,r} + \int_r^t \int_{\mathbb{R}} b(\mu_s^N, x) \partial_x \delta G(m, x) d\mu_s^N(x) ds \\
& + \frac{1}{2} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 \partial_{xx}^2 \delta G(\mu_s^N, x) d\mu_s^N(x) ds \\
& + \frac{1}{2N} \int_r^t \int_{\mathbb{R}} \sigma(\mu_s^N, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(\mu_s^N, y_1, y_2))_{|y_1=y_2=x} d(\mu_s^N(x))^{\otimes 2}(x) ds \\
& + \frac{1}{2} \int_r^t \int_{\mathbb{R}^2} \varsigma(\mu_s^N, x) \varsigma(\mu_s^N, y) \partial_{xy}^2 \delta^2 G(\mu_s^N, x, y) d(\mu_s^N)^{\otimes 2}(x, y) ds \\
& + N \int_r^t \int_{\mathbb{R}} \int_{\mathbb{R}} f(\mu_s^N, y) [G(\mathcal{S}(\mu_s^N, h(\mu_s^N, y, u))) - G(\mu_s^N)] d\mu_s^N(y) d\nu(u) ds,
\end{align*}
$$

with $(M_t^{N,r})_t$ a martingale that depends only on the process $\mu^N$, on the Brownian motions $W, B^k$
$(1 \le k \le N)$ and on the Poisson measures $\pi^l$ $(1 \le l \le N)$ such that it adapted to the union
of the filtrations of these objects. The equality above being true for all $G \in \mathcal{P}^2$, Proposition 3.3
entails that the measure-valued process $(\mu_t^N)_t$ is a Markov process (which was already guaranteed
by Proposition 2.3.3 of ?), but also that the domain of the generator of $\mu^N$ contains the set $\mathcal{P}^2$,
and that the expression of the generator on any $G \in \mathcal{P}^2$ is the one given in the statement of
Theorem 1.7.

4. Proof of the results of Section 1.1

4.1. *Proof of Theorem 1.4*

It is sufficient to prove the result for $n = 1$. Indeed, the general case can be deduced inductively
using the following Markovian properties of the processes: if $(\mu_t)_t$ is an E-valued homogeneous
Now, let us verify Condition 3 of Proposition 3.4. Let $G \in \mathcal{P}^4$, by Theorem 1.5, the generator $\bar{A}$ of $\bar{\mu}$ satisfy: for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ \bar{A}G(m) = \int_{\mathbb{R}} I(m, x)dm(x) + \int_{\mathbb{R}^2} J(m, x, y)dm^{\otimes 2}(x, y), $$

with

$$ I(m, x) = b(m, x)\partial_x\delta G(m, x) + \frac{1}{2}\sigma(m, x)^2\partial_{xx}^2\delta G(m, x) \quad (31) $$

$$ + \frac{1}{2} \left( \int_{\mathbb{R}} u^2 \, d\nu(u) \right) \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) \, dm(z) \right) \partial_{xx}^2 \delta G(m, x), $$

$$ J(m, x, y) = \frac{1}{2} \left( \int_{\mathbb{R}} u^2 \, d\nu(u) \right) \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) \, dm(z) \right) \partial_{xy}^2 \delta^2 G(m, x, y). \quad (32) $$

In particular thanks to Proposition 2.31, we have the following control: there exists $C > 0$ such that for all $G \in \mathcal{P}^4$ and $m \in \mathcal{P}_2(\mathbb{R})$,

$$ |\bar{A}G(m)| \le C \left( 1 + \int_{\mathbb{R}} |x|^2 dm(x) \right) ||G||_2. \quad (33) $$

And, for all $t_0, t \le T$,

$$ \begin{aligned} |\bar{A}G(\bar{\mu}_{t_0}) - \bar{A}G(\bar{\mu}_t)| & \le \left| \int_{\mathbb{R}} I(\bar{\mu}_{t_0}, x) d(\bar{\mu}_{t_0} - \bar{\mu}_t)(x) \right| + \int_{\mathbb{R}} |I(\bar{\mu}_{t_0}, x) - I(\bar{\mu}_t, x)| d\bar{\mu}_t(x) \\ & \quad + \left| \int_{\mathbb{R}^2} J(\bar{\mu}_{t_0}, x, y) d(\bar{\mu}_{t_0}^{\otimes 2} - \bar{\mu}_t^{\otimes 2})(x, y) \right| \\ & \quad + \int_{\mathbb{R}^2} |J(\bar{\mu}_{t_0}, x, y) - J(\bar{\mu}_t, x, y)| d\bar{\mu}_t^{\otimes 2}(x, y) \\ & \le \mathbb{E} \left[ |I(\bar{\mu}_{t_0}, \bar{X}_{t_0}) - I(\bar{\mu}_{t_0}, \bar{X}_t)| \mathcal{W}_T \right] \\ & \quad + \mathbb{E} \left[ |I(\bar{\mu}_{t_0}, \bar{X}_t) - I(\bar{\mu}_t, \bar{X}_t)| \mathcal{W}_T \right] \\ & \quad + \mathbb{E} \left[ |J(\bar{\mu}_{t_0}, \bar{X}_{t_0}^1, \bar{X}_{t_0}^2) - J(\bar{\mu}_{t_0}, \bar{X}_t^1, \bar{X}_t^2)| \mathcal{W}_T \right] \\ & \quad + \mathbb{E} \left[ |J(\bar{\mu}_{t_0}, \bar{X}_t^1, \bar{X}_t^2) - J(\bar{\mu}_t, \bar{X}_t^1, \bar{X}_t^2)| \mathcal{W}_T \right], \end{aligned} $$

with $\bar{X}^1, \bar{X}^2$ solutions to the SDE (4) w.r.t. the same Brownian motion $W$, but w.r.t. two independent Brownian motions $B^1, B^2$ instead of $B$. Let us note that, the functions $I, J$ are locally Lipschitz continuous in the following sense: there exists $C_G > 0$ such that, for all $m_1, m_2 \in \mathcal{P}_2(\mathbb{R})$, and $x_1, x_2, y_1, y_2 \in \mathbb{R}$,

$$ \begin{aligned} & |I(m_1, x_1) - I(m_2, x_2)| + |J(m_1, x_1, y_1) - J(m_2, x_2, y_2)| \\ & \le C_G \left( 1 + |x_1|^2 + |x_2|^2 + |y_1|^2 + |y_2|^2 + \int_{\mathbb{R}} |z|^2 dm_1(z) + \int_{\mathbb{R}} |z|^2 dm_2(z) \right) \\ & \qquad \qquad \qquad \qquad (|x_1 - x_2| + D_{KR}(m_1, m_2)). \end{aligned} $$

In particular, by Cauchy-Schwarz' inequality and Jensen's inequality and (5), denoting $E_m$ the expectation w.r.t. the probability measure under which $\bar{\mu}_0 = m$,

$$ |\bar{P}_{t_0} \bar{A} G(m) - \bar{P}_t \bar{A} G(m)| \le E_m [ |\bar{A} G(\bar{\mu}_{t_0}) - \bar{A} G(\bar{\mu}_t)| ] $$
Markov process with semigroup $(P_t)_t$, then for $t_1 < t_2$, and $G_1, G_2$ “smooth enough”,

$$ \mathbb{E}[G_1(\mu_{t_1})G_2(\mu_{t_2})] = \mathbb{E}[G_1(\mu_{t_1})P_{t_2-t_1}G_2(\mu_{t_1})]. $$

So we only prove the result for $n = 1$. Let us recall that, for each $N \in \mathbb{N}^*$, the processes $X^{N,k}$ ($1 \le k \le N$) are defined as the (strong) solutions of (3), and $\bar{X}$ as the (strong) solution of (4). And, by definition, for all $t \ge 0$,

$$ \mu_t^N = \frac{1}{N} \sum_{k=1}^{N} \delta_{X_t^{N,k}} \quad \text{and} \quad \bar{\mu}_t = \mathcal{L}(\bar{X}_t | \mathcal{W}_t). $$

Let us denote

$$ \mathcal{E}_N = \left\{ m_w = \frac{1}{N} \sum_{k=1}^{N} \delta_{w_k} : w_1, \dots, w_N \in \mathbb{R} \right\} \subseteq \mathcal{P}_{\infty}(\mathbb{R}). $$

It is classical that for any $t \ge 0$, there exists $C_t > 0$ such that for all $p \in \mathbb{N}^*$, $m \in \mathcal{P}_p(\mathbb{R})$ and $w_1, \dots, w_N \in \mathbb{R}$

$$ \mathbb{E}_m \left[ \sup_{s \le t} |\bar{X}_s|^p \right] \le C_t \left( 1 + \int_{\mathbb{R}} |x|^p dm(x) \right), \quad (34) $$

$$ \frac{1}{N} \sum_{k=1}^{N} \mathbb{E}_{m_w} \left[ \sup_{s \le t} |X_s^{N,k}|^p \right] \le C_t \left( 1 + \frac{1}{N} \sum_{k=1}^{N} |w_k|^p \right), \quad (35) $$

where $m_w = N^{-1} \sum_{k=1}^{N} w_k$, $\mathbb{E}_m$ (resp. $\mathbb{E}_{m_w}$) denotes the expectation w.r.t. the probability measure under which the Markov process $\bar{\mu}$ (resp. $\mu^N$) has $m$ (resp. $m_w$) as initial condition.

*Step 1.* The first step of the proof consists in showing that, for any fixed $N \in \mathbb{N}^*$, the $\mathcal{P}_1(\mathbb{R})$-valued Markov processes $\mu^N$ and $\bar{\mu}$ satisfy the following Trotter-Kato formula: for all $G \in \mathcal{P}^4$, $m \in \mathcal{E}_N$ and $t \ge 0$,

$$ (\bar{P}_t - P_t^N) G(m) = \int_0^t P_{t-s}^N (\bar{A} - A^N) \bar{P}_s G(m) ds, $$

with $P^N, \bar{P}$ (resp. $A^N, \bar{A}$) the semigroups (resp. generators) of $\mu^N, \bar{\mu}$, in the sense of Definition 3.1.

To prove this, it is sufficient to check that $\mu^N, \bar{\mu}$ satisfy the four conditions of Proposition 3.4. Condition 1 holds true for $\mathcal{A} = \mathcal{P}^4$ by Proposition 3.7. Besides, Condition 2 is a direct consequence of Theorems 1.5 and 1.7.

Now, let us verify Condition 3 of Proposition 3.4. Let $G \in \mathcal{P}^4$, by Theorem 1.5, the generator $\bar{A}$ of $\bar{\mu}$ satisfy: for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ \bar{A}G(m) = \int_{\mathbb{R}} I(m, x)dm(x) + \int_{\mathbb{R}^2} J(m, x, y)dm^{\otimes 2}(x, y), $$

with

$$ I(m, x) = b(m, x) \partial_x \delta G(m, x) + \frac{1}{2} \sigma(m, x)^2 \partial_{xx}^2 \delta G(m, x) \quad (36) $$

$$ + \frac{1}{2} \left( \int_{\mathbb{R}} u^2 \, d\nu(u) \right) \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) \, dm(z) \right) \partial_{xx}^2 \delta G(m, x), $$
$$
\begin{align*}
&\le C_G \mathbb{E}_m \left[ \left( 1 + \sup_{s \le T} \mathbb{E} \left[ |\bar{X}_s|^2 | \mathcal{W}_T \right] \right) \mathbb{E} \left[ |\bar{X}_{t_0} - \bar{X}_t| | \mathcal{W}_T \right] \right] \\
&\le C_G \mathbb{E}_m \left[ |\bar{X}_{t_0} - \bar{X}_t|^2 \right]^{1/2}.
\end{align*}
$$

Recalling that $\bar{X}$ is solution to (4), it is classical (using for example Burkholder-Davis-Gundy's inequality with (5)) that the RHS of the above inequality vanishes as $t$ goes to $t_0$. This implies that, for any $m \in \mathcal{P}_2(\mathbb{R})$ and $G \in \mathcal{P}^4$, the function

$t \mapsto \bar{P}_t \bar{A} G(m)$

is continuous.

The continuity of the function

$t \mapsto P_t^N A^N G(m),$

for $m \in \mathcal{E}_N$ follows from the same reasoning provided Lemma 4.1 below to deal with the jump term that depends on a shift operator $\mathcal{S}$. Note that it is important to work with an initial condition belonging to $\mathcal{E}_N$ in order to interpret $\mu_t^N$ as the empirical measure of some $N$-particle system satisfying the SDEs (3).

**Lemma 4.1.** Let $m_1, m_2 \in \mathcal{P}_1(\mathbb{R})$ and $h_1, h_2 \in \mathbb{R}$. Then

$$
D_{KR}(\mathcal{S}(m_1, h_1), \mathcal{S}(m_2, h_2)) \le D_{KR}(m_1, m_2) + |h_1 - h_2|.
$$

*Proof.* Let $\varphi : \mathbb{R} \to \mathbb{R}$ be any Lipschitz continuous function with Lipschitz constant non-greater than one.

$$
\begin{align*}
\left| \int_{\mathbb{R}} \varphi(x) d(\mathcal{S}(m_1, h_1))(x) - \int_{\mathbb{R}} \varphi(x) d(\mathcal{S}(m_2, h_2))(x) \right| \\
&= \left| \int_{\mathbb{R}} \varphi(x+h_1) dm_1(x) - \int_{\mathbb{R}} \varphi(x+h_2) dm_2(x) \right| \\
&\le \left| \int_{\mathbb{R}} \varphi(x+h_1) d(m_1-m_2)(x) \right| + \int_{\mathbb{R}} |\varphi(x+h_1) - \varphi(x+h_2)| dm_2(x) \\
&\le D_{KR}(m_1, m_2) + |h_1 - h_2|.
\end{align*}
$$

Taking the supremum over all such functions $\varphi$ proves the result.

<figure><img src="image_5.png" /></figure>

Indeed, with the same computation, we obtain

$$
\begin{align*}
|P_{t_0}^N A^N G(m) - P_t^N A^N G(m)| &\le \mathbb{E}_m [|A^N G(\mu_{t_0}^N) - A^N G(\mu_t^N)|] \\
&\le C_G \mathbb{E}_m \left[ \left( 1 + \sup_{s \le T} \frac{1}{N} \sum_{k=1}^N |X_s^{N,k}|^2 \right) \frac{1}{N} \sum_{k=1}^N |X_{t_0}^{N,k} - X_t^{N,k}| \right] \\
&\le C_G \left( \frac{1}{N} \sum_{k=1}^N \mathbb{E}_m \left[ |X_{t_0}^{N,k} - X_t^{N,k}|^2 \right] \right)^{1/2} .
\end{align*}
$$

Once again, this proves that for any $m \in \mathcal{E}_N$ and $G \in \mathcal{P}^4$, the function

$t \mapsto P_t^N A^N G(m)$
$$J(m, x, y) = \frac{1}{2} \left( \int_{\mathbb{R}} u^2 \, d\nu(u) \right) \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) \, dm(z) \right) \partial_{xy}^2 \delta^2 G(m, x, y). \quad (37)$$

In particular thanks to Proposition 2.32, we have the following control: there exists $C > 0$ such that for all $G \in \mathcal{P}^4$ and $m \in \mathcal{P}_2(\mathbb{R})$,

$$|\bar{A}G(m)| \le C \left( 1 + \int_{\mathbb{R}} |x|^2 dm(x) \right) ||G||_2. \quad (38)$$

And, for all $t_0, t \le T$,

$$\begin{aligned} |\bar{A}G(\bar{\mu}_{t_0}) - \bar{A}G(\bar{\mu}_t)| & \le \left| \int_{\mathbb{R}} I(\bar{\mu}_{t_0}, x) d(\bar{\mu}_{t_0} - \bar{\mu}_t)(x) \right| + \int_{\mathbb{R}} |I(\bar{\mu}_{t_0}, x) - I(\bar{\mu}_t, x)| d\bar{\mu}_t(x) \\ & \quad + \left| \int_{\mathbb{R}^2} J(\bar{\mu}_{t_0}, x, y) d(\bar{\mu}_{t_0}^{\otimes 2} - \bar{\mu}_t^{\otimes 2})(x, y) \right| \\ & \quad + \int_{\mathbb{R}^2} |J(\bar{\mu}_{t_0}, x, y) - J(\bar{\mu}_t, x, y)| d\bar{\mu}_t^{\otimes 2}(x, y) \\ & \le \mathbb{E} \left[ |I(\bar{\mu}_{t_0}, \bar{X}_{t_0}) - I(\bar{\mu}_{t_0}, \bar{X}_t)| |W_T| \right] \\ & \quad + \mathbb{E} \left[ |I(\bar{\mu}_{t_0}, \bar{X}_t) - I(\bar{\mu}_t, \bar{X}_t)| |W_T| \right] \\ & \quad + \mathbb{E} \left[ |J(\bar{\mu}_{t_0}, \bar{X}_{t_0}^1, \bar{X}_{t_0}^2) - J(\bar{\mu}_{t_0}, \bar{X}_t^1, \bar{X}_t^2)| |W_T| \right] \\ & \quad + \mathbb{E} \left[ |J(\bar{\mu}_{t_0}, \bar{X}_t^1, \bar{X}_t^2) - J(\bar{\mu}_t, \bar{X}_t^1, \bar{X}_t^2)| |W_T| \right], \end{aligned}$$

with $\bar{X}^1, \bar{X}^2$ solutions to the SDE (4) w.r.t. the same Brownian motion $W$, but w.r.t. two independent Brownian motions $B^1, B^2$ instead of $B$. Let us note that, the functions $I, J$ are locally Lipschitz continuous in the following sense: there exists $C_G > 0$ such that, for all $m_1, m_2 \in \mathcal{P}_2(\mathbb{R})$, and $x_1, x_2, y_1, y_2 \in \mathbb{R}$,

$$\begin{aligned} & |I(m_1, x_1) - I(m_2, x_2)| + |J(m_1, x_1, y_1) - J(m_2, x_2, y_2)| \\ & \le C_G \left( 1 + |x_1|^2 + |x_2|^2 + |y_1|^2 + |y_2|^2 + \int_{\mathbb{R}} |z|^2 dm_1(z) + \int_{\mathbb{R}} |z|^2 dm_2(z) \right) \\ & \qquad (|x_1 - x_2| + D_{KR}(m_1, m_2)). \end{aligned}$$

In particular, by Cauchy-Schwarz' inequality and Jensen's inequality and (5), denoting $E_m$ the expectation w.r.t. the probability measure under which $\bar{\mu}_0 = m$,

$$\begin{aligned} |\bar{P}_{t_0} \bar{A} G(m) - \bar{P}_t \bar{A} G(m)| & \le E_m [ |\bar{A} G(\bar{\mu}_{t_0}) - \bar{A} G(\bar{\mu}_t)| ] \\ & \le C_G E_m \left[ \left( 1 + \sup_{s \le T} \mathbb{E} [ |\bar{X}_s|^2 |W_T| ] \right) \mathbb{E} [ |\bar{X}_{t_0} - \bar{X}_t| |W_T| ] \right] \\ & \le C_G E_m [ |\bar{X}_{t_0} - \bar{X}_t|^2 ]^{1/2}. \end{aligned}$$

Recalling that $\bar{X}$ is solution to (4), it is classical (using for example Burkholder-Davis-Gundy's inequality with (5)) that the RHS of the above inequality vanishes as $t$ goes to $t_0$. This implies that, for any $m \in \mathcal{P}_2(\mathbb{R})$ and $G \in \mathcal{P}^4$, the function

$$t \mapsto \bar{P}_t \bar{A} G(m)$$
is continuous. The continuity of $t \mapsto P_t^N \bar{A}G(m)$ is guaranteed by the same arguments. So Condition 3 of Proposition 3.4 is verified.

To end the *Step 1* of the proof, let us now check Condition 4. Let us fix in the rest of *Step 1* some $G \in \mathcal{P}^4$, $0 \le t < T$ and some sequence $(s_n)_n$ converging to $t$ such that $s_n \le T$ for all $n \in \mathbb{N}$. For $M > 0$, let us introduce

$$ \mathcal{K}_M = \{m \in \mathcal{P}_1(\mathbb{R}) : \text{Supp}(m) \subseteq [-M, M]\}. $$

For any $M > 0$, $\mathcal{K}_M$ is a compact set of $\mathcal{P}_p(\mathbb{R})$, for all $p \ge 1$ (see Remark 6.19 of Villani (2009)). Let us denote, for every $n \in \mathbb{N}$,

$$ G_n = \bar{P}_t G - \bar{P}_{s_n} G. $$

Thanks to Proposition 3.7, we know that each $G_n$ belongs to $\mathcal{P}^4$. Hence, by Proposition 2.31, we can use Corollaries 2.26 and 2.27 (and Lemma 2.18 to differentiate the jump term from the generator $A^N$) to compute $\partial(\bar{A}G_n)(m, x)$ and $\partial(A^N g_n)(m, x)$ (recalling the expressions of $\bar{A}$ and $A^N$ given in Theorems 1.5 and 1.7) to prove that, for all $M > 0$,

$$ \sup_{\substack{n \in \mathbb{N} \\ m \in \mathcal{K}_M \\ |x| \le M}} \left| \partial_x \delta(\bar{A}G_n)(m, x) \right| + \sup_{\substack{n \in \mathbb{N} \\ m \in \mathcal{K}_M \\ |x| \le M}} \left| \partial_x \delta(A^N G_n)(m, x) \right| \le C_G (1 + M^2). $$

Note that the fact that the control above is uniform in $n$ is guaranteed by Proposition 3.7.(b).

Then, by the Mean Value Theorem for measure-variable functions (i.e. Proposition 2.21) and since $\mathcal{K}_M$ ($M > 0$) are convex sets, we know that the families of functions $(\bar{A}G_n)_n$ and $(A^N G_n)_n$ are both uniformly equicontinuous on each compact set $\mathcal{K}_M$ (for any $M > 0$). In addition, let us recall that it is possible to write the generator $\bar{A}$ and $A^N$ in the following form: for any $m \in \mathcal{P}_2(\mathbb{R})$,

$$ \begin{aligned} \bar{A}G_n(m) &= \int_{\mathbb{R}} I_n(m, x)dm(x) + \int_{\mathbb{R}^2} J_n(m, x, y)dm^{\otimes 2}(x, y) \\ A^N G_n(m) &= \int_{\mathbb{R}} I_n^N(m, x)dm(x) + \int_{\mathbb{R}^2} J_n^N(m, x, y)dm^{\otimes 2}(x, y), \end{aligned} $$

with $I_n, J_n, I_n^N$ and $J_n^N$ defined similarly as $I, J$ in (31) and (32), for the function $G_n$ instead of $G$.

And, by Proposition 3.7.(c), for any $x, y \in \mathbb{R}$ and $m \in \mathcal{P}_2(\mathbb{R})$, the integrands above vanish as $n$ goes to infinity, and are bounded by

$$ C_G \left( 1 + x^2 + y^2 + \int |z|^2 dm(z) \right), $$

where $C_G > 0$ does not depend on $(m, x, y, n)$. Then, by the Dominated Convergence Theorem, for any $m \in \mathcal{K}_M$ (for any $M > 0$), both $\bar{A}G_n(m)$ and $A^N G_n(m)$ vanish as $n$ goes to infinity. Since these two sequences are uniformly equicontinuous on each compact set $\mathcal{K}_M$, this implies that, for all $M > 0$,

$$ \sup_{m \in \mathcal{K}_M} |\bar{A}G_n(m)| + \sup_{m \in \mathcal{K}_M} |A^N G_n(m)| \xrightarrow{n \to \infty} 0. \quad (34) $$

Then, let us introduce, for $M > 0$ the following event

$$ D_M = \left\{ \sup_{1 \le k \le N} |X_{r}^{N,k}| \le M \right\}, $$
is continuous.

The continuity of the function

$$t \mapsto P_t^N A^N G(m),$$

for $m \in \mathcal{E}_N$ follows from the same reasoning provided Lemma 4.1 below to deal with the jump term that depends on a shift operator $\mathcal{S}$. Note that it is important to work with an initial condition belonging to $\mathcal{E}_N$ in order to interpret $\mu_t^N$ as the empirical measure of some $N$-particle system satisfying the SDEs (3).

**Lemma 4.1.** Let $m_1, m_2 \in \mathcal{P}_1(\mathbb{R})$ and $h_1, h_2 \in \mathbb{R}$. Then

$$D_{KR}(\mathcal{S}(m_1, h_1), \mathcal{S}(m_2, h_2)) \le D_{KR}(m_1, m_2) + |h_1 - h_2|.$$

*Proof.* Let $\varphi : \mathbb{R} \to \mathbb{R}$ be any Lipschitz continuous function with Lipschitz constant non-greater than one.

$$\begin{aligned} & \left| \int_{\mathbb{R}} \varphi(x) d(\mathcal{S}(m_1, h_1))(x) - \int_{\mathbb{R}} \varphi(x) d(\mathcal{S}(m_2, h_2))(x) \right| \\ &= \left| \int_{\mathbb{R}} \varphi(x+h_1) dm_1(x) - \int_{\mathbb{R}} \varphi(x+h_2) dm_2(x) \right| \\ &\le \left| \int_{\mathbb{R}} \varphi(x+h_1) d(m_1-m_2)(x) \right| + \int_{\mathbb{R}} |\varphi(x+h_1) - \varphi(x+h_2)| dm_2(x) \\ &\le D_{KR}(m_1, m_2) + |h_1 - h_2|. \end{aligned}$$

Taking the supremum over all such functions $\varphi$ proves the result.

<figure><img src="image_4.png" /></figure>

Indeed, with the same computation, we obtain

$$\begin{aligned} & |P_{t_0}^N A^N G(m) - P_t^N A^N G(m)| \le \mathbb{E}_m [|A^N G(\mu_{t_0}^N) - A^N G(\mu_t^N)|] \\ & \le C_G \mathbb{E}_m \left[ \left( 1 + \sup_{s \le T} \frac{1}{N} \sum_{k=1}^N |X_s^{N,k}|^2 \right) \frac{1}{N} \sum_{k=1}^N |X_{t_0}^{N,k} - X_t^{N,k}| \right] \\ & \le C_G \left( \frac{1}{N} \sum_{k=1}^N \mathbb{E}_m \left[ |X_{t_0}^{N,k} - X_t^{N,k}|^2 \right] \right)^{1/2}. \end{aligned}$$

Once again, this proves that for any $m \in \mathcal{E}_N$ and $G \in \mathcal{P}^4$, the function

$$t \mapsto P_t^N A^N G(m)$$

is continuous. The continuity of $t \mapsto P_t^N \bar{A}G(m)$ is guaranteed by the same arguments. So Condition 3 of Proposition 3.4 is verified.

To end the *Step 1* of the proof, let us now check Condition 4. Let us fix in the rest of *Step 1* some $G \in \mathcal{P}^4$, $0 \le t < T$ and some sequence $(s_n)_n$ converging to $t$ such that $s_n \le T$ for all $n \in \mathbb{N}$. For $M > 0$, let us introduce

$$\mathcal{K}_M = \{m \in \mathcal{P}_1(\mathbb{R}) : \text{Supp}(m) \subseteq [-M, M]\}.$$

For any $M > 0$, $\mathcal{K}_M$ is a compact set of $\mathcal{P}_p(\mathbb{R})$, for all $p \ge 1$ (see Remark 6.19 of?). Let us denote, for every $n \in \mathbb{N}$,

$$G_n = \bar{P}_t G - \bar{P}_{s_n} G.$$
and $D_M^c$ its complementary. We have, by Cauchy-Schwarz' inequality and Markov's inequality, for any $m \in \mathcal{E}_N$,

$$ \begin{aligned} \sup_{r \le T} |P_r^N \bar{A} G_n(m)| &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| \right] \\ &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| 1_{D_M} \right] + \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N) 1_{D_M^c}| \right] \\ &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| 1_{D_M} \right] + C_{T,G,m} M^{-1/2}, \end{aligned} $$

where we have used the controls given at (30), (33) and Proposition 3.7.(b).

In particular, for any $\varepsilon > 0$ it is possible to fix some $M_\varepsilon > 0$ (whose also depends on $T, G, m$, but not on $n$) such that

$$ \sup_{r \le T} |P_r^N \bar{A} G_n(m)| \le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| 1_{D_{M_\varepsilon}} \right] + \varepsilon. $$

Noticing that,

$$ D_{M_\varepsilon} \subseteq \{\forall r \le T, \mu_r^N \in \mathcal{K}_{M_\varepsilon}\}, $$

we know, by (34), that, almost surely,

$$ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| 1_{D_{M_\varepsilon}} \xrightarrow{n \to \infty} 0, $$

and, by the Dominated Convergence Theorem (and (30), (33) and Proposition 3.7.(b)), this entails that, for $m \in \mathcal{E}_N$,

$$ \sup_{r \le T} |P_r^N \bar{A} G_n(m)| \xrightarrow{n \to \infty} 0. $$

With exactly the same reasoning, we also have

$$ \sup_{r \le T} |P_r^N A^N G_n(m)| \xrightarrow{n \to \infty} 0. $$

So Condition 4 of Proposition 3.4 is satisfied.

*Step 2.* We have proved that for all $G \in \mathcal{P}^4$, $m \in \mathcal{E}_N$ and $t \ge 0$,

$$ (\bar{P}_t - P_t^N) G(m) = \int_0^t P_{t-s}^N (\bar{A} - A^N) \bar{P}_s G(m) ds. $$

This formula allows to obtain a convergence speed for the semigroups provided a convergence speed of the generators. So, the goal of *Step 2* is to prove the following assertion: for $G \in \mathcal{P}^3$ and $m \in \mathcal{P}_4(\mathbb{R})$,

$$ |(A^N - \bar{A}) G(m)| \le C ||G||_3 \frac{1}{\sqrt{N}} \left( 1 + \int_{\mathbb{R}} |x|^4 dm(x) \right). \quad (35) $$

To this end, let us admit the following lemma, whose proof consists in using three times the Taylor-Lagrange's inequality. This lemma allows to simplify computation in our proof.
Thanks to Proposition 3.7, we know that each $G_n$ belongs to $\mathcal{P}^4$. Hence, by Proposition 2.32, we can use Corollaries 2.26 and 2.27 (and Lemma 2.18 to differentiate the jump term from the generator $A^N$) to compute $\partial(\bar{A}G_n)(m, x)$ and $\partial(A^N g_n)(m, x)$ (recalling the expressions of $\bar{A}$ and $A^N$ given in Theorems 1.5 and 1.7) to prove that, for all $M > 0$,

$$ \sup_{\substack{n \in \mathbb{N} \\ m \in \mathcal{K}_M \\ |x| \le M}} |\partial_x \delta(\bar{A}G_n)(m, x)| + \sup_{\substack{n \in \mathbb{N} \\ m \in \mathcal{K}_M \\ |x| \le M}} |\partial_x \delta(A^N G_n)(m, x)| \le C_G(1 + M^2). $$

Note that the fact that the control above is uniform in $n$ is guaranteed by Proposition 3.7.(b).

Then, by the Mean Value Theorem for measure-variable functions (i.e. Proposition 2.21) and since $\mathcal{K}_M$ ($M > 0$) are convex sets, we know that the families of functions $(\bar{A}G_n)_n$ and $(A^N G_n)_n$ are both uniformly equicontinuous on each compact set $\mathcal{K}_M$ (for any $M > 0$). In addition, let us recall that it is possible to write the generator $\bar{A}$ and $A^N$ in the following form: for any $m \in \mathcal{P}_2(\mathbb{R})$,

$$ \begin{aligned} \bar{A}G_n(m) &= \int_{\mathbb{R}} I_n(m, x)dm(x) + \int_{\mathbb{R}^2} J_n(m, x, y)dm^{\otimes 2}(x, y) \\ A^N G_n(m) &= \int_{\mathbb{R}} I_n^N(m, x)dm(x) + \int_{\mathbb{R}^2} J_n^N(m, x, y)dm^{\otimes 2}(x, y), \end{aligned} $$

with $I_n, J_n, I_n^N$ and $J_n^N$ defined similarly as $I, J$ in (36) and (37), for the function $G_n$ instead of $G$.

And, by Proposition 3.7.(c), for any $x, y \in \mathbb{R}$ and $m \in \mathcal{P}_2(\mathbb{R})$, the integrands above vanish as $n$ goes to infinity, and are bounded by

$$ C_G \left( 1 + x^2 + y^2 + \int |z|^2 dm(z) \right), $$

where $C_G > 0$ does not depend on $(m, x, y, n)$. Then, by the Dominated Convergence Theorem, for any $m \in \mathcal{K}_M$ (for any $M > 0$), both $\bar{A}G_n(m)$ and $A^N G_n(m)$ vanish as $n$ goes to infinity. Since these two sequences are uniformly equicontinuous on each compact set $\mathcal{K}_M$, this implies that, for all $M > 0$,

$$ \sup_{m \in \mathcal{K}_M} |\bar{A}G_n(m)| + \sup_{m \in \mathcal{K}_M} |A^N G_n(m)| \underset{n \to \infty}{\longrightarrow} 0. \quad (39) $$

Then, let us introduce, for $M > 0$ the following event

$$ D_M = \left\{ \sup_{\substack{r \le T \\ 1 \le k \le N}} |X_r^{N,k}| \le M \right\}, $$

and $D_M^c$ its complementary. We have, by Cauchy-Schwarz' inequality and Markov's inequality, for any $m \in \mathcal{E}_N$,

$$ \begin{aligned} \sup_{r \le T} |P_r^N \bar{A}G_n(m)| &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A}G_n(\mu_r^N)| \right] \\ &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A}G_n(\mu_r^N)| 1_{D_M} \right] + \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A}G_n(\mu_r^N) 1_{D_M^c}| \right] \\ &\le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A}G_n(\mu_r^N)| 1_{D_M} \right] + C_{T,G,m} M^{-1/2}, \end{aligned} $$
**Lemma 4.2.** Let $g \in C_b^3(\mathbb{R}^2)$. Then, for all $x, y, \lambda \in \mathbb{R}$,

$$ \left| g(x + \lambda, y + \lambda) - g(x + \lambda, y) - g(x, y + \lambda) + g(x, y) - \frac{\lambda^2}{2} \partial_{xy}^2 g(x, y) \right| \le \frac{|\lambda|^3}{2} \sum_{|\alpha|=3} \|\partial_\alpha g\|_{\infty}. $$

Let us denote $\zeta > 0$ the standard deviation of the probability measure $\nu$:

$$ \zeta = \left( \int_{\mathbb{R}} u^2 d\nu(u) \right)^{1/2}. $$

Then, by Theorems 1.5 and 1.7, for all $G \in \mathcal{P}^2$ and $m \in \mathcal{P}_2(\mathbb{R})$,

$$ \begin{aligned} \bar{A}G(m) = & \int_{\mathbb{R}} \left[ b(m, x) \partial_x \delta G(m, x) + \frac{1}{2} \sigma(m, x)^2 \partial_{xx}^2 \delta G(m, x) \right] dm(x) \\ & + \frac{1}{2} \zeta^2 \int_{\mathbb{R}} \left( \int_{\mathbb{R}} h(m, y)^2 f(m, y) dm(y) \right) \partial_{xx}^2 \delta G(m, x) dm(x) \\ & + \frac{1}{2} \zeta^2 \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) dm(z) \right) \partial_{xy}^2 \delta^2 G(m, x, y) dm^{\otimes 2}(x, y), \end{aligned} $$

$$ \begin{aligned} A^N G(m) = & \int_{\mathbb{R}} \left[ b(m, x) \partial_x \delta G(m, x) + \frac{1}{2} \sigma(m, x)^2 \partial_{xx}^2 \delta G(m, x) \right] dm(x) \\ & + \frac{1}{2N} \int_{\mathbb{R}} \sigma(m, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(m, y_1, y_2))_{|y_1=y_2=x} dm(x) \\ & + N \int_{\mathbb{R}} \int_{\mathbb{R}} f(m, x) \left[ G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, x) \right) \right) - G(m) \right] d\nu(u) dm(x). \end{aligned} $$

So

$$ (A^N - \bar{A}) G(m) = \frac{1}{2N} \int_{\mathbb{R}} \sigma(m, x)^2 (\partial_{y_1 y_2}^2 \delta^2 G(m, y_1, y_2))_{|y_1=y_2=x} dm(x) \quad (36) $$

$$ + N \int_{\mathbb{R}} f(m, z) \int_{\mathbb{R}} \left[ G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) \right) - G(m) \right] dm \quad (37) $$

$$ - \frac{u^2}{2N} h(m, z)^2 \int_{\mathbb{R}} \partial_{xx}^2 \delta G(m, x) dm(x) $$

$$ - \frac{u^2}{2N} h(m, z)^2 \int_{\mathbb{R}^2} \partial_{xy}^2 \delta^2 G(m, x, y) dm^{\otimes 2}(x, y) \right] d\nu(u) dm(z). $$

The absolute value of the term at (36) is bounded by

$$ C \frac{1}{N} \|G\|_2 \left( 1 + \int_{\mathbb{R}} x^2 dm(x) \right). $$

For the rest of the proof, let us introduce, for $u, z \in \mathbb{R}$, the signed measure

$$ m_{z,u}^N = S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) - m. \quad (38) $$
where we have used the controls given at (35), (38) and Proposition 3.7.(b).

In particular, for any $\varepsilon > 0$ it is possible to fix some $M_\varepsilon > 0$ (whose also depends on $T, G, m$, but not on $n$) such that

$$ \sup_{r \le T} |P_r^N \bar{A} G_n(m)| \le \mathbb{E}_m \left[ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| \mathbb{1}_{D_{M_\varepsilon}} \right] + \varepsilon. $$

Noticing that,

$$ D_{M_\varepsilon} \subseteq \{\forall r \le T, \mu_r^N \in \mathcal{K}_{M_\varepsilon}\}, $$

we know, by (39), that, almost surely,

$$ \sup_{r \le T} |\bar{A} G_n(\mu_r^N)| \mathbb{1}_{D_{M_\varepsilon}} \xrightarrow{n \to \infty} 0, $$

and, by the Dominated Convergence Theorem (and (35), (38) and Proposition 3.7.(b)), this entails that, for $m \in \mathcal{E}_N$,

$$ \sup_{r \le T} |P_r^N \bar{A} G_n(m)| \xrightarrow{n \to \infty} 0. $$

With exactly the same reasoning, we also have

$$ \sup_{r \le T} |P_r^N A^N G_n(m)| \xrightarrow{n \to \infty} 0. $$

So Condition 4 of Proposition 3.4 is satisfied.

*Step 2.* We have proved that for all $G \in \mathcal{P}^4$, $m \in \mathcal{E}_N$ and $t \ge 0$,

$$ (\bar{P}_t - P_t^N) G(m) = \int_0^t P_{t-s}^N (\bar{A} - A^N) \bar{P}_s G(m) ds. $$

This formula allows to obtain a convergence speed for the semigroups provided a convergence speed of the generators. So, the goal of *Step 2* is to prove the following assertion: for $G \in \mathcal{P}^3$ and $m \in \mathcal{P}_4(\mathbb{R})$,

$$ |(A^N - \bar{A}) G(m)| \le C ||G||_3 \frac{1}{\sqrt{N}} \left( 1 + \int_{\mathbb{R}} |x|^4 dm(x) \right). \quad (40) $$

To this end, let us admit the following lemma, whose proof consists in using three times the Taylor-Lagrange's inequality. This lemma allows to simplify computation in our proof.

**Lemma 4.2.** Let $g \in C_b^3(\mathbb{R}^2)$. Then, for all $x, y, \lambda \in \mathbb{R}$,

$$ \left| g(x + \lambda, y + \lambda) - g(x + \lambda, y) - g(x, y + \lambda) + g(x, y) - \frac{\lambda^2}{2} \partial_{xy}^2 g(x, y) \right| \le \frac{|\lambda|^3}{2} \sum_{|\alpha|=3} ||\partial_\alpha g||_\infty. $$

Let us denote $\zeta > 0$ the standard deviation of the probability measure $\nu$:

$$ \zeta = \left( \int_{\mathbb{R}} u^2 d\nu(u) \right)^{1/2}. $$

Then, by Theorems 1.5 and 1.7, for all $G \in \mathcal{P}^2$ and $m \in \mathcal{P}_2(\mathbb{R})$,

$$ \bar{A}G(m) = \int_{\mathbb{R}} \left[ b(m, x) \partial_x \delta G(m, x) + \frac{1}{2} \sigma(m, x)^2 \partial_{xx}^2 \delta G(m, x) \right] dm(x) $$
To control the term at (37), by using Theorem 2.24, we need to compute the integrals of the k-th order derivative of G w.r.t. the power of $m_{z,u}^N$ to k, for $k \in \{1, 2\}$.

$$ \int_{\mathbb{R}} \int_{\mathbb{R}} \delta G(m, x) dm_{z,u}^N(x) d\nu(u) = \int_{\mathbb{R}} \int_{\mathbb{R}} \left[ \delta G \left( m, x + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta G(m, x) - \frac{u}{\sqrt{N}} h(m, z) \partial_x \delta G(m, x) \right] d\nu(u) dm(x), $$

where the term at the second line above has been added artificially since $\nu$ is a centered measure. In addition,

$$ \begin{aligned} & \int_{\mathbb{R}} \int_{\mathbb{R}^2} \delta^2 G(m, x, y) d(m_{z,u}^N)^{\otimes 2} (x, y) d\nu(u) \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} \left[ \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y + \frac{u}{\sqrt{N}} h(m, z) \right) \right. \\ & \qquad \left. - \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y \right) - \delta^2 G \left( m, x, y \frac{u}{\sqrt{N}} h(m, z) \right) \right. \\ & \qquad \left. + \delta^2 G(m, x, y) \right] d\nu(u) dm(x) dm(y). \end{aligned} $$

Whence, the absolute value of the term within the brackets at (37) is non-greater than the sum of the three following quantities:

$$ \left| G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) \right) - G(m) - \int_{\mathbb{R}} \delta G(m, x) dm_{z,u}^{N}(x) dm(x) \right| \quad (39) $$

$$ \int_{\mathbb{R}} \left| \delta G \left( m, x + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta G(m, x) - \frac{u}{\sqrt{N}} h(m, z) \partial_x \delta G(m, x) \right| dm(x), \quad (40) $$

$$ \int_{\mathbb{R}^2} \left| \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y \right) \right. \\ \left. - \delta^2 G \left( m, x, y \frac{u}{\sqrt{N}} h(m, z) \right) + \delta^2 G(m, x, y) - \frac{u^2}{2N} h(m, z)^2 \partial_{xy}^2 \delta^2 G(m, x, y) \right| dm^{\otimes 2}(x, y). \quad (41) $$

Each of the three terms above is bounded by

$$ C \|G\|_3 \frac{1}{N\sqrt{N}} \left( 1 + |z|^3 + \int_{\mathbb{R}} |x|^3 dm(x) \right) (1 + |u|^3). $$

Indeed, for (39), it is a consequence of the Taylor-Lagrange's inequality for measure-variable functions (i.e. Theorem 2.24), and the fact that (by Lemma 2.18)

$$ D_{KR} \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right), m \right) \leq \frac{|u|}{\sqrt{N}} |h(m, z)|, $$
$$ 
\begin{aligned}
& + \frac{1}{2} \zeta^2 \int_{\mathbb{R}} \left( \int_{\mathbb{R}} h(m, y)^2 f(m, y) dm(y) \right) \partial_{xx}^2 \delta G(m, x) dm(x) \\
& + \frac{1}{2} \zeta^2 \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}} h(m, z)^2 f(m, z) dm(z) \right) \partial_{xy}^2 \delta^2 G(m, x, y) dm^{\otimes 2}(x, y), \\
A^N G(m) = & \int_{\mathbb{R}} \left[ b(m, x) \partial_x \delta G(m, x) + \frac{1}{2} \sigma(m, x)^2 \partial_{xx}^2 \delta G(m, x) \right] dm(x) \\
& + \frac{1}{2N} \int_{\mathbb{R}} \sigma(m, x)^2 \left( \partial_{y_1 y_2}^2 \delta^2 G(m, y_1, y_2) \right)_{|y_1=y_2=x} dm(x) \\
& + N \int_{\mathbb{R}} \int_{\mathbb{R}} f(m, x) \left[ G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, x) \right) \right) - G(m) \right] d\nu(u) dm(x).
\end{aligned}
$$

So

$$ (A^N - \bar{A}) G(m) = \frac{1}{2N} \int_{\mathbb{R}} \sigma(m, x)^2 \left( \partial_{y_1 y_2}^2 \delta^2 G(m, y_1, y_2) \right)_{|y_1=y_2=x} dm(x) \quad (41) $$

$$ 
\begin{aligned}
& + N \int_{\mathbb{R}} f(m, z) \int_{\mathbb{R}} \left[ G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) \right) - G(m) \right] \quad (42) \\
& - \frac{u^2}{2N} h(m, z)^2 \int_{\mathbb{R}} \partial_{xx}^2 \delta G(m, x) dm(x) \\
& - \frac{u^2}{2N} h(m, z)^2 \int_{\mathbb{R}^2} \partial_{xy}^2 \delta^2 G(m, x, y) dm^{\otimes 2}(x, y) \Big] d\nu(u) dm(z).
\end{aligned}
$$

The absolute value of the term at (41) is bounded by

$$ C \frac{1}{N} \|G\|_2 \left( 1 + \int_{\mathbb{R}} x^2 dm(x) \right). $$

For the rest of the proof, let us introduce, for $u, z \in \mathbb{R}$, the signed measure

$$ m_{z,u}^N = S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) - m. \quad (43) $$

To control the term at (42), by using Theorem 2.24, we need to compute the integrals of the k-th order derivative of G w.r.t. the power of $m_{z,u}^N$ to k, for $k \in \{1, 2\}$.

$$ 
\begin{aligned}
\int_{\mathbb{R}} \int_{\mathbb{R}} \delta G(m, x) dm_{z,u}^N(x) d\nu(u) = & \int_{\mathbb{R}} \int_{\mathbb{R}} \left[ \delta G \left( m, x + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta G(m, x) \right. \\
& \left. - \frac{u}{\sqrt{N}} h(m, z) \partial_x \delta G(m, x) \right] d\nu(u) dm(x),
\end{aligned}
$$

where the term at the second line above has been added artificially since $\nu$ is a centered measure. In addition,

$$ 
\begin{aligned}
& \int_{\mathbb{R}} \int_{\mathbb{R}^2} \delta^2 G(m, x, y) d(m_{z,u}^N)^{\otimes 2}(x, y) d\nu(u) \\
= & \int_{\mathbb{R}} \int_{\mathbb{R}} \int_{\mathbb{R}} \left[ \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y + \frac{u}{\sqrt{N}} h(m, z) \right) \right]
\end{aligned}
$$
for (40), it comes from the Taylor-Lagrange's inequality for real-variable functions, and the control of (41) is obtained by Lemma 4.2.

Finally using this last bound to control (37) is sufficient to prove (35).

Step 3. From (35), we deduce a control between the semigroups of $\mu^N, \bar{\mu}$. Let $w_1, \dots, w_N \in \mathbb{R}$, and

$$m_w = \frac{1}{N} \sum_{k=1}^{N} w_k.$$

Then, thanks to Steps 1 and 2,

$$ \begin{aligned} |(\bar{P}_t - P_t^N) G(m_w)| &\le \int_0^t \mathbb{E}_{m_w} [|(\bar{A} - A^N) \bar{P}_s G(\mu_{t-s}^N)|] ds \\ &\le C \frac{1}{\sqrt{N}} \int_0^t ||\bar{P}_s G||_3 \left( 1 + \frac{1}{N} \sum_{k=1}^N \mathbb{E}_{m_w} [|X_{t-s}^{N,k}|^4] \right) ds \\ &\le C_t \frac{1}{\sqrt{N}} ||G||_3 \left( 1 + \frac{1}{N} \sum_{k=1}^N |w_k|^4 \right), \end{aligned} \quad (42) $$

where we have used Proposition 3.7.

Step 4. Then end of the proof is now quite classical. For $G \in \mathcal{P}^4$,

$$ \begin{aligned} |\mathbb{E}[G(\bar{\mu}_t)] - \mathbb{E}[G(\mu_t^N)]| &\le \left| \int_{\mathcal{P}_1(\mathbb{R})} \bar{P}_t G(m) d(\delta_{\bar{\mu}_0} - \mathcal{L}(\mu_0^N))(m) \right| \\ &\quad + \int_{\mathcal{P}_1(\mathbb{R})} |(\bar{P}_t - P_t^N) G(m)| d(\mathcal{L}(\mu_0^N))(m) \\ &\le C_t ||G||_1 \mathcal{D}_{KR}(\delta_{\bar{\mu}_0}, \mathcal{L}(\mu_0^N)) + C_t ||G||_3 \frac{1}{\sqrt{N}}, \end{aligned} $$

where the last line has been obtained using Proposition 3.7 (and the definition of the Kantorovich-Rubinstein metric) for the first term of the sum, and (42) for the second one.

Finally, by Theorem 1 of Fournier and Guillin (2015) (or using Theorem 3.2 of Bobkov and Ledoux (2019) and the discussion thereafter),

$$ \mathcal{D}_{KR}(\delta_{\bar{\mu}_0}, \mathcal{L}(\mu_0^N)) = \mathbb{E} \left[ D_{KR} \left( \bar{\mu}_0, \frac{1}{N} \sum_{k=1}^N \delta_{X_0^{N,k}} \right) \right] \le C N^{-1/2}. $$

This ends the proof of Theorem 1.4.

## 4.2. Proof of Corollary 1.1

The proof of Corollary 1.1 consists in remarking that the SDEs (1) and (2) are particular cases of (3) and (4) with: for $m \in \mathcal{P}_1(\mathbb{R})$ and $x \in \mathbb{R}$,

$$b(m, x) = \tilde{b}(x); \quad \sigma(m, x) = \tilde{\sigma}(x); \quad f(m, x) = \tilde{f}(x); \quad h(m, x) = 1;$$
$$-\delta^2 G \left(m, x + \frac{u}{\sqrt{N}} h(m, z), y\right) - \delta^2 G \left(m, x, y \frac{u}{\sqrt{N}} h(m, z)\right) + \delta^2 G(m, x, y) \right] d\nu(u) dm(x) dm(y).$$

Whence, the absolute value of the term within the brackets at (42) is non-greater than the sum of the three following quantities:

$$\left| G \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right) \right) - G(m) - \int_{\mathbb{R}} \delta G(m, x) dm_{z,u}^{N}(x) dm(x) \right. \qquad (44) \\ \left. - \frac{1}{2} \int_{\mathbb{R}^2} \delta^2 G(m, x, y) d(m_{z,u}^{N})^{\otimes 2}(x, y) \right|,$$

$$\int_{\mathbb{R}} \left| \delta G \left( m, x + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta G(m, x) - \frac{u}{\sqrt{N}} h(m, z) \partial_x \delta G(m, x) \right| dm(x), \qquad (45)$$

$$\int_{\mathbb{R}^2} \left| \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y + \frac{u}{\sqrt{N}} h(m, z) \right) - \delta^2 G \left( m, x + \frac{u}{\sqrt{N}} h(m, z), y \right) \right| dm(x), \qquad (46)$$ $$- \delta^2 G \left( m, x, y \frac{u}{\sqrt{N}} h(m, z) \right) + \delta^2 G(m, x, y) - \frac{u^2}{2N} h(m, z)^2 \partial_{xy}^2 \delta^2 G(m, x, y) \bigg| dm^{\otimes 2}(x, y).$$

Each of the three terms above is bounded by

$$C \|G\|_3 \frac{1}{N \sqrt{N}} \left( 1 + |z|^3 + \int_{\mathbb{R}} |x|^3 dm(x) \right) (1 + |u|^3).$$

Indeed, for (44), it is a consequence of the Taylor-Lagrange's inequality for measure-variable functions (i.e. Theorem 2.24), and the fact that (by Lemma 2.18)

$$D_{KR} \left( S \left( m, \frac{u}{\sqrt{N}} h(m, z) \right), m \right) \leq \frac{|u|}{\sqrt{N}} |h(m, z)|,$$

for (45), it comes from the Taylor-Lagrange's inequality for real-variable functions, and the control of (46) is obtained by Lemma 4.2.

Finally using this last bound to control (42) is sufficient to prove (40).

Step 3. From (40), we deduce a control between the semigroups of $\mu^N, \bar{\mu}$. Let $w_1, \dots, w_N \in \mathbb{R}$, and

$$m_w = \frac{1}{N} \sum_{k=1}^{N} w_k.$$

Then, thanks to Steps 1 and 2,

$$\begin{aligned} |(\bar{P}_t - P_t^N) G(m_w)| & \leq \int_0^t \mathbb{E}_{m_w} \left[ |(\bar{A} - A^N) \bar{P}_s G(\mu_{t-s}^N)| \right] ds \\ & \leq C \frac{1}{\sqrt{N}} \int_0^t \|\bar{P}_s G\|_3 \left( 1 + \frac{1}{N} \sum_{k=1}^N \mathbb{E}_{m_w} \left[ |X_{t-s}^{N,k}|^4 \right] \right) ds \end{aligned}$$
$$ \zeta(m, x) = \left( \int_{\mathbb{R}} u^2 d\nu(u) \right)^{1/2} \left( \int_{\mathbb{R}} \tilde{f}(y) dm(y) \right)^{1/2} . $$

Then, the result of Corollary 1.1 comes from the fact that, for any $\varphi \in C_b^4(\mathbb{R})$, the function

$$ G : m \in \mathcal{P}_1(\mathbb{R}) \mapsto \int_{\mathbb{R}} \varphi(x) dm(x) $$

belongs to $\mathcal{P}^4$, and that

$$ \mathbb{E}[G(\bar{\mu}_t)] = G(\bar{\mu}_t) = \mathbb{E}[\varphi(\bar{Y}_t)] \quad \text{and} \quad \mathbb{E}[G(\mu_t^N)] = \mathbb{E}\left[\frac{1}{N}\sum_{k=1}^N \varphi(Y_t^{N,k})\right] = \mathbb{E}[\varphi(Y_t^{N,1})]. $$

## Appendix A: Separating class on spaces of probability measures

The goal of this section is to prove Proposition 2.30. It means that the set $\mathcal{P}^\infty$ is a separating class for $\mathcal{P}_p(\mathbb{R})$ (for any $p \ge 1$). Before proving this statement, let us introduce some definitions and useful results about separating classes. This section relies strongly on the content of Section 3.4 of Ethier and Kurtz (2005).

In all this section, $E$ denotes some Polish space, and $\mathcal{P}(E)$ the space of probability measures on $E$ endowed with Prohorov metric (i.e. the topology of the weak convergence).

**Definition A.1.** Some set $\mathcal{A}$ of measurable functions $f : E \to \mathbb{R}$ is said to separate $E$ if, for all $x, y \in E$,

$$ (\forall f \in \mathcal{A}, f(x) = f(y)) \implies x = y. $$

A set $\mathcal{A}$ of measurable functions $f : E \to \mathbb{R}$ is called a separating class for $E$ if, for all $m, \mu \in \mathcal{P}(E)$,

$$ (\forall f \in \mathcal{A}, \int_E f(x) dm(x) = \int_E f(x) d\mu(x)) \implies m = \mu. $$

In other words, $\mathcal{A}$ is a separating class for $E$ means that

$$ \{m \in \mathcal{P}(E) \mapsto \int_E f(x) dm(x) : f \in \mathcal{A}\} $$

separates $\mathcal{P}(E)$. The interesting point of a separating class for $E$ is that it allows to identify the laws on $E$ in the following sense (it is a mere rephrasing of the definition of a separating class).

**Lemma A.2.** If $\mathcal{A}$ is a separating class for $E$, and $X, Y$ are two $E$-valued random variables such that: for all $f \in \mathcal{A}$,

$$ \mathbb{E}[f(X)] = \mathbb{E}[f(Y)]. $$

Then $X$ and $Y$ have the same law.

The lemma below should be classical, but since we have not found a proof in the literature, we provide one for self-completeness.

**Lemma A.3.** The set $C_c^\infty(\mathbb{R})$ is a separating class for $\mathbb{R}$.
$$ \leq C_t \frac{1}{\sqrt{N}} ||G||_3 \left( 1 + \frac{1}{N} \sum_{k=1}^{N} |w_k|^4 \right), \quad (47) $$

where we have used Proposition 3.7.

Step 4. Then end of the proof is now quite classical. For $G \in \mathcal{P}^4$,

$$ \begin{aligned} |\mathbb{E}[G(\bar{\mu}_t)] - \mathbb{E}[G(\mu_t^N)]| &\le \left| \int_{\mathcal{P}_1(\mathbb{R})} \bar{P}_t G(m) d(\delta_{\bar{\mu}_0} - \mathcal{L}(\mu_0^N))(m) \right| \\ &\quad + \int_{\mathcal{P}_1(\mathbb{R})} |(\bar{P}_t - P_t^N) G(m)| d(\mathcal{L}(\mu_0^N))(m) \\ &\le C_t ||G||_1 \mathcal{D}_{KR}(\delta_{\bar{\mu}_0}, \mathcal{L}(\mu_0^N)) + C_t ||G||_3 \frac{1}{\sqrt{N}}, \end{aligned} $$

where the last line has been obtained using Proposition 3.7 (and the definition of the Kantorovich-Rubinstein metric) for the first term of the sum, and (47) for the second one.

Finally, by Theorem 1 of ? (or using Theorem 3.2 of ? and the discussion thereafter),

$$ \mathcal{D}_{KR}(\delta_{\bar{\mu}_0}, \mathcal{L}(\mu_0^N)) = \mathbb{E} \left[ D_{KR} \left( \bar{\mu}_0, \frac{1}{N} \sum_{k=1}^{N} \delta_{X_0^{N,k}} \right) \right] \le C N^{-1/2}. $$

This ends the proof of Theorem 1.4.

## 4.2. Proof of Corollary 1.1

The proof of Corollary 1.1 consists in remarking that the SDEs (1) and (2) are particular cases of (3) and (4) with: for $m \in \mathcal{P}_1(\mathbb{R})$ and $x \in \mathbb{R}$,

$$ b(m, x) = \tilde{b}(x); \quad \sigma(m, x) = \tilde{\sigma}(x); \quad f(m, x) = \tilde{f}(x); \quad h(m, x) = 1; $$

$$ \varsigma(m, x) = \left( \int_{\mathbb{R}} u^2 d\nu(u) \right)^{1/2} \left( \int_{\mathbb{R}} \tilde{f}(y) dm(y) \right)^{1/2}. $$

Then, the result of Corollary 1.1 comes from the fact that, for any $\varphi \in C_b^4(\mathbb{R})$, the function

$$ G : m \in \mathcal{P}_1(\mathbb{R}) \mapsto \int_{\mathbb{R}} \varphi(x) dm(x) $$

belongs to $\mathcal{P}^4$, and that

$$ \mathbb{E}[G(\bar{\mu}_t)] = G(\bar{\mu}_t) = \mathbb{E}[\varphi(\bar{Y}_t)] \quad \text{and} \quad \mathbb{E}[G(\mu_t^N)] = \mathbb{E}\left[\frac{1}{N}\sum_{k=1}^{N}\varphi(Y_t^{N,k})\right] = \mathbb{E}[\varphi(Y_t^{N,1})]. $$

## Appendix A: Separating class on spaces of probability measures

The goal of this section is to prove Proposition 2.31. It means that the set $\mathcal{P}^\infty$ is a separating class for $\mathcal{P}_p(\mathbb{R})$ (for any $p \ge 1$). Before proving this statement, let us introduce some definitions and
Proof. Let $m, \mu \in \mathcal{P}(\mathbb{R})$ such that, for all $g \in C_c^\infty(\mathbb{R})$,

$$ \int_{\mathbb{R}} g(x) dm(x) = \int_{\mathbb{R}} g(x) d\mu(x). \qquad (43) $$

Let $a < b$. Let $(g_n)_n$ be a $C_c^\infty(\mathbb{R})$-valued sequence such that

$$ \begin{cases} (g_n)_n \text{ converges pointwise to } 1|_{[a,b[}, \\ \forall (x,n) \in \mathbb{R} \times \mathbb{N}, \ 0 \le g_n(x) \le 1. \end{cases} $$

Then, rewriting (43) with $g_n$ instead of $g$ (for all $n \in \mathbb{N}^*$), and using the Dominated Convergence Theorem,

$$ m([a, b[) = \mu([a, b[). $$

The previous equality being true for any $a < b$, this proves that $m = \mu$. $\square$

The proof that the polynomials form a separating class for $\mathcal{P}_p(\mathbb{R})$ relies on the following criterion.

**Theorem A.4** (Theorem 3.4.5.(a) of Ethier and Kurtz (2005)). Any algebra that separates E is a separating class for E.

Besides, since $\mathcal{P}_p(\mathbb{R})$ is Polish (see Theorem 6.18 of Villani (2009)), Theorem A.4 above can be used with $E = \mathcal{P}_p(\mathbb{R})$. Now we can prove that the polynomials form a separating class of $\mathcal{P}_p(\mathbb{R})$.

**Proof of Proposition 2.30.** By Lemma A.3, the set

$$ \{ m \in \mathcal{P}(\mathbb{R}) \mapsto \int_{\mathbb{R}} h(x) dm(x) : h \in C_c^\infty(\mathbb{R}) \} $$

separates $\mathcal{P}(\mathbb{R})$, hence it also separates $\mathcal{P}_p(\mathbb{R})$. The set $\mathcal{P}^\infty$ being an algebra containing the set above, it is a separating class by Theorem A.4. $\square$

# Appendix B: Proof of Lemma 3.6

The proof of Lemma 3.6 uses the following classical lemma whose proof is omitted.

**Lemma B.1.** Let $(f_n)_n$ be a sequence of $C^1(\mathbb{R})$ such that:

* $(f_n)_n$ converges point-wisely to some function $f$,

* $(f'_n)_n$ converges uniformly on every compact set to some function $g$.

Then, $f \in C^1(\mathbb{R})$ and $f' = g$.

Let us recall that, by definition, for all $x \in \mathbb{R}, t \ge 0$,

$$ \bar{X}_t^{(x)} = x + \int_0^t b(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) ds + \int_0^t \sigma(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) dB_s + \int_0^t \zeta(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) dW_s, $$

where $\bar{\mu}_s^{(x)} = \mathcal{L}(\bar{X}_s^{(x)}|W_s)$.

In order to prove Lemma 3.6, we use the Banach-Picard iteration scheme related to the above equation. Namely, for all $x \in \mathbb{R}, t \ge 0, n \in \mathbb{N}$,

$$ \bar{X}_t^{(x), [0]} = x, $$
useful results about separating classes. This section relies strongly on the content of Section 3.4 of ?.

In all this section, $E$ denotes some Polish space, and $\mathcal{P}(E)$ the space of probability measures on $E$ endowed with Prohorov metric (i.e. the topology of the weak convergence).

**Definition A.1.** Some set $\mathcal{A}$ of measurable functions $f: E \to \mathbb{R}$ is said to separate $E$ if, for all $x, y \in E$,

$$ (\forall f \in \mathcal{A}, f(x) = f(y)) \implies x = y. $$

A set $\mathcal{A}$ of measurable functions $f: E \to \mathbb{R}$ is called a separating class for $E$ if, for all $m, \mu \in \mathcal{P}(E)$,

$$ (\forall f \in \mathcal{A}, \int_E f(x) dm(x) = \int_E f(x) d\mu(x)) \implies m = \mu. $$

In other words, $\mathcal{A}$ is a separating class for $E$ means that

$$ \{m \in \mathcal{P}(E) \mapsto \int_E f(x) dm(x) : f \in \mathcal{A}\} $$

separates $\mathcal{P}(E)$. The interesting point of a separating class for $E$ is that it allows to identify the laws on $E$ in the following sense (it is a mere rephrasing of the definition of a separating class).

**Lemma A.2.** If $\mathcal{A}$ is a separating class for $E$, and $X, Y$ are two $E$-valued random variables such that: for all $f \in \mathcal{A}$,

$$ \mathbb{E}[f(X)] = \mathbb{E}[f(Y)]. $$

Then $X$ and $Y$ have the same law.

The lemma below should be classical, but since we have not found a proof in the literature, we provide one for self-completeness.

**Lemma A.3.** The set $C_c^\infty(\mathbb{R})$ is a separating class for $\mathbb{R}$.

*Proof.* Let $m, \mu \in \mathcal{P}(\mathbb{R})$ such that, for all $g \in C_c^\infty(\mathbb{R})$,

$$ \int_{\mathbb{R}} g(x) dm(x) = \int_{\mathbb{R}} g(x) d\mu(x). \qquad (48) $$

Let $a < b$. Let $(g_n)_n$ be a $C_c^\infty(\mathbb{R})$-valued sequence such that

$$ \begin{cases} (g_n)_n \text{ converges pointwise to } 1_{]a,b[}, \\ \forall (x, n) \in \mathbb{R} \times \mathbb{N}, \ 0 \le g_n(x) \le 1. \end{cases} $$

Then, rewriting (48) with $g_n$ instead of $g$ (for all $n \in \mathbb{N}^*$), and using the Dominated Convergence Theorem,

$$ m([a, b]) = \mu([a, b]). $$

The previous equality being true for any $a < b$, this proves that $m = \mu$. $\square$

The proof that the polynomials form a separating class for $\mathcal{P}_p(\mathbb{R})$ relies on the following criterion.

**Theorem A.4** (Theorem 3.4.5.(a) of?). Any algebra that separates $E$ is a separating class for $E$.
$$
\begin{align*}
\bar{\mu}_t^{(x), [0]} &= \delta_x, \\
\bar{X}_t^{(x), [n+1]} &= x + \int_0^t b(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) ds + \int_0^t \sigma(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) dB_s \\
&\quad + \int_0^t \varsigma(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) dW_s, \\
\bar{\mu}_t^{(x), [n+1]} &= \mathcal{L}(\bar{X}_t^{(x), [n+1]} | \mathcal{W}_t).
\end{align*}
$$

Step 1. In this first step, we prove the almost sure convergence of $\bar{X}_t^{(x), [n]}$ to $\bar{X}_t^{(x)}$ as $n$ goes to infinity, for any $t$ belonging to some sufficiently small interval, and locally uniformly w.r.t. $x$.
For $n \in \mathbb{N}, p \in \mathbb{N}^*, t \ge 0$ and $M > 0$,

$$
u_t^{[n],0}(M,p) = \mathbb{E} \left[ \sup_{s \le t, |x| \le M} \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right|^p \right].
$$

With classical computation (using Burkholder-Davis-Gundy's inequality and the assumption that $b, \sigma, \varsigma$ are Lipschitz continuous), we have that, for all $T > 0$, $p \in \mathbb{N}^*, n \in \mathbb{N}$,

$$
u_T^{[n+1],0}(M,p) \le C_p (T^p + T^{p/2}) u_T^{[n],0}(M,p),
$$

with $C_p > 0$ independent of $T$, $M$ and $n$. In particular, let us fix some small enough $T_p > 0$ such that, for all $n \in \mathbb{N}, p \in \mathbb{N}, M > 0$,

$$
u_{T_p}^{[n+1],0}(M,p) \le \frac{1}{2} u_{T_p}^{[n],0}(M,p),
$$

whence

$$
u_{T_p}^{[n],0}(M,p) \le C_p \frac{1}{2^n} (1+M^p). \quad (44)
$$

and

$$
\mathbb{E} \left[ \sup_{s \le T_p, |x| \le M} \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right| \right] \le u_{T_p}^{[n],0} (M,p)^{1/p} \le C_p \frac{1}{2^{n/p}} (1+M).
$$

As a consequence, for all $M > 0, p \in \mathbb{N}^*$,

$$
\mathbb{E} \left[ \sup_{t \le T_p, |x| \le M} \left| \sum_{n=0}^{+\infty} (\bar{X}_t^{(x), [n+1]} - \bar{X}_t^{(x), [n]}) \right| \right] < \infty,
$$

and, almost surely,

$$
\sup_{t \le T_p, |x| \le M} \left| \sum_{n=0}^{+\infty} (\bar{X}_t^{(x), [n+1]} - \bar{X}_t^{(x), [n]}) \right| < \infty.
$$

This proves that, almost surely, for all $t \in [0, T_p]$, $\bar{X}_t^{(x), [n]}$ converges as $n$ goes to infinity to $\bar{X}_t^{(x)}$ (recalling that $\bar{X}^{(x)}$ is solution to (25)) locally uniformly w.r.t. $x$.

In addition, with similar technics and Fatou's lemma, we can prove that, for all $p \in \mathbb{N}^*, M > 0$, $T > 0$,

$$
\mathbb{E} \left[ \sup_{t \le T, |x| \le M} \left| \bar{X}_t^{(x)} \right|^p \right] < \infty. \quad (45)
$$
Besides, since $\mathcal{P}_p(\mathbb{R})$ is Polish (see Theorem 6.18 of ?), Theorem A.4 above can be used with $E = \mathcal{P}_p(\mathbb{R})$. Now we can prove that the polynomials form a separating class of $\mathcal{P}_p(\mathbb{R})$.

**Proof of Proposition 2.31.** By Lemma A.3, the set

$$ \left\{ m \in \mathcal{P}(\mathbb{R}) \mapsto \int_{\mathbb{R}} h(x) dm(x) : h \in C_c^{\infty}(\mathbb{R}) \right\} $$

separates $\mathcal{P}(\mathbb{R})$, hence it also separates $\mathcal{P}_p(\mathbb{R})$. The set $\mathcal{P}^{\infty}$ being an algebra containing the set above, it is a separating class by Theorem A.4. □

# Appendix B: Proof of Lemma 3.6

The proof of Lemma 3.6 uses the following classical lemma whose proof is omitted.

**Lemma B.1.** Let $(f_n)_n$ be a sequence of $C^1(\mathbb{R})$ such that:

* $(f_n)_n$ converges point-wisely to some function $f$,

* $(f'_n)_n$ converges uniformly on every compact set to some function $g$.

Then, $f \in C^1(\mathbb{R})$ and $f' = g$.

Let us recall that, by definition, for all $x \in \mathbb{R}, t \ge 0$,

$$ \bar{X}_t^{(x)} = x + \int_0^t b(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) ds + \int_0^t \sigma(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) dB_s + \int_0^t \varsigma(\bar{\mu}_s^{(x)}, \bar{X}_s^{(x)}) dW_s, $$

where $\bar{\mu}_s^{(x)} = \mathcal{L}(\bar{X}_s^{(x)}|W_s)$.

In order to prove Lemma 3.6, we use the Banach-Picard iteration scheme related to the above equation. Namely, for all $x \in \mathbb{R}, t \ge 0, n \in \mathbb{N}$,

$$ \bar{X}_t^{(x), [0]} = x, $$

$$ \bar{\mu}_t^{(x), [0]} = \delta_x, $$

$$ \bar{X}_t^{(x), [n+1]} = x + \int_0^t b(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) ds + \int_0^t \sigma(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) dB_s \\ + \int_0^t \varsigma(\bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]}) dW_s, $$

$$ \bar{\mu}_t^{(x), [n+1]} = \mathcal{L}(\bar{X}_t^{(x), [n+1]} | W_t). $$

**Step 1.** In this first step, we prove the almost sure convergence of $\bar{X}_t^{(x), [n]}$ to $\bar{X}_t^{(x)}$ as $n$ goes to infinity, for any $t$ belonging to some sufficiently small interval, and locally uniformly w.r.t. $x$.

For $n \in \mathbb{N}, p \in \mathbb{N}^*, t \ge 0$ and $M > 0$,

$$ u_t^{[n], 0}(M, p) = \mathbb{E} \left[ \sup_{s \le t, |x| \le M} \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right|^p \right]. $$
Step 2. Now we study the regularity of the functions

$$x \mapsto \bar{X}_t^{(x), [n]}.$$

Let us prove by induction on $n \in \mathbb{N}$ that there exists $T > 0$ (independent of $n$) such that, for all $t \in [0, T]$ the function $x \mapsto \bar{X}_t^{(x), [n]} C^1$ and that, for all $M > 0$,

$$\mathbb{E} \left[ \sup_{t \le T, |x| \le M} \left| \partial_x \bar{X}_t^{(x), [n]} \right| \right] < \infty \quad \text{and} \quad \sup_{x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \le T} \left| \partial_x \bar{X}_t^{(x), [n]} \right| \right] < \infty.$$

For $n = 0$,

$$\partial_x \bar{X}_t^{(x), [0]} = 1.$$

Then fix some $n \in \mathbb{N}$ and assume that the induction hypothesis holds true for this $n$. Then, by Lemma 2.19 (and Remark 2.20), and assuming that $b, \sigma, \zeta$ admits bounded second order mixed derivatives (what implies that the first order mixed derivatives are continuous w.r.t. all their variables, by Remark 2.13),

$$\begin{aligned}
\partial_x \bar{X}_t^{(x), [n+1]} = & 1 + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) ds \quad (46) \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( b_{\bar{X}_s^{(x), [n]}} \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \right) \middle| \mathcal{W}_s \right] ds \\
& + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} \sigma \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) dB_s \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( \sigma_{\bar{X}_s^{(x), [n]}} \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \right) \middle| \mathcal{W}_s \right] dB_s \\
& + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} \zeta \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) dW_s \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( \zeta_{\bar{X}_s^{(x), [n]}} \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \right) \middle| \mathcal{W}_s \right] dW_s,
\end{aligned}$$

where $\tilde{X}$ is defined in the same way as $\bar{X}$ w.r.t. the same Brownian motion $W$, but w.r.t. a Brownian motion $\tilde{B}$ independent of $(B, W)$, and $\tilde{\mathbb{E}}$ is the expectation w.r.t. the law of $\tilde{B}$. One can note that, the expression above of $\partial_x \bar{X}_t^{(x), [n+1]}$ is closely related to the ones of Proposition 3.1 of Crisan and McMurray (2018) (we actually use their notation for $\tilde{\mathbb{E}}$ and $\bar{X}$).

Let us remark an important point: the functions appearing as coefficients in the expression above are exactly the first order mixed derivatives of the functions $b, \sigma, \zeta$.

And, with similar computation as in Step 1, we can prove that, for all $p \in \mathbb{N}^*, M > 0$,

$$\sup_{n \in \mathbb{N}} \mathbb{E} \left[ \sup_{t \le T_p, |x| \le M} \left| \partial_x \bar{X}_t^{(x), [n]} \right|^p \right] < \infty \quad \text{and} \quad \sup_{n \in \mathbb{N}, x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \le T_p} \left| \partial_x \bar{X}_t^{(x), [n]} \right|^p \right] < \infty, \quad (47)$$

for some $T_p > 0$ independent of $n, M$.

Let

$$u_t^{[n],1}(M, p) = \mathbb{E} \left[ \sup_{s \le t, |x| \le M} \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right|^p \right].$$
With classical computation (using Burkholder-Davis-Gundy's inequality and the assumption that $b, \sigma, \zeta$ are Lipschitz continuous), we have that, for all $T > 0$, $p \in \mathbb{N}^*$, $n \in \mathbb{N}$,

$$u_{T}^{[n+1],0}(M,p) \leq C_{p} (T^{p} + T^{p/2}) u_{T}^{[n],0}(M,p),$$

with $C_p > 0$ independent of $T$, $M$ and $n$. In particular, let us fix some small enough $T_p > 0$ such that, for all $n \in \mathbb{N}$, $p \in \mathbb{N}$, $M > 0$,

$$u_{T_p}^{[n+1],0}(M,p) \leq \frac{1}{2} u_{T_p}^{[n],0}(M,p),$$

whence

$$u_{T_p}^{[n],0}(M,p) \leq C_p \frac{1}{2^n} (1 + M^p). \qquad (49)$$

and

$$\mathbb{E} \left[ \sup_{s \leq T_p, |x| \leq M} \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right| \right] \leq u_{T_p}^{[n],0}(M,p)^{1/p} \leq C_p \frac{1}{2^{n/p}} (1 + M).$$

As a consequence, for all $M > 0$, $p \in \mathbb{N}^*$,

$$\mathbb{E} \left[ \sup_{t \leq T_p, |x| \leq M} \left| \sum_{n=0}^{+\infty} \left( \bar{X}_t^{(x), [n+1]} - \bar{X}_t^{(x), [n]} \right) \right| \right] < \infty,$$

and, almost surely,

$$\sup_{t \leq T_p, |x| \leq M} \left| \sum_{n=0}^{+\infty} \left( \bar{X}_t^{(x), [n+1]} - \bar{X}_t^{(x), [n]} \right) \right| < \infty.$$

This proves that, almost surely, for all $t \in [0, T_p]$, $\bar{X}_t^{(x), [n]}$ converges as $n$ goes to infinity to $\bar{X}_t^{(x)}$ (recalling that $\bar{X}^{(x)}$ is solution to (30)) locally uniformly w.r.t. $x$.

In addition, with similar technics and Fatou's lemma, we can prove that, for all $p \in \mathbb{N}^*$, $M > 0$, $T > 0$,

$$\mathbb{E} \left[ \sup_{t \leq T, |x| \leq M} \left| \bar{X}_t^{(x)} \right|^p \right] < \infty. \qquad (50)$$

*Step 2.* Now we study the regularity of the functions

$$x \mapsto \bar{X}_t^{(x), [n]}.$$

Let us prove by induction on $n \in \mathbb{N}$ that there exists $T > 0$ (independent of $n$) such that, for all $t \in [0, T]$ the function $x \mapsto \bar{X}_t^{(x), [n]} C^1$ and that, for all $M > 0$,

$$\mathbb{E} \left[ \sup_{t \leq T, |x| \leq M} \left| \partial_x \bar{X}_t^{(x), [n]} \right| \right] < \infty \quad \text{and} \quad \sup_{x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \leq T} \left| \partial_x \bar{X}_t^{(x), [n]} \right| \right] < \infty.$$

For $n = 0$,

$$\partial_x \bar{X}_t^{(x), [0]} = 1.$$
To control the above quantity in a similar way as in Step 1, it is required to control two kind of terms. In order to simplify the reading we only handle the drift terms (the Brownian terms can be treated in the exact same way after using Burkholder-Davis-Gundy's inequality). The first term is of the following form: for $|x| \le M$,

$$ \left| \left( \partial_x \bar{X}_s^{(x), [n+1]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n+1]}, \bar{X}_s^{(x), [n+1]} \right) - \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) \right| \\ \le \left( \sup_{m \in \mathcal{P}_1(\mathbb{R}), |y| \le M} \left| \partial_{(0,1)} b(m, y) \right| \right) \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right| \\ + C \left| \partial_x X_s^{(x), [n]} \right| \left( \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right| + D_{KR} \left( \bar{\mu}_s^{(x), [n+1]}, \bar{\mu}_s^{(x), [n]} \right) \right). $$

In particular, thanks to (44), (47) and Cauchy-Schwarz' inequality, for any $p \in \mathbb{N}^*$, for $t \le T_p$ (where $T_p$ has possibly been reduced compared to the previous one),

$$ \mathbb{E} \left[ \left( \int_0^t \left| \left( \partial_x \bar{X}_s^{(x), [n+1]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n+1]}, \bar{X}_s^{(x), [n+1]} \right) - \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) \right| ds \right)^p \right] \\ \le C_{M,p} T^p u_t^{[n],1}(M,p) + C_{M,p} T^p \frac{1}{2^{n/2}}. \qquad (48) $$

The second term involves the following term

$$ \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial \left( b_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right), $$

and can be handled exactly as (48), since the function $(m, x, y) \mapsto \partial_y \delta(b_x)(m, y)$ is assumed to be Lipschitz continuous and bounded (recalling that this function belongs to the set of first order mixed derivatives of $b$). Whence, for $T \le T_p$,

$$ u_T^{[n+1],1}(M,p) \le C_{M,p} (T^p + T^{p/2}) \left( u_T^{[n],1}(M,p) + \frac{1}{2^{n/2}} \right). $$

And, possibly by reducing $T_p$,

$$ u_T^{[n+1],1}(M,p) \le \frac{1}{2} u_T^{[n],1}(M,p) + \frac{1}{2^{1+n/2}}. $$

Consequently, for $T \le T_p$,

$$ u_T^{[n],1}(M,p) \le C_{M,p} 2^{-n/2}, \qquad (49) $$

and

$$ \mathbb{E} \left[ \sup_{s \le T_p, |x| \le M} \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right| \right] \le u_T^{[n],1}(M,p)^{1/p} \le C_{M,p} 2^{-n/(2p)}. $$

The quantities above being the terms of a convergent series, we can conclude as in Step 1 that, almost surely, for all $t \in [0, T_p]$, $\partial_x \bar{X}_t^{(x), [n]}$ converges as $n$ goes to infinity to some function locally uniformly w.r.t. $x$. Then, by Lemma B.1, we know that, almost surely, for all $t \in [0, T_p]$, the function
Then fix some $n \in \mathbb{N}$ and assume that the induction hypothesis holds true for this $n$. Then, by Lemma 2.19 (and Remark 2.20), and assuming that $b, \sigma, \zeta$ admits bounded second order mixed derivatives (what implies that the first order mixed derivatives are continuous w.r.t. all their variables, by Remark 2.13),

$$ 
\begin{aligned}
\partial_x \bar{X}_t^{(x), [n+1]} = & 1 + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) ds \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( b_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \middle| \mathcal{W}_s \right] ds \\
& + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} \sigma \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) dB_s \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( \sigma_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \middle| \mathcal{W}_s \right] dB_s \\
& + \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} \zeta \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) dW_s \\
& + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( \zeta_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \middle| \mathcal{W}_s \right] dW_s,
\end{aligned}
\quad (51) $$

where $\tilde{X}$ is defined in the same way as $\bar{X}$ w.r.t. the same Brownian motion $W$, but w.r.t. a Brownian motion $\tilde{B}$ independent of $(B, W)$, and $\tilde{\mathbb{E}}$ is the expectation w.r.t. the law of $\tilde{B}$. One can note that, the expression above of $\partial_x \bar{X}_t^{(x), [n+1]}$ is closely related to the ones of Proposition 3.1 of ? (we actually use their notation for $\tilde{\mathbb{E}}$ and $\tilde{X}$).

Let us remark an important point: the functions appearing as coefficients in the expression above are exactly the first order mixed derivatives of the functions $b, \sigma, \zeta$.

And, with similar computation as in Step 1, we can prove that, for all $p \in \mathbb{N}^*, M > 0$,

$$ \sup_{n \in \mathbb{N}} \mathbb{E} \left[ \sup_{t \le T_p, |x| \le M} \left| \partial_x \bar{X}_t^{(x), [n]} \right|^p \right] < \infty \quad \text{and} \quad \sup_{n \in \mathbb{N}, x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \le T_p} \left| \partial_x \bar{X}_t^{(x), [n]} \right| \right] < \infty, \quad (52) $$

for some $T_p > 0$ independent of $n, M$.

Let

$$ u_t^{[n],1}(M,p) = \mathbb{E} \left[ \sup_{s \le t, |x| \le M} \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right|^p \right]. $$

To control the above quantity in a similar way as in Step 1, it is required to control two kind of terms. In order to simplify the reading we only handle the drift terms (the Brownian terms can be treated in the exact same way after using Burkholder-Davis-Gundy's inequality). The first term is of the following form: for $|x| \le M$,

$$ 
\begin{aligned}
& \left| \left( \partial_x \bar{X}_s^{(x), [n+1]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n+1]}, \bar{X}_s^{(x), [n+1]} \right) - \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) \right| \\
& \le \left( \sup_{m \in \mathcal{P}_1(\mathbb{R}), |y| \le M} \left| \partial_{(0,1)} b(m, y) \right| \right) \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right| \\
& \quad + C \left| \partial_x X_s^{(x), [n]} \right| \left( \left| \bar{X}_s^{(x), [n+1]} - \bar{X}_s^{(x), [n]} \right| + D_{KR} \left( \bar{\mu}_s^{(x), [n+1]}, \bar{\mu}_s^{(x), [n]} \right) \right).
\end{aligned}
$$
$x \mapsto \bar{X}_t^{(x)}$ is $C^1$, and that its derivative is the limit of $\partial_x \bar{X}_t^{(x),[n]}$. In particular, by Fatou's lemma and (47),

$$ \mathbb{E} \left[ \sup_{t \le T, |x| \le M} \left| \partial_x \bar{X}_t^{(x)} \right|^p \right] < \infty \quad \text{and} \quad \sup_{x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \le T} \left| \partial_x \bar{X}_t^{(x)} \right|^p \right] < \infty. \quad (50) $$

Step 3. The proof that the function $x \mapsto \bar{X}_t^{(x)}$ is $C^2$ uses the same arguments as the ones used in Step 2. In order to make it clear, we just write the terms of the dynamics of $\partial_{xx}^2 \bar{X}_t^{(x),[n+1]}$ coming from the two first lines of (46).

$$ \begin{aligned} \partial_{xx}^2 \bar{X}_t^{(x),[n+1]} = & \int_0^t \left( \partial_{xx}^2 \bar{X}_s^{(x),[n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x),[n]}, \bar{X}_s^{(x),[n]} \right) ds \\ & + \int_0^t \left( \partial_x \bar{X}_s^{(x),[n]} \right)^2 \partial_{(0,2)} b \left( \bar{\mu}_s^{(x),[n]}, \bar{X}_s^{(x),[n]} \right) ds \\ & + \int_0^t \left( \partial_x \bar{X}_s^{(x),[n]} \right) \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x),[n]} \right) \partial \left( \left( \partial_{(0,1)} b \right) \bar{X}_s^{(x),[n]} \right) \left( \bar{\mu}_s^{(x),[n]}, \tilde{X}^{(x),[n]} \right) \middle| \mathcal{W}_s \right] ds \\ & + \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_{xx}^2 \tilde{X}_s^{(x),[n]} \right) \partial \left( b_{\bar{X}_s^{(x),[n]}} \right) \left( \bar{\mu}_s^{(x),[n]}, \tilde{X}_s^{(x),[n]} \right) \middle| \mathcal{W}_s \right] ds \\ & + \int_0^t \left( \partial_x \tilde{X}_s^{(x),[n]} \right) \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x),[n]} \right) \left( \partial_{\bar{x}} \partial \left( b_{\bar{x}} \right) \left( \bar{\mu}_s^{(x),[n]}, \tilde{X}_s^{(x),[n]} \right) \right) \middle|_{\bar{x}=\bar{X}_s^{(x),[n]}} \middle| \mathcal{W}_s \right] ds \\ & + \int_0^t \tilde{\mathbb{E}} \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x),[n]} \right) \left( \partial_x \tilde{X}_s^{(x),[n]} \right) \partial^2 \left( b_{\tilde{X}_s^{(x),[n]}} \right) \left( \bar{\mu}_s^{(x),[n]}, \tilde{X}_s^{(x),[n]}, \tilde{X}_s^{(x),[n]} \right) \middle| \mathcal{W}_s \right] ds \\ & + \dots \end{aligned} $$

where the omitted terms (hidden in the ellipsis) are the same as the ones written replacing the function $b$ respectively by $\sigma$ and $\zeta$, and $ds$ by $dB_s$ and $dW_s$. As previously, $\tilde{X}$ and $\tilde{X}$ are defined as $\tilde{X}$ w.r.t. the same Brownian motion $W$, and w.r.t. respective Brownian motions $\tilde{B}$ and $\tilde{B}$ such that $W, \tilde{B}$ and $\tilde{B}$ are independent. The expectations $\tilde{\mathbb{E}}$ and $\tilde{\mathbb{E}}$ are the expectation w.r.t. the laws of $\tilde{B}$ and $\tilde{B}$.

Once again, it can be noted that the functions appearing in the expression of $\partial_{xx}^2 \bar{X}_t^{(x),[n+1]}$ are the mixed-derivatives of $b, \sigma, \zeta$ up to order two. Hence the same reasoning as in Step 2 allows to conclude. Finally, the proof that $x \mapsto \bar{X}_t^{(x)}$ is $C^j$ (for any $j \le k - 1$) is the same as previously remarking that, calculating the expression of $\partial_{x_j}^j \bar{X}_t^{(x),[n+1]}$ makes appear the mixed-derivatives of $b, \sigma, \zeta$ up to order $j$. Since it is assumed that all the mixed-derivatives up to order $k$ are bounded, we know by, Remark 2.20, that the mixed-derivatives up to order $k - 1$ are Lipschitz continuous. So the end of the proof follows the same arguments as Step 2, and similar controls as (50) can be proved for the $j$-th order derivatives, for any $j \le k - 1$.

## Appendix C: Proofs of some technical results

*Proof of Lemma 2.3.* Let us consider $m_0 \in \mathcal{P}_1(\mathbb{R})$ and $h_1, h_2 : \mathbb{R} \to \mathbb{R}$ such that, for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ \int_{\mathbb{R}} h_1(x) d(m - m_0)(x) = \int_{\mathbb{R}} h_2(x) d(m - m_0)(x) + \varepsilon_{m_0}(m), $$
In particular, thanks to (49), (52) and Cauchy-Schwarz' inequality, for any $p \in \mathbb{N}^*$, for $t \le T_p$ (where $T_p$ has possibly been reduced compared to the previous one),

$$ \mathbb{E} \left[ \left( \int_0^t \left| \left( \partial_x \bar{X}_s^{(x), [n+1]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n+1]}, \bar{X}_s^{(x), [n+1]} \right) - \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) \right| ds \right)^p \right] \le C_{M,p} T^p u_t^{[n],1}(M,p) + C_{M,p} T^p \frac{1}{2^{n/2}}. \quad (53) $$

The second term involves the following term

$$ \left( \partial_x \bar{X}_s^{(x), [n]} \right) \partial \left( b_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right), $$

and can be handled exactly as (53), since the function $(m, x, y) \mapsto \partial_y \delta(b_x)(m, y)$ is assumed to be Lipschitz continuous and bounded (recalling that this function belongs to the set of first order mixed derivatives of $b$). Whence, for $T \le T_p$,

$$ u_T^{[n+1],1}(M,p) \le C_{M,p} (T^p + T^{p/2}) \left( u_T^{[n],1}(M,p) + \frac{1}{2^{n/2}} \right). $$

And, possibly by reducing $T_p$,

$$ u_T^{[n+1],1}(M,p) \le \frac{1}{2} u_T^{[n],1}(M,p) + \frac{1}{2^{1+n/2}}. $$

Consequently, for $T \le T_p$,

$$ u_T^{[n],1}(M,p) \le C_{M,p} 2^{-n/2}, \quad (54) $$

and

$$ \mathbb{E} \left[ \sup_{s \le T_p, |x| \le M} \left| \partial_x \bar{X}_s^{(x), [n+1]} - \partial_x \bar{X}_s^{(x), [n]} \right| \right] \le u_T^{[n],1}(M,p)^{1/p} \le C_{M,p} 2^{-n/(2p)}. $$

The quantities above being the terms of a convergent series, we can conclude as in Step 1 that, almost surely, for all $t \in [0, T_p]$, $\partial_x \bar{X}_t^{(x), [n]}$ converges as $n$ goes to infinity to some function locally uniformly w.r.t. $x$. Then, by Lemma B.1, we know that, almost surely, for all $t \in [0, T_p]$, the function $x \mapsto \bar{X}_t^{(x)}$ is $C^1$, and that its derivative is the limit of $\partial_x \bar{X}_t^{(x), [n]}$. In particular, by Fatou's lemma and (52),

$$ \mathbb{E} \left[ \sup_{t \le T, |x| \le M} \left| \partial_x \bar{X}_t^{(x)} \right|^p \right] < \infty \quad \text{and} \quad \sup_{x \in \mathbb{R}} \mathbb{E} \left[ \sup_{t \le T} \left| \partial_x \bar{X}_t^{(x)} \right|^p \right] < \infty. \quad (55) $$

Step 3. The proof that the function $x \mapsto \bar{X}_t^{(x)}$ is $C^2$ uses the same arguments as the ones used in Step 2. In order to make it clear, we just write the terms of the dynamics of $\partial_{xx}^2 \bar{X}_t^{(x), [n+1]}$ coming from the two first lines of (51).

$$ \partial_{xx}^2 \bar{X}_t^{(x), [n+1]} = \int_0^t \left( \partial_{xx}^2 \bar{X}_s^{(x), [n]} \right) \partial_{(0,1)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) ds $$
where $\varepsilon_{m_0}(m)/D_{KR}(m, m_0)$ vanishes as $m$ converges to $m_0$.

Let us define

$$ \Phi : m \in \mathcal{P}_1(\mathbb{R}) \mapsto \int_{\mathbb{R}} (h_1(x) - h_2(x)) d(m - m_0)(x). $$

So, by hypothesis, $\Phi(m)/D_{KR}(m, m_0)$ vanishes as $m$ goes to $m_0$.

*Step 1.* In a first time, we prove that $\Phi$ is the zero function. Let us fix $m \in \mathcal{P}_1(\mathbb{R})$ and introduce, for any $n \in \mathbb{N}^*$,

$$ \mu_n = \frac{n-1}{n}m_0 + \frac{1}{n}m \in \mathcal{P}_1(\mathbb{R}). $$

We have that, for all $n \in \mathbb{N}^*$,

$$ D_{KR}(m_0, \mu_n) \leq \frac{1}{n} D_{KR}(m_0, m), $$

hence, $\mu_n$ converges to $m_0$ as $n$ goes to infinity.

On the other hand,

$$ \Phi(m) = n\Phi(\mu_n) = n \frac{\Phi(\mu_n)}{D_{KR}(m_0, \mu_n)} D_{KR}(m_0, \mu_n) \leq \frac{\Phi(\mu_n)}{D_{KR}(m_0, \mu_n)} D_{KR}(m_0, m) \underset{n \to \infty}{\longrightarrow} 0. $$

This proves that, for all $m \in \mathcal{P}_1(\mathbb{R})$, $\Phi(m) = 0$.

*Step 2.* To conclude the proof, it is then sufficient to notice that, thanks to *Step 1*, for any $y \in \mathbb{R}$, $\Phi(\delta_y) = 0$. Whence, for any $y \in \mathbb{R}$,

$$ h_1(y) - h_2(y) = \int_{\mathbb{R}} (h_1(x) - h_2(x)) dm_0(x). $$

In particular, the function $h_1 - h_2$ is constant, and the lemma is proved. $\square$

*Proof of Corollary 2.26.* Let $m, m_0 \in \mathcal{P}_1(\mathbb{R})$ be fixed in all the proof.

$$ \begin{aligned} F(m) - F(m_0) &= \int_{\mathbb{R}} H(m, x)d(m - m_0)(x) + \int_{\mathbb{R}} (H(m, x) - H(m_0, x)) dm_0(x) \\ &= \int_{\mathbb{R}} H(m_0, x)d(m - m_0)(x) + \int_{\mathbb{R}} (H(m, x) - H(m_0, x)) dm_0(x) \\ &\quad + \int_{\mathbb{R}} (H(m, x) - H(m_0, x)) d(m - m_0)(x). \end{aligned} $$

Consequently,

$$ \begin{aligned} & \left| F(m) - F(m_0) - \int_{\mathbb{R}} \left( H(m_0, x) - \int_{\mathbb{R}} \delta H_y(m_0, x) dm_0(y) \right) d(m - m_0)(x) \right| \\ & \leq \int_{\mathbb{R}} \left| H_x(m) - H_x(m_0) - \int_{\mathbb{R}} \delta H_x(m_0, y) d(m - m_0)(y) \right| dm_0(x) \end{aligned} \quad (51) $$

$$ + \left| \int_{\mathbb{R}} (H(m, x) - H(m_0, x)) d(m - m_0)(x) \right|. \quad (52) $$

Then, by the hypothesis (i) of the corollary and Theorem 2.24, the integrand of (51) is bounded by

$$ C(1 + |x|) D_{KR}(m, m_0)^2 $$
$$
\begin{align*}
&+ \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right)^2 \partial_{(0,2)} b \left( \bar{\mu}_s^{(x), [n]}, \bar{X}_s^{(x), [n]} \right) ds \\
&+ \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial \left( \left( \partial_{(0,1)} b \right) \bar{X}_s^{(x), [n]} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}^{(x), [n]} \right) \middle| \mathcal{W}_s \right] ds \\
&+ \int_0^t \tilde{\mathbb{E}} \left[ \left( \partial_{xx}^2 \tilde{X}_s^{(x), [n]} \right) \partial \left( b_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \middle| \mathcal{W}_s \right] ds \\
&+ \int_0^t \left( \partial_x \bar{X}_s^{(x), [n]} \right) \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \left( \partial_{\bar{x}} \partial \left( b_{\bar{x}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \right) \middle|_{\bar{x}=\bar{X}_s^{(x), [n]}} \middle| \mathcal{W}_s \right] ds \\
&+ \int_0^t \tilde{\mathbb{E}} \tilde{\mathbb{E}} \left[ \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \left( \partial_x \tilde{X}_s^{(x), [n]} \right) \partial^2 \left( b_{\bar{X}_s^{(x), [n]}} \right) \left( \bar{\mu}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]}, \tilde{X}_s^{(x), [n]} \right) \middle| \mathcal{W}_s \right] ds \\
&+ \dots
\end{align*}
$$

where the omitted terms (hidden in the ellipsis) are the same as the ones written replacing the function b respectively by σ and ς, and ds by dBs and dWs. As previously, ¯X and ˜X are defined as ¯X w.r.t. the same Brownian motion W, and w.r.t. respective Brownian motions ˜B and ˜B such that W, ˜B and ˜B are independent. The expectations ˜E and ˜E are the expectation w.r.t. the laws of ˜B and ˜B.

Once again, it can be noted that the functions appearing in the expression of $\partial_{xx}^2 \bar{X}_t^{(x), [n+1]}$ are the mixed-derivatives of $b, \sigma, \zeta$ up to order two. Hence the same reasoning as in Step 2 allows to conclude. Finally, the proof that $x \mapsto \bar{X}_t^{(x)}$ is $C^3$ (resp. $C^4$) is the same as previously remarking that, calculating the expression of $\partial_{xxx}^3 \bar{X}_t^{(x), [n+1]}$ (resp. $\partial_{xxxx}^4 \bar{X}_t^{(x), [n+1]}$) makes appear the mixed-derivatives of $b, \sigma, \zeta$ up to order three (resp. four). Since it is assumed that all the mixed-derivatives up to order five are bounded, we know by, Remark 2.20, that the mixed-derivatives up to order four are Lipschitz continuous. So the end of the proof follows the same arguments as Step 2, and similar controls as (55) can be proved for the third and fourth derivatives.

Appendix C: Proofs of some technical lemmas on spaces of probability measures

*Proof of Lemma 2.3.* Let us consider $m_0 \in \mathcal{P}_1(\mathbb{R})$ and $h_1, h_2 : \mathbb{R} \to \mathbb{R}$ such that, for all $m \in \mathcal{P}_1(\mathbb{R})$,

$$ \int_{\mathbb{R}} h_1(x) d(m - m_0)(x) = \int_{\mathbb{R}} h_2(x) d(m - m_0)(x) + \varepsilon_{m_0}(m), $$

where $\varepsilon_{m_0}(m)/D_{KR}(m, m_0)$ vanishes as $m$ converges to $m_0$.

Let us define

$$ \Phi : m \in \mathcal{P}_1(\mathbb{R}) \mapsto \int_{\mathbb{R}} (h_1(x) - h_2(x)) d(m - m_0)(x). $$

So, by hypothesis, $\Phi(m)/D_{KR}(m, m_0)$ vanishes as $m$ goes to $m_0$.

*Step 1.* In a first time, we prove that $\Phi$ is the zero function. Let us fix $m \in \mathcal{P}_1(\mathbb{R})$ and introduce, for any $n \in \mathbb{N}^*$,

$$ \mu_n = \frac{n-1}{n} m_0 + \frac{1}{n} m \in \mathcal{P}_1(\mathbb{R}). $$

We have that, for all $n \in \mathbb{N}^*$,

$$ D_{KR}(m_0, \mu_n) \leq \frac{1}{n} D_{KR}(m_0, m), $$
for any $x \in \mathbb{R}$ (with $C > 0$ independent of $x, m, m_0$), whence the quantity at (51) is bounded by

$$C \left( 1 + \int_{\mathbb{R}} |x| dm_0(x) \right) D_{KR}(m, m_0)^2$$

which is negligible compared to $D_{KR}(m, m_0)$ when $m$ converges to $m_0$ (for a fixed $m_0$).

In addition, the term (52) is non-greater than

$$D_{KR}(m, m_0) \cdot \sup_{x \in \mathbb{R}} |\partial_x (H(m, x) - H(m_0, x))|.$$

And, by the Mean Value Theorem (i.e. Proposition 2.21) and the hypothesis (ii) of the corollary, for all $x \in \mathbb{R}$,

$$|\partial_x (H(m, x) - H(m_0, x))| = |\partial_x H_x(m) - \partial_x H_x(m_0)| \\ \le D_{KR}(m, m_0) \cdot \sup_{\substack{m \in \mathcal{P}_1(\mathbb{R}) \\ x, y \in \mathbb{R}}} |\partial_y \delta(\partial_x H_x)(m, y)|,$$

which entails that the term at (52) is bounded by

$$CD_{KR}(m, m_0)^2$$

for some constant $C > 0$ independent of $m, m_0$.

So we have proved that $F$ is differentiable on $\mathcal{P}_1(\mathbb{R})$, and that the following function is one version of its derivative:

$$(m, x) \in \mathcal{P}_1(\mathbb{R}) \times \mathbb{R} \mapsto H(m, x) + \int_{\mathbb{R}} \delta H_y(m, x) dm(y).$$

It is then sufficient to subtract $F(m)$ to the quantity above to obtain the canonical derivative of $F$. $\square$

The proof of Corollary 2.27 requires the following lemma.

**Lemma C.1.** Let $d \in \mathbb{N}^*$ and $m_1, \dots, m_d, \mu_1, \dots, \mu_d \in \mathcal{P}_1(\mathbb{R})$. Then,

$$D_{KR} \left( \bigotimes_{k=1}^{d} m_k, \bigotimes_{k=1}^{d} \mu_k \right) \le \sum_{k=1}^{d} D_{KR}(m_k, \mu_k).$$

*Proof.* For each $1 \le k \le d$, there exists a probability space $(\Omega_k, \mathcal{F}_k, \mathbb{P}_k)$ and two random variables on this space $X_k, Y_k$ of respective laws $m_k, \mu_k$ such that

$$D_{KR}(m_k, \mu_k) = \mathbb{E}_k [|X_k - Y_k|],$$

with $\mathbb{E}_k$ the expectation w.r.t. $\mathbb{P}_k$ (the existence of these random variables is guaranteed by Theorem 4.1 of Villani (2009)). Let us assume that all the probability spaces $(\Omega_k, \mathcal{F}_k, \mathbb{P}_k)$ are disjoint, and consider the product probability space of these spaces, denoted by $(\Omega, \mathcal{F}, \mathbb{P})$. In particular, the variables $X_k$ (resp. $Y_k$) ($1 \le k \le d$) are independent, whence

$$\mathcal{L}(X_1, \dots, X_d) = \bigotimes_{k=1}^{d} m_k \quad \text{and} \quad \mathcal{L}(Y_1, \dots, Y_d) = \bigotimes_{k=1}^{d} \mu_k.$$
hence, $\mu_n$ converges to $m_0$ as $n$ goes to infinity.

On the other hand,

$$ \Phi(m) = n\Phi(\mu_n) = n\frac{\Phi(\mu_n)}{D_{KR}(m_0, \mu_n)}D_{KR}(m_0, \mu_n) \le \frac{\Phi(\mu_n)}{D_{KR}(m_0, \mu_n)}D_{KR}(m_0, m) \underset{n\to\infty}{\longrightarrow} 0. $$

This proves that, for all $m \in \mathcal{P}_1(\mathbb{R})$, $\Phi(m) = 0$.

*Step 2.* To conclude the proof, it is then sufficient to notice that, thanks to *Step 1*, for any $y \in \mathbb{R}$, $\Phi(\delta_y) = 0$. Whence, for any $y \in \mathbb{R}$,

$$ h_1(y) - h_2(y) = \int_{\mathbb{R}} (h_1(x) - h_2(x)) dm_0(x). $$

In particular, the function $h_1 - h_2$ is constant, and the lemma is proved. $\square$

*Proof of Lemma 2.28.* For each $1 \le k \le d$, there exists a probability space $(\Omega_k, \mathcal{F}_k, \mathbb{P}_k)$ and two random variables on this space $X_k, Y_k$ of respective laws $m_k, \mu_k$ such that

$$ D_{KR}(m_k, \mu_k) = \mathbb{E}_k [|X_k - Y_k|], $$

with $\mathbb{E}_k$ the expectation w.r.t. $\mathbb{P}_k$ (the existence of these random variables is guaranteed by Theorem 4.1 of?). Let us assume that all the probability spaces $(\Omega_k, \mathcal{F}_k, \mathbb{P}_k)$ are disjoint, and consider the product probability space of these spaces, denoted by $(\Omega, \mathcal{F}, \mathbb{P})$. In particular, the variables $X_k$ (resp. $Y_k$) ($1 \le k \le d$) are independent, whence

$$ \mathcal{L}(X_1, \dots, X_d) = \bigotimes_{k=1}^{d} m_k \quad \text{and} \quad \mathcal{L}(Y_1, \dots, Y_d) = \bigotimes_{k=1}^{d} \mu_k. $$

Consequently,

$$ D_{KR} \left( \bigotimes_{k=1}^{d} m_k, \bigotimes_{k=1}^{d} \mu_k \right) \le \mathbb{E} \left[ \sum_{k=1}^{d} |X_k - Y_k| \right] = \sum_{k=1}^{d} \mathbb{E}_k [|X_k - Y_k|] = \sum_{k=1}^{d} D_{KR}(m_k, \mu_k), $$

which proves the result. $\square$

## Appendix D: Some technical lemmas about Ito's integrals

Let $(X_t)_{t \ge 0}$ be some càdlàg $\mathbb{R}$-valued process that is locally $L^2$: for all $t \ge 0$,

$$ \int_0^t \mathbb{E}[X_s^2] ds < \infty. \qquad (56) $$

Let $W, B$ be two standard Brownian motions of dimension one, and $(\mathcal{F}_t)_t$ be a filtration such that $\mathcal{F}, B$ and $W$ are independent. Let us denote $(W_t)_{t \text{ resp. } (B_t)_t)$ the filtration of $W$ (resp. $B$), meaning

$$ W_t = \sigma(W_s : s \le t) \text{ and } B_t = \sigma(B_s : s \le t), $$

and define $(\mathcal{G}_t)_t$ the union (in the filtration sense) of $\mathcal{F}, W$ and $B$: for all $t \ge 0$,

$$ \mathcal{G}_t = \mathcal{F}_t \vee W_t \vee \mathcal{B}_t. $$
Consequently,

$$D_{KR}\left(\bigotimes_{k=1}^{d} m_k, \bigotimes_{k=1}^{d} \mu_k\right) \leq \mathbb{E}\left[\sum_{k=1}^{d} |X_k - Y_k|\right] = \sum_{k=1}^{d} \mathbb{E}_k\left[|X_k - Y_k|\right] = \sum_{k=1}^{d} D_{KR}(m_k, \mu_k),$$

which proves the result.

Proof of Corollary 2.27. Let us fix some $m, m_0 \in \mathcal{P}_1(\mathbb{R})$ in the proof.

<figure><img src="image_6.png" /></figure>

$$ \begin{aligned} F(m) - F(m_0) &= \int_{\mathbb{R}^2} H(m, x) d(m^{\otimes 2} - m_0^{\otimes 2}) (x) \\ &\quad + \int_{\mathbb{R}^2} (H(m, x) - H(m_0, x)) dm_0^{\otimes 2} (x) \\ &= \int_{\mathbb{R}^2} H(m_0, x_1, x_2) dm_0(x_1) d(m - m_0)(x_2) \\ &\quad + \int_{\mathbb{R}^2} H(m_0, x_1, x_2) d(m - m_0)(x_1) dm_0(x_2) \\ &\quad + \int_{\mathbb{R}^2} (H(m, x) - H(m_0, x)) dm_0^{\otimes 2} (x) \\ &\quad + \int_{\mathbb{R}^2} (H(m, x) - H(m_0, x)) d(m^{\otimes 2} - m_0^{\otimes 2}) (x). \end{aligned} $$

Then,

$$ \left| F(m) - F(m_0) - \int_{\mathbb{R}} \int_{\mathbb{R}^2} \delta(H_x)(m_0, y) dm_0^{\otimes 2}(x) d(m - m_0)(y) - \int_{\mathbb{R}} \left( \int_{\mathbb{R}} H(m_0, x_1, y) dm_0(x_1) + \int_{\mathbb{R}} H(m_0, y, x_2) dm_0(x_2) \right) d(m - m_0)(y) \right| \le \left| \int_{\mathbb{R}^2} H(m_0, x_1, x_2) d(m - m_0)^{\otimes 2}(x_1, x_2) \right| \quad (53) $$

$$ + \left| \int_{\mathbb{R}^2} \left( H(m, x) - H(m_0, x) - \int_{\mathbb{R}} \delta(H_x)(m_0, y) d(m - m_0)(y) \right) dm_0^{\otimes 2}(x) \right| \quad (54) $$

$$ + \left| \int_{\mathbb{R}^2} (H(m, x) - H(m_0, x)) d(m^{\otimes 2} - m_0^{\otimes 2}) (x) \right| \quad (55) $$

By Lemma 2.23 and thanks to the hypothesis (iii) of the corollary, the term at (53) is bounded by

$$ D_{KR}(m, m_0)^2 \sup_{\mu \in \mathcal{P}_1(\mathbb{R}), x_1, x_2 \in \mathbb{R}} \left| \partial_{x_1 x_2}^2 H(\mu, x) \right|. $$

According to Taylor-Lagrange's inequality (i.e. Theorem 2.24) and using Hypothesis (i), we know that the term at (54) is non-greater than

$$ C \cdot D_{KR}(m, m_0)^2 \left( 1 + \left( \int_{\mathbb{R}} |x| dm_0(x) \right)^2 \right), $$
**Lemma D.1.** Assume that X is G-adapted. Then, for all $t \ge 0$,

$$ \mathbb{E} \left[ \int_0^t X_s dW_s \mid \mathcal{W}_t \right] = \int_0^t \mathbb{E} [X_s \mid \mathcal{W}_s] dW_s. $$

*Proof.* Let us fix $t \ge 0$. By definition of Ito's integral,

$$ \int_0^t X_s dW_s = \lim_{n \to \infty} \sum_{k=0}^{n-1} X_{s_k} (W_{s_{k+1}} - W_{s_k}) $$

in probability, where, for any $0 \le k \le n$, $s_k = t \cdot k/n$.

Then, thanks to (56), the (conditional) Vitali's convergence theorem implies that, almost surely,

$$ \begin{aligned} \mathbb{E} \left[ \int_0^t X_s dW_s \mid \mathcal{W}_t \right] &= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} (W_{s_{k+1}} - W_{s_k}) \mid \mathcal{W}_t \right] \\ &= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} \mid \mathcal{W}_t \right] (W_{s_{k+1}} - W_{s_k}) \\ &= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} \mid \mathcal{W}_{s_k} \right] (W_{s_{k+1}} - W_{s_k}) \\ &= \int_0^t \mathbb{E} [X_s \mid \mathcal{W}_s] dW_s, \end{aligned} $$

where the before last equality above comes from the fact that X is G-adapted and that $W_t$ can be written as the union (in the filtration sense) of $W_{s_k}$ and $\sigma(W_r - W_{s_k} : s_k < r \le t)$ which are independent (whence, $X_{s_k}$ is also necessarily independent of $\sigma(W_r - W_{s_k} : s_k < r \le t)$). $\square$

**Lemma D.2.** Assume that X is G-adapted. Then, for all $t \ge 0$,

$$ \mathbb{E} \left[ \int_0^t X_s dW_s \mid \mathcal{B}_t \right] = 0. $$

*Proof.* With the same reasoning (and using the same notation) as in the beginning of the proof of Lemma D.1, for all $t \ge 0$,

$$ \mathbb{E} \left[ \int_0^t X_s dB_s \mid \mathcal{W}_t \right] = \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} (B_{s_{k+1}} - B_{s_k}) \mid \mathcal{W}_t \right]. \quad (57) $$

Let us recall that X is G-adapted and that B, W are independent. Let us consider any $W_t$-measurable random variable Z. Then, both Z and $X_{s_k}$ are independent of $B_{s_{k+1}} - B_{s_k}$. So,

$$ \mathbb{E} [X_{s_k} (B_{s_{k+1}} - B_{s_k}) Z] = \mathbb{E} [X_{s_k} Z] \mathbb{E} [B_{s_{k+1}} - B_{s_k}] = 0, $$

implying

$$ \mathbb{E} [X_{s_k} (B_{s_{k+1}} - B_{s_k}) \mid \mathcal{W}_t] = 0. $$

Combining the equation above with (57) proves the result. $\square$
for some $C > 0$ independent of $(m, m_0)$.

For the last term (i.e. (55)), by definition of Kantorovich-Rubinstein's metric, is bounded by

$$D_{KR}(m^{\otimes 2}, m_0^{\otimes 2}) \cdot \sup_{x_1, x_2 \in \mathbb{R}} |\partial_{x_1} H_x(m) - \partial_{x_1} H_x(m_0)| + |\partial_{x_2} H_x(m) - \partial_{x_2} H_x(m_0)|.$$

As in the end of the proof of the previous corollary, we have

$$\sup_{x_1, x_2 \in \mathbb{R}} |\partial_{x_1} H_x(m) - \partial_{x_1} H_x(m_0)| + |\partial_{x_2} H_x(m) - \partial_{x_2} H_x(m_0)| \le C \cdot D_{KR}(m, m_0).$$

On the other hand, by Lemma C.1, we have that

$$D_{KR}(m^{\otimes 2}, m_0^{\otimes 2}) \le 2D_{KR}(m, m_0).$$

So the term at (55) is bounded by $C \cdot D_{KR}(m, m_0)^2$. This proves finally that the function $F$ is differentiable on $P_1(\mathbb{R})$ and that the function

$$(m, y) \in P_1(\mathbb{R}) \times \mathbb{R} \mapsto \int_{\mathbb{R}^2} \delta(H_x)(m, y) dm^{\otimes 2}(x) + \int_{\mathbb{R}} H(m, x_1, y) dm(x_1) \\ + \int_{\mathbb{R}} H(m, y, x_2) dm(x_2)$$

is one version of the derivative of $F$. It is then sufficient to subtract $2F(m)$ to the quantity above to obtain the canonical derivative of $F$. $\square$

## Appendix D: Some technical lemmas about Ito's integrals

Let $(X_t)_{t \ge 0}$ be some càdlàg $\mathbb{R}$-valued process that is locally $L^2$: for all $t \ge 0$,

$$\int_0^t \mathbb{E}[X_s^2] ds < \infty. \qquad (56)$$

Let $W, B$ be two standard Brownian motions of dimension one, and $(F_t)_t$ be a filtration such that $F, B$ and $W$ are independent. Let us denote $(W_t)_t$ (resp. $(B_t)_t$) the filtration of $W$ (resp. $B$), meaning

$$W_t = \sigma(W_s : s \le t) \text{ and } B_t = \sigma(B_s : s \le t),$$

and define $(G_t)_t$ the union (in the filtration sense) of $F, W$ and $B$: for all $t \ge 0$,

$$G_t = F_t \vee W_t \vee B_t.$$

**Lemma D.1.** Assume that $X$ is $G$-adapted. Then, for all $t \ge 0$,

$$\mathbb{E} \left[ \int_0^t X_s dW_s \mid W_t \right] = \int_0^t \mathbb{E} [X_s | W_s] dW_s.$$

*Proof.* Let us fix $t \ge 0$. By definition of Ito's integral,

$$\int_0^t X_s dW_s = \lim_{n \to \infty} \sum_{k=0}^{n-1} X_{s_k} (W_{s_{k+1}} - W_{s_k})$$
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in probability, where, for any $0 \le k \le n$, $s_k = t \cdot k/n$.

Then, thanks to (56), the (conditional) Vitali's convergence theorem implies that, almost surely,

$$ 
\begin{aligned}
\mathbb{E} \left[ \int_0^t X_s dW_s \middle| \mathcal{W}_t \right] &= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} (W_{s_{k+1}} - W_{s_k}) \middle| \mathcal{W}_t \right] \\
&= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} \middle| \mathcal{W}_t \right] (W_{s_{k+1}} - W_{s_k}) \\
&= \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} \middle| \mathcal{W}_{s_k} \right] (W_{s_{k+1}} - W_{s_k}) \\
&= \int_0^t \mathbb{E} \left[ X_s \middle| \mathcal{W}_s \right] dW_s,
\end{aligned}
$$

where the before last equality above comes from the fact that $X$ is $\mathcal{G}$-adapted and that $W_t$ can be written as the union (in the filtration sense) of $W_{s_k}$ and $\sigma(W_r - W_{s_k} : s_k < r \le t)$ which are independent (whence, $X_{s_k}$ is also necessarily independent of $\sigma(W_r - W_{s_k} : s_k < r \le t)$). $\square$

**Lemma D.2.** Assume that $X$ is $\mathcal{G}$-adapted. Then, for all $t \ge 0$,

$$ \mathbb{E} \left[ \int_0^t X_s dW_s \middle| \mathcal{B}_t \right] = 0. $$

*Proof.* With the same reasoning (and using the same notation) as in the beginning of the proof of Lemma D.1, for all $t \ge 0$,

$$ \mathbb{E} \left[ \int_0^t X_s dB_s \middle| \mathcal{W}_t \right] = \lim_{n \to \infty} \sum_{k=0}^{n-1} \mathbb{E} \left[ X_{s_k} (B_{s_{k+1}} - B_{s_k}) \middle| \mathcal{W}_t \right]. \quad (57) $$

Let us recall that $X$ is $\mathcal{G}$-adapted and that $B, W$ are independent. Let us consider any $W_t$-measurable random variable $Z$. Then, both $Z$ and $X_{s_k}$ are independent of $B_{s_{k+1}} - B_{s_k}$. So,

$$ \mathbb{E} [X_{s_k} (B_{s_{k+1}} - B_{s_k}) Z] = \mathbb{E} [X_{s_k} Z] \mathbb{E} [B_{s_{k+1}} - B_{s_k}] = 0, $$

implying

$$ \mathbb{E} [X_{s_k} (B_{s_{k+1}} - B_{s_k}) | \mathcal{W}_t] = 0. $$

Combining the equation above with (57) proves the result. $\square$

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FOURNIER, N. and LÖCHERBACH, E. (2016). On a toy model of interacting neurons. *Annales de l'Institut Henri Poincaré, Probabilités et Statistiques* **52** 1844–1876. Publisher: Institut Henri Poincaré.
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GUO, X., PHAM, H. and WEI, X. (2023). Itô's formula for flows of measures on semimartingales. *Stochastic Processes and their Applications* **159** 350–390.

JOURDAIN, B. and TSE, A. (2021). Central limit theorem over non-linear functionals of empirical measures with applications to the mean-field fluctuation of interacting diffusions. *Electronic Journal of Probability* **26** 1–34. Publisher: Institute of Mathematical Statistics and Bernoulli Society.

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LU, X. and ABERGEL, F. (2018). High-dimensional Hawkes processes for limit order books: modelling, empirical analysis and numerical calibration. *Quantitative Finance* **18** 249–264.

MEYN, S. P. and TWEEDIE, R. L. (1993). Stability of Markovian Processes III: Foster-Lyapunov Criteria for Continuous-Time Processes. *Advances in Applied Probability* **25** 518–548. Publisher: Applied Probability Trust.

MISCHLER, S., MOUHOT, C. and WENNBERG, B. (2015). A new approach to quantitative propagation of chaos for drift, diffusion and jump processes. *Probability Theory and Related Fields* **161** 1–59.

MITCHELL, L. and CATES, M. E. (2009). Hawkes process as a model of social interactions: a view on video dynamics. *Journal of Physics A: Mathematical and Theoretical* **43** 045101.

REYNAUD-BOURET, P. and SCHBATH, S. (2010). Adaptive Estimation for Hawkes Processes; Application to Genome Analysis. *The Annals of Statistics* **38** 2781–2822. Publisher: Institute of Mathematical Statistics.

VEDENYAPIN, V., SINITSYN, A. and DULOV, E. (2011). Boltzmann Equation. In *Kinetic Boltzmann, Vlasov and Related Equations* 165–182. Elsevier, London.

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