Here, we recall that $\eta_0^\delta$ is deterministic because $m$ and $\tilde{m}$ are deterministic, see Remark 16. Finally, using inequalities (33) and (44), we obtain

$$ (47) \qquad \begin{aligned} & \kappa_1 \mathbb{E}_0 [|\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|] \le e^{-ct} |\mathbb{E}_1[X_0^\delta] - \mathbb{E}_1[\tilde{X}_0^\delta]| + \frac{h(\delta)}{c} \\ & \qquad + C_b \int_0^t e^{-c(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d}) \mathbb{E}_0 \left[ |A_s^\delta - \tilde{A}_s^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \mathrm{d}s, \end{aligned} $$
where $C_b := C_G + C_F$.

Step 5. We now focus on the (normed) difference process $(|A_t^\delta - \tilde{A}_t^\delta|)_{t \ge 0}$. Thanks to the properties of $F$, as stated in Assumption 2, we are here able to show that it decreases exponentially fast to zero.

Recalling from Step 2 that $A_t^\delta = X_t^\delta - \mathbb{E}_1[X_t^\delta]$ for $t \ge 0$, we have

$$ dA_t^\delta = (G(X_t^\delta) - \mathbb{E}_1[G(X_t^\delta)]) dt + (F(X_t^\delta - \mathbb{E}_1[X_t^\delta])dt - \mathbb{E}_1[F(X_t^\delta - \mathbb{E}_1[X_t^\delta])] dt + \sigma dB_t, $$

from which we obtain

$$ \frac{d}{dt}(A_t^\delta - \tilde{A}_t^\delta) = [G(X_t^\delta) - G(\tilde{X}_t^\delta)] + [F(A_t^\delta) - F(\tilde{A}_t^\delta)] \\ - [\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]] - [\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]] $$

Then,

$$ (48) \qquad \begin{aligned} \frac{d}{dt}|A_t^\delta - \tilde{A}_t^\delta|^2 = & 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta)) \\ & + 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (F(A_t^\delta) - F(\tilde{A}_t^\delta)) \\ & - 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \\ & - 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]) \end{aligned} $$

We can immediately leverage on Assumption 2 to control the second term on the right hand side:

$$ (49) \qquad 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (F(A_t^\delta) - F(\tilde{A}_t^\delta)) \le -2\alpha_F |A_t^\delta - \tilde{A}_t^\delta|^2. $$

Let us now focus on the terms involving $G$. We have

$$ \begin{aligned} & (A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \\ &= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &\quad + (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &\quad - (A_t^\delta - \tilde{A}_t^\delta) \cdot \mathbb{E}_1 \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &\quad - (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \right]. \end{aligned} $$
Here, we recall that $\eta_0^\delta$ is deterministic because $m$ and $\tilde{m}$ are deterministic, see Remark 16. Finally, using inequalities (33) and (44), we obtain

$$ (47) \qquad \begin{aligned} & \kappa_1 \mathbb{E}_0 [|\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|] \le e^{-ct} |\mathbb{E}_1[X_0^\delta] - \mathbb{E}_1[\tilde{X}_0^\delta]| + \frac{h(\delta)}{c} \\ & \qquad + C_b \int_0^t e^{-c(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d}) \mathbb{E}_0 \left[ |A_s^\delta - \tilde{A}_s^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \mathrm{d}s, \end{aligned} $$
where $C_b := C_G + C_F$.

Step 5. We now focus on the (normed) difference process $(|A_t^\delta - \tilde{A}_t^\delta|)_{t \ge 0}$. Thanks to the properties of $F$, as stated in Assumption 2, we are here able to show that it decreases exponentially fast to zero.

Recalling from Step 2 that $A_t^\delta = X_t^\delta - \mathbb{E}_1[X_t^\delta]$ for $t \ge 0$, we have

$$ \begin{aligned} dA_t^\delta = & (G(X_t^\delta) - \mathbb{E}_1[G(X_t^\delta)]) dt + (F(X_t^\delta - \mathbb{E}_1[X_t^\delta])dt - \mathbb{E}_1[F(X_t^\delta - \mathbb{E}_1[X_t^\delta])] dt + \sigma dB_t, \\ & \text{from which we obtain} \end{aligned} $$

$$ \frac{d}{dt}(A_t^\delta - \tilde{A}_t^\delta) = [G(X_t^\delta) - G(\tilde{X}_t^\delta)] + [F(A_t^\delta) - F(\tilde{A}_t^\delta)] \\ - [\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]] - [\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]] $$

Then,

$$ (48) \qquad \begin{aligned} \frac{d}{dt}|A_t^\delta - \tilde{A}_t^\delta|^2 = & 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta)) \\ & + 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (F(A_t^\delta) - F(\tilde{A}_t^\delta)) \\ & - 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \\ & - 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]) \end{aligned} $$

We can immediately leverage on Assumption 2 to control the second term on the right hand side:

$$ (49) \qquad 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (F(A_t^\delta) - F(\tilde{A}_t^\delta)) \le -2\alpha_F |A_t^\delta - \tilde{A}_t^\delta|^2. $$

Let us now focus on the terms involving $G$. We have

$$ \begin{aligned} & (A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \\ &= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &+ (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &- (A_t^\delta - \tilde{A}_t^\delta) \cdot \mathbb{E}_1 \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\ &- (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \right]. \end{aligned} $$
By combining the second and fourth term in the right hand side, we obtain

$$
\begin{align*}
& (A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta)) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)] \\
& = (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\
& \quad + (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (A_t^\delta - \tilde{A}_t^\delta) \\
& \quad - (A_t^\delta - \tilde{A}_t^\delta) \cdot \mathbb{E}_1 \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\
& =: \mathbf{\Delta}_1(t) + \mathbf{\Delta}_2(t) + \mathbf{\Delta}_3(t).
\end{align*}
$$

Then, if we perform the following decomposition of the first term on the right-hand side, we obtain

$$
(50) \qquad \mathbf{\Delta}_1(t) = \mathbf{\Delta}\mathbf{\Delta}_1(t) + \mathbf{\Delta}\mathbf{\Delta}_2(t),
$$

with

$$
\begin{align*}
\overline{\mathbf{A}\mathbf{A}}_1(t) &:= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \left\{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \right\} d\rho \right) (\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \right] \\
\overline{\mathbf{A}\mathbf{A}}_2(t) &:= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 \left\{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \right\} d\rho \right) (A_t^\delta - \tilde{A}_t^\delta).
\end{align*}
$$

Then, combining the terms $\overline{\mathbf{A}}_2(t)$ and $\overline{\mathbf{A}}_2(t)$, we get

$$
\begin{align}
(A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \tag{51} \\
&= \overline{\mathbf{A}}_1(t) + [\overline{\mathbf{A}}_2(t) + \overline{\mathbf{A}}_2(t)] + \overline{\mathbf{A}}_3(t) \nonumber \\
&=: \overline{\mathbf{A}}_1(t) + \mathbf{\Pi}_2(t) + \overline{\mathbf{A}}_3(t), \nonumber
\end{align}
$$

where $\overline{\mathbf{A}}_1(t)$ and $\overline{\mathbf{A}}_3(t)$ have been already defined and $\mathbf{\Pi}_2(t)$ is defined by

$$
\mathbf{\Pi}_2(t) := (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) d\rho \right) (A_t^\delta - \tilde{A}_t^\delta).
$$

We now control $\mathbf{\Pi}_2(t)$ as we handled Equation (49). We get

$$
(52) \qquad \mathbf{\Pi}_2(t) \le m_G |A_t^\delta - \tilde{A}_t^\delta|^2,
$$

where we recall from (6) that $m_G \in \mathbb{R}$ stands for the upper bound of the function $r \mapsto [-\sigma_0^2 \kappa(r)/2]$ introduced in Assumption 1. Indeed, as a consequence of 1, one can show that for any $x, y \in \mathbb{R}^d$,

$$
y \cdot (DG(x)y) \leq m_G |y|^2.
$$
By combining the second and fourth term in the right hand side, we obtain

$$
\begin{align*}
& (A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta)) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)] \\
&= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\
&\quad + (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) d\rho \right) (A_t^\delta - \tilde{A}_t^\delta) \\
&\quad - (A_t^\delta - \tilde{A}_t^\delta) \cdot \mathbb{E}_1 \left[ \left( \int_0^1 \{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \} d\rho \right) (X_t^\delta - \tilde{X}_t^\delta) \right] \\
&=: \mathbf{\Delta}_1(t) + \mathbf{\Delta}_2(t) + \mathbf{\Delta}_3(t).
\end{align*}
$$

Then, if we perform the following decomposition of the first term on the right-hand side, we obtain

$$
(50) \qquad \mathbf{\Delta}_1(t) = \mathbf{\Delta}\mathbf{\Delta}_1(t) + \mathbf{\Delta}\mathbf{\Delta}_2(t),
$$

with

$$
\begin{align*}
\overline{\mathbf{A}\mathbf{A}}_1(t) &:= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left[ \left( \int_0^1 \left\{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \right\} d\rho \right) (\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \right] \\
\overline{\mathbf{A}\mathbf{A}}_2(t) &:= (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 \left\{ DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) - DG(\rho \mathbb{E}_1[X_t^\delta] + (1-\rho)\mathbb{E}_1[\tilde{X}_t^\delta]) \right\} d\rho \right) (A_t^\delta - \tilde{A}_t^\delta).
\end{align*}
$$

Then, combining the terms $\overline{\mathbf{A}}_2(t)$ and $\overline{\mathbf{A}}_2(t)$, we get

$$
\begin{align}
(A_t^\delta - \tilde{A}_t^\delta) \cdot (G(X_t^\delta) - G(\tilde{X}_t^\delta) - \mathbb{E}_1[G(X_t^\delta)] + \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \tag{51} \\
&= \overline{\mathbf{A}}_1(t) + [\overline{\mathbf{A}}_2(t) + \overline{\mathbf{A}}_2(t)] + \overline{\mathbf{A}}_3(t) \nonumber \\
&:= \overline{\mathbf{A}}_1(t) + \underline{\mathbf{\Pi}}_2(t) + \overline{\mathbf{A}}_3(t), \nonumber
\end{align}
$$

where $\overline{\mathbf{A}}_1(t)$ and $\overline{\mathbf{A}}_3(t)$ have been already defined and $\underline{\mathbf{\Pi}}_2(t)$ is defined by

$$
\underline{\Pi}_2(t) := (A_t^\delta - \tilde{A}_t^\delta) \cdot \left( \int_0^1 DG(\rho X_t^\delta + (1-\rho)\tilde{X}_t^\delta) d\rho \right) (A_t^\delta - \tilde{A}_t^\delta).
$$

We now control $\underline{\Pi}_2(t)$ as we handled Equation (49). We get

$$
(52) \qquad \underline{\Pi}_2(t) \le m_G |A_t^\delta - \tilde{A}_t^\delta|^2,
$$

where we recall from (6) that $m_G \in \mathbb{R}$ stands for the upper bound of the function $r \mapsto [-\sigma_0^2 \kappa(r)/2]$ introduced in Assumption 1. Indeed, as a consequence of 1, one can show that for any $x, y \in \mathbb{R}^d$,

$$
y \cdot (DG(x)y) \le m_G |y|^2.
$$
Then, inserting (49), (51) and (52) into (48), we obtain

$$ \frac{d}{dt} |A_t^\delta - \tilde{A}_t^\delta|^2 \\ \le -2(\alpha_F - m_G) |A_t^\delta - \tilde{A}_t^\delta|^2 + \underline{\mathbf{AA}}_1(t) + \underline{\mathbf{A}}_3(t) + 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]). $$

By taking expectation under $\mathbb{P}_1$ and by using the fact that $\mathbb{E}_1[A_t^\delta] = \mathbb{E}_1[\tilde{A}_t^\delta] = 0$, we get rid the last two terms in the above right-hand side. We obtain

$$ (53) \qquad \frac{d}{dt} \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \le -2(\alpha_F - m_G) \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] + 2 \mathbb{E}_1 \left[ \underline{\mathbf{AA}}_1(t) \right]. $$

In order to control $\underline{\mathbf{AA}}_1(t)$, we come back to (50) and use the fact that $\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]$ is measurable with respect to the common noise. By Cauchy-Schwarz' inequality, we obtain

$$ \mathbb{E}_1[\underline{\mathbf{AA}}_1(t)] \le C_G \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \eta_t^\delta |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

where we recall that $\eta_t^\delta = \mathbb{E}_1 [|A_t^\delta|^2 + |\tilde{A}_t^\delta|^2]^{1/2}$. Then, from Proposition 10, we get

$$ \mathbb{E}_1[\underline{\mathbf{AA}}_1(t)] \le \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \left( \eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

and (53) rewrites

$$ (54) \qquad \frac{d}{dt} \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \le -2(\alpha_F - m_G) \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \\ \qquad \qquad + 2 \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \left( \eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|. $$

Then, we get

$$ (55) \qquad \frac{d}{dt} \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \le -(\alpha_F - m_G) \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \\ \qquad \qquad + C_G (\eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

see Appendix A.2 for details. Using Grönwall's Lemma, we obtain

$$ \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \le \mathbb{E}_1 [|A_0^\delta - \tilde{A}_0^\delta|^2]^{1/2} \exp(-(\alpha_F - m_G)t) \\ + C_G \int_0^t e^{-(\alpha_F - m_G)(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| ds. $$

Finally, taking again expectation $\mathbb{E}_0$,

$$ (56) \qquad \mathbb{E}_0 \left[ \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \right] \le \exp(-(\alpha_F - m_G)t) \mathbb{E}_1 [|A_0^\delta - \tilde{A}_0^\delta|^2]^{1/2} \\ \qquad \qquad + C_G \int_0^t e^{-(\alpha_F - m_G)(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) \mathbb{E}_0 \left[ |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \right] ds. $$
Then, inserting (49), (51) and (52) into (48), we obtain

$$ \frac{d}{dt} |A_t^\delta - \tilde{A}_t^\delta|^2 \\ \le -2(\alpha_F - m_G) |A_t^\delta - \tilde{A}_t^\delta|^2 + \underline{\mathbf{AA}}_1(t) + \underline{\mathbf{A}}_3(t) + 2(A_t^\delta - \tilde{A}_t^\delta) \cdot (\mathbb{E}_1[F(A_t^\delta)] - \mathbb{E}_1[F(\tilde{A}_t^\delta)]). $$

By taking expectation under $\mathbb{P}_1$ and by using the fact that $\mathbb{E}_1[A_t^\delta] = \mathbb{E}_1[\tilde{A}_t^\delta] = 0$, we get rid the last two terms in the above right-hand side. We obtain

$$ (53) \qquad \frac{d}{dt} \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \le -2(\alpha_F - m_G) \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] + 2 \mathbb{E}_1 \left[ \underline{\mathbf{AA}}_1(t) \right]. $$

In order to control $\underline{\mathbf{AA}}_1(t)$, we come back to (50) and use the fact that $\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]$ is measurable with respect to the common noise. By Cauchy-Schwarz' inequality, we obtain

$$ \mathbb{E}_1[\underline{\mathbf{AA}}_1(t)] \le C_G \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \eta_t^\delta |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

where we recall that $\eta_t^\delta = \mathbb{E}_1 [|A_t^\delta|^2 + |\tilde{A}_t^\delta|^2]^{1/2}$. Then, from Proposition 10, we get

$$ \mathbb{E}_1[\underline{\mathbf{AA}}_1(t)] \le \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \left( \eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

and (53) rewrites

$$ (54) \qquad \frac{d}{dt} \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \le -2(\alpha_F - m_G) \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right] \\ \qquad \qquad + 2 \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \left( \eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|. $$

Then, we get

$$ (55) \qquad \frac{d}{dt} \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \le -(\alpha_F - m_G) \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \\ \qquad \qquad + C_G (\eta_0^\delta e^{-c_{\alpha_F} t} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|, $$

see Appendix A.2 for details. Using Grönwall's Lemma, we obtain

$$ \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \le \mathbb{E}_1 [|A_0^\delta - \tilde{A}_0^\delta|^2]^{1/2} \exp(-(\alpha_F - m_G)t) \\ + C_G \int_0^t e^{-(\alpha_F - m_G)(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| ds. $$

Finally, taking again expectation $\mathbb{E}_0$,

$$ (56) \qquad \mathbb{E}_0 \left[ \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} \right] \le \exp(-(\alpha_F - m_G)t) \mathbb{E}_1 [|A_0^\delta - \tilde{A}_0^\delta|^2]^{1/2} \\ \qquad \qquad + C_G \int_0^t e^{-(\alpha_F - m_G)(t-s)} (\eta_0^\delta e^{-c_{\alpha_F} s} + \sigma c_{\alpha_F}^{-1/2} \sqrt{d}) \mathbb{E}_0 \left[ |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \right] ds. $$
Step 6. Now, as (47) remains true replacing the rate c by $\tilde{c} \le c$, we can assume without loss of generality that $\alpha_F > m_G + c$. We can then combine Equations (47) and (56) to obtain, for all $t \ge 0$,

$$ (57) \quad \begin{aligned} & \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]| \right] \\ & \le \left( \mathbb{E}_1 \left[ |A_0^\delta - \tilde{A}_0^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_0^\delta] - \mathbb{E}_1[\tilde{X}_0^\delta]| \right) e^{-ct} + \frac{h(\delta)}{c} \\ & \quad + c_3^b \int_0^t e^{-c(t-s)} \left( \eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |A_s^\delta - \tilde{A}_s^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \right] ds, \end{aligned} $$

for some constant $c_3^b > 0$ that depends on F and G.

With the notation $\Theta_t^\delta := \mathbb{E}_1 [|A_t^\delta - \tilde{A}_t^\delta|^2]^{1/2} + |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|$, Equation (57) reads

$$ \mathbb{E}_0[\Theta_t^\delta] \le \Theta_0^\delta e^{-ct} + \frac{h(\delta)}{c} + c_3^b \int_0^t e^{-c(t-s)} \left( \eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) \mathbb{E}_0[\Theta_s^\delta] ds. $$

Then, using Grönwall Lemma, we obtain

$$ (58) \quad \begin{aligned} e^{ct} \mathbb{E}_0[\Theta_t^\delta] \le & \left( \Theta_0 + \frac{h(\delta)}{c} \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} t \right) \\ & + h(\delta) \int_0^t e^{cs} \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} (t-s) \right) ds. \end{aligned} $$

We emphasize here that we write $\Theta_0$ and $\eta_0$ without reference to $\delta$ because these quantities are predetermined and independent of $\delta$. Now, let us notice that creating the couplage we defined two Brownian motions adapted to the filtration $\mathbb{F}_0$:

$$ \beta_t^\delta = \int_0^t \pi_\delta(E_s^\delta)(\mathrm{Id} - 2e_s^\delta(e_s^\delta)^\top)dB_s^0 + \int_0^t \lambda_\delta(E_s^\delta)d\tilde{B}_s^0 $$

$$ \tilde{\beta}_t^\delta = \int_0^t \pi_\delta(E_s^\delta)dB_s^0 + \int_0^t \lambda_\delta(E_s^\delta)d\tilde{B}_s^0. $$

Setting $m_t^\delta = \mathcal{L}_1(X_t^\delta)$ et $\tilde{m}_t^\delta = \mathcal{L}_1(\tilde{X}_t^\delta)$, we get

$$ d_t m_t^\delta = \nabla \cdot \left( \frac{\sigma^2 + \sigma_0^2}{2} \nabla m_t^\delta - m_t^\delta (G + F(\cdot - \mu_1(m_t^\delta))) \right) dt - \sigma_0 \nabla m_t^\delta \cdot d\beta_t^\delta, $$

$$ d_t \tilde{m}_t^\delta = \nabla \cdot \left( \frac{\sigma^2 + \sigma_0^2}{2} \nabla \tilde{m}_t^\delta - \tilde{m}_t^\delta (G + F(\cdot - \mu_1(\tilde{m}_t^\delta))) \right) dt - \sigma_0 \nabla \tilde{m}_t^\delta \cdot d\tilde{\beta}_t^\delta. $$

Then, one has $\mathcal{P}_t\phi(m) = \mathbb{E}_0[\phi(m_t^\delta)|m_0^\delta = m]$ and $\mathcal{P}_t\phi(\tilde{m}) = \mathbb{E}_0[\phi(\tilde{m}_t^\delta)|\tilde{m}_0^\delta = \tilde{m}]$. Then, we obtain that for any $\delta > 0$,

$$ (59) \quad \begin{aligned} |\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| & \le \|\phi\|_{\mathrm{lip}} \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |X_t^\delta - \tilde{X}_t^\delta| \right] \right] \\ & \le \|\phi\|_{\mathrm{lip}} \mathbb{E}_0[\Theta_t^\delta] \end{aligned} $$
Step 6. Now, as (47) remains true replacing the rate c by $\tilde{c} \le c$, we can assume without loss of generality that $\alpha_F > m_G + c$. We can then combine Equations (47) and (56) to obtain, for all $t \ge 0$,

$$ (57) \quad \begin{aligned} & \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |A_t^\delta - \tilde{A}_t^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]| \right] \\ & \le \left( \mathbb{E}_1 \left[ |A_0^\delta - \tilde{A}_0^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_0^\delta] - \mathbb{E}_1[\tilde{X}_0^\delta]| \right) e^{-ct} + \frac{h(\delta)}{c} \\ & \quad + c_3^b \int_0^t e^{-c(t-s)} \left( \eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |A_s^\delta - \tilde{A}_s^\delta|^2 \right]^{1/2} + |\mathbb{E}_1[X_s^\delta] - \mathbb{E}_1[\tilde{X}_s^\delta]| \right] ds, \end{aligned} $$

for some constant $c_3^b > 0$ that depends on F and G.

With the notation $\Theta_t^\delta := \mathbb{E}_1 [ |A_t^\delta - \tilde{A}_t^\delta|^2 ]^{1/2} + |\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|$, Equation (57) reads

$$ \mathbb{E}_0[\Theta_t^\delta] \le \Theta_0^\delta e^{-ct} + \frac{h(\delta)}{c} + c_3^b \int_0^t e^{-c(t-s)} \left( \eta_0^\delta e^{-c_{\alpha_F} s} + 2\sigma c_{\alpha_F}^{-1/2} \sqrt{d} \right) \mathbb{E}_0[\Theta_s^\delta] ds. $$

Then, using Grönwall Lemma, we obtain

$$ (58) \quad \begin{aligned} e^{ct} \mathbb{E}_0[\Theta_t^\delta] \le & \left( \Theta_0 + \frac{h(\delta)}{c} \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} t \right) \\ & + h(\delta) \int_0^t e^{cs} \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} (t-s) \right) ds. \end{aligned} $$

We emphasize here that we write $\Theta_0$ and $\eta_0$ without reference to $\delta$ because these quantities are predetermined and independent of $\delta$. Now, let us notice that creating the couplage we defined two Brownian motions adapted to the filtration $\mathbb{F}_0$:

$$ \beta_t^\delta = \int_0^t \pi_\delta(E_s^\delta)(\mathrm{Id} - 2e_s^\delta(e_s^\delta)^\top)dB_s^0 + \int_0^t \lambda_\delta(E_s^\delta)d\tilde{B}_s^0 $$

$$ \tilde{\beta}_t^\delta = \int_0^t \pi_\delta(E_s^\delta)dB_s^0 + \int_0^t \lambda_\delta(E_s^\delta)d\tilde{B}_s^0. $$

Setting $m_t^\delta = \mathcal{L}_1(X_t^\delta)$ et $\tilde{m}_t^\delta = \mathcal{L}_1(\tilde{X}_t^\delta)$, we get

$$ d_t m_t^\delta = \nabla \cdot \left( \frac{\sigma^2 + \sigma_0^2}{2} \nabla m_t^\delta - m_t^\delta (G + F(\cdot - \mu_1(m_t^\delta))) \right) dt - \sigma_0 \nabla m_t^\delta \cdot d\beta_t^\delta, $$

$$ d_t \tilde{m}_t^\delta = \nabla \cdot \left( \frac{\sigma^2 + \sigma_0^2}{2} \nabla \tilde{m}_t^\delta - \tilde{m}_t^\delta (G + F(\cdot - \mu_1(\tilde{m}_t^\delta))) \right) dt - \sigma_0 \nabla \tilde{m}_t^\delta \cdot d\tilde{\beta}_t^\delta. $$

Then, one has $\mathcal{P}_t\phi(m) = \mathbb{E}_0[\phi(m_t^\delta)|m_0^\delta = m]$ and $\mathcal{P}_t\phi(\tilde{m}) = \mathbb{E}_0[\phi(\tilde{m}_t^\delta)|\tilde{m}_0^\delta = \tilde{m}]$. Then, we obtain that for any $\delta > 0$,

$$ (59) \quad \begin{aligned} |\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| & \le \|\phi\|_{\mathrm{lip}} \mathbb{E}_0 \left[ \mathbb{E}_1 \left[ |X_t^\delta - \tilde{X}_t^\delta| \right] \right] \\ & \le \|\phi\|_{\mathrm{lip}} \mathbb{E}_0[\Theta_t^\delta] \end{aligned} $$
Plugging the estimate given by Equation (58) into Equation (59) gives

$$|\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| \le \|\phi\|_{\text{lip}} \left(\Theta_0 + \frac{h(\delta)}{c}\right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \left( \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} - c \right) t \right) \\ + \|\phi\|_{\text{lip}} h(\delta) \int_0^t e^{-c(t-s)} \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} (t-s) \right) ds.$$

The latter being true for all $\delta > 0$, we can consider the limit $\delta \to 0$ to finally obtain

$$|\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| \le \|\phi\|_{\text{lip}} \Theta_0 \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} t - ct \right).$$

Then, for $\sigma$ and $\alpha$ satisfying $c := \sigma c_3^b c_{\alpha_F}^{-1/2} - c < 0$, we get the expected exponential rate of convergence. Moreover, writing

$$\Theta_0 \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 \right) = \left( \mathbb{E}_1 \left[ |A_0 - \tilde{A}_0|^2 \right]^{1/2} + |\mathbb{E}_1[X_0] - \mathbb{E}_1[\tilde{X}_0]| \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 \right) \\ \le C_2 \left( \mu_2(m)^{1/2} + \mu_2(\tilde{m})^{1/2} \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} (v(m)^{1/2} + v(\tilde{m})^{1/2}) \right),$$

for a new constant positive $C_2$, we complete the proof of Proposition 7. $\square$

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

**Acknowledgments.** François Delarue acknowledges the financial support of the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program (AdG ELISA project, Grant agreement No. 101054746). Raphaël Maillet wishes to thank the Université Côte d'Azur for its hospitality during the preparation of this work, along with Pierre Cardaliaguet for fruitful discussions.

## REFERENCES

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[3] L. Angeli, J. Barré, M. Kolodziejczyk, and M. Ottobre, Well-posedness and stationary solutions of McKean-Vlasov (S)PDEs, J. Math. Anal. Appl. 526 (2023), no. 2, Paper No. 127301, 49, doi. MR-4579326

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Plugging the estimate given by Equation (58) into Equation (59) gives

$$ |\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| \le \|\phi\|_{\text{lip}} \left(\Theta_0 + \frac{h(\delta)}{c}\right) \exp\left(\frac{c_3^b}{c_{\alpha_F}}\eta_0 + \left(\frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} - c\right)t\right) \\ + \|\phi\|_{\text{lip}}h(\delta) \int_0^t e^{-c(t-s)} \exp\left(\frac{c_3^b}{c_{\alpha_F}}\eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} (t-s)\right) ds. $$

The latter being true for all $\delta > 0$, we can consider the limit $\delta \to 0$ to finally obtain

$$ |\mathcal{P}_t\phi(m) - \mathcal{P}_t\phi(\tilde{m})| \le \|\phi\|_{\text{lip}} \Theta_0 \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 + \frac{c_3^b \sigma \sqrt{d}}{\sqrt{c_{\alpha_F}}} t - ct \right). $$

Then, for $\sigma$ and $\alpha$ satisfying $c := \sigma c_3^b c_{\alpha_F}^{-1/2} - c < 0$, we get the expected exponential rate of convergence. Moreover, writing

$$ \Theta_0 \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 \right) = \left( \mathbb{E}_1 \left[ |A_0 - \tilde{A}_0|^2 \right]^{1/2} + |\mathbb{E}_1[X_0] - \mathbb{E}_1[\tilde{X}_0]| \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} \eta_0 \right) \\ \le C_2 \left( \mu_2(m)^{1/2} + \mu_2(\tilde{m})^{1/2} \right) \exp \left( \frac{c_3^b}{c_{\alpha_F}} (v(m)^{1/2} + v(\tilde{m})^{1/2}) \right), $$

for a new constant positive $C_2$, we complete the proof of Proposition 7. $\square$

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

**Acknowledgments.** François Delarue acknowledges the financial support of the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program (AdG ELISA project, Grant agreement No. 101054746). Raphaël Maillet wishes to thank the Université Côte d'Azur for its hospitality during the preparation of this work, along with Pierre Cardaliaguet for fruitful discussions.

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[50] ——, The geometry of dissipative evolution equations: the porous medium equation, Comm. Partial Differential Equations 26 (2001), no. 1-2, 101–174, doi. MR-1842429

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## APPENDIX A.

### A.1. Proof of Equation (38).

*Proof.* Using Itô's formula, and recalling that $E^\delta = X^\delta - \tilde{X}^\delta$, we get that for any $\delta > 0$,

$$ 
\begin{aligned}
d|\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|^2 = & \ 2(\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot (\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \, dt \\
& + 2(\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot (\mathbb{E}_1[F(X_t^\delta - \mathbb{E}_1[X_t^\delta])] - \mathbb{E}_1[G(F(\tilde{X}_t^\delta - \mathbb{E}_1[\tilde{X}_t^\delta]))]) \, dt \\
& + 4\sigma_0\pi_\delta(E_t^\delta)((\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot e_t^\delta)e_t^\delta \cdot dB_t^0 \\
& + 4\sigma_0^2\pi_\delta^2(E_t^\delta)dt.
\end{aligned}
$$
[35] S. Herrmann and J. Tugaut, Non-uniqueness of stationary measures for self-stabilizing processes, Stochastic Process. Appl. 120 (2010), no. 7, 1215-1246, doi. MR-2639745

[36] B. Jianhai and W. Jian, Long time behavior of one-dimensional mckean-vlasov sdes with common noise, arXiv:2401.07665 (2024).

[37] R. Jordan, D. Kinderlehrer, and F. Otto, The variational formulation of the Fokker-Planck equation, SIAM J. Math. Anal. 29 (1998), no. 1, 1-17, doi. MR-1617171

[38] M. Kac, Foundations of kinetic theory, Proceedings of the Third Berkeley Symposium on Mathematical Statistics and Probability, 1954-1955, vol. III, Univ. California Press, Berkeley-Los Angeles, Calif., 1956, pp. 171-197. MR-84985

[39] T.G. Kurtz and J. Xiong, Particle representations for a class of nonlinear SPDEs, Stochastic Process. Appl. 83 (1999), no. 1, 103-126, doi. MR-1705602

[40] D. Lacker, M. Shkolnikov, and J. Zhang, Superposition and mimicking theorems for conditional McKean-Vlasov equations, J. Eur. Math. Soc. (JEMS) 25 (2023), no. 8, 3229-3288, doi. MR-4612111

[41] T. Lindvall and L. C. G. Rogers, Coupling of multidimensional diffusions by reflection, Ann. Probab. 14 (1986), no. 3, 860-872, https://www.jstor.org/stable/pdf/2244138.pdf. MR-841588

[42] P.-L. Lions, B. Perthame, and P.E. Souganidis, Scalar conservation laws with rough (stochastic) fluxes, Stoch. Partial Differ. Equ. Anal. Comput. 1 (2013), no. 4, 664-686, doi. MR-3327520

[43] P.-L Lions, B. Perthame, and P.E. Souganidis, Scalar conservation laws with rough (stochastic) fluxes: the spatially dependent case, Stoch. Partial Differ. Equ. Anal. Comput. 2 (2014), no. 4, 517-538, doi. MR-3274890

[44] R. Maillet, A note on the Long-Time behaviour of Stochastic McKean-Vlasov Equations with common noise, arXiv:2306.16130 (2023).

[45] F. Malrieu, Logarithmic Sobolev inequalities for some nonlinear PDE's, Stochastic Process. Appl. 95 (2001), no. 1, 109-132, doi. MR-1847094

[46] ——, Convergence to equilibrium for granular media equations and their Euler schemes, Ann. Appl. Probab. 13 (2003), no. 2, 540-560, doi. MR-1970276

[47] H. P. McKean, A class of Markov processes associated with nonlinear parabolic equations, Proc. Nat. Acad. Sci. U.S.A. 56 (1966), 1907-1911, doi. MR-221595

[48] S. Méléard, Asymptotic behaviour of some interacting particle systems; McKean-Vlasov and Boltzmann models, Probabilistic models for nonlinear partial differential equations (Montecatini Terme, 1995), Lecture Notes in Math., vol. 1627, Springer, Berlin, 1996, pp. 42-95, doi. MR-1431299

[49] F. Otto, Evolution of microstructure in unstable porous media flow: a relaxational approach, Comm. Pure Appl. Math. 52 (1999), no. 7, 873-915, doi. MR-1682800

[50] ——, The geometry of dissipative evolution equations: the porous medium equation, Comm. Partial Differential Equations 26 (2001), no. 1-2, 101-174, doi. MR-1842429

[51] D. Revuz and M. Yor, Continuous martingales and brownian motion, vol. 293, Springer Science & Business Media, 2013.

[52] A.-S. Sznitman, Topics in propagation of chaos, École d'Été de Probabilités de Saint-Flour XIX-1989, Lecture Notes in Math., vol. 1464, Springer, Berlin, 1991, pp. 165-251, doi. MR-1108185

## APPENDIX A.

### A.1. Proof of Equation (38).

*Proof.* Using Itô's formula, and recalling that $E^\delta = X^\delta - \tilde{X}^\delta$, we get that for any $\delta > 0$,

$$ 
\begin{aligned}
d|\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]|^2 = & \ 2(\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot (\mathbb{E}_1[G(X_t^\delta)] - \mathbb{E}_1[G(\tilde{X}_t^\delta)]) \, dt \\
& + 2(\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot (\mathbb{E}_1[F(X_t^\delta - \mathbb{E}_1[X_t^\delta])] - \mathbb{E}_1[G(F(\tilde{X}_t^\delta - \mathbb{E}_1[\tilde{X}_t^\delta]))]) \, dt \\
& + 4\sigma_0\pi_\delta(E_t^\delta)((\mathbb{E}_1[X_t^\delta] - \mathbb{E}_1[\tilde{X}_t^\delta]) \cdot e_t^\delta)e_t^\delta \cdot dB_t^0 \\
& + 4\sigma_0^2\pi_\delta^2(E_t^\delta)dt.
\end{aligned}
$$
For any $\varepsilon > 0$, let us introduce $\psi_{\varepsilon} : [0, +\infty] \in r \mapsto (r + \varepsilon)^{1/2}$. Since this function is twice continuously differentiable, we can write

$$ (60) \qquad \begin{aligned} \mathrm{d}\psi_{\varepsilon} \left( |\mathbb{E}_1[X_t^{\delta}] - \mathbb{E}_1[\tilde{X}_t^{\delta}|]^2 \right) = & \ 2\psi'_{\varepsilon}(|E_t^{\delta}|^2)E_t^{\delta} \cdot \left( \mathbb{E}_1[G(X_t^{\delta})] - \mathbb{E}_1[G(\tilde{X}_t^{\delta})] \right) \mathrm{d}t \\ & + 2\psi'_{\varepsilon}(|E_t^{\delta}|^2)E_t^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) \mathrm{d}t \\ & + 4\sigma_0\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}(E_t^{\delta})(E_t^{\delta} \cdot e_t^{\delta})e_t^{\delta} \cdot \mathrm{d}B_t^0 \\ & + 4\sigma_0^2\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}^2(E_t^{\delta})\mathrm{d}t \\ & + 8\sigma_0^2\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2\pi_{\delta}(E_t^{\delta})\mathrm{d}t. \end{aligned} $$

We now want to take the limit $\varepsilon \to 0$. Using dominated convergence theorem and stochastic dominated convergence theorem as stated in [51, Theorem 2.12], combined with the bound $4r\psi'_{\varepsilon}(r^2) \le 1$, we can deal with the first two lines and get that for all $t \ge 0$,

$$ (61) \quad \lim_{\varepsilon \to 0} \int_0^t 2\psi'_{\varepsilon}(|E_s^{\delta}|^2)E_s^{\delta} \cdot \left( \mathbb{E}_1[G(X_s^{\delta})] - \mathbb{E}_1[G(\tilde{X}_s^{\delta})] \right) \mathrm{d}s = \int_0^t e_s^{\delta} \cdot \left( \mathbb{E}_1[G(X_s^{\delta})] - \mathbb{E}_1[G(\tilde{X}_s^{\delta})] \right) \mathrm{d}s, $$

$$ (62) \quad \begin{aligned} & \lim_{\varepsilon \to 0} \int_0^t 2\psi'_{\varepsilon}(|E_s^{\delta}|^2)E_s^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) \mathrm{d}s \\ & = \int_0^t e_s^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) \mathrm{d}s, \end{aligned} $$

and

$$ (63) \quad \lim_{\varepsilon \to 0} \int_0^t 4\sigma_0\psi'_{\varepsilon}(|E_s^{\delta}|^2)\pi_{\delta}(E_s^{\delta})(E_s^{\delta} \cdot e_s^{\delta})e_s^{\delta} \cdot \mathrm{d}B_s^0 = \int_0^t 2\sigma_0\pi_{\delta}(E_s^{\delta})e_s^{\delta} \cdot \mathrm{d}B_s^0, $$

almost surely. For the last two terms in Equation (60), let us remark that for all $r \ge 0$,

$$ (64) \quad \psi'_{\varepsilon}(r) + 2\psi''(r) = \frac{1}{2(r + \varepsilon)^{1/2}} - \frac{r}{2(r + \varepsilon)^{3/2}} \to 0, $$

when $\varepsilon$ tends to 0. In order to we need to properly take the limit $\varepsilon \to 0$ in Equation (60), we need to take advantage of the presence of the function $\pi_{\delta}$ for $\delta > 0$. In Indeed, we have

$$ \begin{aligned} & |4\sigma_0^2\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}^2(E_t^{\delta}) + 8\sigma_0^2\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2\pi_{\delta}(E_t^{\delta})| \\ & \le |\pi_{\delta}^2(E_t^{\delta})\sigma_0^2(4\psi'_{\varepsilon}(|E_t^{\delta}|^2) + 8\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2)|. \end{aligned} $$

Moreover, we know that $\psi'_{\varepsilon}(r^2) + 2\psi''_{\varepsilon}(r^2)r^2 \le r^{-3}$, for all $r > 0$ and $\varepsilon \le 1$. Using the presence of $\pi_{\delta}$, we have that the integrand is null near 0. Then, we can once again apply dominated convergence theorem and obtain

$$ (65) \quad \lim_{\varepsilon \to 0} \int_0^t 4 \{ \sigma_0^2 \psi'_{\varepsilon} (|E_s^{\delta}|^2) \pi_{\delta} (E_s^{\delta})^2 + 8 \sigma_0^2 \psi''_{\varepsilon} (|E_s^{\delta}|^2) |E_s^{\delta}|^2 \psi_{\delta} (E_s^{\delta}) \} \mathrm{d}s = 0. $$

Finally, combining Equations (61), (62) (63) and (65), we get the expected result. $\square$

## A.2. Proof of Equation (55).

*Proof.* To prove Equation, let us introduce the following technical Lemma:
For any $\varepsilon > 0$, let us introduce $\psi_{\varepsilon} : [0, +\infty] \in r \mapsto (r + \varepsilon)^{1/2}$. Since this function is twice continuously differentiable, we can write

$$ (60) \qquad \begin{aligned} d\psi_{\varepsilon} \left( |\mathbb{E}_1[X_t^{\delta}] - \mathbb{E}_1[\tilde{X}_t^{\delta}|^2] \right) = & \ 2\psi'_{\varepsilon}(|E_t^{\delta}|^2)E_t^{\delta} \cdot \left( \mathbb{E}_1[G(X_t^{\delta})] - \mathbb{E}_1[G(\tilde{X}_t^{\delta})] \right) dt \\ & + 2\psi'_{\varepsilon}(|E_t^{\delta}|^2)E_t^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) dt \\ & + 4\sigma_0\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}(E_t^{\delta})(E_t^{\delta} \cdot e_t^{\delta})e_t^{\delta} \cdot dB_t^0 \\ & + 4\sigma_0^2\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}^2(E_t^{\delta})dt \\ & + 8\sigma_0^2\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2\pi_{\delta}(E_t^{\delta})dt. \end{aligned} $$

We now want to take the limit $\varepsilon \to 0$. Using dominated convergence theorem and stochastic dominated convergence theorem as stated in [51, Theorem 2.12], combined with the bound $4r\psi'_{\varepsilon}(r^2) \le 1$, we can deal with the first two lines and get that for all $t \ge 0$,

$$ (61) \quad \lim_{\varepsilon \to 0} \int_0^t 2\psi'_{\varepsilon}(|E_s^{\delta}|^2)E_s^{\delta} \cdot \left( \mathbb{E}_1[G(X_s^{\delta})] - \mathbb{E}_1[G(\tilde{X}_s^{\delta})] \right) ds = \int_0^t e_s^{\delta} \cdot \left( \mathbb{E}_1[G(X_s^{\delta})] - \mathbb{E}_1[G(\tilde{X}_s^{\delta})] \right) ds, $$

$$ (62) \qquad \begin{aligned} & \lim_{\varepsilon \to 0} \int_0^t 2\psi'_{\varepsilon}(|E_s^{\delta}|^2)E_s^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) ds \\ & = \int_0^t e_s^{\delta} \cdot \left( \mathbb{E}_1[F(X_t^{\delta} - \mathbb{E}_1[X_t^{\delta}])] - \mathbb{E}_1[F(\tilde{X}_t^{\delta} - \mathbb{E}_1[\tilde{X}_t^{\delta}])] \right) ds, \end{aligned} $$

and

$$ (63) \quad \lim_{\varepsilon \to 0} \int_0^t 4\sigma_0\psi'_{\varepsilon}(|E_s^{\delta}|^2)\pi_{\delta}(E_s^{\delta})(E_s^{\delta} \cdot e_s^{\delta})e_s^{\delta} \cdot dB_s^0 = \int_0^t 2\sigma_0\pi_{\delta}(E_s^{\delta})e_s^{\delta} \cdot dB_s^0, $$

almost surely. For the last two terms in Equation (60), let us remark that for all $r \ge 0$,

$$ (64) \qquad \psi'_{\varepsilon}(r) + 2\psi''(r) = \frac{1}{2(r + \varepsilon)^{1/2}} - \frac{r}{2(r + \varepsilon)^{3/2}} \to 0, $$

when $\varepsilon$ tends to 0. In order to we need to properly take the limit $\varepsilon \to 0$ in Equation (60), we need to take advantage of the presence of the function $\pi_{\delta}$ for $\delta > 0$. In Indeed, we have

$$ \begin{aligned} & |4\sigma_0^2\psi'_{\varepsilon}(|E_t^{\delta}|^2)\pi_{\delta}^2(E_t^{\delta}) + 8\sigma_0^2\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2\pi_{\delta}(E_t^{\delta})| \\ & \le |\pi_{\delta}^2(E_t^{\delta})\sigma_0^2(4\psi'_{\varepsilon}(|E_t^{\delta}|^2) + 8\psi''_{\varepsilon}(|E_t^{\delta}|^2)|E_t^{\delta}|^2)|. \end{aligned} $$

Moreover, we know that $\psi'_{\varepsilon}(r^2) + 2\psi''_{\varepsilon}(r^2)r^2 \le r^{-3}$, for all $r > 0$ and $\varepsilon \le 1$. Using the presence of $\pi_{\delta}$, we have that the integrand is null near 0. Then, we can once again apply dominated convergence theorem and obtain

$$ (65) \quad \lim_{\varepsilon \to 0} \int_0^t 4 \{ \sigma_0^2 \psi'_{\varepsilon} (|E_s^{\delta}|^2) \pi_{\delta} (E_s^{\delta})^2 + 8 \sigma_0^2 \psi''_{\varepsilon} (|E_s^{\delta}|^2) |E_s^{\delta}|^2 \psi_{\delta} (E_s^{\delta}) \} ds = 0. $$

Finally, combining Equations (61), (62) (63) and (65), we get the expected result. $\square$

## A.2. Proof of Equation (55).

*Proof.* To prove Equation, let us introduce the following technical Lemma:
**Lemma 17.** For any positive and differentiable function $u : [0, +\infty) \to [0, +\infty)$ such that

$$ \frac{d}{dt} u(t) \le 2k(t) \sqrt{u(t)}, $$

for some function $k : [0, +\infty) \to \mathbb{R}$, we have

$$ \frac{d}{dt} \sqrt{u(t)} \le k(t). $$

*Proof of Lemma 17.* We consider for any $\varepsilon > 0$ the function $\psi_{\varepsilon} : [0, +\infty) \ni r \mapsto (r + \varepsilon)^{1/2}$. The function $\psi_{\varepsilon}$ being differentiable with positive derivative, we get

$$ \frac{d}{dt} \psi_{\varepsilon} (u(t)) \le 2k(t) \sqrt{u(t)} \psi'_{\varepsilon} (u(t)). $$

Using the fact that $2r\psi'_{\varepsilon}(r^2) = r(r^2 + \varepsilon)^{-1/2} \le 1$, we use dominated convergence theorem to conclude the proof. $\square$

Such a lemma is classical and applying it to Equation (54) concludes the proof of Equation (55). $\square$

FRANÇOIS DELARUE, UNIVERSITÉ CÔTE D'AZUR, LABORATOIRE J.A.DIEUDONNÉ, 06108 NICE, FRANCE

Email address: francois.delarue@univ-cotedazur.fr

RAPHAËL MAILLET, UNIVERSITÉ PARIS-DAUPHINE & PSL, CNRS, CEREMADE, 75016 PARIS, FRANCE

Email address: maillet@ceremade.dauphine.fr

ETIENNE TANRÉ, UNIVERSITÉ CÔTE D'AZUR, INRIA, CNRS, LJAD, FRANCE

Email address: Etienne.Tanre@inria.fr
**Lemma 17.** For any positive and differentiable function $u : [0, +\infty) \to [0, +\infty)$ such that

$$ \frac{d}{dt} u(t) \le 2k(t) \sqrt{u(t)}, $$

for some function $k : [0, +\infty) \to \mathbb{R}$, we have

$$ \frac{d}{dt} \sqrt{u(t)} \le k(t). $$

*Proof of Lemma 17.* We consider for any $\varepsilon > 0$ the function $\psi_{\varepsilon} : [0, +\infty) \ni r \mapsto (r + \varepsilon)^{1/2}$. The function $\psi_{\varepsilon}$ being differentiable with positive derivative, we get

$$ \frac{d}{dt} \psi_{\varepsilon} (u(t)) \le 2k(t) \sqrt{u(t)} \psi'_{\varepsilon} (u(t)). $$

Using the fact that $2r\psi'_{\varepsilon}(r^2) = r(r^2 + \varepsilon)^{-1/2} \le 1$, we use dominated convergence theorem to conclude the proof. □

Such a lemma is classical and applying it to Equation (54) concludes the proof of Equation (55). □

FRANÇOIS DELARUE, UNIVERSITÉ CÔTE D'AZUR, LABORATOIRE J.A.DIEUDONNÉ, 06108 NICE, FRANCE

Email address: francois.delarue@univ-cotedazur.fr

RAPHAËL MAILLET, UNIVERSITÉ PARIS-DAUPHINE & PSL, CNRS, CEREMADE, 75016 PARIS, FRANCE

Email address: maillet@ceremade.dauphine.fr

ETIENNE TANRÉ, UNIVERSITÉ CÔTE D'AZUR, INRIA, CNRS, LJAD, FRANCE

Email address: Etienne.Tanre@inria.fr