By continuity of $z_\varepsilon$ and $y$, the last term of the right-hand side converges to 0 when $\delta \to 0$. Hence the left-hand side is positive for all $\delta$ small enough. Since $z_\varepsilon(\vartheta_\varepsilon) = y(\vartheta_\varepsilon)$, this implies that

$$z_{\varepsilon}(\vartheta_{\varepsilon} - \delta) - y(\vartheta_{\varepsilon} - \delta) < 0$$

for such $\delta$. This contradicts the definition of $\vartheta_\varepsilon$ as the first time $t$ such that $z_\varepsilon(t) \le y(t)$. $\square$

**Lemma 4.2.** There exists $K_0 \in \mathbb{N}^*$ large enough and two constants $C_1, C_2 > 0$ such that for all $K \ge K_0$ and $t \ge 0$,

$$\mathbb{E} \left( M_6 \left( \mu_t^K \right) 1_{t \le \check{\tau}^K} \right) \le C_1 \exp \left( - \frac{9\underline{b}}{28K^2 \sigma_K^2} t \right) + C_2$$

*Proof. Step 1. Pathwise inequality.* Note that for all $t \ge 0$,

$$M_7 \left( \mu_{t \wedge \check{\tau}^K}^K \right) \le \mathrm{Diam} \left( \mathrm{Supp} \, \mu_{t \wedge \check{\tau}^K}^K \right) M_6 \left( \mu_{t \wedge \check{\tau}^K}^K \right) \le \frac{M_6 \left( \mu_{t \wedge \check{\tau}^K}^K \right)}{\sigma_K K^{\frac{3+\varepsilon}{2}}}. \quad (32)$$

Introducing for all $t \ge 0$, $A_t := \frac{3}{4} M_6(\mu_t^K) + 7 M_4(\mu_t^K) M_2(\mu_t^K) - 4 \tilde{M}_3^2(\mu_t^K) + \frac{3}{2} M_2^3(\mu_t^K)$, we obtain from Lemma 3.7, (32) and Assumptions (A) that there exists a constant $C > 0$ such that for all $t > s \ge 0$,

$$A_{t \wedge \check{\tau}^K} \le A_{s \wedge \check{\tau}^K} - \frac{1}{K^2 \sigma_K^2} \int_{s \wedge \check{\tau}^K}^{t \wedge \check{\tau}^K} \left\{ \left( \underline{b} - \frac{C}{K^{\frac{\varepsilon}{2}}} \right) \left[ 3M_6(\mu_r^K) + \frac{9}{2}M_4(\mu_r^K)M_2(\mu_r^K) + 6M_2^3(\mu_r^K) \right] \right. \\ \left. - \bar{\theta} \bar{m}_2 \left[ \frac{91}{4}M_4(\mu_r^K) + 42M_2^2(\mu_r^K) \right] \right\} dr + \mathrm{Mart}_{t \wedge \check{\tau}^K} - \mathrm{Mart}_{s \wedge \check{\tau}^K}$$

where $(\mathrm{Mart}_t)_{t \ge 0}$ is a martingale. Note that, choosing $K_0$ large enough, for all $K \ge K_0$, $\underline{b} - CK^{-\varepsilon/2} > \underline{b}/2$. From the inequalities $M_2^2(\mu) \le M_4(\mu)$ and $M_4(\mu) \le M_6^{2/3}(\mu) \le \alpha M_6(\mu) + 1/\alpha^2$ with $\alpha := 2\underline{b}/259\bar{\theta}\bar{m}_2$, we deduce that

$$\frac{3\underline{b}}{2} M_6(\mu) - \bar{\theta} \bar{m}_2 \left[ \frac{91}{4} M_4(\mu) + 42 M_2^2(\mu) \right] \ge \frac{3\underline{b}}{2} M_6(\mu) - \frac{259 \bar{\theta} \bar{m}_2}{4} M_4(\mu) \ge \underline{b} M_6(\mu) - \frac{259^3 \bar{\theta}^3 \bar{m}_2^3}{16 \underline{b}^2}.$$

Thus, for all $t > s \ge 0$,

$$A_{t \wedge \check{\tau}^K} \le A_{s \wedge \check{\tau}^K} + \mathrm{Mart}_{t \wedge \check{\tau}^K} - \mathrm{Mart}_{s \wedge \check{\tau}^K} \\ - \frac{\underline{b}}{2K^2 \sigma_K^2} \int_{s \wedge \check{\tau}^K}^{t \wedge \check{\tau}^K} \left[ 2M_6(\mu_r^K) + \frac{9}{2}M_4(\mu_r^K)M_2(\mu_r^K) + 6M_2^3(\mu_r^K) - \left( \frac{259\bar{\theta}\bar{m}_2}{2\underline{b}} \right)^3 \right] dr \\ \le A_{s \wedge \check{\tau}^K} - \frac{9\underline{b}}{28K^2 \sigma_K^2} \int_{s \wedge \check{\tau}^K}^{t \wedge \check{\tau}^K} \left( A_r - \frac{14}{9} \left( \frac{259\bar{\theta}\bar{m}_2}{2\underline{b}} \right)^3 \right) dr + \mathrm{Mart}_{t \wedge \check{\tau}^K} - \mathrm{Mart}_{s \wedge \check{\tau}^K}$$

*Step 2. Conclusion.* Let us consider the stochastic process $(N(\mu_t^K))_{t \ge 0}$ defined by

$$N(\mu_t^K) := \begin{cases} A_t & \text{if } s < t \le \check{\tau}^K \\ (A_{\check{\tau}^K} - C_0) \exp \left( -\frac{3\underline{b}}{4K^2\sigma_K^2} (t - \check{\tau}^K) \right) + C_0 & \text{if } s < \check{\tau}^K < t \\ 0 & \text{if } \check{\tau}^K < s < t \end{cases}$$
Proof. Let us consider $\mu_0^K := \frac{1}{K} \sum_{k=1}^K \delta_{x_k}$ and $y = (y_k)_{1\le k\le K}$ defined by

$$ y = \begin{cases} (x_1, \dots, x_{i-1}, x_i + \frac{H}{\sqrt{K}}, x_{i+1}, \dots, x_K) & \text{on } E_1, \\ (x_1, \dots, x_{j-1}, x_i, x_{j+1}, \dots, x_K) & \text{on } E_2, \end{cases} $$

where $H$ follows the mutation law $m(x_i, dh)$, $E_1$ is the event “a mutation occurs at time $\vartheta_1$” and $E_2$ is the event “a resampling between $i$ and $j$ occurs at time $\vartheta_1$”. Note that $\mu_{\vartheta_1}^K = \tau_{-\langle \text{id}, \frac{1}{K} \sum_{k=1}^K \delta_{y_k} \rangle} \sharp(\frac{1}{K} \sum_{k=1}^K \delta_{y_k})$. Hence, noting that for all $k \in \{1, \dots, K\}$,

$$ |x_k| \le \sqrt{K} (1 + M_2(\mu_0^K)) $$

and using Assumption (A2), for $K$ large enough we have that

$$ \begin{aligned} M_2(\mu_{\vartheta_1}^K) \le M_2 \left( \frac{1}{K} \sum_{i=1}^K \delta_{y_i} \right) &= \frac{1}{K} \sum_{i=1}^K y_i^2 \le \begin{cases} \frac{1}{K} \sum_{k=1}^K x_k^2 + \frac{2A_m x_k}{K^{\frac{3}{2}}} + \frac{A_m^2}{K^2} & \text{on } E_1, \\ \frac{1}{K} \sum_{k=1}^K x_k^2 + \frac{x_j^2 - x_i^2}{K} & \text{on } E_2, \end{cases} \\ &\le \begin{cases} M_2(\mu_0^K) + \frac{2A_m (1 + M_2(\mu_0^K))}{K} + \frac{A_m^2}{K^2} & \text{on } E_1, \\ 2M_2(\mu_0^K) & \text{on } E_2, \end{cases} \\ &\le 2M_2(\mu_0^K) + 1. \end{aligned} \quad \square $$

Each of the previous steps will be called *transitions of $M_2(\mu_t^K)$*. The idea of the proof of Lemma 4.8 is to estimate the transition probabilities of the sequence $(L_k^K)_{k \in \mathbb{N}}$ in order to construct a coupling between $(L_k^K)_{k \in \mathbb{N}}$ and a biased random walk on $\mathbb{N}^*$ and reflected in 1.

At time $\hat{\tau}^K$, $M_2(\mu_{\hat{\tau}^K}^K) \in \mathcal{I}_{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}$ where $\lfloor x \rfloor$ is the lower integer part of $x$. So, the number of transitions of $M_2(\mu_t^K)$ before $\hat{\tau}^K$ is greater than or equal to $T_{k_0}^K$ where $k_0$ is the first integer such that $L_{k_0}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$.

By using estimates on the number of steps that a biased random walk takes to reach $\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ and estimates on the durations between two transitions $T_{k+1}^K - T_k^K$, we will deduce a lower bound, exponential in $K$, on $\hat{\tau}^K$. An additional difficulty comes from the fact that the previous coupling argument is only valid up to time $\hat{\tau}^K$. Hence we will construct a coupling that takes into account the possibility that $\hat{\tau}^K$ happens during each transition step. The proof is divided into four steps: in Section 4.4.1, we characterise the behavior of the first transition step; in Section 4.4.2, the proposed coupling is constructed; in Section 4.4.3, we give estimates on the first exit time from an attractive domain for random walks; finally, we conclude in Section 4.4.4.

### 4.4.1 One-step transitions

In this section we look at only one transition: we suppose that $M_2(\mu_0^K) \in [2u_{\ell-1} + 1, 2u_\ell + 1)$ for $\ell \in \mathbb{N}^*$ fixed and we look for bounds on the first transition probabilities of $M_2(\mu_t^K)$ and on time $T_1^K$. The main result is the following.
where $C_0 := \frac{14}{9} \left( \frac{259\theta\overline{m_2}}{2b} \right)^3$. From Step 1, for all $t > s \ge 0$,

$$N(\mu_t^K) \le N(\mu_s^K) - \frac{9b}{28K^2\sigma_K^2} \int_s^t [N(\mu_r^K) - C_0] dr + \text{Mart}_t - \text{Mart}_s$$

which entails

$$y(t) \le y(s) - \frac{9b}{28K^2\sigma_K^2} \int_s^t (y(r) - C_0) dr$$

where $y(t) := \mathbb{E}(N(\mu_t^K))$. From Lemma 4.1, we deduce that for all $t \ge 0$, $y(t) \le z(t)$ where $z(t)$ is the unique solution of

$$z(t) = z(s) - \frac{9b}{28K^2\sigma_K^2} \int_s^t (z(r) - C_0) dr$$

such that $z(0) = A_0$, i.e. $z(t) = (z(0) - C_0) \exp(-\frac{9b}{28K^2\sigma_K^2}t) + C_0$. From (7) then (4), note that

$$M_6(\mu_0^K) = \langle \mathrm{id}^6, \mu_0^K \rangle = \frac{1}{K^3\sigma_K^6} \langle (\mathrm{id} - x_0)^6, \nu_0^K \rangle \le C_6^*.$$

From Cauchy-Schwarz's inequality $\tilde{M}_3^2(\mu) \le M_4(\mu)M_2(\mu)$, note that $z(0) \le 21C_6^*/4$ and so, for all $t \ge 0$,

$$\begin{aligned} \mathbb{E}(M_6(\mu_t^K) \mathbb{1}_{t \le \hat{\tau}^K}) &\le \frac{4}{3} \mathbb{E} \left( \left[ \frac{3}{4} M_6(\mu_t^K) + 3M_4(\mu_t^K) M_2(\mu_t^K) + \frac{3}{2} M_2^3(\mu_t^K) \right] \mathbb{1}_{t \le \hat{\tau}^K} \right) \\ &\le \frac{4}{3} z(t) \le \frac{4}{3} \left( \frac{21}{4} (C_6^* - C_0) \exp \left( -\frac{9b}{28K^2\sigma_K^2} t \right) + C_0 \right) \end{aligned}$$

and the announced result follows. □

The following lemma will be useful in the proof of Theorem 5.1.

**Lemma 4.3.** For all $t \ge 0$,

$$\sup_{K \in \mathbb{N}^*} \mathbb{E} \left( \int_0^{t \wedge \hat{\tau}^K} M_6(\mu_s^K) ds \right) < \infty.$$

*Proof.* Let $t \ge 0$ be fixed. By FUBINI's theorem and Lemma 4.2, there exists two constants $C_1, C_2 > 0$ such that

$$\mathbb{E} \left( \int_0^t M_6 (\mu_s^K) \mathbb{1}_{s \wedge \hat{\tau}^K} ds \right) \le C_1 \int_0^t \exp \left( -\frac{9b}{28K^2\sigma_K^2} s \right) ds + C_2t \le (C_1 + C_2)t$$

which ends the proof. □

The last lemma has the following consequence, used in Sections 5 and 6.
**Lemma 4.10.** For all $\ell \in \mathbb{N}$, for all $\mu_0^K \in \mathcal{M}_{1,K}^{c,2}(\mathbb{R})$ such that $M_2(\mu_0^K) \in [2u_{\ell-1}+1, 2u_\ell+1)$, there exist a constant $C > 0$, $K_0 \in \mathbb{N}^*$ such that for all $K \ge K_0$, we have

$$ \mathbb{P}_{\mu_0^K} \left( \left[ \{L_1^K = \ell - 1\} \cap \{\check{\tau}^K > T_1^K \ge K\sigma^2\} \right] \cup \{T_1^K \ge \check{\tau}^K\} \right) \ge \frac{1}{2} + \eta_K, \quad (31) $$

where $\eta_K := \frac{1}{2} - \frac{C}{K^{\epsilon/2}}$.

The proof of the previous result is based on Corollary 4.11 which is obtained as a straightforward consequence of Lemma 4.4.

**Corollary 4.11.** Let $\ell \in \mathbb{N}^*$ be fixed. There exists $K_0 \in \mathbb{N}^*$ large enough such that for all $K \ge K_0$, for all $\mu_0^K \in \mathcal{M}_{1,K}^{c,2}(\mathbb{R})$ such that $M_2(\mu_0^K) \in [2u_{\ell-1}+1, 2u_\ell+1)$, we have for all $t \le T_1^K$ that

$$ 2u_{\ell-1} + 1 - \frac{5\bar{b}u_{\ell+1} + \theta m_2}{2K^2\sigma^2} t \wedge \check{\tau}^K + M_{t \wedge \check{\tau}^K}^{K, P_{\text{id}^2,1}} \le M_2(\mu_{t \wedge \check{\tau}^K}^K) \le 2u_\ell + 1 - \frac{bu_{\ell-1}}{K^2\sigma^2} t \wedge \check{\tau}^K + M_{t \wedge \check{\tau}^K}^{K, P_{\text{id}^2,1}}. $$

*Proof of Lemma 4.10.* By passing on the complementary of (31), it is equivalent to prove that

$$ \mathbb{P}_{\mu_0^K} \left( \left[ \{L_1^K = \ell + 1\} \cup \{T_1^K < K\sigma^2\} \right] \cap \{T_1^K < \check{\tau}^K\} \right) < \frac{1}{2} - \eta_K. $$

As for all events $A, B$, $\mathbb{P}(A \cup B) = \mathbb{P}(A \cap B^c) + \mathbb{P}(B)$, we have

$$ \begin{align} \mathbb{P}_{\mu_0^K} \left( \left[ \{L_1^K = \ell + 1\} \cup \{T_1^K < K\sigma^2\} \right] \cap \{T_1^K < \check{\tau}^K\} \right) & \nonumber \\ &= \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\} \cap \{T_1^K \ge K\sigma^2\} \right) \nonumber \\ & \quad + \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{T_1^K < K\sigma^2\} \right) \tag{32} \\ & \le \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\} \right) + \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{T_1^K < K\sigma^2\} \right). \nonumber \end{align} $$

**Step 1.** Control of the first right-hand term of (32). Let us consider the martingale $(\text{Mart}_t^{K,+})_{t \ge 0}$ defined by

$$ \text{Mart}_t^{K,+} := \begin{cases} M_t^{K, P_{\text{id}^2,1}} & \text{if } t \le \check{\tau}^K \\ M_{\check{\tau}^K}^{K, P_{\text{id}^2,1}} + \frac{1}{K\sigma} B_{t-\check{\tau}^K} & \text{if } t > \check{\tau}^K \end{cases} $$

where $(B_t)_{t \ge 0}$ is a standard Brownian motion independent of $(\nu_t^K)_{t \ge 0}$. Note that from Lemma 4.6, there exists a constant $C > 0$ such that for all $t \ge 0$ we have

$$ \mathbb{E} \left( \langle \text{Mart}^{K,+} \rangle_t \right) \le \frac{C}{K^2 \sigma^2} t. \quad (33) $$

Let us consider the process $(M_2^{K,+}(t))_{t \ge 0}$ defined by

$$ M_2^{K,+}(t) := \begin{cases} M_2(\mu_t^K) & \text{if } t \le \check{\tau}^K \\ M_2(\mu_{\check{\tau}^K}^K) - \frac{bu_{\ell-1}}{K^2\sigma^2} (t - \check{\tau}^K) + (\text{Mart}_t^{K,+} - \text{Mart}_{\check{\tau}^K}^{K,+}) & \text{if } t > \check{\tau}^K \end{cases}. $$
**Corollary 4.4.** For all $T > 0$, the family

$$ \left( \int_{0}^{t \wedge \hat{\tau}^{K}} M_{5}\left(\mu_{s}^{K}\right) \mathrm{d}s \right)_{t \in\left[0, T\right], K \in \mathbb{N}^{*}} $$

is uniformly integrable.

*Proof.* Note that for all $K \in \mathbb{N}^*$, for all $t \in [0, T]$, for all $A \in (0, +\infty)$,

$$ \mathbf{1}_{\int_{0}^{t} M_{5}(\mu_{s}^{K}) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \ge A} \int_{0}^{t} M_{5}\left(\mu_{s}^{K}\right) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \le \frac{1}{\sqrt[5]{A}}\left(\int_{0}^{t} M_{5}\left(\mu_{s}^{K}\right) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s\right)^{6/5}, $$

we deduce from HÖLDER' inequality that, for all $t \in [0, T]$, for all $A \in (0, +\infty)$,

$$ \begin{aligned} & \sup_{K \in \mathbb{N}^*} \mathbb{E} \left( \mathbf{1}_{\int_{0}^{t} M_{5}(\mu_{s}^{K}) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \ge A} \int_{0}^{t} M_{5}\left(\mu_{s}^{K}\right) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \right) \\ & \le \frac{T}{\sqrt[5]{A}} \sup_{K \in \mathbb{N}^*} \mathbb{E} \left( \int_{0}^{t} M_{5}^{6/5}\left(\mu_{s}^{K}\right) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \right) \\ & \le \frac{T}{\sqrt[5]{A}} \sup_{K \in \mathbb{N}^*} \mathbb{E} \left( \int_{0}^{t} M_{6}\left(\mu_{s}^{K}\right) \mathbf{1}_{s \le \hat{\tau}^{K}} \mathrm{d}s \right). \end{aligned} $$

From Lemma 4.3, we deduce that the right hand side of the previous inequality goes to 0 when $A \to +\infty$ and so the announced result. □

## 4.2 Some inequalities on the moment of order 2

**Lemma 4.5.** There exists $K_0 \in \mathbb{N}^*$ large enough such that for all $K \ge K_0$, the process $M_2(\mu^K)$ satisfies for all $t \ge 0$ the following inequalities:

$$ \begin{aligned} & M_2(\mu_{s \wedge \hat{\tau}^K}^K) - \frac{5\bar{b}}{2K^2\sigma_K^2} \int_{s \wedge \hat{\tau}^K}^{t \wedge \hat{\tau}^K} \left[ M_2(\mu_r^K) - \frac{2\theta\bar{m}_2}{5\bar{b}} + \frac{2\bar{\theta}\bar{m}_2}{5\bar{b}K} \right] \mathrm{d}r + M_{t \wedge \hat{\tau}^K}^{K, P_{\mathrm{id}}^{2,1}} - M_{s \wedge \hat{\tau}^K}^{K, P_{\mathrm{id}}^{2,1}} \\ & \le M_2(\mu_{t \wedge \hat{\tau}^K}^K) \le M_2(\mu_{s \wedge \hat{\tau}^K}^K) - \frac{b}{K^2\sigma_K^2} \int_{s \wedge \hat{\tau}^K}^{t \wedge \hat{\tau}^K} \left[ M_2(\mu_r^K) - \frac{\bar{\theta}\bar{m}_2}{b} \right] \mathrm{d}r + M_{t \wedge \hat{\tau}^K}^{K, P_{\mathrm{id}}^{2,1}} - M_{s \wedge \hat{\tau}^K}^{K, P_{\mathrm{id}}^{2,1}} \end{aligned} \quad (33) $$

where $M_{t \wedge \hat{\tau}^K}^{K, P_{\mathrm{id}}^{2,1}}$ is defined in Proposition 3.5. Moreover, for $t \ge 0$,

$$ \mathbb{E} \left( M_2 \left( \mu_t^K \right) \mathbf{1}_{t \le \hat{\tau}^K} \right) \le \mathbb{E} \left( M_2 \left( \mu_0^K \right) - \frac{\bar{\theta} \bar{m}_2}{b} \right) \exp \left( - \frac{b}{K^2 \sigma_K^2} t \right) + \frac{\bar{\theta} \bar{m}_2}{b}. $$

*Proof.* Note that for all $t \ge 0$,

$$ M_3(\mu_{t \wedge \hat{\tau}^K}^K) \le \mathrm{Diam} \left( \mathrm{Supp} \, \mu_{t \wedge \hat{\tau}^K}^K \right) M_2(\mu_{t \wedge \hat{\tau}^K}^K) \le \frac{M_2(\mu_{t \wedge \hat{\tau}^K}^K)}{\sigma_K K^{\frac{3+\varepsilon}{2}}}. \quad (34) $$
Note that from Corollary 4.11, for all $t \ge 0$,

$$2u_{\ell-1} + 1 - \frac{5\bar{b}u_{\ell+1} + \theta m_2}{2K^2\sigma^2}t + \mathrm{Mart}_t^{K,+} \le M_2^{K,+}(t) \le 2u_{\ell} + 1 - \frac{bu_{\ell-1}}{K^2\sigma^2}t + \mathrm{Mart}_t^{K,+}. \quad (34)$$

By GIRSANOV'S theorem, the process $(E_t^{K,+})_{t \ge 0}$ defined for all $t \ge 0$ by

$$E_t^{K,+} := \exp \left( \mathrm{Mart}_t^{K,+} - \frac{1}{2} \langle \mathrm{Mart}^{K,+} \rangle_t \right)$$

is a local martingale. Let $t_0^K := \frac{6K^2\sigma^2}{b}$ be fixed and let us consider

$$T_1^{K,+} := \inf \left\{ t \ge 0 \mid M_2^{K,+}(t) \notin \mathcal{I}_\ell \right\} \quad \text{and} \quad L_1^{K,+} := \min \left\{ \ell \in \mathbb{N} \mid M_2^{K,+}(T_1^{K,+}) \le 2u_\ell - 1 \right\}.$$

From (34) and denoting $\Gamma := \{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} \le \frac{2bu_{\ell-1}}{K^2\sigma^2} T_1^{K,+} \wedge t_0^K \}$, we have

$$\begin{aligned} 1 &= \mathbb{E}_{\mu_0^K} \left( E_{T_1^{K,+} \wedge t_0^K}^{K,+} \right) \ge \mathbb{E}_{\mu_0^K} \left( E_{T_1^{K,+} \wedge t_0^K}^{K,+} \mathbf{1}_{\Gamma} \right) \\ &\ge \exp(-2u_\ell - 1) \mathbb{E}_{\mu_0^K} \left( \exp \left( M_2^{K,+} (T_1^{K,+} \wedge t_0^K) + \frac{bu_{\ell-1}}{K^2\sigma^2} T_1^{K,+} \wedge t_0^K - \frac{1}{2} \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} \right) \mathbf{1}_{\Gamma} \right) \\ &\ge \exp(-2u_\ell - 1) \mathbb{E}_{\mu_0^K} \left( \exp \left( M_2^{K,+} (T_1^{K,+} \wedge t_0^K) \right) \mathbf{1}_{\Gamma} \right) \\ &\ge \exp(-2u_\ell - 1) \left[ \exp(u_{\ell+1}) \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell+1\} \cap \Gamma \cap \{T_1^{K,+} \le t_0^K\} \right) \right. \\ &\qquad \left. + \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell-1\} \cap \Gamma \cap \{T_1^{K,+} \le t_0^K\} \right) \right]. \end{aligned}$$

As for all events $A, B, C$, $\mathbb{P}(A \cap B \cap C) \ge \mathbb{P}(A) - \mathbb{P}(A \cap B^c) - \mathbb{P}(A \cap C^c)$, and denoting $p := \mathbb{P}_{\mu_0^K} (L_1^{K,+} = \ell + 1)$ we have that

$$\begin{aligned} 1 &\ge \exp(u_\ell - 1) \left[ p - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell + 1\} \cap \Gamma^c \right) - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell + 1\} \cap \{T_1^{K,+} > t_0^K\} \right) \right] \\ &\quad + \exp(-2u_\ell - 1) \left[ 1 - p - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell - 1\} \cap \Gamma^c \right) - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell - 1\} \cap \{T_1^{K,+} > t_0^K\} \right) \right] \\ &\ge (\exp(u_\ell - 1) - \exp(-2u_\ell - 1)) p + \exp(-2u_\ell - 1) \\ &\quad - (\exp(u_\ell - 1) + \exp(-2u_\ell - 1)) \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right]. \end{aligned}$$

Then, considering $s_0^K := \frac{u_{\ell-1}K^2\sigma^2}{5\bar{b}u_{\ell+1}+\theta m_2} < t_0^K$ be fixed, we deduce that

$$\begin{aligned} p &\le \frac{1 - \exp(-2u_\ell - 1)}{\exp(u_\ell - 1) - \exp(-2u_\ell - 1)} + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right] \\ &\le \exp(-u_\ell + 1) \times \frac{1 - \exp(-2u_\ell - 1)}{1 - \exp(-u_{\ell+1})} + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \\ &\quad \times \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c \cap \{T_1^{K,+} \ge s_0^K\}) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right]. \end{aligned}$$
From Lemma 3.6, (34) and Assumptions (A), there exists two constants $C, \tilde{C} > 0$ such that for all $t > s \ge 0$

$$
\begin{align*}
M_2(\mu_{s \wedge \tilde{\tau}^K}^K) - \frac{1}{K^2 \sigma_K^2} \int_{s \wedge \tilde{\tau}^K}^{t \wedge \tilde{\tau}^K} \left( \left[ 2\bar{b} + \frac{C}{K^{\frac{\epsilon}{2}}} \right] M_2(\mu_r^K) - \underline{\theta m}_2 + \frac{\bar{\theta} \bar{m}_2}{K} \right) dr &+ M_{t \wedge \tilde{\tau}^K}^{K, P_{id}^{2,1}} - M_{s \wedge \tilde{\tau}^K}^{K, P_{id}^{2,1}} \\
&\le M_2(\mu_{t \wedge \tilde{\tau}^K}^K) \\
&\le M_2(\mu_{s \wedge \tilde{\tau}^K}^K) - \frac{1}{K^2 \sigma_K^2} \int_{s \wedge \tilde{\tau}^K}^{t \wedge \tilde{\tau}^K} \left( \left[ 2\underline{b} - \frac{\tilde{C}}{K^{\frac{\epsilon}{2}}} \right] M_2(\mu_r^K) - \bar{\theta} \bar{m}_2 + \frac{\underline{\theta m}_2}{K} \right) dr &+ M_{t \wedge \tilde{\tau}^K}^{K, P_{id}^{2,1}} - M_{s \wedge \tilde{\tau}^K}^{K, P_{id}^{2,1}}.
\end{align*}
$$

Noting that for all $K \ge K_0$, $2\underline{b} - \tilde{C}K^{-\epsilon/2} > \underline{b}$ and $2\bar{b} + CK^{-\epsilon/2} < 5\bar{b}/2$, the announced first result follows. In similar way to Step 2 of the proof of Lemma 4.2, we obtain the second part of the announced result. $\square$

**Corollary 4.6.** For all $t \ge 0$, $\mathbb{E}(M_2(\mu_t^K) 1_{t \le \tilde{\tau}^K}) \le \max\{C_2^*, \bar{\theta}\bar{m}_2/\underline{b}\}$.

*Proof.* From (7) and (4), note that

$$M_2(\mu_0^K) = \langle \mathrm{id}^2, \mu_0^K \rangle \le C_2^*.$$

The announced result follows from Lemma 4.5. $\square$

The goal of the following result is to bound the martingale bracket $\langle M^{K, P_{id}^{2,1}} \rangle_{t \wedge \tilde{\tau}^K}$. To do this, we exploit the martingale approximation of the moment of order 6 in order to control the dominant term $M_4(\mu_s^K) - M_2^2(\mu_s^K)$ in (28).

**Lemma 4.7.** There exists a constant $C > 0$ such that for all $K \in \mathbb{N}^*$, for all $t \ge 0$ the martingale bracket $\langle M^{K, P_{id}^{2,1}} \rangle_t$ satisfies

$$\mathbb{E} \left( \langle M^{K, P_{id}^{2,1}} \rangle_{t \wedge \tilde{\tau}^K} \right) \le \frac{C}{K^2 \sigma_K^2} \mathbb{E} \left( t \wedge \tilde{\tau}^K \right).$$

*Proof.* From Lemma 3.6 and Hölder's inequalities $M_2^2(\mu) \le M_4(\mu) \le M_6^{2/3} \le 1 + M_6(\mu)$ and $M_5(\mu) \le M_6(\mu)^{5/6} \le 1 + M_6(\mu)$, we deduce that there exists a constant $C > 0$ such that for all $t \ge 0$

$$\langle M^{K, P_{id}^{2,1}} \rangle_{t \wedge \tilde{\tau}^K} \le \frac{C}{K^2 \sigma_K^2} \int_0^{t \wedge \tilde{\tau}^K} \left(1 + M_6(\mu_s^K)\right) \left(1 + \frac{1}{K}\right) ds.$$

The announced result follows from FUBINI's theorem and Lemma 4.2. $\square$

## 4.3 Convergence to $+\infty$ of the stopping time $\tau^K$ and support concentration property

The main result of this section is the following convergence result for the stopping time $\tau^K$, defined by (13), when $K \to +\infty$.

**Proposition 4.8.** Under Assumption (5), $\tau^K$ converges in probability to $+\infty$ when $K \to +\infty$.
Denoting ⌈a⌋ the upper integer part of a ∈ ℝ, we obtain on the one hand that there exists a constant C₁ > 0 such that

$$
\begin{align*}
& \mathbb{P}_{\mu_0^K} (\Gamma^c \cap \{T_1^{K,+} \ge s_0^K\}) \\
&= \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma^2} T_1^{K,+} \wedge t_0^K \right\} \cap \{T_1^{K,+} \ge s_0^K\} \right) \\
&= \sum_{k=1}^{\lceil t_0^K/s_0^K \rceil - 1} \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma^2} T_1^{K,+} \wedge t_0^K \right\} \cap \{ks_0^K \le T_1^{K,+} \le (k+1)s_0^K\} \right) \\
&\quad + \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma^2} T_1^{K,+} \wedge t_0^K \right\} \cap \{T_1^{K,+} \ge \lceil t_0^K/s_0^K \rceil s_0^K\} \right) \\
&\le \sum_{k=1}^{\lceil t_0^K/s_0^K \rceil - 1} \mathbb{P}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{(k+1)s_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma^2} ks_0^K \right) + \mathbb{P}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma^2} t_0^K \right) \\
&\le K^2 \sigma^2 \left( \sum_{k=1}^{\lceil t_0^K/s_0^K \rceil - 1} \frac{\mathbb{E} \left( \langle \mathrm{Mart}^{K,+} \rangle_{(k+1)s_0^K} \right)}{2bu_{\ell-1}ks_0^K} + \frac{\mathbb{E} \left( \langle \mathrm{Mart}^{K,+} \rangle_{t_0^K} \right)}{2bu_{\ell-1}t_0^K} \right) \\
&\le \frac{C_1}{K^{\frac{\epsilon}{2}}}
\end{align*}
$$

where we use MARKOV'S inequality in the second inequality and (34) in the last inequality. On the other hand, we control $\mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K)$ as follows:

$$
\mathbb{P}_{\mu_0^K} (T_1^{K,+} < s_0^K) \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \sup_{0 \le t \le s_0^K} M_2^{K,+}(t) \ge u_{\ell+1} \right\} \right) \\
\qquad + \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le t \le s_0^K} M_2^{K,+}(t) < u_{\ell-1} \right\} \right).
$$

Let us consider $K_1 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_1^{\epsilon/2} \ge 2$. There exists a constant $C_1 > 0$ such that for all $K \ge K_1$, we have that

$$
\begin{align*}
& \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^{K,+} < s_0^K \right\} \cap \left\{ \sup_{0 \le t \le s_0^K} M_2^{K,+}(t) \ge u_{\ell+1} \right\} \right) \\
&\le \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^{K,+} < s_0^K \right\} \cap \left\{ \sup_{0 \le t \le s_0^K} \left\{ 2u_\ell + 1 - \frac{bu_{\ell-1}}{K^2 \sigma^2} t + \mathrm{Mart}_{t_\ell}^{K,+} \right\} \ge u_{\ell+1} \right\} \right) \\
&\le \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^{K,+} < s_0^K \right\} \cap \left\{ \sup_{0 \le t \le s_0^K} \mathrm{Mart}_{t_\ell}^{K,+} \ge u_\ell - 1 \right\} \right) \\
&\le \mathbb{P}_{\mu_0^K} \left( \sup_{0 \le t \le s_0^K} \left| \mathrm{Mart}_{t_\ell}^{K,+} \right| \ge \frac{3^\ell K^{\frac{\epsilon}{2}}}{2} \right) \\
&\le 4E_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{s_0^K} \right) \times \frac{4}{3^{2\ell} K^\epsilon} \\
&\le \frac{C_1}{K^\epsilon}
\end{align*}
$$
The proof is based on the next lemma proved in Section 4.4 below. Thanks to Proposition 4.8, we deduce the support concentration property of $(\nu_{t/K\sigma_K^2}^K)$ given by (6). Indeed, with probability which tends to 1 when $K \to +\infty$, $\text{Diam}(\mu_t^K) \le \frac{1}{\sigma_K K^{\frac{3+\varepsilon}{2}}}$ for all $t \in [0, T]$. Hence,

$$ \text{Diam} (\nu_{t/K\sigma_K^2}^K) = \sigma_K \sqrt{K} \text{Diam} (\mu_t^K) \le \frac{1}{K^{1+\frac{\varepsilon}{2}}} $$

and (6) follows.

**Lemma 4.9.** Under Assumption (5), for all $T \ge 0$,

$$ \lim_{K \to +\infty} \mathbb{P} (\hat{\tau}^K \le \check{\tau}^K \wedge T) = \lim_{K \to +\infty} \mathbb{P} \left( \sup_{0 \le t \le T \wedge \check{\tau}^K} M_2 (\mu_t^K) \ge K^\varepsilon \right) = 0. $$

*Proof of Proposition 4.8.* Note that for all $T \ge 0$,

$$ \mathbb{P} (\check{\tau}^K < T \wedge \hat{\tau}^K) \le \mathbb{P} \left( \exists t < T \wedge \hat{\tau}^K, \text{Diam} (\text{Supp } \mu_t^K) > \frac{1}{\sigma_K K^{\frac{3+\varepsilon}{2}}} \right). $$

As for all $t \ge 0$, for all $K \in \mathbb{N}^*$, $\text{Diam} (\text{Supp } \mu_t^K)^2 \le (2 \max \{|x| \mid x \in \text{Supp } \mu_t^K\})^2 \le 4KM_2(\mu_t^K)$, we deduce that

$$ \begin{aligned} & \mathbb{P} \left( \exists t < T \wedge \hat{\tau}^K, \text{Diam} (\text{Supp } \mu_t^K) > \frac{1}{\sigma_K K^{\frac{3+\varepsilon}{2}}} \right) \\ & \le \mathbb{P} \left( \exists t < T \wedge \hat{\tau}^K M_2 (\mu_t^K) > \frac{1}{4\sigma_K^2 K^{4+\varepsilon}} \right). \end{aligned} $$

According to (5), we deduce that the following inequality $4\sigma_K^2 K^{4+\varepsilon} \le 1/K^\varepsilon$ holds true when $K \to +\infty$ and $\sigma_K \to 0$. Hence, we deduce that

$$ \mathbb{P} \left( \exists t < T \wedge \hat{\tau}^K, M_2 (\mu_t^K) > \frac{1}{4\sigma_K^2 K^{4+\varepsilon}} \right) \le \mathbb{P} \left( \exists t < T \wedge \hat{\tau}^K, M_2 (\mu_t^K) > K^\varepsilon \right) = 0. $$

Now, for all $T > 0$, $\{\check{\tau}^K \ge T \wedge \hat{\tau}^K\} = \{\check{\tau}^K \ge \hat{\tau}^K\} \cup \{\check{\tau}^K \ge T\}$, and so

$$ \begin{aligned} & \mathbb{P} (\{\check{\tau}^K \ge T \wedge \hat{\tau}^K\} \cap \{\tau^K < T\}) \le \mathbb{P} ([\{\hat{\tau}^K < T\} \cap \{\hat{\tau}^K \le \check{\tau}^K\}] \cup \{\hat{\tau}^K < T \le \check{\tau}^K\}) \\ & \le \mathbb{P} (\hat{\tau}^K \le T \wedge \check{\tau}^K). \end{aligned} $$

Note that for all $T > 0$,

$$ \mathbb{P} (\tau^K < T) \le \mathbb{P} (\check{\tau}^K < T \wedge \hat{\tau}^K) + \mathbb{P} (\hat{\tau}^K \le T \wedge \check{\tau}^K). $$

Hence, from Lemma 4.9, we have for all $T \ge 0$, $\lim_{K \to +\infty} \mathbb{P} (\tau^K < T) = 0$ and the announced result follows. $\square$
where we use (34) in the first inequality, DOOB's maximal inequality in the fourth inequality and (33) in the last inequality. Now, let us consider $K_2 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_2^{\varepsilon/2} \ge 4$. Using again (34), DOOB's maximal inequality, (33), and the definition of $s_0^K$, we deduce that there exists a constant $C_2 > 0$ such that for all $K \ge K_2$, we have that

$$
\begin{aligned}
& \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le t \le s_0^K} M_2^{K,+}(t) < u_{\ell-1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le t \le s_0^K} \left\{ 2u_{\ell-1} - 1 - \frac{5\bar{b}u_{\ell+1} + \frac{\theta m_2}{2K^2\sigma^2}t + \text{Mart}_t^{K,+}}{2K^2\sigma^2} \right\} < u_{\ell-1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le t \le s_0^K} \text{Mart}_t^{K,+} < -\frac{u_{\ell-1}}{2} + 1 \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \sup_{0 \le t \le s_0^K} \left| \text{Mart}_t^{K,+} \right| > \frac{u_{\ell-1}}{2} - 1 \right) \\
& \le \frac{C_2}{K^{\varepsilon}}.
\end{aligned}
$$

Hence, for all $K \ge \max\{K_1, K_2\}$, $\mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K) \le \frac{C_1+C_2}{K^\varepsilon}$. Finally, from (34) and MARKOV'S inequality and the definition of $t_0^K$, note that there exists a constant $C_3 > 0$ such that

$$
\begin{aligned}
\mathbb{P}_{\mu_0^K} (T_1^{K,+} > t_0^K) &\le \mathbb{P}_{\mu_0^K} \left( \left\{ \sup_{0 \le t \le t_0^K} M_2^{K,+}(t) < u_{\ell+1} \right\} \cap \left\{ \inf_{0 \le t \le t_0^K} M_2^{K,+}(t) \ge u_{\ell-1} \right\} \right) \\
&\le \mathbb{P}_{\mu_0^K} \left( \text{Mart}_{t_0^K}^{K,+} \ge u_{\ell-1} - 2u_\ell - 1 + \frac{bu_{\ell-1}}{K^2\sigma^2} t_0^K \right) \\
&\le \frac{\mathbb{E}_{\mu_0^K} \left( \langle \text{Mart}^{K,+} \rangle_{t_0^K} \right)}{(u_{\ell-1} - 1)^2} \\
&\le \frac{C_3}{K^\varepsilon}.
\end{aligned}
$$

Hence, for all $K \ge \max\{K_1, K_2\}$, we deduce that $p \le \varepsilon_K$ where

$$
\varepsilon_K := \exp(-u_\ell + 1) \times \frac{1 - \exp(-2u_\ell - 1)}{1 - \exp(-u_{\ell+1})} + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \times \frac{C_1}{K^{\frac{\varepsilon}{2}}}
$$

and thus, for all $K \ge \max\{K_1, K_2\}$,

$$
\mathbb{P}_{\mu_0^K} \left( \{T_1^K < \tilde{\tau}^K\} \cap \{L_1^K = \ell + 1\} \right) = \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < \tilde{\tau}^K\} \cap \{L_1^K = \ell + 1\} \right) \le p \le \varepsilon_K.
$$

**Step 2.** Control of the second right-hand term of (32). Similarly, let us consider $K_3 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_3^{\varepsilon/2} \ge 2$. Using Corollary 4.11, DOOB's maximal inequality and Lemma 4.6 in similar way to Step 1, there exists a constant $C_4 > 0$ such that for all $K \ge K_3$ we have that

$$
\mathbb{P}_{\mu_0^K} \left( \{T_1^K < \tilde{\tau}^K\} \cap \left\{ \sup_{0 \le t \le K \sigma^2 \wedge \tilde{\tau}^K} M_2(\mu_t^K) \ge u_{\ell+1} \right\} \right) \le \frac{16 \mathbb{E}_{\mu_0^K} \left( \left\langle M^{K,P_{\text{id}2,1}} \right\rangle_{K \sigma^2 \wedge \tilde{\tau}^K} \right)}{3^{2\ell} K^{\varepsilon}} \le \frac{C_4}{K^{1+\varepsilon}}.
$$
## 4.4 Proof of Lemma 4.9

Note that if $M_2(\mu_s^K)$ becomes larger than $\bar{\theta}\bar{m}_2/\underline{b}$, the drift term in the right-hand side of (33) is then negative preventing $M_2(\mu_s^K)$ to become excessively large unless the martingale $M^{K,P_{id}^2,1}$ has an exceptional path with large increments on a small time interval. We then expect to have large deviations estimates and bounds on the time of exit from attractive domains for $M_2(\mu^K)$. However, we cannot consider establishing directly a large deviation principle on $M_2(\mu^K)$, in particular because of the slow-fast limit. The approach considered below is based on coupling arguments between $M_2(\mu^K)$ and a simpler process for which large deviations estimates are known.

Let us introduce $u_0 := 0$ and for all $\ell \in \mathbb{N}^*$ the real number $u_\ell := 3^\ell K^{\frac{\ell}{2}}$ and the interval $\mathcal{I}_\ell := [u_{\ell-1}, u_{\ell+1})$. We will look at the process $M_2(\mu^K)$ at successive exit times of $(\mathcal{I}_\ell)_{\ell \in \mathbb{N}^*}$. We set $L_0^K := 1$ when $M_2(\mu_0^K) \in [0, 2u_1+1]$ and $L_0^K := \ell_0 \in \mathbb{N}^*$ when $M_2(\mu_0^K) \in (2u_{\ell_0-1}+1, 2u_{\ell_0}+1]$. We also set $T_0^K := 0$ and by induction on $k \in \mathbb{N}^*$, if $(L_i^K)_{1 \le i \le k}$ and $(T_i^K)_{1 \le i \le k}$ are constructed,

$$T_{k+1}^{K} := \inf \left\{ t \ge T_{k}^{K} \mid M_{2}(\mu_{t}^{K}) \notin \mathcal{I}_{L_{k}^{K}} \right\}$$

and

$$L_{k+1}^{K} := \min \left\{ \ell \in \mathbb{N} \mid M_{2} \left( \mu_{T_{k+1}^{K}}^{K} \right) \le 2u_{\ell} + 1 \right\}.$$

Note that from (4), for $K$ large enough, $L_0^K = 1$ because $M_2(\mu_0^K) \le C_2^*$. Since $(\mu_t^K)_{t \ge 0}$ is a pure jump process, note that $(M_2(\mu_t^K))_{t \ge 0}$ is also a pure jump process. The definition of $u_\ell$ is motivated by the fact that, for all $k \ge 0$, $L_{k+1}^K \le L_k^K + 1$, which follows from the next lemma.

**Lemma 4.10.** Let $\vartheta_1$ be the first jump time of $\mu^K$. Then, for $K$ large enough

$$M_2(\mu_{\vartheta_1}^K) \le 2M_2(\mu_0^K) + 1.$$

In particular, for all $t > 0$, $\Delta M_2(\mu_t^K) := M_2(\mu_t^K) - M_2(\mu_{t-}^K) \le M_2(\mu_{t-}^K) + 1$.

*Proof.* Let us consider $\mu_0^K := \frac{1}{K} \sum_{k=1}^K \delta_{x_k}$ and $y = (y_k)_{1 \le k \le K}$ defined by

$$y = \begin{cases} (x_1, \dots, x_{i-1}, x_i + \frac{H}{\sqrt{K}}, x_{i+1}, \dots, x_K) & \text{on } E_1, \\ (x_1, \dots, x_{j-1}, x_i, x_{j+1}, \dots, x_K) & \text{on } E_2, \end{cases}$$

where $H$ follows the mutation law $m(x_i, dh)$, $E_1$ is the event “a mutation occurs at time $\vartheta_1$” and $E_2$ is the event “a resampling between $i$ and $j$ occurs at time $\vartheta_1$”. Note that $\mu_{\vartheta_1}^K = \tau_{-\langle \mathrm{id}, \frac{1}{K} \sum_{k=1}^K \delta_{y_k} \rangle} \sharp (\frac{1}{K} \sum_{k=1}^K \delta_{y_k})$. Hence, noting that for all $k \in \{1, \dots, K\}$,

$$|x_k| \le \sqrt{K} (1 + M_2(\mu_0^K))$$
Let us consider $K_4 \in \mathbb{N}^*$ independent of $\ell$ satisfying the relation $u_{\ell-1} > 1 + \frac{5\bar{b}u_{\ell+1}+\theta m_2}{2K_4}$. As previously, we establish that there exists a constant $C_5 > 0$ such that for all $K \ge K_4$ we have that

$$ \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^K < \check{\tau}^K \right\} \cap \left\{ \inf_{0 \le t \le K \sigma^2 \wedge \check{\tau}^K} M_2 (\mu_t^K) < u_{\ell-1} \right\} \right) \le \frac{4 \mathbb{E}_{\mu_0^K} \left( \left\langle M^K, P_{\mathrm{id}2,1} \right\rangle_{K \sigma^2 \wedge \check{\tau}^K} \right)}{\left( u_{\ell-1} - 1 - \frac{5\bar{b}u_{\ell+1}+\theta m_2}{2K} \right)^2} \le \frac{C_5}{K^{1+\varepsilon}}. $$

To conclude this proof, there exists a constant $C > 0$ such that for all $K \ge K_0$ where $K_0 := \max_{i \in \{1, \dots, 4\}} K_i$,

$$ \mathbb{P}_{\mu_0^K} \left( \left[ \left\{ L_1^K = \ell+1 \right\} \cup \left\{ T_1^K < K\sigma^2 \right\} \right] \cap \left\{ T_1^K < \check{\tau}^K \right\} \right) < \frac{1}{2} - \eta_K $$

for any $\eta_K \ge \frac{1}{2} - \epsilon_K - \frac{C_4+C_5}{K^{1+\varepsilon}}$. A convenient choice is given by $\eta_K := \frac{1}{2} - \frac{C}{K^{\varepsilon/2}}$ for $K$ large enough independent of $\ell$ which completes the proof. $\square$

### 4.4.2 Construction of the coupling

The goal of this section is to construct a coupling between $(L_k^K)_{k \in \mathbb{N}}$ and a biased random walk on $\mathbb{N}^*$ and reflected in 1. To do this, we will construct a sequence $(\xi_k^K)_{k \in \mathbb{N}}$ of i.i.d. random variables with values in $\{-1, 1\}$ as follows.

**Step 1.** First of all, thanks to Lemma 4.10 we can construct the random variable $\xi_0^K \in \{-1, 1\}$ such that

$$ \{\xi_0^K = -1\} \subset \left[ \left\{ L_1^K = L_0^K - 1 \right\} \cap \left\{ K\sigma^2 \le T_1^K < \check{\tau}^K \right\} \right] \cup \left\{ T_1^K \ge \hat{\tau}^K \right\} $$

$$ \text{and } \mathbb{P}(\xi_0^K = -1) = \frac{1}{2} + \eta_K. $$

**Step 2.** Then, thanks to Lemma 4.10 again and after having applied the MARKOV property at time $T_1^K$, we can construct the random variable $\xi_1^K \in \{-1, 1\}$ such that, conditionally to $\mathcal{F}_{T_1^K} \lor \sigma(\xi_0^K)$,

$$ \{\xi_1^K = -1\} \subset \left[ \left\{ L_2^K = L_1^K - 1 \right\} \cap \left\{ K\sigma^2 + T_1^K \le T_2^K < \check{\tau}^K \right\} \right] \cup \left\{ T_2^K \ge \hat{\tau}^K \right\} $$

and $\mathbb{P}(\xi_1^K = -1 | \mathcal{F}_{T_1^K} \lor \sigma(\xi_0^K)) = \frac{1}{2} + \eta_K$. Note that this implies in particular that $\xi_1^K$ is independent of $\xi_0^K$.

**Step 3.** By induction on $k \in \mathbb{N}$, if $(\xi_k^K)_{0 \le i \le k}$ are constructed, then thanks to Lemma 4.10 again and after having applied the MARKOV property at time $T_{k+1}^K$, we can construct the random variable $\xi_{k+1}^K \in \{-1, 1\}$ such that, conditionally to $\mathcal{F}_{T_{k+1}^K} \lor \sigma(\xi_0^K, \dots, \xi_k^K)$,

$$ \{\xi_{k+1}^K = -1\} \subset \left[ \left\{ L_{k+2}^K = L_{k+1}^K - 1 \right\} \cap \left\{ K\sigma^2 + T_{k+1}^K \le T_{k+2}^K < \check{\tau}^K \right\} \right] \cup \left\{ T_{k+2}^K \ge \hat{\tau}^K \right\} $$

$$ \text{and } \mathbb{P}(\xi_1^K = -1 | \mathcal{F}_{T_{k+1}^K} \lor \sigma(\xi_0^K, \dots, \xi_k^K)) = \frac{1}{2} + \eta_K. $$
and using Assumption (A2), for $K$ large enough we have that

$$
\begin{align*}
M_2(\mu_{\vartheta_1}^K) &\le M_2\left(\frac{1}{K}\sum_{i=1}^K \delta_{y_i}\right) = \frac{1}{K}\sum_{i=1}^K y_i^2 \\
&\le \begin{cases}
\frac{1}{K}\sum_{k=1}^K x_k^2 + \frac{2A_m x_k}{K^{\frac{3}{2}}} + \frac{A_m^2}{K^2} & \text{on } E_1, \\
\frac{1}{K}\sum_{k=1}^K x_k^2 + \frac{x_j^2 - x_i^2}{K} & \text{on } E_2,
\end{cases} \\
&\le \begin{cases}
M_2(\mu_0^K) + \frac{2A_m(1+M_2(\mu_0^K))}{K} + \frac{A_m^2}{K^2} & \text{on } E_1, \\
2M_2(\mu_0^K) & \text{on } E_2,
\end{cases} \\
&\le 2M_2(\mu_0^K) + 1.
\end{align*}
$$

□

Each of the previous steps will be called *transitions of $M_2(\mu_t^K)$*. The idea of the proof of Lemma 4.9 is to estimate the transition probabilities of the sequence $(L_k^K)_{k \in \mathbb{N}}$ in order to construct a coupling between $(L_k^K)_{k \in \mathbb{N}}$ and a biased random walk on $\mathbb{N}^*$ reflected in 1.

At time $\hat{\tau}^K$, $M_2(\mu_{\hat{\tau}^K}^K) \in \mathcal{I}_{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}$ where $\lfloor x \rfloor$ is the lower integer part of $x$. So, the number of transitions of $M_2(\mu_t^K)$ before $\hat{\tau}^K$ is greater than or equal to $T_{k_0}^K$ where $k_0$ is the first integer such that $L_{k_0}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$.

By using estimates on the number of steps that a biased random walk takes to reach $\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ and estimates on the durations between two transitions $T_{k+1}^K - T_k^K$, we will deduce a lower bound, exponential in $K$, on $\hat{\tau}^K$. An additional difficulty comes from the fact that the previous coupling argument is only valid up to time $\check{\tau}^K$. Hence we will construct a coupling that takes into account the possibility that $\check{\tau}^K$ happens during each transition step. The proof is divided into four steps: in Section 4.4.1, we characterise the behaviour of the first transition step; in Section 4.4.2, the proposed coupling is constructed; in Section 4.4.3, we give estimates on the first exit time from an attractive domain for random walks; finally, we conclude in Section 4.4.4.

### 4.4.1 One-step transitions

In this section we look at only one transition: we suppose that $M_2(\mu_0^K) \in [2u_{\ell-1} + 1, 2u_\ell + 1)$ for $\ell \in \mathbb{N}^*$ fixed and we look for bounds on the first transition probabilities of $M_2(\mu_t^K)$ and on time $T_1^K$. The main result is the following.

**Lemma 4.11.** For all $\ell \in \mathbb{N}$, for all $\mu_0^K \in \mathcal{M}_{1,K}^{c,2}(\mathbb{R})$ such that

$$M_2(\mu_0^K) \in [2u_{\ell-1} + 1, 2u_\ell + 1),$$

there exist a constant $C > 0$, $K_0 \in \mathbb{N}^*$ such that for all $K \ge K_0$, we have

$$\mathbb{P}_{\mu_0^K} \left( \left[ \left\{ L_1^K = \ell - 1 \right\} \cap \left\{ \check{\tau}^K > T_1^K \ge K \sigma_K^2 \right\} \right] \cup \left\{ T_1^K \ge \check{\tau}^K \right\} \right) \ge \frac{1}{2} + \eta_K, \quad (35)$$
To conclude, we set

$$ \forall K \in \mathbb{N}^*, \quad \forall k \in \mathbb{N}, \quad Z_{k+1}^K - Z_k^K := \xi_k^K \quad \text{and} \quad Z_0^K = L_0^K \quad (35) $$

and the sequence $(Z_k^K)_{k \in \mathbb{N}}$ satisfies by construction the following lemma:

**Lemma 4.12.** The sequence $(Z_k^K)_{k \in \mathbb{N}}$, given by (35) is a biased simple random walk on $\mathbb{N}^*$, reflected in 1 and for all $k \in \mathbb{N}$ as long as $T_{k+1}^K < \tilde{\tau}^K$, that

$$ (1) \quad Z_{k+1}^K - Z_k^K \ge L_{k+1}^K - L_k^K $$

$$ (2) \quad T_{k+1}^K - T_k^K \ge K\sigma^2 \quad \text{when} \quad Z_{k+1}^K - Z_k^K = -1 $$

On the one hand, note that for all $k \in \mathbb{N}$, $L_k^K \le Z_k^K$. On the other hand, note that Lemma 4.9 implies for all $k \in \mathbb{N}$, $L_{k+1}^K - L_k^K \in \{\dots, -3, -2, -1, 1\}$ which justifies (1) of Lemma 4.12. In Figure 4.2, we illustrate the coupling between $(L_k^K)_{0 \le k \le 7}$ and $(Z_k^K)_{0 \le k \le 7}$. In this figure, we observe that $\xi_k^K = L_{k+1}^K - L_k^K$ for all $k \in \{0, 1, 2, 4, 5, 6\}$ and $\xi_3^K = -1 \ge -2 = L_4^K - L_3^K$. This illustrates that $Z_{k+1}^K - Z_k^K$ is always 1 when $L_{k+1}^K - L_k^K$ is equal to 1. Note that $Z_{k+1}^K - Z_k^K$ can be 1 when $L_{k+1}^K - L_k^K$ is equal to $-1$.

<figure><img src="image_6.png" /><figcaption>Figure 4.2: For $K$ large enough, coupling between $(L_k^K)_{0 \le k \le 7}$ and $(Z_k^K)_{0 \le k \le 7}$ up to time $T$ and before $\tilde{\tau}^K$ and where $M_2(\mu_0^K) \in [0, 2u_1 + 1)$ so that $L_0^K = 1$ and $\tilde{Z}_k^K := u_{Z_k^K} = 3^{Z_k^K} K^{\epsilon/2}$ for all $k \in \mathbb{N}$.</figcaption></figure>
where $\eta_K := \frac{1}{2} - \frac{C}{K^{\epsilon/2}}$.

The proof of the previous result is based on Corollary 4.12 which is obtained as a straightforward consequence of Lemma 4.5.

**Corollary 4.12.** Let $\ell \in \mathbb{N}^*$ be fixed. There exists $K_0 \in \mathbb{N}^*$ large enough such that for all $K \ge K_0$, for all $\mu_0^K \in \mathcal{M}_{1,K}^{c,2}(\mathbb{R})$ such that $M_2(\mu_0^K) \in [2u_{\ell-1} + 1, 2u_\ell + 1)$, we have for all $t \le T_1^K$ that

$$2u_{\ell-1} + 1 - \frac{5\bar{b}u_{\ell+1} + \frac{\theta m_2}{2K^2\sigma_K^2}}{2K^2\sigma_K^2} t \wedge \check{\tau}^K + M_{t\wedge\check{\tau}^K}^{K,P_{\text{id}2,1}} \\ \le M_2(\mu_{t\wedge\check{\tau}^K}^K) \le 2u_\ell + 1 - \frac{\bar{b}u_{\ell-1}}{K^2\sigma_K^2} t \wedge \check{\tau}^K + M_{t\wedge\check{\tau}^K}^{K,P_{\text{id}2,1}}.$$

*Proof of Lemma 4.11.* By passing on the complementary of (35), it is equivalent to prove that

$$\mathbb{P}_{\mu_0^K} \left( \left[ \{L_1^K = \ell + 1\} \cup \{T_1^K < K\sigma_K^2\} \right] \cap \{T_1^K < \check{\tau}^K\} \right) < \frac{1}{2} - \eta_K.$$

As for all events $A, B$, $\mathbb{P}(A \cup B) = \mathbb{P}(A \cap B^c) + \mathbb{P}(B)$, we have

$$\begin{align}
\mathbb{P}_{\mu_0^K} \left( \left[ \{L_1^K = \ell + 1\} \cup \{T_1^K < K\sigma_K^2\} \right] \cap \{T_1^K < \check{\tau}^K\} \right) & \nonumber \\
&= \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\} \cap \{T_1^K \ge K\sigma_K^2\} \right) \nonumber \\
&\quad + \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{T_1^K < K\sigma_K^2\} \right) \tag{36} \\
&\le \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\} \right) + \mathbb{P}_{\mu_0^K} \left( \{T_1^K < \check{\tau}^K\} \cap \{T_1^K < K\sigma_K^2\} \right). \nonumber
\end{align}$$

**Step 1.** Control of the first right-hand term of (36). Let us consider the martingale $(\text{Mart}_t^{K,+})_{t \ge 0}$ defined by

$$\text{Mart}_t^{K,+} := \begin{cases} M_t^{K,P_{\text{id}2,1}} & \text{if } t \le \check{\tau}^K \\ M_{\check{\tau}^K}^{K,P_{\text{id}2,1}} + \frac{1}{K\sigma_K} B_{t-\check{\tau}^K} & \text{if } t > \check{\tau}^K \end{cases}$$

where $(B_t)_{t \ge 0}$ is a standard Brownian motion independent of $(\mu_t^K)_{t \ge 0}$. Note that from Lemma 4.7, there exists a constant $C > 0$ such that for all $t \ge 0$ we have

$$\mathbb{E} \left( \langle \text{Mart}^{K,+} \rangle_t \right) \le \frac{C}{K^2 \sigma_K^2} t. \quad (37)$$

Let us consider the process $(M_2^{K,+}(t))_{t \ge 0}$ defined by

$$M_2^{K,+}(t) := \begin{cases} M_2(\mu_t^K) & \text{if } t \le \check{\tau}^K \\ M_2(\mu_{\check{\tau}^K}^K) - \frac{bu_{\ell-1}}{K^2\sigma_K^2}(t - \check{\tau}^K) + (\text{Mart}_t^{K,+} - \text{Mart}_{\check{\tau}^K}^{K,+}) & \text{if } t > \check{\tau}^K \end{cases}.$$

Note that from Corollary 4.12, for all $t \ge 0$,

$$2u_{\ell-1} + 1 - \frac{5\bar{b}u_{\ell+1} + \frac{\theta m_2}{2K^2\sigma_K^2}}{2K^2\sigma_K^2} t + \text{Mart}_t^{K,+} \le M_2^{K,+}(t) \le 2u_\ell + 1 - \frac{\bar{b}u_{\ell-1}}{K^2\sigma_K^2} t + \text{Mart}_t^{K,+}. \quad (38)$$
### 4.4.3 Estimates of exit from an attractive domain for random walks

Let us consider $N^K$ the number of transitions before reaching $\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ for the random walk $(Z_k^K)_{k \in \mathbb{N}}$.

**Remark 4.13.** By Lemma 4.12, $N^K \le k_0$ where $k_0$ is the first integer such that $L_{k_0}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$, so $\hat{\tau}^K \ge T_{k_0}^K \ge T_{N^K}^K$.

The following lemma gives an estimate on the problem of exit from a domain for $Z^K$. For all $k \in \mathbb{N}^*$, we denote by $\mathbf{P}_k$ the law of the MARKOV chain $Z^K$ given $Z_0^K = L_0^K = k$.

**Lemma 4.14.** We have,

$$ \lim_{K \to +\infty} \mathbf{P}_1 \left( N^K \ge \exp \left( \frac{\varepsilon^2}{16 \log(3)} \log^2(K) \right) \right) = 1. $$

*Proof.* For all $k \in \mathbb{N}^*$, let us consider the stopping time $\mathcal{T}_k^K := \inf \{n \in \mathbb{N} \mid Z_n^K = k\}$ and we set $v_k^K := \mathbf{P}_k \left( \mathcal{T}_1^K < \mathcal{T}_{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}^K \right)$. Note that $v_1^K = 1$. By a classical approach and setting $r_K := \frac{1+2\eta_K}{1-2\eta_K} > 1$, we prove that

$$ v_k^K = \frac{1 - r_K^{-\left(\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - k\right)}}{1 - r_K^{-\left(\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1\right)}} = 1 - (r_K^{k-1} - 1) \exp \left( - \left( \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1 \right) \log(r_K) \right). $$

Note that $N^K$ is greater than the number of transitions from 2 to 1 that occur before $\mathcal{T}_{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}^K$.

Hence, under $\mathbf{P}_1$, $N^K \ge X$ where $X$ is random variable with geometric law of parameter $p_K := (r_K - 1) \exp \left( - \left( \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1 \right) \log(r_K) \right)$. Since $r_K \sim_{K \to +\infty} \frac{K^{\varepsilon/2}}{C}$, with $C > 0$ given by Lemma 4.10, $p_K \to +\infty$ when $K \to +\infty$ and using that for all $x > 0$ small enough, $\log(1-x) \ge -2x$ we deduce that for all $K$ large enough,

$$ \begin{aligned} \mathbf{P}_1 \left( X \ge \exp \left( \frac{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1}{2} \log(r_K) \right) \right) \\ = \exp \left( \exp \left( \frac{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1}{2} \log(r_K) \right) \log(1-p_K) \right) \\ \ge \exp \left( -2(r_K - 1) \exp \left( -\frac{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1}{2} \log(r_K) \right) \right). \end{aligned} $$

Using again that $r_K \sim_{K \to +\infty} \frac{K^{\varepsilon/2}}{C}$, we deduce that for $K$ large enough,

$$ \frac{\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor - 1}{2} \log(r_K) \ge \frac{\varepsilon^2}{16 \log(3)} \log^2(K) $$

and the announced result follows. $\square$
By GIRSANOV'S theorem, the process $(E_t^{K,+})_{t \ge 0}$ defined for all $t \ge 0$ by

$$E_t^{K,+} := \exp \left( \mathrm{Mart}_t^{K,+} - \frac{1}{2} \langle \mathrm{Mart}^{K,+} \rangle_t \right)$$

is a local martingale. Let $t_0^K := \frac{6K^2\sigma_K^2}{b}$ be fixed and let us consider

$$ 
\begin{aligned}
T_1^{K,+} &:= \inf \left\{ t \ge 0 \mid M_2^{K,+}(t) \notin \mathcal{I}_\ell \right\}, \\
L_1^{K,+} &:= \min \left\{ \ell \in \mathbb{N} \mid M_2^{K,+}(T_1^{K,+}) \le 2u_\ell - 1 \right\}.
\end{aligned}
$$

From (38) and denoting $\Gamma := \{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} \le \frac{2bu_{\ell-1}}{K^2\sigma_K^2} T_1^{K,+} \wedge t_0^K \}$, we have

$$ 
\begin{aligned}
1 &= \mathbb{E}_{\mu_0^K} (E_{T_1^{K,+} \wedge t_0^K}^{K,+}) \ge \mathbb{E}_{\mu_0^K} (E_{T_1^{K,+} \wedge t_0^K}^{K,+} 1_{\Gamma}) \\
&\ge \exp(-2u_\ell - 1) \mathbb{E}_{\mu_0^K} \left( \exp \left( M_2^{K,+} (T_1^{K,+} \wedge t_0^K) + \frac{bu_{\ell-1}}{K^2\sigma_K^2} T_1^{K,+} \wedge t_0^K \right) - \frac{1}{2} \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} \right) 1_{\Gamma} \\
&\ge \exp(-2u_\ell - 1) \mathbb{E}_{\mu_0^K} \left( \exp \left( M_2^{K,+} (T_1^{K,+} \wedge t_0^K) \right) 1_{\Gamma} \right) \\
&\ge \exp(-2u_\ell - 1) \left[ \exp(u_{\ell+1}) \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell + 1\} \cap \Gamma \cap \{T_1^{K,+} \le t_0^K\} \right) \right. \\
&\qquad \left. + \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell - 1\} \cap \Gamma \cap \{T_1^{K,+} \le t_0^K\} \right) \right].
\end{aligned}
$$

As for all events $A, B, C$, $\mathbb{P}(A \cap B \cap C) \ge \mathbb{P}(A) - \mathbb{P}(A \cap B^c) - \mathbb{P}(A \cap C^c)$, and denoting $q := \mathbb{P}_{\mu_0^K} (L_1^{K,+} = \ell + 1)$ we have that

$$ 
\begin{aligned}
1 &\ge \exp(u_\ell - 1) \left[ q - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell + 1\} \cap \Gamma^c \right) \right. \\
&\qquad \left. - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} = \ell + 1\} \cap \{T_1^{K,+} > t_0^K\} \right) \right] \\
&\quad + \exp(-2u_\ell - 1) \left[ 1 - q - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell - 1\} \cap \Gamma^c \right) \right. \\
&\qquad \left. - \mathbb{P}_{\mu_0^K} \left( \{L_1^{K,+} \le \ell - 1\} \cap \{T_1^{K,+} > t_0^K\} \right) \right] \\
&\ge (\exp(u_\ell - 1) - \exp(-2u_\ell - 1)) q + \exp(-2u_\ell - 1) \\
&\quad - (\exp(u_\ell - 1) + \exp(-2u_\ell - 1)) \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right].
\end{aligned}
$$

Then, considering $s_0^K := \frac{u_{\ell-1}K^2\sigma_K^2}{5bu_{\ell+1}+\theta m_2} < t_0^K$ be fixed, we deduce that

$$ 
\begin{aligned}
q &\le \frac{1 - \exp(-2u_\ell - 1)}{\exp(u_\ell - 1) - \exp(-2u_\ell - 1)} \\
&\quad + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right] \\
&\le \exp(-u_\ell + 1) \times \frac{1 - \exp(-2u_\ell - 1)}{1 - \exp(-u_{\ell+1})} + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \\
&\quad \times \left[ \mathbb{P}_{\mu_0^K}(\Gamma^c \cap \{T_1^{K,+} \ge s_0^K\}) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K) + \mathbb{P}_{\mu_0^K}(T_1^{K,+} > t_0^K) \right].
\end{aligned}
$$
It is in the next Corollary 4.15 that we see the importance of Assumption (5) where $\sigma$ must not be too small.

**Corollary 4.15.** For all $T > 0$,

$$ \lim_{K \to +\infty} \mathbb{P} \left( \frac{K\sigma^2}{2} \left( N^K - \frac{\varepsilon}{2\log(3)} \log(K) \right) < T \right) = 0. $$

*Proof.* From Lemma 4.14 and Assumption (5) we deduce in a straightforward manner that

$$ \lim_{K \to +\infty} \mathbf{P}_1 \left( \frac{K\sigma^2}{2} \left( N^K - \frac{\varepsilon}{2\log(3)} \log(K) \right) < T \right) = 0. $$

As $L_0^K$ depends on $\mu_0^K$ note that $L_0^K = Z_0^K$, is function of the initial condition $\mu_0^K$. However, by Assumptions (4) and MARKOV'S inequality, we have that

$$ \mathbb{P} (L_0^K > 1) \le \mathbb{P} (M_2 (\mu_0^K) \ge K^{\frac{\varepsilon}{2}}) \le \frac{C_2^*}{K^{\varepsilon}} $$

which tends to 0 when $K \to +\infty$ and the announced result follows. $\square$

### 4.4.4 Conclusion

We denote respectively by $N_+^K$ and $N_-^K$ the number of upward transitions and downward transitions for $(Z_k^K)_{k \in \mathbb{N}}$ before reaching $\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$. Let $T \ge 0$ be fixed. Note that

$$ \mathbb{P} (\hat{\tau}^K \le \check{\tau}^K \wedge T) \le \mathbb{P} (\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K > T_{NK}^K\}) + \mathbb{P} (\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K \le T_{NK}^K\}). \quad (36) $$

**Step 1.** Control of the first right-hand term of (36). As $Z_0^K = 1$ with probability converging to 1 and $Z_{NK}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$, we have $N_+^K - N_-^K \le \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ and $N_+^K + N_-^K = N^K$. Thus,

$$ N_-^K \ge \frac{N^K - \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}{2}. $$

Hence, if $\check{\tau}^K > T_{NK}^K$, then by Lemma 4.12 (2),

$$ \hat{\tau}^K \ge T_{NK}^K \ge \sum_{i=1}^{N^K} (T_i^K - T_{i-1}^K) \ge N_-^K K \sigma^2 \ge \frac{K \sigma^2}{2} \left( N^K - \frac{\varepsilon}{2 \log(3)} \log(K) \right). $$

Therefore,

$$ \mathbb{P} (\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K > T_{NK}^K\}) \le \mathbb{P} \left( \frac{K\sigma^2}{2} \left( N^K - \frac{\varepsilon}{2\log(3)} \log(K) \right) \le T \right) $$

which tends to 0 when $K \to +\infty$ according to Corollary 4.15.
Denoting ⌈a⌋ the upper integer part of a ∈ ℝ, we obtain on the one hand that there exists a constant C₁ > 0 such that

$$
\begin{align*}
& \mathbb{P}_{\mu_0^K} \left( \Gamma^c \cap \left\{ T_1^{K,+} \ge s_0^K \right\} \right) \\
&= \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma_K^2} T_1^{K,+} \wedge t_0^K \right\} \cap \left\{ T_1^{K,+} \ge s_0^K \right\} \right) \\
& \quad \left[ \frac{t_0^K}{s_0^K} \right]^{-1} \\
&= \sum_{k=1} \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma_K^2} T_1^{K,+} \wedge t_0^K \right\} \cap \left\{ ks_0^K \le T_1^{K,+} \le (k+1)s_0^K \right\} \right) \\
& \quad + \mathbb{P}_{\mu_0^K} \left( \left\{ \langle \mathrm{Mart}^{K,+} \rangle_{T_1^{K,+} \wedge t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma_K^2} T_1^{K,+} \wedge t_0^K \right\} \cap \left\{ T_1^{K,+} \ge \lceil t_0^K / s_0^K \rceil s_0^K \right\} \right) \\
& \le \sum_{k=1} \mathbb{P}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{(k+1)s_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma_K^2} ks_0^K \right) + \mathbb{P}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{t_0^K} > \frac{2bu_{\ell-1}}{K^2 \sigma_K^2} t_0^K \right) \\
& \le K^2 \sigma_K^2 \left( \sum_{k=1}^{\lceil t_0^K / s_0^K \rceil - 1} \frac{\mathbb{E} \left( \langle \mathrm{Mart}^{K,+} \rangle_{(k+1)s_0^K} \right)}{2bu_{\ell-1} ks_0^K} + \frac{\mathbb{E} \left( \langle \mathrm{Mart}^{K,+} \rangle_{t_0^K} \right)}{2bu_{\ell-1} t_0^K} \right) \\
& \le \frac{C_1}{K^{\frac{\epsilon}{2}}}
\end{align*}
$$

where we use MARKOV'S inequality in the second inequality and (38) in the last inequality. On the other hand, we control $\mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K)$ as follows:

$$
\mathbb{P}_{\mu_0^K} (T_1^{K,+} < s_0^K) \le \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^{K,+} < s_0^K \right\} \cap \left\{ \sup_{0 \le t \le s_0^K} M_2^{K,+}(t) \ge u_{\ell+1} \right\} \right) \\
+ \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^{K,+} < s_0^K \right\} \cap \left\{ \inf_{0 \le t \le s_0^K} M_2^{K,+}(t) < u_{\ell-1} \right\} \right).
$$

Let us consider $K_1 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_1^{\epsilon/2} \ge 2$. There exists a constant $C_1 > 0$
**Step 2.** Control of the second right-hand term of (36). If $\hat{\tau}^K \le T_{NK}^K$, let us consider $\tilde{N}^K$ the last index $k$ such that $T_k^K < \check{\tau}^K$. We have $\tilde{N}^K < N^K$ and so $\tilde{N}^K + 1 \le \tilde{N}^K$ which implies, thanks to Lemma 4.9, that for all $t \le T_{\tilde{N}^{K+1}}$ and $K$ large enough,

$$M_2(\mu_t^K) \le 2u_{\lfloor \frac{\epsilon}{2\log(3)} \rfloor \log(K)} + 1 \le 2K^{\frac{\epsilon}{2}} + 1 < K^{\epsilon}.$$

As the jump times of $\nu^K$ are isolated, we deduce that $\hat{\tau}^K > T_{\tilde{N}^{K+1}}^K$. Now, by definition of $\tilde{N}^K$, we have $T_{\tilde{N}^{K+1}}^K \ge \check{\tau}^K$ and thus $\hat{\tau}^K > \check{\tau}^K$. Hence, we obtain that

$$\mathbb{P}(\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K \le T_{NK}^K\}) = \mathbb{P}(\emptyset) = 0.$$

Therefore, $\lim_{K \to +\infty} \mathbb{P}(\hat{\tau}^K \le \check{\tau}^K \wedge T) = 0$ which concludes the proof of Lemma 4.8.

# 5 Tightness on the torus

The main result of this section is given by Theorem 5.1. This is a stochastic averaging result inspired by KURTZ [49, Theorem 2.1] establishing a tightness result, in the torus case, of the sequence of laws of $\{(z^K, \Gamma^K)\}_{K \in \mathbb{N}^*}$ where $\Gamma^K$ is the occupation measure of the fast component $\mu^K$. We will use criteria proposed by ETHIER-KURTZ in [34, Theorems 3.9.1 and 3.9.4]. In Sections 6 and 7 we identify the limit and we prove its uniqueness.

Let introduce the torus $\mathbb{T}_R := [x_0 - 2R, x_0 + 2R]$ of length $4R$ with $R > 0$ fixed and $x_0$ is the value of the mean trait of $\nu_0^K$ as in Theorem 2.1. We define $M_1(\mathbb{T}_R)$ the set of probability measure on $\mathbb{T}_R$. We denote by $M_m(M_1(\mathbb{R}))$ the set of measures $\Gamma$ on $\mathbb{R}_+ \times M_1(\mathbb{R})$ such that for all $t \ge 0$, $\Gamma([0, t] \times M_1(\mathbb{R})) = t$. For any $t \ge 0$, we denote by $M_m^t(M_1(\mathbb{R}))$ the set of measures $\Gamma \in M_m(M_1(\mathbb{R}))$ restricted to $[0, t] \times M_1(\mathbb{R})$.

Let $b_R \in \mathscr{C}_b^2(\mathbb{T}_R^2, \mathbb{R})$ and $\theta_R \in \mathscr{C}_b^2(\mathbb{T}_R, \mathbb{R})$ be two functions satisfying $b_R = b$ on $[x_0 - R, x_0 + R]^2$ and $\theta_R = \theta$ on $[x_0 - R, x_0 + R]$. Let $T > 0$ be fixed. We define on $\mathbb{T}_R$, in similar way as in Section 2, the processes $\nu^{K,R} := (\nu_t^{K,R})_{t \in [0,T]}$, $z^{K,R} := (z_t^{K,R})_{t \in [0,T]}$ and $\mu^{K,R} = (\mu_t^{K,R})_{t \in [0,T]}$ as follows

$$\nu_t^{K,R} := \frac{1}{K} \sum_{i=1}^{K} \delta_{x_i^R(t)}, \quad z_t^{K,R} := \langle \mathrm{id}, \nu_{t/K\sigma^2}^{K,R} \rangle \quad \text{and} \quad \mu_t^{K,R} := \left(h_{\frac{1}{\sigma\sqrt{K}}} \circ \tau_{-z_t^K}\right) \sharp \nu_{t/K\sigma^2}^K$$

where for all $i \in \{1, \dots, K\}$, $\nu_t^{K,R}$ is defined on $M_1(\mathbb{T}_R)$ from $b_R$ and $\theta_R$ as $\nu_t^K$ was defined from $b$ and $\theta$ and $z_t^{K,R} \in \mathbb{T}_R$. Note that $\mu_t^{K,R}$ takes values in $M_1(\mathbb{T}_R^{K,\sigma})$ where $\mathbb{T}_R^{K,\sigma}$ is the torus corresponding to the interval $[\frac{1}{\sigma\sqrt{K}}(x_0 - 2R - z_t^K), \frac{1}{\sigma\sqrt{K}}(x_0 + 2R - z_t^K)]$. However, we will identify in the sequel $M_1(\mathbb{T}_R^{K,\sigma})$ as a subset of $M_1(\mathbb{R})$ using the natural embedding of $\mathbb{T}_R^{K,\sigma}$ in $\mathbb{R}$. So, $\mu_t^{K,R} \in M_1(\mathbb{R})$. We define also the stopping times

$$\hat{\tau}^{K,R} := \inf \left\{ t \ge 0 \mid M_2(\mu_t^{K,R}) \ge K^\epsilon \right\},$$

$$\check{\tau}^{K,R} := \inf \left\{ t \ge 0 \mid \mathrm{Diam} \left( \mathrm{Supp} \, \nu_t^{K,R} \right) > \frac{1}{\sigma K^{\frac{3+\epsilon}{2}}} \right\},$$
such that for all $K \ge K_1$, we have that

$$
\begin{align*}
& \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \sup_{0 \le t \le s_0^K} M_2^{K,+}(t) \ge u_{\ell+1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \sup_{0 \le t \le s_0^K} \left\{ 2u_\ell + 1 - \frac{bu_{\ell-1}}{K^2 \sigma_K^2} t + \mathrm{Mart}_t^{K,+} \right\} \ge u_{\ell+1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \sup_{0 \le t \le s_0^K} \mathrm{Mart}_t^{K,+} \ge u_\ell - 1 \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \sup_{0 \le t \le s_0^K} \left| \mathrm{Mart}_t^{K,+} \right| \ge \frac{3^\ell K^{\frac{\ell}{2}}}{2} \right) \\
& \le 4 \mathbb{E}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{s_0^K} \right) \times \frac{4}{3^{2\ell} K^\varepsilon} \\
& \le \frac{C_1}{K^\varepsilon}
\end{align*}
$$

where we use (38) in the first inequality, DOOB's maximal inequality in the fourth inequality and (37) in the last inequality. Now, let us consider $K_2 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_2^{\varepsilon/2} \ge 4$. Using again (38), DOOB's maximal inequality, (37), and the definition of $s_0^K$, we deduce that there exists a constant $C_2 > 0$ such that for all $K \ge K_2$, we have that

$$
\begin{align*}
& \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le l \le s_0^K} M_2^{K,+}(t) < u_{l-1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le l \le s_0^K} \left\{ 2u_{l-1} - 1 - \frac{5\bar{b}u_{l-1} + \theta m_2}{2K^2 \sigma_K^2} t + \mathrm{Mart}_t^{K,+} \right\} < u_{l-1} \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \{T_1^{K,+} < s_0^K\} \cap \left\{ \inf_{0 \le l \le s_0^K} \mathrm{Mart}_t^{K,+} < -\frac{u_{l-1}}{2} + 1 \right\} \right) \\
& \le \mathbb{P}_{\mu_0^K} \left( \sup_{0 \le l \le s_0^K} \left| \mathrm{Mart}_t^{K,+} \right| > \frac{u_{l-1}}{2} - 1 \right) \\
& \le \frac{C_2}{K^{\varepsilon}}.
\end{align*}
$$

Hence, for all $K \ge \max\{K_1, K_2\}$, $\mathbb{P}_{\mu_0^K}(T_1^{K,+} < s_0^K) \le \frac{C_1+C_2}{K^\varepsilon}$. Finally, from (38) and MARKOV'S inequality and the definition of $t_0^K$, note that there exists a constant $C_3 > 0$ such that

$$
\begin{align*}
\mathbb{P}_{\mu_0^K} (T_1^{K,+} > t_0^K) &\le \mathbb{P}_{\mu_0^K} \left( \left\{ \sup_{0 \le t \le t_0^K} M_2^{K,+}(t) < u_{\ell+1} \right\} \cap \left\{ \inf_{0 \le t \le t_0^K} M_2^{K,+}(t) \ge u_{\ell-1} \right\} \right) \\
&\le \mathbb{P}_{\mu_0^K} \left( \mathrm{Mart}_{t_0^K}^{K,+} \ge u_{\ell-1} - 2u_\ell - 1 + \frac{bu_{\ell-1}}{K^2 \sigma_K^2} t_0^K \right) \\
&\le \frac{\mathbb{E}_{\mu_0^K} \left( \langle \mathrm{Mart}^{K,+} \rangle_{t_0^K} \right)}{(u_{\ell-1} - 1)^2} \\
&\le \frac{C_3}{K^\varepsilon}.
\end{align*}
$$
and $\tau^{K,R} := \tilde{\tau}^{K,R} \wedge \hat{\tau}^{K,R}$. We define on $M_m^T(M_1(\mathbb{R}))$ the sequence of random measures $(\Gamma^{K,R})_{K \in \mathbb{N}^*}$ as follows

$$ \Gamma^{K,R}(dt, d\mu) = \delta_{\mu_t^{K,R}}(d\mu) dt. $$

In the sequel, we study the limit as $K$ tends to $+\infty$ of the sequence of laws of $((z^{K,R}, \Gamma^{K,R}))_{K \in \mathbb{N}^*}$ in $M_1(\mathbb{D}([0,T], \mathbb{T}_R) \times M_m^T(M_1(\mathbb{R})))$.

**Theorem 5.1.** Let $T, R > 0$. The sequence of laws of $((z^{K,R}, \Gamma^{K,R}))_{K \in \mathbb{N}^*}$ is tight in the set of probability measures on $\mathbb{D}([0,T], \mathbb{T}_R) \times M_m^T(M_1(\mathbb{R}))$ and for any limiting value $\mathbb{Q}$ of this sequence, the canonical process $(\zeta^R, \Gamma^R)$ on $\mathbb{D}([0,T], \mathbb{T}_R) \times M_m^T(M_1(\mathbb{R}))$ satisfies that for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$,

$$ N_t^{f,R} := f(\zeta_t^R) - f(\zeta_0^R) - \int_0^t \int_{M_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\zeta_s^R, \mu) \Gamma^R(ds, d\mu) = 0 \quad (37) $$

$\mathbb{Q}$-a.s. Moreover, for all $t \in [0, T]$,

$$ \mathbb{E}_{\mathbb{Q}} \left( \int_0^t \int_{M_1(\mathbb{R})} M_5(\mu) \Gamma^R(ds, d\mu) \right) < \infty, \quad (38) $$

so that the definition (37) makes sense.

This result is based on Propositions 3.1 and 4.7, Lemma 4.2. The proof is divided in eight steps. In Step 1, we establish the tightness of the family of laws of the stopped slow component $(z_{\bullet \wedge \tau^{K,R}}^{K,R})_{K \in \mathbb{N}^*}$. In Step 2, we establish a compact containment condition for the stopped fast component $(\mu_{\bullet \wedge \tau^{K,R}}^{K,R})_{K \in \mathbb{N}^*}$. In Step 3, we prove the tightness of the family of laws of the occupation measure. In Step 4, we deduce the tightness of the family of laws of the couple (slow, occupation measure fast). In Step 5, we prove uniform integrability results for a family $(N_t^{f,K,R})_{t \in [0,T], K \in \mathbb{N}^*}$ constructed from $z^{K,R}$ and $\Gamma^{K,R}$ similarly as $N_t^{f,R}$ in (37). Step 6 is devoted to establish the convergence in distribution of $(N_{\bullet \wedge \tau^{K,R}}^{f,K,R})_{K \in \mathbb{N}^*}$ to $N^{f,R}$. In Steps 7 and 8, we prove that the limit $N^{f,R}$ is null $\mathbb{Q}$-a.s.

The main modification of KURTZ's setting of [49, Theorem 2.1] is that we have to work with stopped times and need to be careful with moment estimates and uniform integrability properties. This led us to rewrite the proof.

**Proof of Theorem 5.1. Step 1. Tightness of the family of laws of $\{z_{\bullet \wedge \tau^{K,R}}^{K,R}\}_{K \in \mathbb{N}^*}$ on $\mathbb{D}([0,T], \mathbb{T}_R)$**. Let $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$. For all $K \in \mathbb{N}^*$ and for all $t \in [0, T]$, let us consider the two processes $Y_t^{K,R} := f(z_{t \wedge \tau^{K,R}}^{K,R})$ and $Z_t^{K,R}$ defined by the relation

$$ Y_t^{K,R} = Y_0^{K,R} + \int_0^t Z_s^{K,R} ds + M_{t \wedge \tau^{K,R}}^{f,K,R} $$

where $M_{t \wedge \tau^{K,R}}^{f,K,R}$ is the martingale given by (18) in the torus case. Note that, for all $s > \tau^{K,R}$, $Z_s^{K,R} = 0$. From Proposition 3.2, (30) and Lemma 4.2, note that there exists a constant $C > 0$
Hence, for all $K \ge \max\{K_1, K_2\}$, we deduce that $q \le \varepsilon_K$ where

$$ \varepsilon_K := \exp(-u_\ell + 1) \times \frac{1 - \exp(-2u_\ell - 1)}{1 - \exp(-u_\ell + 1)} + \frac{\exp(u_\ell) + \exp(-2u_\ell)}{\exp(u_\ell) - \exp(-2u_\ell)} \times \frac{C_1}{K^{\frac{\varepsilon}{2}}} $$

and thus, for all $K \ge \max\{K_1, K_2\}$,

$$ \mathbb{P}_{\mu_0^K} (\{T_1^K < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\}) = \mathbb{P}_{\mu_0^K} (\{T_1^{K,+} < \check{\tau}^K\} \cap \{L_1^K = \ell + 1\}) \le q \le \varepsilon_K. $$

**Step 2.** Control of the second right-hand term of (36). Similarly, let us consider $K_3 \in \mathbb{N}^*$ independent of $\ell$ satisfying $K_3^{\varepsilon/2} \ge 2$. Using Corollary 4.12, Doob's maximal inequality and Lemma 4.7 in similar way to Step 1, there exists a constant $C_4 > 0$ such that for all $K \ge K_3$ we have that

$$ \begin{aligned} \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^K < \check{\tau}^K \right\} \cap \left\{ \sup_{0 \le t \le K \sigma_K^2 \wedge \check{\tau}^K} M_2(\mu_t^K) \ge u_{\ell+1} \right\} \right) &\le \frac{16 \mathbb{E}_{\mu_0^K} \left( \left\langle M^{K, P_{\mathrm{id}2,1}} \right\rangle_{K \sigma_K^2 \wedge \check{\tau}^K} \right)}{3^{2\ell} K^{\varepsilon}} \\ &\le \frac{C_4}{K^{1+\varepsilon}}. \end{aligned} $$

Let us consider $K_4 \in \mathbb{N}^*$ independent of $\ell$ satisfying the relation $u_{\ell-1} > 1 + \frac{5bu_{\ell+1}+\theta m_2}{2K_4}$. As previously, we establish that there exists a constant $C_5 > 0$ such that for all $K \ge K_4$ we have that

$$ \begin{aligned} \mathbb{P}_{\mu_0^K} \left( \left\{ T_1^K < \check{\tau}^K \right\} \cap \left\{ \inf_{0 \le t \le K \sigma_K^2 \wedge \check{\tau}^K} M_2(\mu_t^K) < u_{\ell-1} \right\} \right) &\le \frac{4 \mathbb{E}_{\mu_0^K} \left( \left\langle M^{K, P_{\mathrm{id}2,1}} \right\rangle_{K \sigma_K^2 \wedge \check{\tau}^K} \right)}{\left( u_{\ell-1} - 1 - \frac{5bu_{\ell+1}+\theta m_2}{2K} \right)^2} \\ &\le \frac{C_5}{K^{1+\varepsilon}}. \end{aligned} $$

To conclude this proof, there exists a constant $C > 0$ such that for all $K \ge K_0$ where $K_0 := \max_{i \in \{1, \dots, 4\}} K_i$,

$$ \mathbb{P}_{\mu_0^K} \left( \left[ \left\{ L_1^K = \ell + 1 \right\} \cup \left\{ T_1^K < K \sigma_K^2 \right\} \right] \cap \left\{ T_1^K < \check{\tau}^K \right\} \right) < \frac{1}{2} - \eta_K $$

for any $\eta_K \ge \frac{1}{2} - \epsilon_K - \frac{C_4+C_5}{K^{1+\varepsilon}}$. A convenient choice is given by $\eta_K := \frac{1}{2} - \frac{C}{K^{\varepsilon/2}}$ for $K$ large enough independent of $\ell$ which completes the proof. $\square$

### 4.4.2 Construction of the coupling

The goal of this section is to construct a coupling between $(L_k^K)_{k \in \mathbb{N}}$ and a biased random walk on $\mathbb{N}^*$ and reflected in 1. To do this, we will construct a sequence $(\xi_k^K)_{k \in \mathbb{N}}$ of i.i.d. random variables with values in $\{-1, 1\}$ as follows.

**Step 1.** First of all, thanks to Lemma 4.11 we can construct the random variable $\xi_0^K \in \{-1, 1\}$ such that

$$ \{\xi_0^K = -1\} \subset \left[ \left\{ L_1^K = L_0^K - 1 \right\} \cap \left\{ K \sigma_K^2 \le T_1^K < \check{\tau}^K \right\} \right] \cup \left\{ T_1^K \ge \hat{\tau}^K \right\} $$
such that

$$
\begin{align*}
\sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ \left( \int_0^T |Z_t^{K,R}|^2 dt \right)^{\frac{1}{2}} \right] & \\
&\le \sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ 1 + \int_0^T |Z_t^{K,R}|^2 dt \right] & \\
&\le C \left\{ 1 + \sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ \int_0^{T \wedge \tau^{K,R}} \left( M_2^2 (\mu_t^{K,R}) + \frac{1}{K^2} [1 + M_2^2 (\mu_t^{K,R})] \right) dt \right] \right\} & \\
&< \infty. &
\end{align*}
$$

Hence, from [34, Theorem 3.9.4], the family of laws of $\{Y^{K,R}\}_{K \in \mathbb{N}^*}$, on $\mathbb{D}([0, T], \mathbb{T}_R)$, is tight. Let us observe that the compact containment condition is satisfied by the stopped slow component $(z_{\bullet \wedge \tau^{K,R}}^{K,R})_{K \in \mathbb{N}^*}$ since $\mathbb{T}_R$ is compact and

$$
\forall T > 0, \quad \inf_{K \in \mathbb{N}^*} \mathbb{P} \left( \sup_{0 \le t \le T} z_{t \wedge \tau^{K,R}}^{K,R} \in \mathbb{T}_R \right) = 1. \quad (39)
$$

As $\mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$ is a dense subset of $\mathcal{C}_b^0(\mathbb{T}_R, \mathbb{T}_R)$ in the topology of the uniform norm, we deduce from (39) and [34, Theorem 3.9.1] that the family of laws of $\{z_{\bullet \wedge \tau^{K,R}}^{K,R}\}_{K \in \mathbb{N}^*}$ is tight on $\mathbb{D}([0, T], \mathbb{T}_R)$.

**Step 2. Compact containment condition.**

**Lemma 5.2.** Let $R > 0$. For all $T \ge 0$, the family of laws of the marginal random variables of the stopped fast process $\{\mu_{t \wedge \tau^{K,R}}^{K,R}\}_{t \in [0,T], K \in \mathbb{N}^*}$ is tight on $M_1(\mathbb{R})$ i.e.

$$
\forall \eta > 0, \exists D_{\varepsilon, \eta}^{R,T} \subset \mathcal{M}_1(\mathbb{R}) \text{ compact, } \forall t \in [0,T], \forall K \in \mathbb{N}^*, \mathbb{P}(\mu_{t \wedge \tau^{K,R}}^{K,R} \in D_{\varepsilon, \eta}^{R,T}) \ge 1 - \eta.
$$

*Proof.* Let $\eta > 0$ be fixed. From Proposition 4.7, there exists $K_0 \in \mathbb{N}^*$ large enough such that for all $K \ge K_0$, for all $t \in [0, T]$, $\mathbb{P}(t > \tau^{K,R}) \le \frac{\eta}{2}$. We consider the $\mathbb{R}$-valued sequence $(a_{p,\eta})_{p \in \mathbb{N}}$ satisfying for all $p \in \mathbb{N}$, $a_{p,\eta} > 1$, and $\sum_{p \in \mathbb{N}^*} \frac{p}{a_{p,\eta}^2} < \frac{\eta}{2M_2}$ where $\bar{M}_2 := \max\{C_2^*, \bar{\theta}\bar{m}_2\}$ is a uniform upper bound of $\mathbb{E}(M_2(\mu_{t \wedge \tau^{K,R}}^{K,R}))$ given by Corollary 4.5. Let $(\kappa_{p,\eta})_{p \in \mathbb{N}}$ be a sequence of compact on $\mathbb{R}$, increasing for inclusion, of the form $[-a_{p,\eta}, a_{p,\eta}]$. Let $t \in [0, T]$. For all $p \in \mathbb{N}^*$, and $K \ge K_0$, note that

$$
M_2(\mu_t^{K,R}) \mathbb{1}_{t \le \tau^{K,R}} \ge \mathbb{1}_{t \le \tau^{K,R}} \int_{\kappa_{p,\eta}^c} x^2 \mu_t^{K,R}(dx) \ge a_{p,\eta}^2 \mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{p,\eta}^c) \mathbb{1}_{t \le \tau^{K,R}},
$$

and we deduce that $\{\mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{p,\eta}^c) > \frac{1}{p}\} \cap \{t \le \tau^{K,R}\} \subset \{M_2 (\mu_{t \wedge \tau^{K,R}}^{K,R}) > \frac{a_{p,\eta}^2}{p}\} \cap \{t \le \tau^{K,R}\}$.
and $\mathbb{P}(\xi_0^K = -1) = \frac{1}{2} + \eta_K$.

Step 2. Then, thanks to Lemma 4.11 again and after having applied the MARKOV property at time $T_1^K$, we can construct the random variable $\xi_1^K \in \{-1, 1\}$ such that, conditionally to $\mathcal{F}_{T_1^K} \lor \sigma(\xi_0^K)$,

$$ \{\xi_1^K = -1\} \subset \left[ \{L_2^K = L_1^K - 1\} \cap \{K\sigma_K^2 + T_1^K \le T_2^K < \check{\tau}^K\} \right] \cup \{T_2^K \ge \hat{\tau}^K\} $$

and $\mathbb{P}(\xi_1^K = -1 | \mathcal{F}_{T_1^K} \lor \sigma(\xi_0^K)) = \frac{1}{2} + \eta_K$. Note that this implies in particular that $\xi_1^K$ is independent of $\xi_0^K$.

Step 3. By induction on $k \in \mathbb{N}$, if $(\xi_i^K)_{0 \le i \le k}$ are constructed, then thanks to Lemma 4.11 again and after having applied the MARKOV property at time $T_{k+1}^K$, we can construct the random variable $\xi_{k+1}^K \in \{-1, 1\}$ such that, conditionally to $\mathcal{F}_{T_{k+1}^K} \lor \sigma(\xi_0^K, \dots, \xi_k^K)$,

$$ \{\xi_{k+1}^K = -1\} \subset \left[ \{L_{k+2}^K = L_{k+1}^K - 1\} \cap \{K\sigma_K^2 + T_{k+1}^K \le T_{k+2}^K < \check{\tau}^K\} \right] \cup \{T_{k+2}^K \ge \hat{\tau}^K\} $$

and $\mathbb{P}(\xi_1^K = -1 | \mathcal{F}_{T_{k+1}^K} \lor \sigma(\xi_0^K, \dots, \xi_k^K)) = \frac{1}{2} + \eta_K$.

To conclude, we set

$$ \forall K \in \mathbb{N}^*, \quad \forall k \in \mathbb{N}, \quad Z_{k+1}^K - Z_k^K := \xi_k^K \quad \text{and} \quad Z_0^K = L_0^K \quad (39) $$

and the sequence $(Z_k^K)_{k \in \mathbb{N}}$ satisfies by construction the following lemma:

**Lemma 4.13.** The sequence $(Z_k^K)_{k \in \mathbb{N}}$, given by (39) is a biased simple random walk on $\mathbb{N}^*$, reflected in 1 and for all $k \in \mathbb{N}$ as long as $T_{k+1}^K < \check{\tau}^K$, that

$$ (1) \quad Z_{k+1}^K - Z_k^K \ge L_{k+1}^K - L_k^K $$

$$ (2) \quad T_{k+1}^K - T_k^K \ge K\sigma_K^2 \quad \text{when} \quad Z_{k+1}^K - Z_k^K = -1. $$

On the one hand, note that for all $k \in \mathbb{N}$, $L_k^K \le Z_k^K$. On the other hand, note that Lemma 4.10 implies for all $k \in \mathbb{N}$, $L_{k+1}^K - L_k^K \in \{\dots, -3, -2, -1, 1\}$ which justifies (1) of Lemma 4.13. In Figure 4.7, we illustrate the coupling between $(L_k^K)_{0 \le k \le 7}$ and $(Z_k^K)_{0 \le k \le 7}$. In this figure, we observe that $\xi_k^K = L_{k+1}^K - L_k^K$ for all $k \in \{0, 1, 2, 4, 5, 6\}$ and $\xi_3^K = -1 \ge -2 = L_4^K - L_3^K$. This illustrates that $Z_{k+1}^K - Z_k^K$ is always 1 when $L_{k+1}^K - L_k^K$ is equal to 1. Note that $Z_{k+1}^K - Z_k^K$ can be 1 when $L_{k+1}^K - L_k^K$ is equal to $-1$.

### 4.4.3 Estimates of exit from an attractive domain for random walks

Let us consider $N^K$ the number of transitions before reaching $\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ for the random walk $(Z_k^K)_{k \in \mathbb{N}}$.

**Remark 4.14.** By Lemma 4.13, $N^K \le k_0$ where $k_0$ is the first integer such that $L_{k_0}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$, so $\hat{\tau}^K \ge T_{k_0}^K \ge T_{N^K}^K$.
Hence,

$$
\begin{align*}
\mathbb{P}\left(\left\{\exists p \in \mathbb{N}^*, \mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{p,\eta}^c) > \frac{1}{p}\right\}\right) &\le \mathbb{P}\left(\tau^{K,R} < t\right) + \sum_{p \in \mathbb{N}^*} \mathbb{P}\left(\left\{\mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{p,\eta}^c) > \frac{1}{p}\right\} \cap \left\{t \le \tau^{K,R}\right\}\right) \\
&\le \mathbb{P}\left(\tau^{K,R} < t\right) + \sum_{p \in \mathbb{N}^*} \mathbb{P}\left(M_2\left(\mu_t^{K,R}\right) \mathbb{1}_{t \le \tau^{K,R}} > \frac{a_{p,\eta}^2}{p}\right) \\
&\le \mathbb{P}\left(\tau^{K,R} < t\right) + \sum_{p \in \mathbb{N}^*} \frac{\mathbb{E}\left(M_2\left(\mu_t^{K,R}\right) \mathbb{1}_{t \le \tau^{K,R}}\right)}{\frac{a_{p,\eta}^2}{p}} \\
&\le \eta.
\end{align*}
$$

Therefore, we have proved that

$$
\forall \eta > 0, \forall t \in [0, T], \forall K \geq K_0, \quad \mathbb{P} \left( \mu_{t \wedge \tau^{K,R}}^{K,R} \in \mathcal{K}_{\eta} \right) \geq 1 - \eta,
$$

where $\mathcal{K}_\eta := \{\mu \in \mathcal{M}_1(\mathbb{R}) \mid \forall p \in \mathbb{N}^*, \mu(\kappa_{p,\eta}^c) \le \frac{1}{p}\}$ which is compact by PROHOROV'S theorem. $\square$

**Step 3. Tightness of the family of laws of {Γ<sup>K,R</sup>}<sub>K∈N<sup>★</sup></sub> on M<sup>T</sup><sub>m</sub>(M<sub>1</sub>(R)).** Let η > 0 be fixed. From Proposition 4.7, there exists K<sub>0</sub> ∈ N<sup>★</sup> such that for all K ≥ K<sub>0</sub>, for all t ∈ [0, T], P(t ≥ τ<sup>K,R</sup>) ≤ η/2. Consider D<sup>R,T</sup><sub>ε,η/2</sub> the compact set in Lemma 5.2. It follows that for all t ∈ [0, T], K ≥ K<sub>0</sub>,

$$
\begin{align*}
\mathbb{E} \left( \Gamma^{K,R} \left( [0,t] \times D_{\varepsilon, \frac{\eta}{2}}^{R,T} \right) \right) &\ge \mathbb{E} \left( \Gamma^{K,R} \left( [0,t] \times D_{\varepsilon, \frac{\eta}{2}}^{R,T} \right) \mathbb{1}_{t < \tau^{K,R}} \right) \\
&= \int_0^t \mathbb{P} \left( \left\{ \mu_{s \wedge \tau^{K,R}}^{K,R} \in D_{\varepsilon, \frac{\eta}{2}}^{R,T} \right\} \cap \left\{ t < \tau^{K,R} \right\} \right) \mathrm{d}s \\
&\ge \int_0^t \mathbb{P} \left( \mu_{s \wedge \tau^{K,R}}^{K,R} \in D_{\varepsilon, \frac{\eta}{2}}^{R,T} \right) \mathrm{d}s - \int_0^t \mathbb{P} \left( t \ge \tau^{K,R} \right) \mathrm{d}s \\
&\ge t (1 - \eta).
\end{align*}
$$

Therefore, the tightness of the family of laws of {Γ<sup>K,R</sup>}<sub>K∈N<sup>★</sup></sub> follows from [49, Lemma 1.3].

**Step 4. Tightness of the family of laws of {⟨z<sup>K,R</sup>, Γ<sup>K,R</sup>⟩}<sub>K∈N<sup>★</sup></sub> on D([0, T], T<sub>R</sub>) × M<sup>T</sup><sub>m</sub>(M<sub>1</sub>(R)).** From Steps 1 and 3 and PROHOROV'S theorem, we deduce that the family of laws of the couple {⟨z<sup>K,R</sup><sub>Λ&tau;K,R</sub>, Γ<sup>K,R</sup>⟩}<sub>K∈N<sup>★</sup></sub> is relatively compact in M<sub>1</sub>(D([0, T], T<sub>R</sub>) × M<sup>T</sup><sub>m</sub>(M<sub>1</sub>(R))). Hence, there exists a probability measure Q on the canonical space D([0, T], T<sub>R</sub>) × M<sup>T</sup><sub>m</sub>(M<sub>1</sub>(R)) and an increasing function n: N<sup>★</sup> → N<sup>★</sup> such that the subsequence of laws of {⟨z<sup>n(K),R</sup><sub>Λ&tau; n(K),R</sub>, Γ<sup>n(K),R</sup>⟩}<sub>K∈N<sup>★</sup></sub> converges weakly to the limiting value Q when K → +∞. Thanks to Proposition 4.7, we deduce that the family of laws of {⟨z<sup>n(K),R</sup>, Γ<sup>n(K),R</sup>⟩}<sub>K∈N<sup>★</sup></sub> converges weakly to Q when K → +∞ and therefore that the family of laws of {⟨z<sup>K,R</sup>, Γ<sup>K,R</sup>⟩}<sub>K∈N<sup>★</sup></sub> is relatively compact, thus tight on D([0, T], T<sub>R</sub>) × M<sup>T</sup><sub>m</sub>(M<sub>1</sub>(R)) by PROHOROV'S theorem.
<figure><img src="image_9.png" /><figcaption>Figure 4.7: For $K$ large enough, coupling between $(L_k^K)_{0 \le k \le 7}$ and $(Z_k^K)_{0 \le k \le 7}$ up to time $T$ and before $\tilde{\tau}^K$ and where $M_2(\mu_0^K) \in [0, 2u_1 + 1)$ so that $L_0^K = 1$ and $\tilde{Z}_k^K := u_{Z_k^K} = 3^{Z_k^K} K^{\epsilon/2}$ for all $k \in \mathbb{N}$.</figcaption></figure>

The following lemma gives an estimate on the problem of exit from a domain for $Z^K$. For all $k \in \mathbb{N}^*$, we denote by $\mathbf{P}_k$ the law of the MARKOV chain $Z^K$ given $Z_0^K = L_0^K = k$.

**Lemma 4.15.** We have,

$$ \lim_{K \to +\infty} \mathbf{P}_1 \left( N^K \ge \exp \left( \frac{\varepsilon^2}{16 \log(3)} \log^2(K) \right) \right) = 1. $$

*Proof.* For all $k \in \mathbb{N}^*$, let us consider the stopping time $\mathcal{T}_k^K := \inf \{n \in \mathbb{N} \mid Z_n^K = k\}$ and we set $v_k^K := \mathbf{P}_k \left( \mathcal{T}_1^K < \mathcal{T}_{\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor}^K \right)$. Note that $v_1^K = 1$. By a classical approach and setting $r_K := \frac{1+2\eta_K}{1-2\eta_K} > 1$, we prove that

$$ \begin{aligned} v_k^K &= \frac{1 - r_K^{-\left(\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - k\right)}}{1 - r_K^{-\left(\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1\right)}} \\ &= 1 - (r_K^{k-1} - 1) \exp \left( - \left( \left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1 \right) \log(r_K) \right). \end{aligned} $$

Note that $N^K$ is greater than the number of transitions from 2 to 1 that occur before $\mathcal{T}_{\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor}^K$.

Hence, under $\mathbf{P}_1$, $N^K \ge X$ where $X$ is random variable with geometric law of parameter $q_K := (r_K - 1) \exp \left( - \left( \left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1 \right) \log(r_K) \right)$. Since $r_K \sim_{K \to +\infty} \frac{K^{\epsilon/2}}{C}$, with $C > 0$ given by Lemma 4.11, $q_K \to 0$ when $K \to +\infty$ and using that for all $x > 0$ small enough,
**Step 5. Uniform integrability.** For all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, let us consider $(N_t^{f,K,R})_{t \in [0,T]}$ the stochastic process defined by

$$N_t^{f,K,R} := f(z_t^{K,R}) - f(z_0^{K,R}) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(z_s^{K,R}, \mu) \Gamma^{K,R}(ds, d\mu).$$

From (9), we have for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, there exists a constant $C > 0$ such that

$$\forall t \in [0, T], \forall K \in \mathbb{N}^*, \quad |N_{t \wedge \tau^{K,R}}^{f,K,R}| \le C \left(1 + \int_0^t M_2(\mu_s^{K,R}) 1_{s \le \tau^{K,R}} ds\right),$$

Hence, the uniform integrability of $(N_{\bullet \wedge \tau^{K,R}}^{f,K,R})_{K \in \mathbb{N}^*}$ follows from Corollary 4.3.

**Step 6. Proof of (38) and almost sure convergence of $(\tilde{N}_{\bullet \wedge \tau^{K,R}}^{f,K,R})_{K \in \mathbb{N}^*}$ to $\tilde{N}^{f,R}$**

From Step 4, Proposition 4.7 and SKOROHOD's representation theorem, there exists an increasing function $\tilde{n}: \mathbb{N}^* \to \mathbb{N}^*$ and a probability space on which we define, the random variable $\tilde{\tau}^{\tilde{n}(K),R}$, the families $\{\tilde{z}^{\tilde{n}(K),R}\}_{K \in \mathbb{N}^*}$ and $\{\tilde{\Gamma}^{\tilde{n}(K),R}\}_{K \in \mathbb{N}^*}$ and $\tilde{\zeta}^R, \tilde{\Gamma}^R$ copies of $\{z^{\tilde{n}(K),R}\}_{K \in \mathbb{N}^*}$, $\{\tilde{\Gamma}^{\tilde{n}(K),R}\}_{K \in \mathbb{N}^*}, \tilde{\zeta}^R, \tilde{\Gamma}^R$ under $\mathbb{Q}$ such that for all $K \in \mathbb{N}^*$, the sequence $(\tilde{z}^{\tilde{n}(K),R}, \tilde{\Gamma}^{\tilde{n}(K),R}, \tilde{\tau}^{\tilde{n}(K),R})_{K \in \mathbb{N}^*}$ converges a.s. to $(\tilde{\zeta}^R, \tilde{\Gamma}^R, +\infty)$ when $K \to +\infty$. Note that for all $t \in [0, T]$,

$$\tilde{\Gamma}^{\tilde{n}(K),R} ([0, t \wedge \tilde{\tau}^{\tilde{n}(K),R}] \times \mathcal{M}_1(\mathbb{R})) = t \wedge \tilde{\tau}^{\tilde{n}(K),R} \xrightarrow{K \to +\infty} t = \tilde{\Gamma} ([0, t] \times \mathcal{M}_1(\mathbb{R})).$$

From [49, Lemma 1.5 (b)(c)(d)] and Corollary 4.3, we have

$$\int_0^{t \wedge \tilde{\tau}^{K,R}} \int_{\mathcal{M}_1(\mathbb{R})} M_5(\mu) \tilde{\Gamma}^{K,R}(ds, d\mu) \xrightarrow{K \to +\infty} \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} M_5(\mu) \tilde{\Gamma}(ds, d\mu)$$

and (38) follows. From [49, Lemma 1.5 (b)(c)(d)], we also deduce that for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, for all $t \in [0, T]$, a.s.

$$\lim_{K \to +\infty} \int_0^{t \wedge \tilde{\tau}^{\tilde{n}(K),R}} \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{z}_s^{\tilde{n}(K),R}, \mu) \tilde{\Gamma}^{\tilde{n}(K),R}(ds, d\mu) \\ = \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{\zeta}^R, \mu) \tilde{\Gamma}^R(ds, d\mu)$$

and thus the sequence $(\tilde{N}_{\bullet \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f,\tilde{n}(K),R})_{K \in \mathbb{N}^*}$ converges a.s. to the process $(\tilde{N}_t^{f,K,R})_{t \in [0,T]}$ where $\tilde{N}_t^{f,K,R}$ and $\tilde{N}_t^{f,R}$ are respectively defined for all $K \in \mathbb{N}^*$ by

$$\tilde{N}_t^{f,K,R} := f(\tilde{z}_t^{K,R}) - f(\tilde{z}_0^{K,R}) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{z}_t^{K,R}, \mu) \tilde{\Gamma}^{K,R}(ds, d\mu), \quad t \le T \wedge \tilde{\tau}^{K,R}$$

$$\tilde{N}_t^{f,R} := f(\tilde{\zeta}_t^R) - f(\tilde{\zeta}_0^R) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{\zeta}_t^R, \mu) \tilde{\Gamma}^R (ds, d\mu), \quad t \le T.$$

**Step 7. $N^{f,R}$ is a martingale.** Let us consider the filtration $(\tilde{\mathcal{F}}_t^{K,R})_{t \in [0,T]}$ defined by $\tilde{\mathcal{F}}_t^{K,R} := \sigma(\tilde{z}_s^{K,R}, \tilde{\Gamma}^{K,R}([0, s] \times H) | s \le t, H \in \mathcal{B}(\mathcal{M}_1(\mathbb{R}))), (f_i)_{1 \le i \le p}, p \in \mathbb{N}^*$ bounded LIPS-CHITZ functions from $\mathbb{T}_R$ to $\mathbb{T}_R$ and $0 < t_1 \le \dots \le t_p \le s < t$. Let us denote for all $K \in \mathbb{N}^*$, for
$\log(1-x) \ge -2x$ we deduce that for all $K$ large enough,

$$
\begin{aligned}
& \mathbf{P}_1 \left( X \ge \exp \left( \frac{\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1}{2} \log(r_K) \right) \right) \\
& = \exp \left( \exp \left( \frac{\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1}{2} \log(r_K) \right) \log(1 - q_K) \right) \\
& \ge \exp \left( -2(r_K - 1) \exp \left( -\frac{\left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor - 1}{2} \log(r_K) \right) \right).
\end{aligned}
$$

Using again that $r_K \sim_{K \to +\infty} \frac{K^{\varepsilon/2}}{C}$, we deduce that for $K$ large enough,

$$
\frac{\left\lfloor \frac{\varepsilon}{2 \log(3)} \log(K) \right\rfloor - 1}{2} \log(r_K) \ge \frac{\varepsilon^2}{16 \log(3)} \log^2(K)
$$

and the announced result follows.

<figure><img src="image_10.png" /></figure>

It is in the next Corollary 4.16 that we see the importance of Assumption (5) where $\sigma_K$ must not be too small.

**Corollary 4.16.** For all $T > 0$,

$$
\lim_{K \to +\infty} \mathbb{P} \left( \frac{K \sigma_K^2}{2} \left( N^K - \frac{\varepsilon}{2 \log(3)} \log(K) \right) < T \right) = 0.
$$

*Proof.* From Lemma 4.15 and Assumption (5) we deduce in a straightforward manner that

$$
\lim_{K \to +\infty} \mathbf{P}_1 \left( \frac{K \sigma_K^2}{2} \left( N^K - \frac{\varepsilon}{2 \log(3)} \log(K) \right) < T \right) = 0.
$$

As $L_0^K$ depends on $\mu_0^K$ note that $L_0^K = Z_0^K$, is function of the initial condition $\mu_0^K$. However, by
Assumption (4) and MARKOV'S inequality, we have that

$$
\mathbb{P} (L_0^K > 1) \le \mathbb{P} (M_2 (\mu_0^K) \ge K^{\frac{\epsilon}{2}}) \le \frac{C_2^*}{K^{\epsilon}}
$$

which tends to 0 when $K \to +\infty$ and the announced result follows.

<figure><img src="image_11.png" /></figure>

**4.4.4 Conclusion**

We denote respectively by $N_+^K$ and $N_-^K$ the number of upward transitions and downward tran-
sitions for $(Z_k^K)_{k \in \mathbb{N}}$ before reaching $\lfloor \frac{\varepsilon}{2 \log(3)} \log(K) \rfloor$. Let $T \ge 0$ be fixed. Note that

$$
\begin{align}
\mathbb{P} (\hat{\tau}^K \le \check{\tau}^K \wedge T) \le \mathbb{P} & (\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K > T_{NK}^K\}) \nonumber \\
& + \mathbb{P} (\{\hat{\tau}^K \le \check{\tau}^K \wedge T\} \cap \{\check{\tau}^K \le T_{NK}^K\}). \tag{40}
\end{align}
$$
all $t \le \tilde{\tau}^{K,R} \wedge T$, $\tilde{M}_t^{f,\tilde{n}(K),R} := \tilde{N}_t^{f,\tilde{n}(K),R} + \tilde{\mathcal{E}}_t^{f,\tilde{n}(K),R}$ where $\tilde{M}^{f,\tilde{n}(K),R}$ is constructed from $\tilde{z}^{K,R}$ and $\tilde{\mu}^{K,R}$ as in (18) and $\tilde{\mathcal{E}}_t^{f,\tilde{n}(K),R}$ is an error term. Note that $\tilde{M}_t^{f,\tilde{n}(K),R}$ is a $(\tilde{\mathcal{F}}_t^{K,R})_{t \in [0,T]}$-martingale as in Proposition 3.1. Hence,

$$ \mathbb{E} \left( \tilde{M}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \prod_{i=1}^{p} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) \right) = \mathbb{E} \left( \tilde{M}_{s \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \prod_{i=1}^{p} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) \right). $$

From Proposition 3.2 and (30), $\tilde{\mathcal{E}}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f,\tilde{n}(K),R} = \frac{1}{K} O(\int_0^t [1 + M_2(\tilde{\mu}_s^{K,R})] 1_{s \le \tilde{\tau}^{\tilde{n}(K),R}} ds)$ and then satisfy the condition $\lim_{K \to +\infty} \mathbb{E}(\sup_{0 \le t \le T} |\mathcal{E}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f,\tilde{n}(K),R}|) = 0$. As for all $i \in \{1, \dots, p\}$, $f_i$ is LIPSCHITZ, there exists a constant $C > 0$ such that

$$ \mathbb{E} \left( \left| f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} - \mathcal{E}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) - f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) \right| \right) \\ \le \frac{C}{\tilde{n}(K)} \mathbb{E} \left( \int_0^{t_i} [1 + M_2(\tilde{\mu}_r^{\tilde{n}(K),R})] 1_{r \le \tilde{\tau}^{\tilde{n}(K),R}} dr \right) $$

where the term of the right hand side of the previous inequality goes to 0 when $K \to +\infty$ according to Lemma 4.2. We deduce that for all $u \in \{s, t\}$,

$$ \lim_{K \to +\infty} \mathbb{E} \left( \tilde{M}_{u \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \prod_{i=1}^{p} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) \right) = \lim_{K \to +\infty} \mathbb{E} \left( \tilde{N}_{u \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \prod_{i=1}^{p} f_i \left( \tilde{N}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K), R} \right) \right). $$

From Steps 5 and 6, we deduce letting $K \to +\infty$ that

$$ \mathbb{E} \left( \tilde{N}_t^{f,R} \prod_{i=1}^{p} f_i \left( \tilde{N}_{t_i}^{f,R} \right) \right) = \mathbb{E} \left( \tilde{N}_s^{f,R} \prod_{i=1}^{p} f_i \left( \tilde{N}_{t_i}^{f,R} \right) \right). $$

Hence,

$$ \mathbb{E} \left( N_t^{f,R} \prod_{i=1}^{p} f_i \left( N_{t_i}^{f,R} \right) \right) = \mathbb{E} \left( N_s^{f,R} \prod_{i=1}^{p} f_i \left( N_{t_i}^{f,R} \right) \right). $$

Since the last property is true for all $p \in \mathbb{N}^*$, $t_1 \le \dots \le t_p \le s < t$ and for all bounded LIPSCHITZ functions $(f_i)_{1 \le i \le p}$, the monotone class theorem ensures us that

$$ \mathbb{E} \left( N_t^{f,R} \mid \sigma \left( N_u^{f,R} \mid u \le s \right) \right) = N_s^{f,R}. $$

Hence the announced result.

**Step 8. Nullity of $N^{f,R}$**. On the one hand, from Itô's formula [71, Theorem 32 of Chapter II], for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$ and $t \in [0, T]$

$$ \begin{aligned} f^2(\zeta_t^R) = f^2(\zeta_0^R) &+ 2 \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} f(\zeta_s^R) \mathcal{L}_{\text{SLOW}} f(\zeta_s^R, \mu) \Gamma^R (ds, d\mu) \\ &+ 2 \int_0^t f(\zeta_s^R) dN_s^{f,R} + \langle N^{f,R} \rangle_t + \sum_{0<s \le t} \{f(\zeta_s^R) - f(\zeta_{s-}^R)\}^2. \end{aligned} \quad (40) $$
Step 1. Control of the first right-hand term of (40). As $Z_0^K = 1$ with probability converging to 1 and $Z_{NK}^K = \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$, we have $N_+^K - N_-^K \le \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor$ and $N_+^K + N_-^K = N^K$. Thus,

$$N_-^K \ge \frac{N^K - \lfloor \frac{\varepsilon}{2\log(3)} \log(K) \rfloor}{2}.$$

Hence, if $\check{\tau}^K > T_{NK}^K$, then by Lemma 4.13 (2),

$$\hat{\tau}^K \ge T_{NK}^K \ge \sum_{i=1}^{NK} (T_i^K - T_{i-1}^K) \ge N_-^K K \sigma_K^2 \ge \frac{K \sigma_K^2}{2} \left( N^K - \frac{\varepsilon}{2\log(3)} \log(K) \right).$$

Therefore,

$$\mathbb{P} \left( \left\{ \hat{\tau}^K \le \check{\tau}^K \wedge T \right\} \cap \left\{ \check{\tau}^K > T_{NK}^K \right\} \right) \le \mathbb{P} \left( \frac{K \sigma_K^2}{2} \left( N^K - \frac{\varepsilon}{2\log(3)} \log(K) \right) \le T \right)$$

which tends to 0 when $K \to +\infty$ according to Corollary 4.16.

Step 2. Control of the second right-hand term of (40). If $\check{\tau}^K \le T_{NK}^K$, let us consider $\tilde{N}^K$ the last index $k$ such that $T_k^K < \check{\tau}^K$. We have $\tilde{N}^K < N^K$ and so $\tilde{N}^K + 1 \le \tilde{N}^K$ which implies, thanks to Lemma 4.10, that for all $t \le T_{\tilde{N}^K+1}$ and $K$ large enough,

$$M_2 \left( \mu_t^K \right) \le 2u \left\lfloor \frac{\varepsilon}{2\log(3)} \log(K) \right\rfloor + 1 \le 2K^{\frac{\varepsilon}{2}} + 1 < K^{\varepsilon}.$$

As the jump times of $\nu^K$ are isolated, we deduce that $\hat{\tau}^K > T_{\tilde{N}^K+1}^K$. Now, by definition of $\tilde{N}^K$, we have $T_{\tilde{N}^K+1}^K \ge \check{\tau}^K$ and thus $\hat{\tau}^K > \check{\tau}^K$. Hence, we obtain that

$$\mathbb{P} \left( \left\{ \hat{\tau}^K \le \check{\tau}^K \wedge T \right\} \cap \left\{ \check{\tau}^K \le T_{NK}^K \right\} \right) = \mathbb{P}(\emptyset) = 0.$$

Therefore, $\lim_{K \to +\infty} \mathbb{P} (\hat{\tau}^K \le \check{\tau}^K \wedge T) = 0$ which concludes the proof of Lemma 4.9.

# 5 Tightness on the torus

The main result of this section is given by Theorem 5.1. This is a stochastic averaging result inspired by KURTZ [57, Theorem 2.1] establishing a tightness result, in the torus case, of the sequence of laws of $((z^K, \Gamma^K))_{K \in \mathbb{N}^*}$ where $\Gamma^K$ is the occupation measure of the fast component $\mu^K$. We will use criteria proposed by ETHIER-KURTZ in [40, Theorems 3.9.1 and 3.9.4]. In Sections 6 and 7 we identify the limit and we prove its uniqueness.

Let introduce the torus $\mathbb{T}_R := [x_0 - 2R, x_0 + 2R]$ of length $4R$ with $R > 0$ fixed and $x_0$ is the value of the mean trait of $\nu_0^K$ as in Theorem 2.1. We define $M_1(\mathbb{T}_R)$ the set of probability measure on $\mathbb{T}_R$. We denote by $M_m(M_1(\mathbb{R}))$ the set of measures $\Gamma$ on $\mathbb{R}_+ \times M_1(\mathbb{R})$ such that for all $t \ge 0$, $\Gamma([0, t] \times M_1(\mathbb{R})) = t$. For any $t \ge 0$, we denote by $M_m^t(M_1(\mathbb{R}))$ the set of measures $\Gamma \in M_m(M_1(\mathbb{R}))$ restricted to $[0, t] \times M_1(\mathbb{R})$. For all $T > 0$, we denote by $\mathbb{D}([0, T], \mathbb{T}_R)$ the space of càd-làg functions on $[0, T]$ with values in $\mathbb{T}_R$.
On the other hand, applying (37) with $f^2 \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, we obtain that for all $t \in [0, T]$

$$f^2(\zeta_t^R) = f^2(\zeta_0^R) + \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f^2(\zeta_s^R, \mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu) + N_t^{f^2,R}. \quad (41)$$

Comparing (40) and (41) leads for all $t \in [0, T]$ to

$$N_t^{f^2,R} - 2 \int_0^t f(\zeta_s^R) \mathrm{d}N_s^{f,R} = \langle N^{f,R} \rangle_t + \sum_{0<s \le t} \{f(\zeta_s^R) - f(\zeta_{s-}^R)\}^2$$

and thus by [42, Theorem 4.1] that $\mathbb{Q}$-a.s. for all $t \in [0, T]$,

$$\langle N^{f,R} \rangle_t = - \sum_{0<s \le t} \{f(\zeta_s^R) - f(\zeta_{s-}^R)\}^2 \le 0$$

so that $\langle N^{f,R} \rangle_t = 0$. Hence, $\mathbb{Q}$-a.s. $N^{f,R} = 0$ which completes the proof. $\square$

# 6 Characterisation of the occupation measure limit on the torus

Consider a probability measure $\mathbb{Q}$ on $\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ and the canonical process $(\zeta^R, \Gamma^R)$ as in Theorem 5.1. The following lemma gives us a desintegration result of the occupation measure $\Gamma^R$ that we characterise below.

**Lemma 6.1.** Let $T \ge 0$ be fixed. With the notations of Theorem 5.1, there exists a random probability measure-valued process $(\gamma_t^R)_{t \in [0,T]}$ that is predictable in $(\omega, t)$ and such that for all bounded measurable function $\psi: [0, T] \times \mathcal{M}_1(\mathbb{R}) \to \mathbb{T}_R$,

$$\int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \psi(s, \mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu) = \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \psi(s, \mu) \gamma_s^R(\mathrm{d}\mu) \mathrm{d}s. \quad (42)$$

*Proof.* The desintegration result of $\Gamma^R$ follows directly from [49, Lemma 1.4]. $\square$

**Corollary 6.2.** Let $T \ge 0$ be fixed. With the notations of Theorem 5.1, we have that $\zeta^R \in \mathcal{C}^0([0, T], \mathbb{T}_R)$ is differentiable of derivative in $L^1(\mathbb{R})$ $\mathbb{Q}$-a.s.

*Proof.* Applying (37) with $f = \mathrm{id} \in \mathcal{C}_b^1(\mathbb{T}_R, \mathbb{T}_R)$, we deduce from Theorem 5.1, (9) and (42), $\mathbb{Q}$-a.s., for all $t \in [0, T]$

$$\zeta_t^R = \zeta_0^R + \int_0^t \left( \int_{\mathcal{M}_1(\mathbb{R})} M_2(\mu) \gamma_s^R(\mathrm{d}\mu) \right) \partial_1 \mathrm{Fit}(\zeta_s^R, \zeta_{s-}^R) \mathrm{d}s.$$

From Assumptions (A) and (38), the integrand of the previous time integral is in $L^1(\mathbb{R})$ $\mathbb{Q}$-a.s. Hence, the announced result follows from the fundamental theorem of calculus. $\square$

We now want to characterise the limiting value $\Gamma^R(\mathrm{d}t, \mathrm{d}\mu) = \gamma_t^R(\mathrm{d}\mu) \mathrm{d}t$ under $\mathbb{Q}$.
Let $b_R \in \mathcal{C}_b^2(\mathbb{T}_R^2, \mathbb{R})$ and $\theta_R \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{R})$ be two functions satisfying $b_R = b$ on $[x_0 - R, x_0 + R]^2$ and $\theta_R = \theta$ on $[x_0 - R, x_0 + R]$. Let $T > 0$ be fixed. We define on $\mathbb{T}_R$, in similar way as in Section 2, the processes $\nu^{K,R} := (\nu_t^{K,R})_{t \in [0,T]}$, $z^{K,R} := (z_t^{K,R})_{t \in [0,T]}$ and $\mu^{K,R} = (\mu_t^{K,R})_{t \in [0,T]}$ as follows

$$ \nu_t^{K,R} := \frac{1}{K} \sum_{i=1}^{K} \delta_{x_i^R(t)}, \quad z_t^{K,R} := \langle \mathrm{id}, \nu_{t/K \sigma_K^2}^{K,R} \rangle \quad \text{and} \quad \mu_t^{K,R} := \left( h_{\frac{1}{\sigma_K \sqrt{K}}} \circ \tau_{-z_t^K} \right) \sharp \nu_{t/K \sigma_K^2}^K $$

where for all $i \in \{1, \dots, K\}$, $\nu_t^{K,R}$ is defined on $\mathcal{M}_1(\mathbb{T}_R)$ from $b_R$ and $\theta_R$ as $\nu_t^K$ was defined from $b$ and $\theta$ and $z_t^{K,R} \in \mathbb{T}_R$. Note that $\mu_t^{K,R}$ takes values in $\mathcal{M}_1(\mathbb{T}_R^K)$ where $\mathbb{T}_R^K$ is the torus corresponding to the interval $[\frac{1}{\sigma_K \sqrt{K}}(x_0 - 2R - z_t^K), \frac{1}{\sigma_K \sqrt{K}}(x_0 + 2R - z_t^K)]$. However, we will identify in the sequel $\mathcal{M}_1(\mathbb{T}_R^K)$ as a subset of $\mathcal{M}_1(\mathbb{R})$ using the natural embedding of $\mathbb{T}_R^K$ in $\mathbb{R}$. So, $\mu_t^{K,R} \in \mathcal{M}_1(\mathbb{R})$. We define also the stopping times

$$ \begin{aligned} \hat{\tau}^{K,R} &:= \inf \left\{ t \ge 0 \mid M_2 \left( \mu_t^{K,R} \right) \ge K^\epsilon \right\}, \\ \check{\tau}^{K,R} &:= \inf \left\{ t \ge 0 \mid \mathrm{Diam} \left( \mathrm{Supp} \, \nu_t^{K,R} \right) > \frac{1}{\sigma_K K^{\frac{3+\epsilon}{2}}} \right\}, \end{aligned} $$

and $\tau^{K,R} := \check{\tau}^{K,R} \wedge \hat{\tau}^{K,R}$. We define on $\mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ the sequence of random measures $(\Gamma^{K,R})_{K \in \mathbb{N}^*}$ as follows

$$ \Gamma^{K,R}(dt, d\mu) = \delta_{\mu_t^{K,R}}(d\mu) dt. $$

In the sequel, we study the limit as $K$ tends to $+\infty$ of the sequence of laws of

$$ ((z^{K,R}, \Gamma^{K,R}))_{K \in \mathbb{N}^*} $$

in $\mathcal{M}_1(\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R})))$.

**Theorem 5.1.** Let $T, R > 0$. The sequence of laws of $((z^{K,R}, \Gamma^{K,R}))_{K \in \mathbb{N}^*}$ is tight in the set of probability measures on $\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ and for any limiting value $\mathbb{Q}$ of this sequence, the canonical process $(\zeta^R, \Gamma^R)$ on $\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ satisfies that for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$,

$$ N_t^{f,R} := f(\zeta_t^R) - f(\zeta_0^R) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\zeta_s^R, \mu) \Gamma^R(ds, d\mu) = 0 \quad (41) $$

$\mathbb{Q}$-a.s. Moreover, for all $t \in [0, T]$,

$$ \mathbb{E}_{\mathbb{Q}} \left( \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} M_5(\mu) \Gamma^R (ds, d\mu) \right) < \infty, \quad (42) $$

so that the definition (41) makes sense.

This result is based on Propositions 3.1 and 4.8, Lemma 4.3. The proof is divided in eight steps. In Step 1, we establish the tightness of the family of laws of the stopped slow component $(z_{\bullet \wedge \tau^{K,R}}^{K,R})_{K \in \mathbb{N}^*}$. In Step 2, we establish a compact containment condition for the stopped
**Proposition 6.3.** With the notations of Theorem 5.1, for a.e. $t \in [0, T]$, $Q-a.s., \gamma_t^R = \pi^\lambda(\zeta^R)$ where $\pi^\lambda$ is the unique invariant probability measure of the centered FLEMING-VIOT process with resampling rate $\lambda$ (see [16, Section 4]).

It is here that we exploit ergodicity properties for the fast limit component. The proof of Proposition 6.3, given in Section 6.1 is based on the following technical lemma giving a characterisation of $\pi^\lambda$ and proved in Section 6.2. To state this lemma, let us first recall from (21) the definition of polynomials in $\mu$:

$$P_{f,n}(\mu) := \langle f, \mu^n \rangle := \int_{\mathbb{R}} \dots \int_{\mathbb{R}} f(x_1, \dots, x_n) \mu(dx_1) \dots \mu(dx_n)$$

with $n \in \mathbb{N}^*$, $\mu \in \mathcal{M}_1^c(\mathbb{R})$, $f \in \mathcal{C}_b^3(\mathbb{R}^n, \mathbb{R})$. For all $n \in \mathbb{N}^*$, for any function $f: \mathbb{R}^n \to \mathbb{R}$ whose second derivatives exist, we denote by Hess($f$) := $(\partial_{ij}^2 f)_{1 \le i,j \le n}$ the Hessian matrix of $f$. For all $n \in \mathbb{N}^*$, let us denote by $\mathcal{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^n, \mathbb{R})$ the set

$$\{f \in \mathcal{C}^2(\mathbb{R}^n, \mathbb{R}) \mid \exists C > 0, \forall x \in \mathbb{R}^n, |f(x)| + \|\nabla f(x)\|_{\infty} + \|\text{Hess}(f)(x)\|_{\infty} \le C (1 + \|x\|_{\infty}^2)\}.$$

From (25), we can see that if $f \in \mathcal{C}_b^4(\mathbb{R}^n, \mathbb{R})$, $\mathcal{L}_{\text{FVc}}^\lambda P_{f,n}(\mu)$ is a polynomial in $\mu$ of the form $P_{J,n+1}(\mu)$ for some function $J \in \mathcal{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^{n+1}, \mathbb{R})$. We recall from [16, Proposition 2.11] that if $\mu \in \mathcal{M}_1^{c,5}(\mathbb{R})$, then $\sup_{0 \le t \le T} \mathbb{E}_\mu(M_5(X_t)) < \infty$ where $(X_t)_{t \ge 0}$ denotes the centered FLEMING-VIOT process with resampling rate $\lambda$. Since, by (25), for all $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$,

$$|\mathcal{L}_{\text{FVc}} P_{J,n+1}(\mu)| \le C (1 + M_4(\mu)),$$

for some constant $C > 0$, we can apply the martingale problem (26) to the function $P_{J,n+1}(\mu)$ using classical localisation techniques to obtain that the process $(\widehat{M}_t^{P_{J,n+1}})_{t \ge 0}$ defined by

$$\widehat{M}_t^{P_{J,n+1}} := P_{J,n+1}(X_t) - P_{J,n+1}(X_0) - \int_0^t \mathcal{L}_{\text{FVc}}^\lambda P_{J,n+1}(X_s) \, ds \quad (43)$$

is a $\mathbb{P}_\mu$-martingale for all $\mu \in \mathcal{M}_1^{c,5}(\mathbb{R})$.

**Lemma 6.4.** Let be fixed $\lambda \in \mathbb{R}$. If $\gamma \in \mathcal{M}_1(\mathcal{M}_1(\mathbb{R}))$ satisfies $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma(d\mu) < \infty$ and

$$\forall n \in \mathbb{N}^*, \forall f \in \mathcal{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^\lambda P_{f,n}(\mu) \gamma(d\mu) = 0,$$

then, $\gamma = \pi^\lambda$.

## 6.1 Proof of Proposition 6.3

Let us define for all $\ell \in \mathbb{N}$, $\mathcal{C}_K^\ell(\mathbb{R}^n, \mathbb{R})$ the space of real functions of class $\mathcal{C}^\ell(\mathbb{R}^n, \mathbb{R})$ with compact support. From Lemma 3.4 and (12), for all $t \in [0, T]$, for all $n \in \mathbb{N}^*$ we have for all $f \in \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R})$ that

$$K^2 \sigma^2 M_{t \wedge \tau K, R}^{K, P_{f,n}} = \sum_{i=1}^3 (A)_i^{K,R} (t \wedge \tau^{K,R})$$
fast component $(\mu_{\bullet\wedge\tau^{K,R}}^{K,R})_{K\in\mathbb{N}^*}$. In Step 3, we prove the tightness of the family of laws of the occupation measure. In Step 4, we deduce the tightness of the family of laws of the couple (slow, occupation measure fast). In Step 5, we prove uniform integrability results for a family $(N_{t\wedge\tau^{K,R}}^{f,K,R})_{t\in[0,T], K\in\mathbb{N}^*}$ constructed from $z^{K,R}$ and $\Gamma^{K,R}$ similarly as $N_t^{f,R}$ in (41). Step 6 is devoted to establish the convergence in distribution of $(N_{\bullet\wedge\tau^{K,R}}^{f,K,R})_{K\in\mathbb{N}^*}$ to $N^{f,R}$. In Steps 7 and 8, we prove that the limit $N^{f,R}$ is null $\mathbb{Q}$-a.s.

The main modification of KURTZ's setting of [57, Theorem 2.1] is that we have to work with stopped times and need to be careful with moment estimates and uniform integrability properties. This led us to rewrite the proof.

**Proof of Theorem 5.1. Step 1. Tightness of the family of laws of $(z_{\bullet\wedge\tau^{K,R}}^{K,R})_{K\in\mathbb{N}^*}$ on $\mathbb{D}([0,T], \mathbb{T}_R)$**. Let $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$. For all $K \in \mathbb{N}^*$ and for all $t \in [0, T]$, let us consider the two processes $Y_t^{K,R} := f(z_{t\wedge\tau^{K,R}}^{K,R})$ and $Z_t^{K,R}$ defined by the relation

$$Y_t^{K,R} = Y_0^{K,R} + \int_0^t Z_s^{K,R} ds + M_{t\wedge\tau^{K,R}}^{f,K,R}$$

where $M_{t\wedge\tau^{K,R}}^{f,K,R}$ is the martingale given by (19) in the torus case. Note that, for all $s > \tau^{K,R}$, $Z_s^{K,R} = 0$. From Proposition 3.2, (34) and Lemma 4.3, note that there exists a constant $C > 0$ such that

$$\begin{aligned}
\sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ \left( \int_0^T |Z_t^{K,R}|^2 dt \right)^{\frac{1}{2}} \right] & \\
& \le \sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ 1 + \int_0^T |Z_t^{K,R}|^2 dt \right] \\
& \le C \left\{ 1 + \sup_{K \in \mathbb{N}^*} \mathbb{E} \left[ \int_0^{T \wedge \tau^{K,R}} \left( M_2^2(\mu_t^{K,R}) + \frac{1}{K^2} [1 + M_2^2(\mu_t^{K,R})] \right) dt \right] \right\} \\
& < \infty.
\end{aligned}$$

Hence, from [40, Theorem 3.9.4], the family of laws of $(Y^{K,R})_{K\in\mathbb{N}^*}$, on $\mathbb{D}([0,T], \mathbb{T}_R)$, is tight. Let us observe that the compact containment condition is satisfied by the stopped slow component $(z_{\bullet\wedge\tau^{K,R}}^{K,R})_{K\in\mathbb{N}^*}$ since $\mathbb{T}_R$ is compact and

$$\forall T > 0, \quad \inf_{K \in \mathbb{N}^*} \mathbb{P} \left( \sup_{0 \le t \le T} z_{t \wedge \tau^{K,R}, R}^{K,R} \in \mathbb{T}_R \right) = 1. \quad (43)$$

As $\mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$ is a dense subset of $\mathcal{C}_b^0(\mathbb{T}_R, \mathbb{T}_R)$ in the topology of the uniform norm, we deduce from (43) and [40, Theorem 3.9.1] that the family of laws of $(z_{\bullet\wedge\tau^{K,R}}^{K,R})_{K\in\mathbb{N}^*}$ is tight on $\mathbb{D}([0,T], \mathbb{T}_R)$.

### Step 2. Compact containment condition.

**Lemma 5.2.** Let $R > 0$. For all $T \ge 0$, the family of laws of the marginal random variables of the stopped fast process $(\mu_{t\wedge\tau^{K,R}}^{K,R})_{t\in[0,T], K\in\mathbb{N}^*}$ is tight on $\mathcal{M}_1(\mathbb{R})$ i.e.

$$\forall \eta > 0, \exists D_{\eta}^{R,T} \subset \mathcal{M}_1(\mathbb{R}) \text{ compact, } \forall t \in [0, T], \forall K \in \mathbb{N}^*, \mathbb{P}(\mu_{t\wedge\tau^{K,R}}^{K,R} \in D_{\eta}^{R,T}) \ge 1 - \eta.$$
where for all $t \le \tau^{K,R} \wedge T$

$$ (\mathbf{A})_1^{K,R}(t) := K^2 \sigma^2 (P_{f,n}(\mu_t^{K,R}) - P_{f,n}(\mu_0^{K,R})), $$

$$ (\mathbf{A})_2^{K,R}(t) := - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \theta(z_s^{K,R}) m_2(z_s^{K,R}) \mathcal{L}_{\text{FVc}}^{\lambda(z_s^{K,R})} P_{f,n}(\mu) \Gamma^{K,R}(\text{d}s, \text{d}\mu), $$

$$ (\mathbf{A})_3^{K,R}(t) := O\left(\frac{1}{\sqrt{K}} + \sigma K^{\frac{3}{2}+\varepsilon} + \int_0^t \frac{M_3(\mu_s^{K,R})}{K} \mathrm{d}s\right), $$

is a martingale. Since $\sigma K \to 0$ by Assumption (5), $(\mathbf{A})_1^{K,R}(t \wedge \tau^{K,R}) \to 0$ when $K \to +\infty$. From Corollary 4.3, the sequence $((\mathbf{A})_2^{K,R}(t \wedge \tau^K))_{t \in [0,T], K \in \mathbb{N}^*}$ is uniformly integrable and converges in law, when $K \to +\infty$, to

$$ M_t^{P_{f,n}} := - \int_0^t \theta(\zeta_s^R) m_2(\zeta_s^R) \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_s^R)} P_{f,n}(\mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu). $$

Note that, from CAUCHY-SCHWARZ's inequality and Lemma 4.1, there exists a constant $C > 0$ such that

$$ \mathbb{E} \left( \sup_{0 \le t \le T} \int_0^{t \wedge \tau^{K,R}} \frac{M_3(\mu_s^{K,R})}{K} \mathrm{d}s \right)^2 \le \frac{T}{K^2} \mathbb{E} \left( \int_0^{T \wedge \tau^{K,R}} M_6(\mu_s^{K,R}) \mathrm{d}s \right) \le \frac{CT}{K^2}. $$

Hence, $\lim_{K \to +\infty} \sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}) = 0$ in $L^2(\mathbb{R})$. In particular, we deduce that a subsequence of $(\sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}))_{K \in \mathbb{N}^*}$ converges almost surely to 0 and the family $(\sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}))_{K \in \mathbb{N}^*}$ is uniformly integrable along this subsequence [75, Theorem 13.7]. Using the same method based on SKOROHOD'S representation theorem as in the proof of Theorem 5.1, we deduce that the process $(M_t^{P_{f,n}})_{t \ge 0}$ is a $\mathbb{Q}$-martingale. As it is also a continuous and finite variation process by Lemma 6.1, it must hence be $\mathbb{Q}$-almost surely null [42, Theorem 4.1]. Hence, using Lemma 6.1 again, we have proved that

$$ \forall f \in \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R}), \mathbb{Q}-\text{a.s., } dt - \text{a.e., } \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0. $$

The space $\mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R})$ equipped with the norm $\|f\|_{W_0^{2,\infty}} = \|f\|_{\infty} + \|\nabla f\|_{\infty} + \|\text{Hess}(f)\|_{\infty}$ is separable. So, we can choose a dense countable family $\mathcal{B} \subset \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R})$ such that

$$ \forall f \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R}), \exists (f_p)_{p \in \mathbb{N}^*} \in \mathcal{B}^{\mathbb{N}^*}, f_p \xrightarrow{\|\cdot\|_{W_0^{2,\infty}}} f. $$

Then,

$$ \mathbb{Q}-\text{a.s., } dt - \text{a.e., } \forall p \in \mathbb{N}^*, \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f_p,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0. $$

From (38) and Lemma 6.1, we have that $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma_t^R(\mathrm{d}\mu) < \infty$. As for all $p \in \mathbb{N}^*$,

$$ \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f_p,n}(\mu) \le C_1 (1 + M_2(\mu)) \|f_p\|_{W_0^{2,\infty}}. $$
Proof. Let $\eta > 0$ be fixed. From Proposition 4.8, there exists $K_0 \in \mathbb{N}^*$ large enough such that for all $K \ge K_0$, for all $t \in [0, T]$, $\mathbb{P}(t > \tau^{K,R}) \le \frac{\eta}{2}$. We consider the $\mathbb{R}$-valued sequence $(a_{q,\eta})_{q \in \mathbb{N}}$ satisfying for all $q \in \mathbb{N}$, $a_{q,\eta} > 1$, and $\sum_{q \in \mathbb{N}^*} \frac{q}{a_{q,\eta}^2} < \frac{\eta}{2M_2}$ where $\overline{M}_2 := \max \{C_2^*, \overline{\theta} \overline{m}_2 / b\}$ is a uniform upper bound of $\mathbb{E}(M_2(\mu_{t \wedge \tau^{K,R}}^{K,R}))$ given by Corollary 4.6. Let $(\kappa_{q,\eta})_{q \in \mathbb{N}}$ be a sequence of compact intervals on $\mathbb{R}$, increasing for inclusion, of the form $[-a_{q,\eta}, a_{q,\eta}]$. Let $t \in [0, T]$. For all $q \in \mathbb{N}^*$, and $K \ge K_0$, note that

$$M_2(\mu_t^{K,R}) \mathbb{1}_{t \le \tau^{K,R}} \ge \mathbb{1}_{t \le \tau^{K,R}} \int_{\kappa_{q,\eta}^c} x^2 \mu_t^{K,R}(dx) \ge a_{q,\eta}^2 \mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{q,\eta}^c) \mathbb{1}_{t \le \tau^{K,R}},$$

and we deduce that $\{\mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{q,\eta}^c) > \frac{1}{q}\} \cap \{t \le \tau^{K,R}\} \subset \{M_2(\mu_{t \wedge \tau^{K,R}}^{K,R}) > \frac{a_{q,\eta}^2}{q}\} \cap \{t \le \tau^{K,R}\}$. Hence,

$$ \begin{aligned} \mathbb{P}\left(\left\{\exists q \in \mathbb{N}^*, \mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{q,\eta}^c) > \frac{1}{q}\right\}\right) & \le \mathbb{P}(\tau^{K,R} < t) + \sum_{q \in \mathbb{N}^*} \mathbb{P}\left(\left\{\mu_{t \wedge \tau^{K,R}}^{K,R} (\kappa_{q,\eta}^c) > \frac{1}{q}\right\} \cap \{t \le \tau^{K,R}\}\right) \\ & \le \mathbb{P}(\tau^{K,R} < t) + \sum_{q \in \mathbb{N}^*} \mathbb{P}\left(M_2(\mu_t^{K,R}) \mathbb{1}_{t \le \tau^{K,R}} > \frac{a_{q,\eta}^2}{q}\right) \\ & \le \mathbb{P}(\tau^{K,R} < t) + \sum_{q \in \mathbb{N}^*} \frac{\mathbb{E}(M_2(\mu_t^{K,R}) \mathbb{1}_{t \le \tau^{K,R}})}{\frac{a_{q,\eta}^2}{q}} \\ & \le \eta. \end{aligned} $$

Therefore, we have proved that

$$\forall \eta > 0, \forall t \in [0, T], \forall K \ge K_0, \quad \mathbb{P}(\mu_{t \wedge \tau^{K,R}}^{K,R} \in \mathcal{K}_{\eta}) \ge 1 - \eta,$$

where $\mathcal{K}_{\eta} := \{\mu \in \mathcal{M}_1(\mathbb{R}) \mid \forall q \in \mathbb{N}^*, \mu(\kappa_{q,\eta}^c) \le \frac{1}{q}\}$ which is compact by PROHOROV'S theorem. $\square$

Step 3. Tightness of the family of laws of $(\Gamma^{K,R})_{K \in \mathbb{N}^*}$ on $\mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$. Let $\eta > 0$ be fixed. From Proposition 4.8, there exists $K_0 \in \mathbb{N}^*$ such that for all $K \ge K_0$, for all $t \in [0, T]$, $\mathbb{P}(t \ge \tau^{K,R}) \le \frac{\eta}{2}$. Consider $D_{\frac{\eta}{2}}^{R,T}$ the compact set in Lemma 5.2. It follows that for all $t \in [0, T]$, $K \ge K_0$,

$$ \begin{aligned} \mathbb{E}(\Gamma^{K,R}([0,t] \times D_{\frac{\eta}{2}}^{R,T})) & \ge \mathbb{E}(\Gamma^{K,R}([0,t] \times D_{\frac{\eta}{2}}^{R,T}) \mathbb{1}_{t < \tau^{K,R}}) \\ & = \int_0^t \mathbb{P}(\{\mu_{s \wedge \tau^{K,R}}^{K,R} \in D_{\frac{\eta}{2}}^{R,T}\} \cap \{t < \tau^{K,R}\}) \mathrm{d}s \\ & \ge \int_0^t \mathbb{P}(\mu_{s \wedge \tau^{K,R}}^{K,R} \in D_{\frac{\eta}{2}}^{R,T}) \mathrm{d}s - \int_0^t \mathbb{P}(t \ge \tau^{K,R}) \mathrm{d}s \\ & \ge t(1 - \eta). \end{aligned} $$
for some constant $C_1 > 0$, we obtain by the dominated convergence theorem that

$$ \mathbb{Q} - \text{a.s., } dt - \text{a.e., } \forall f \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(d\mu) = 0. \quad (44) $$

Let us consider $f \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^n, \mathbb{R})$ and for all $q \in \mathbb{N}^*$, $x \mapsto \chi_q(x) = \exp\left(-\frac{1}{q^2 - \|x\|_\infty^2}\right) \mathbf{1}_{\|x\|_\infty < q}$ of class $\mathcal{C}^\infty(\mathbb{R}^n, \mathbb{R})$ with compact support. Then, for all $q \in \mathbb{N}^*$, $f\chi_q \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R})$. Noting that, $\left|\int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f\chi_q,n}(\mu) \gamma_t^R(d\mu)\right|$ is dominated by $C_2 \int_{\mathcal{M}_1(\mathbb{R})} (1 + M_4(\mu)) \gamma_t^R(d\mu)$ for some constant $C_2 > 0$, and since $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma_t^R(d\mu) < \infty$, we deduce from (44) applied to $f\chi_q$, by the dominated convergence theorem, that

$$ \mathbb{Q} - \text{a.s., } dt - \text{a.e., } \forall f \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(d\mu) = 0. $$

Therefore, it follows from Lemma 6.4 that $\mathbb{Q}$-a.s., $dt$-a.e., $\gamma_t^R(d\mu) = \pi^{\lambda(\zeta_t^R)}(d\mu)$ which concludes the proof.

## 6.2 Proof of Lemma 6.4

In Section 6.2.1 we extend some duality results for the centered FLEMING-VIOT process, obtained in [16], which are be useful to prove Lemma 6.4 in Section 6.2.2.

### 6.2.1 Extension of the duality result for the centered Fleming-Viot process

Let us recall that the dual process $(\xi_t)_{t \ge 0}$ of the centered FLEMING-VIOT process $(X_t)_{t \ge 0}$ with resampling rate $\lambda$, on the state space $\bigcup_{n \in \mathbb{N}^*} \mathcal{C}^2(\mathbb{R}^n, \mathbb{R})$, obtained in [16, Section 3.2] is defined as below.

Let us consider $M := (M(t))_{t \ge 0}$ a MARKOV'S birth and death process in $\mathbb{N}$ whose transition rates $q_{i,j}$ from $i$ to $j$ are given by:

$$ (1) \quad q_{n,n+1} = \lambda n^2 $$

$$ (2) \quad q_{n,n-1} = \lambda n(n-1) $$

$$ (3) \quad q_{i,j} = 0 \text{ otherwise.} $$

For all $M(0) \in \mathbb{N}^*$, $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ and $\lambda > 0$, we define

$$ \xi_t := T_{\lambda}^{(M(\tau_n))} (t - \tau_n) \Lambda_n T_{\lambda}^{(M(\tau_{n-1}))} (\tau_n - \tau_{n-1}) \Lambda_{n-1} \cdots \Lambda_1 T_{\lambda}^{(M(0))} (\tau_1) \xi_0, \quad \tau_n \le t < \tau_{n+1}, n \in \mathbb{N}, \quad (45) $$

where $(\tau_n)_{n \in \mathbb{N}}$ is the sequence of jump times of $M$ with $\tau_0 = 0$, $(T_{\lambda}^{(n)}(t))_{t \ge 0}$ is the semi-group of operator associated to the generator $B_{\lambda}^{(n)}$ given by (22) and where $(\Lambda_n)_{n \in \mathbb{N}}$ is a sequence of random operators. These are conditionally independent given $M$ and satisfy for all $k \in \mathbb{N}$, $n \ge 1$ and $1 \le i \ne j \le n$,

$$ \mathbb{P} \left( \Lambda_k = \Phi_{i,j} \mid \{ M(\tau_k^-) = n, M(\tau_k) = n-1 \} \right) = \frac{1}{n(n-1)} $$
Therefore, the tightness of the family of laws of $(\Gamma^{K,R})_{K \in \mathbb{N}^*}$ follows from [57, Lemma 1.3].

**Step 4. Tightness of the family of laws of $(z^{K,R}, \Gamma^{K,R})_{K \in \mathbb{N}^*}$ on $D([0,T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$.**
From Steps 1 and 3 and PROHOROV'S theorem, we deduce that the family of laws of the couple $((z^{K,R}_{\bullet\wedge\tau^{K,R}}, \Gamma^{K,R}))_{K \in \mathbb{N}^*}$ is relatively compact in $\mathcal{M}_1(D([0,T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R})))$. Hence, there exists a probability measure $\mathbb{Q}$ on the canonical space $D([0,T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ and an increasing function $n: \mathbb{N}^* \to \mathbb{N}^*$ such that the subsequence of laws of $((z^{n(K,R)}_{\bullet\wedge\tau^{n(K,R)}}, \Gamma^{n(K,R)})_{K \in \mathbb{N}^*})$ converges weakly to the limiting value $\mathbb{Q}$ when $K \to +\infty$. Thanks to Proposition 4.8, we deduce that the family of laws of $((z^{n(K,R)}_{\bullet\wedge\tau^{n(K,R)}}, \Gamma^{n(K,R)})_{K \in \mathbb{N}^*})$ converges weakly to $\mathbb{Q}$ when $K \to +\infty$ and therefore that the family of laws of $((z^{K,R}, \Gamma^{K,R}))_{K \in \mathbb{N}^*}$ is relatively compact, thus tight on $D([0,T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ by PROHOROV'S theorem.

**Step 5. Uniform integrability.** For all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, $K \in \mathbb{N}^*$, let us consider $(N_t^{f,K,R})_{t \in [0,T]}$ the stochastic process defined by

$$N_t^{f,K,R} := f(z_t^{K,R}) - f(z_0^{K,R}) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(z_s^{K,R}, \mu) \Gamma^{K,R}(ds, d\mu).$$

From (9), we have for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, there exists a constant $C > 0$ such that

$$\forall t \in [0, T], \forall K \in \mathbb{N}^*, \quad \left| N_{t \wedge \tau^{K,R}}^{f,K,R} \right| \le C \left( 1 + \int_0^t M_2(\mu_s^{K,R}) \mathbf{1}_{s \le \tau^{K,R}} ds \right),$$

Hence, the uniform integrability of $(N_{\bullet\wedge\tau^{K,R}}^{f,K,R})_{K \in \mathbb{N}^*}$ follows from Corollary 4.4.

**Step 6. Proof of (42) and almost sure convergence of $(\tilde{N}_{\bullet\wedge\tau^{K,R}}^{f,K,R})_{K \in \mathbb{N}^*}$ to $\tilde{N}^{f,R}$.** From Step 4, Proposition 4.8 and SKOROHOD'S representation theorem, there exists an increasing function $\tilde{n}: \mathbb{N}^* \to \mathbb{N}^*$ and a probability space on which we define, the random variable $\tilde{\tau}^{\tilde{n}(K),R}$, the families $(\tilde{z}^{\tilde{n}(K),R})_{K \in \mathbb{N}^*}$ and $(\tilde{\Gamma}^{\tilde{n}(K),R})_{K \in \mathbb{N}^*}$ and $\tilde{\zeta}^R, \tilde{\Gamma}^R$ copies of $(z^{\tilde{n}(K),R})_{K \in \mathbb{N}^*}$, $(\tilde{\Gamma}^{\tilde{n}(K),R})_{K \in \mathbb{N}^*}$, $\zeta^R$, $\Gamma^R$ under $\mathbb{Q}$ such that the sequence $((\tilde{z}^{\tilde{n}(K),R}, \tilde{\Gamma}^{\tilde{n}(K),R}, \tilde{\tau}^{\tilde{n}(K),R}))_{K \in \mathbb{N}^*}$ converges a.s. to $(\tilde{\zeta}^R, \tilde{\Gamma}^R, +\infty)$ when $K \to +\infty$. Note that for all $t \in [0, T]$,

$$\tilde{\Gamma}^{\tilde{n}(K),R} ([0, t \wedge \tilde{\tau}^{\tilde{n}(K),R}] \times \mathcal{M}_1(\mathbb{R})) = t \wedge \tilde{\tau}^{\tilde{n}(K),R} \xrightarrow{\text{a.s.}}_{K \to +\infty} t = \tilde{\Gamma}^R ([0, t] \times \mathcal{M}_1(\mathbb{R})).$$

From [57, Lemma 1.5 (b)(c)(d)] and Corollary 4.4, we have

$$\int_0^{t \wedge \tilde{\tau}^{K,R}} \int_{\mathcal{M}_1(\mathbb{R})} M_5(\mu) \tilde{\Gamma}^{K,R}(ds, d\mu) \xrightarrow{\text{a.s.}}_{K \to +\infty} \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} M_5(\mu) \tilde{\Gamma}^R(ds, d\mu)$$

and (42) follows. From [57, Lemma 1.5 (b)(c)(d)], we also deduce that for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, for all $t \in [0, T]$, a.s.

$$\lim_{K \to +\infty} \int_0^{t \wedge \tilde{\tau}^{\tilde{n}(K),R}} \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{z}_s^{\tilde{n}(K),R}, \mu) \tilde{\Gamma}^{\tilde{n}(K),R}(ds, d\mu) \\ = \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{\zeta}^R, \mu) \tilde{\Gamma}^R(ds, d\mu)$$
and for all $n \ge 1$ and $1 \le i, j \le n$,

$$ \mathbb{P} \left( \Lambda_k = K_{i,j} \mid \{ M(\tau_k^-) = n, M(\tau_k) = n+1 \} \right) = \frac{1}{n^2}, $$

where $\Phi_{i,j}$ and $K_{i,j}$ are respectively defined in (23) and (24). Note that if $M(0) = 1$, the dual process can only jump from $\xi_0$ to $K_{ij}\xi_0$. As for all $t \ge 0$, $\xi_t \in \mathcal{C}^2(\mathbb{R}^{M(t)}, \mathbb{R})$, $K_{i,j}\xi_t$ is well defined. Moreover, the dual process $(\xi_t)_{t \ge 0}$ with initial condition $\xi_0$ is constructed on the same probability space and independently of the centered FLEMING-VIOT process $(X_t)_{t \ge 0}$ with resampling rate $\lambda$ and initial condition $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$. We shall denote by $\mathbb{P}_{(\mu, \xi_0)}$, the law of the couple $(X_t, \xi_t)_{t \ge 0}$ on this probability space.

We denote by $S_t$ the number of jumps of the process $M$ on $[0, t]$. We start with a result giving bounds on the dual process, which is an extension of similar estimates obtained in [16, Lemma 6.4].

**Lemma 6.5.** For all $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ there exists a function $C_0$ from $\bigcup_{k \in \mathbb{N}^*} (0, +\infty)^{k-1} \times \{k\}$ to $\mathbb{R}_+$, locally bounded, such that

$$ C_0(\tau_1, 1) := \|\xi_0\|_{\infty} + \frac{3}{M(0)} \|\text{Hess}(\xi_0)\|_{\infty} $$

and, for all $(t_j)_{j \in \mathbb{N}} \in (0, +\infty)^{\mathbb{N}}$, $k \mapsto C_0((t_i)_{0 \le i \le k-1}, k)$ is non-decreasing and satisfying

$$ \forall k \in \mathbb{N}^*, \forall t \le \tau_k, \forall x \in \mathbb{R}^{M(t)}, \quad |\xi_t(x)| \le C_0((\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k) (1 + \|x\|_{\infty}^{2S_t}). \quad (46) $$

*Proof.* By induction on $k \in \mathbb{N}^*$, we prove the property

$$ (Q_k): \quad \forall x \in \mathbb{R}^{M(\tau_k)}, \quad |\xi_{\tau_k}(x)| \le C_0((\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k) (1 + \|x\|_{\infty}^{2k}). $$

**Step 1. Initial case: computation of $C_0(\tau_1, 1)$.** Let us recall some notations of [16, Theorem 6.1]. For all $n \in \mathbb{N}^*$, we denote by $\mathbf{1} \in \mathbb{R}^n$, the vector whose coordinates are all 1. For all $n \in \mathbb{N}^*$, $t \ge 0$, $x \in \mathbb{R}^n$ and $\lambda > 0$, let us consider $g_{t,x}^{\lambda}$ the density of the Gaussian distribution $N^{(n)}(m_{t,x}^{\lambda}, \Sigma_{t}^{\lambda})$ where $m_{t,x}^{\lambda} := x - \frac{(1-\exp(-2\lambda nt))}{n}(x \cdot \mathbf{1})\mathbf{1}$ and $\Sigma_{t}^{\lambda} := P\sigma_{t}^{\lambda}P^{-1}$ with

$$ \sigma_t^\lambda := \begin{pmatrix} \frac{1-\exp(-4\lambda nt)}{4\lambda n} & 0 & \dots & 0 \\ 0 & t & 0 & \dots \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \dots & \ddots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & t \end{pmatrix} \quad \text{and} \quad P := \begin{pmatrix} \frac{1}{\sqrt{n}} & \frac{1}{\sqrt{2}} & \cdots & \frac{1}{\sqrt{n(n-1)}} \\ \vdots & -\sqrt{\frac{1}{2}} & \ddots & \frac{1}{\sqrt{n(n-1)}} \\ \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \frac{1}{\sqrt{n(n-1)}} \\ \frac{1}{\sqrt{n}} & 0 & \cdots & -\sqrt{\frac{n-1}{n}} \end{pmatrix}. $$

Note that for all $i, j \in \{1, \dots, n\}$, $\partial_{x_i} m_{t,x}^{\lambda} = \epsilon_i - \frac{1-\exp(-2\gamma nt)}{n} \mathbf{1}$ and $\partial_{x_i x_j}^2 m_{t,x}^{\lambda} = 0$ where $(\epsilon_1, \dots, \epsilon_n)$ is the canonical basis of $\mathbb{R}^n$. For all $\lambda > 0$, the key identity for the sequel is

$$ \forall f \in L^{\infty}(\mathbb{R}^n), \forall t \ge 0, \forall x \in \mathbb{R}^n, \quad T_{\lambda}^{(M(0))}(t)f(x) = (f * g_{t,0}^{\lambda})(m_{t,x}^{\lambda}) \quad (47) $$
and thus the sequence $(\tilde{N}_t^{f, \bar{n}(K), R})_{K \in \mathbb{N}^*}$ converges a.s. to the process $(\tilde{N}_t^{f, R})_{t \in [0, T]}$ where $\tilde{N}_t^{f, K, R}$ and $\tilde{N}_t^{f, R}$ are respectively defined for all $K \in \mathbb{N}^*$ by

$$ \tilde{N}_t^{f,K,R} := f(\tilde{z}_t^{K,R}) - f(\tilde{z}_0^{K,R}) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{z}_t^{K,R}, \mu) \tilde{\Gamma}^{K,R}(ds, d\mu), \quad t \le T \wedge \tilde{\tau}^{K,R} $$

$$ \tilde{N}_t^{f,R} := f(\tilde{\zeta}_t^R) - f(\tilde{\zeta}_0^R) - \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\tilde{\zeta}_t^R, \mu) \tilde{\Gamma}^R (ds, d\mu), \quad t \le T. $$

Step 7. $N^{f,R}$ is a martingale. Let us consider the filtration $(\tilde{\mathcal{F}}_t^{K,R})_{t \in [0,T]}$ defined by $\tilde{\mathcal{F}}_t^{K,R} := \sigma (\tilde{z}_s^{K,R}, \tilde{\Gamma}^{K,R} ([0, s] \times H) \mid s \le t, H \in \mathcal{B}(\mathcal{M}_1(\mathbb{R})))$, $(f_i)_{1 \le i \le q}$, $q \in \mathbb{N}^*$ bounded LIPSCHITZ functions from $\mathbb{T}_R$ to $\mathbb{T}_R$ and $0 < t_1 \le \dots \le t_q \le s < t$. Let us denote for all $K \in \mathbb{N}^*$, for all $t \le \tilde{\tau}^{K,R} \wedge T$, $\tilde{M}_t^{f,K,R} := \tilde{N}_t^{f,K,R} + \tilde{\mathcal{E}}_t^{f,K,R}$ where $\tilde{M}^{f,K,R}$ is constructed from $\tilde{z}^{K,R}$ and $\tilde{\mu}^{K,R}$ as in (19) and $\tilde{\mathcal{E}}_t^{f,K,R}$ is an error term. Note that $\tilde{M}_t^{f,K,R}$ is a $(\tilde{\mathcal{F}}_t^{K,R})_{t \in [0,T]}$-martingale as in Proposition 3.1. Hence,

$$ \mathbb{E} \left( \tilde{M}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \prod_{i=1}^{q} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) \right) = \mathbb{E} \left( \tilde{M}_{s \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \prod_{i=1}^{q} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) \right). $$

From Proposition 3.2 and (34), $\tilde{\mathcal{E}}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f,K,R} = \frac{1}{K} \mathcal{O} \left( \int_0^t [1 + M_2(\tilde{\mu}_s^{K,R})] \mathbf{1}_{s \le \tilde{\tau}^{\tilde{n}(K),R}} ds \right)$ and then satisfy the condition $\lim_{K \to +\infty} \mathbb{E} \left( \sup_{0 \le t \le T} \left| \mathcal{E}_{t \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f,K,R} \right| \right) = 0$. As for all $i \in \{1, \dots, q\}$, $f_i$ is LIPSCHITZ, there exists a constant $C > 0$ such that

$$ \mathbb{E} \left( \left| f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} - \mathcal{E}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) - f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) \right| \right) \\ \le \frac{C}{\tilde{n}(K)} \mathbb{E} \left( \int_0^{t_i} [1 + M_2(\tilde{\mu}_r^{\tilde{n}(K),R})] \mathbf{1}_{r \le \tilde{\tau}^{\tilde{n}(K),R}} dr \right) $$

where the term of the right hand side of the previous inequality goes to 0 when $K \to +\infty$ according to Lemma 4.3. We deduce that for all $u \in \{s, t\}$,

$$ \lim_{K \to +\infty} \mathbb{E} \left( \tilde{M}_{u \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \prod_{i=1}^{q} f_i \left( \tilde{M}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) \right) = \lim_{K \to +\infty} \mathbb{E} \left( \tilde{N}_{u \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \prod_{i=1}^{q} f_i \left( \tilde{N}_{t_i \wedge \tilde{\tau}^{\tilde{n}(K),R}}^{f, \tilde{n}(K),R} \right) \right). $$

From Steps 5 and 6, we deduce that

$$ \mathbb{E} \left( \tilde{N}_t^{f,R} \prod_{i=1}^{q} f_i \left( \tilde{N}_{t_i}^{f,R} \right) \right) = \mathbb{E} \left( \tilde{N}_s^{f,R} \prod_{i=1}^{q} f_i \left( \tilde{N}_{t_i}^{f,R} \right) \right). $$

Hence,

$$ \mathbb{E} \left( N_t^{f,R} \prod_{i=1}^{q} f_i \left( N_{t_i}^{f,R} \right) \right) = \mathbb{E} \left( N_s^{f,R} \prod_{i=1}^{q} f_i \left( N_{t_i}^{f,R} \right) \right). $$
where * stands for the convolution product (see (41) in [16, Theorem 6.1]).

On the one hand, note that for all $t < \tau_1$, $M(t) = M(0)$. In this case, for all $x \in \mathbb{R}^{M(0)}$, $\lambda > 0$, $\xi_t(x) = T_{\lambda}^{(M(0))}(t)\xi_0(x)$ and so, from [16, Theorem 6.1], $|\xi_t(x)| \le \|\xi_0\|_{\infty}$. On the other hand, at time $\tau_1$, we make a partition of cases according to whether the dual process loses or gains a variable. Let $i, j \in \{1, \dots, M(0)\}$ be fixed.

• **Case $\Lambda_1 = \Phi_{i,j}$.** In this case, $M(\tau_1) = M(0) - 1$ and we deduce from (45) that

$$ \forall x \in \mathbb{R}^{M(0)-1}, \quad \xi_{\tau_1}(x) = \Phi_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x). $$

By (23), we deduce that for all $x \in \mathbb{R}^{M(0)-1}$, $|\xi_{\tau_1}(x)| \le \|\xi_0\|_{\infty}$.

• **Case $\Lambda_1 = K_{i,j}$.** In this case, $M(\tau_1) = M(0) + 1$ and we deduce from (45) and (24) then from (3) of [16, Theorem 6.1] and properties of the convolution product that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ \begin{aligned} \xi_{\tau_1}(x) = K_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x) &= \partial_{x_i x_j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(\tilde{x}) x_{M(0)+1}^2 \\ &= (\partial_{x_j} m_{\tau_1, \tilde{x}}^{\lambda})^t \left[ (\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda}) (m_{\tau_1, \tilde{x}}^{\lambda}) \partial_{x_i} m_{t, \tilde{x}}^{\lambda} \right] x_{M(0)+1}^2 \end{aligned} $$

where $\tilde{x} = (x_1, \dots, x_{M(0)}) \in \mathbb{R}^{M(0)}$. As,

$$ (\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda}) (m_{\tau_1, \tilde{x}}^{\lambda}) = \left( \int_{\mathbb{R}^{M(0)}} \partial_{ij}^2 \xi_0(u) g_{\tau_1, 0}^{\lambda} (m_{\tau_1, \tilde{x}}^{\lambda} - u) \, du \right)_{1 \le i, j \le M(0)} $$

we obtain that $\|(\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda}) (m_{\tau_1, \tilde{x}}^{\lambda})\|_{\infty} \le \|\partial_{ij}^2 \xi_0\|_{\infty}$. Noting that $\left| (\partial_{x_j} m_{\tau_1, \tilde{x}}^{\lambda})^t \partial_{x_i} m_{t, \tilde{x}}^{\lambda} \right| \le \frac{3}{M(0)}$, we deduce that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ |\xi_{\tau_1}(x)| \le \frac{3}{M(0)} \sup_{1 \le i,j \le M(0)} \|\partial_{ij}^2 \xi_0\|_{\infty} x_{M(0)+1}^2 \le \frac{3}{M(0)} \|\text{Hess}(\xi_0)\|_{\infty} \|x\|_{\infty}^2 $$

and ($Q_1$) follows.

**Step 2. Inductive Step.** We assume that, for $k \in \mathbb{N} \setminus \{0, 1\}$, ($Q_{k-1}$) is satisfied and prove that ($Q_k$) is also. We make again a partition of cases according to whether the dual process loses or gains a variable. Let $i, j \in \{1, \dots, M(\tau_{k-1})\}$ be fixed.

• **Case $\Lambda_k = \Phi_{i,j}$ at the $k^{\text{th}}$ jump.** In this case, $M(\tau_k) = M(\tau_{k-1}) - 1$ and we deduce from (45) that for all $x \in \mathbb{R}^{M(\tau_{k-1})-1}$, $\xi_{\tau_k}(x) = \Phi_{i,j} T_{\lambda}^{(M(\tau_{k-1}))}(\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x)$. By using (23) and ($Q_{k-1}$), we deduce from [16, Corollary 6.2 (1)] that for all $x \in \mathbb{R}^{M(\tau_{k-1})-1}$,

$$ \begin{aligned} & \left| \Phi_{i,j} T_{\lambda}^{(M(\tau_{k-1}))}(\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x) \right| \\ & \le C_2 (\tau_k - \tau_{k-1}, M(\tau_{k-1})) C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) (1 + \|x\|_{\infty}^{2(k-1)}), \end{aligned} \quad (48) $$

where $C_2C_0$ is locally bounded. As $t \mapsto C_2(t, M(\tau_{k-1}))$ is non-decreasing, we deduce ($Q_k$) in that case.
Since the last property is true for all $q \in \mathbb{N}^*$, $t_1 \le \dots \le t_q \le s < t$ and for all bounded LIPSCHITZ functions $(f_i)_{1\le i\le q}$, the monotone class theorem ensures us that

$$ \mathbb{E} \left( N_t^{f,R} \mid \sigma \left( N_u^{f,R} \mid u \le s \right) \right) = N_s^{f,R}. $$

Hence the announced result.

**Step 8. Nullity of $N^{f,R}$**. On the one hand, from It\^a's formula [80, Theorem 32 of Chapter II], for all $f \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$ and $t \in [0, T]$

$$ \begin{aligned} f^2(\zeta_t^R) = f^2(\zeta_0^R) &+ 2 \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} f(\zeta_s^R) \mathcal{L}_{\text{SLOW}} f(\zeta_s^R, \mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu) \\ &+ 2 \int_0^t f(\zeta_s^R) \mathrm{d}N_s^{f,R} + \langle N^{f,R} \rangle_t + \sum_{0<s\le t} (f(\zeta_s^R) - f(\zeta_{s-}^R))^2. \end{aligned} \quad (44) $$

On the other hand, applying (41) with $f^2 \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, we obtain that for all $t \in [0, T]$

$$ f^2(\zeta_t^R) = f^2(\zeta_0^R) + \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f^2(\zeta_s^R, \mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu) + N_t^{f^2,R}. \quad (45) $$

Comparing (44) and (45) leads for all $t \in [0, T]$ to

$$ N_t^{f^2,R} - 2 \int_0^t f(\zeta_s^R) \mathrm{d}N_s^{f,R} = \langle N^{f,R} \rangle_t + \sum_{0<s\le t} (f(\zeta_s^R) - f(\zeta_{s-}^R))^2 $$

and thus by [48, Theorem 4.1] that $\mathbb{Q}$-a.s. for all $t \in [0, T]$,

$$ \langle N^{f,R} \rangle_t = - \sum_{0<s\le t} (f(\zeta_s^R) - f(\zeta_{s-}^R))^2 \le 0 $$

so that $\langle N^{f,R} \rangle_t = 0$. Hence, $\mathbb{Q}$-a.s. $N^{f,R} = 0$ which completes the proof. $\square$

# 6 Characterisation of the occupation measure limit on the torus

Consider a probability measure $\mathbb{Q}$ on $\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ and the canonical process $(\zeta^R, \Gamma^R)$ as in Theorem 5.1. The following lemma gives us a desintegration result of the occupation measure $\Gamma^R$ that we characterise below.

**Lemma 6.1.** Let $T \ge 0$ be fixed. With the notations of Theorem 5.1, there exists a random probability measure-valued process $(\gamma_t^R)_{t \in [0,T]}$ that is predictable in $(\omega, t)$ and such that for all bounded measurable function $\psi : [0, T] \times \mathcal{M}_1(\mathbb{R}) \to \mathbb{T}_R$,

$$ \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \psi(s, \mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu) = \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \psi(s, \mu) \gamma_s^R(\mathrm{d}\mu) \mathrm{d}s. \quad (46) $$

*Proof.* The desintegration result of $\Gamma^R$ follows directly from [57, Lemma 1.4]. $\square$
• **Case $\Lambda_k = K_{i,j}$ at the $k^{\text{th}}$ jump.** In this case, $M(\tau_k) = M(\tau_{k-1}) + 1$. From (45), (24) and [16, Theorem 6.1 (3)], we have for all $x \in \mathbb{R}^{M(\tau_{k-1})+1}$,

$$ \begin{aligned} |\xi_{\tau_k}(x)| &= |K_{i,j} T_{\lambda}^{(M(\tau_{k-1})+1)} (\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x)| \\ &= \left| \left( \partial_{x_j} m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda} \right)^t \left[ \left( \xi_{\tau_{k-1}} * \text{Hess} \left( g_{\tau_k - \tau_{k-1}, 0}^{\lambda} \right) \right) (m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda}) \partial_{x_i} m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda} \right] \right| x_{M(\tau_{k-1})+1}^2, \end{aligned} $$

where $\tilde{x} = (x_1, \dots, x_{M(\tau_{k-1})})^t \in \mathbb{R}^{M(\tau_{k-1})}$. From $(\mathcal{Q}_{k-1})$ and [16, Corollary 6.2 (2)], we deduce that

$$ \begin{aligned} & |K_{i,j} T_{\lambda}^{(M(\tau_{k-1})+1)} (\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x)| \\ & \le C_3 (\tau_k - \tau_{k-1}, M(\tau_{k-1})) C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \left( 1 + \|x\|_{\infty}^{2k} \right), \end{aligned} \quad (49) $$

where $C_3C_0$ is locally bounded and $(\mathcal{Q}_k)$ follows in that case.

We conclude by the principle of induction.

**Step 3.** Proof of (46) for $t < \tau_k$. Note that from (45), for all $k \in \mathbb{N}$ and $t \in ]\tau_k, \tau_{k+1}]$,

$$ \forall x \in \mathbb{R}^{M(t)}, \quad \xi_t(x) = T_{\lambda}^{(M(\tau_k))} (t - \tau_k) \xi_{\tau_k}(x), $$

so the announced result follows from [16, Corollary 6.2 (1)]. □

Let us consider for all $k \in \mathbb{N}^*$, $\ell, m \in \mathbb{N}$ the stopping times

$$ \vartheta_{k,\ell} := \inf \left\{ t \ge 0 \mid \begin{array}{l} S_t \ge k \quad \text{or} \quad \exists s \in [0,t], \langle \xi_s, X_{t-s}^{M(s)} \rangle \ge \ell \\ \phantom{\text{or}} \end{array} \right\}, $$

$$ \vartheta'_{k,m} := \inf \left\{ t \ge 0 \mid \begin{array}{l} S_t \ge k \quad \text{or} \quad C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \ge m \\ \phantom{\text{or}} \end{array} \right\}, $$

$$ \vartheta'_{k,\ell,m} := \inf \left\{ t \ge 0 \mid \begin{array}{l} S_t \ge k \quad \text{or} \quad \exists s \in [0,t], \langle \xi_s, X_{t-s}^{M(s)} \rangle \ge \ell \\ \phantom{\text{or}} \text{or} \quad C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \ge m \end{array} \right\}. $$

As $(M(t), \xi_t)_{t \ge 0}$ is independent of $(X_t)_{t \ge 0}$, note that for all $k \in \mathbb{N}^*$, $m \in \mathbb{N}$, $\vartheta'_{k,m}$ is independent of $(X_t)_{t \ge 0}$. Recall that for all $t \ge 0$, $\xi_t \in \mathcal{C}^2(\mathbb{R}^{M(t)}, \mathbb{R})$. For all $n \ge M(t)$, for all $x \in \mathbb{R}^n$, we denote by $\xi_t^{(n)}(x) = \xi_t^{(n)}(x_1, \dots, x_n) := \xi_t(x_1, \dots, x_{M(t)})$ so that for all $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$,

$$ \langle \xi_t^{(n)}, \mu^n \rangle = \langle \xi_t, \mu^{M(t)} \rangle. $$

The next lemma is an extension of the duality identity proved in [16].

**Lemma 6.6.** Given $(X_t)_{t \ge 0}$, $(\xi_t)_{t \ge 0}$ as above with $X_0 := \mu \in \mathcal{M}_1^{c,2k}(\mathbb{R})$, $k \in \mathbb{N}^*$ and $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$, we have that $\mathbb{P}_{(\mu, \xi_0)}$-a.s., for all $t \ge 0$, for all $m \in \mathbb{N}$,

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k,m}}^{M(0)} \right\rangle \right) = \left\langle \mathbb{E}_{\xi_0} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle. \quad (50) $$
**Corollary 6.2.** Let $T \ge 0$ be fixed. With the notations of Theorem 5.1, we have that $\zeta^R \in \mathscr{C}^0([0, T], \mathbb{T}_R)$ is differentiable of derivative in $L^1(\mathbb{R})$ Q-a.s.

**Proof.** Applying (41) with $f = \text{id} \in \mathscr{C}_b^1(\mathbb{T}_R, \mathbb{T}_R)$, we deduce from Theorem 5.1, (9) and (46), Q-a.s., for all $t \in [0, T]$

$$ \zeta_t^R = \zeta_0^R + \int_0^t \left( \int_{\mathcal{M}_1(\mathbb{R})} M_2(\mu) \gamma_s^R (d\mu) \right) \partial_1 \text{Fit} \left( \zeta_s^R, \zeta_s^R \right) \mathrm{d}s. $$

From Assumptions (A) and (42), the integrand of the previous time integral is in $L^1(\mathbb{R})$ Q-a.s. Hence, the announced result follows from the fundamental theorem of calculus. $\square$

We now want to characterise the limiting value $\Gamma^R (dt, d\mu) = \gamma_t^R (d\mu) dt$ under Q.

**Proposition 6.3.** With the notations of Theorem 5.1, for a.e. $t \in [0, T]$, Q-a.s., $\gamma_t^R = \pi^\lambda(\zeta^R)$ where $\pi^\lambda$ is the unique invariant probability measure of the centered FLEMING-VIOT process with resampling rate $\lambda$ (see [18, Section 4]).

It is here that we exploit ergodicity properties for the fast limit component. The proof of Proposition 6.3, given in Section 6.1 is based on the following technical lemma giving a characterisation of $\pi^\lambda$ and proved in Section 6.2. To state this lemma, let us first recall from (22) the definition of polynomials in $\mu$:

$$ P_{f,n}(\mu) := \langle f, \mu^n \rangle := \int_{\mathbb{R}} \dots \int_{\mathbb{R}} f(x_1, \dots, x_n) \mu(dx_1) \dots \mu(dx_n) $$

with $n \in \mathbb{N}^*$, $\mu \in \mathcal{M}_1^c(\mathbb{R})$, $f \in \mathscr{C}_b^3(\mathbb{R}^n, \mathbb{R})$. For all $n \in \mathbb{N}^*$, for any function $f: \mathbb{R}^n \to \mathbb{R}$ whose second derivatives exist, we denote by Hess($f$) := $(\partial_{ij}^2 f)_{1 \le i,j \le n}$ the Hessian matrix of $f$. For all $n \in \mathbb{N}^*$, let us denote by $\mathscr{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^n, \mathbb{R})$ the set

$$ \left\{ f \in \mathscr{C}^2(\mathbb{R}^n, \mathbb{R}) \mid \exists C > 0, \forall x \in \mathbb{R}^n, \left| f(x) \right| + \| \nabla f(x) \|_{\infty} + \| \text{Hess}(f)(x) \|_{\infty} \le C (1 + \| x \|_{\infty}^2) \right\}. $$

From (26), we can see that if $f \in \mathscr{C}_b^4(\mathbb{R}^n, \mathbb{R})$, $\mathcal{L}_{\text{FVc}}^\lambda P_{f,n}(\mu)$ is a polynomial in $\mu$ of the form $P_{J,n+1}(\mu)$ for some function $J \in \mathscr{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^{n+1}, \mathbb{R})$. We recall from [18, Proposition 2.11] that if $\mu \in \mathcal{M}_1^{c,5}(\mathbb{R})$, then $\sup_{0 \le t \le T} \mathbb{E}_\mu (M_5(X_t)) < \infty$ where $(X_t)_{t \ge 0}$ denotes the centered FLEMING-VIOT process with resampling rate $\lambda$. Since, by (26), for all $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$,

$$ |\mathcal{L}_{\text{FVc}} P_{J,n+1}(\mu)| \le C (1 + M_4(\mu)), $$

for some constant $C > 0$, we can apply the martingale problem (27) to the function $P_{J,n+1}(\mu)$ using classical localisation techniques to obtain that the process $(\widehat{M}_t^{P_{J,n+1}})_{t \ge 0}$ defined by

$$ \widehat{M}_t^{P_{J,n+1}} := P_{J,n+1}(X_t) - P_{J,n+1}(X_0) - \int_0^t \mathcal{L}_{\text{FVc}}^\lambda P_{J,n+1}(X_s) \mathrm{d}s \quad (47) $$

is a $\mathbb{P}_\mu$-martingale for all $\mu \in \mathcal{M}_1^{c,5}(\mathbb{R})$.

**Lemma 6.4.** Let $\lambda \in \mathbb{R}$ be fixed. If $\gamma \in \mathcal{M}_1(\mathcal{M}_1(\mathbb{R}))$ satisfies $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma (d\mu) < \infty$ and

$$ \forall n \in \mathbb{N}^*, \forall f \in \mathscr{C}_{\parallel \cdot \parallel}^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^\lambda P_{f,n}(\mu) \gamma (d\mu) = 0, $$

then, $\gamma = \pi^\lambda$.
Compared to the duality identity of [16], the key point of this lemma is that the right-hand side of the last equation, is a polynomial in $\mu$.

*Proof.* Let us consider $t \ge 0$, $k \in \mathbb{N}^*$, $\ell, m \in \mathbb{N}$. As $\vartheta'_{k,\ell,m} \le \vartheta_{k,\ell}$, the weak duality identity of [16, Theorem 3.4] implies that:

$$ 
\begin{aligned}
\mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, \ell, m}}^{M(0)} \right\rangle \right) &= \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_{t \wedge \vartheta'_{k, \ell, m}}, \mu^{M(t \wedge \vartheta'_{k, \ell, m})} \right\rangle \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right) \\
&= \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)}, \mu^{M(0)+k} \right\rangle \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right).
\end{aligned}
$$

Now, from Lemma 6.5 and the definition of $\vartheta'_{k,\ell,m}$, note that for all $x \in \mathbb{R}^{M(0)+k}$,

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right) \le m (1 + \|x\|_{\infty}^{2k}) \exp \left( \lambda t (M(0) + k)^2 \right). \quad (51) $$

Since $\mu \in \mathcal{M}_1^{c,2k}(\mathbb{R})$, we deduce from FUBINI's theorem that

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, \ell, m}}^{M(0)} \right\rangle \right) = \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right) \right\rangle, \mu^{M(0)+k} \right\rangle. $$

On the one hand, from [16, Section 6.2.1], $\lim_{\ell \to +\infty} \vartheta'_{k,\ell,m} = \vartheta'_{k,m} \mathbb{P}_{(\mu, \xi_0)}$-a.s., and since $(X_t)_{t \ge 0}$ has continuous paths for the topology of weak convergence, we have $\mathbb{P}_{(\mu, \xi_0)}$-a.s.

$$ \lim_{\ell \to +\infty} \left\langle \xi_0, X_{t \wedge \vartheta'_{k, \ell, m}}^{M(0)} \right\rangle = \left\langle \xi_0, X_{t \wedge \vartheta'_{k, m}}^{M(0)} \right\rangle. $$

Therefore, we deduce from the dominated convergence theorem, that for all $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$,

$$ \lim_{\ell \to +\infty} \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, \ell, m}}^{M(0)} \right\rangle \right) = \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, m}}^{M(0)} \right\rangle \right). $$

On the other hand, Lemma 6.5 and the dominated convergence theorem imply that for all $x \in \mathbb{R}^{M(0)+k}$,

$$ 
\begin{aligned}
\lim_{\ell \to +\infty} \mathbb{E}_{(\mu, \xi_0)} & \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right) \\
&= \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, m}} M^2(u) du \right) \right)
\end{aligned}
$$

and the limit satisfies the inequality (51). As $\mu \in \mathcal{M}_1^{c,2k}(\mathbb{R})$, we deduce again from the dominated convergence theorem that

$$ 
\begin{aligned}
\lim_{\ell \to +\infty} \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle \\
= \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle
\end{aligned}
$$
## 6.1 Proof of Proposition 6.3

Let us define for all $\ell \in \mathbb{N}$, $\mathcal{C}_K^\ell(\mathbb{R}^n, \mathbb{R})$ the space of real functions of class $\mathcal{C}^\ell(\mathbb{R}^n, \mathbb{R})$ with compact support. From Lemma 3.4 and (13), for all $t \in [0, T]$, for all $n \in \mathbb{N}^*$ we have for all $f \in \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R})$ that

$$K^2 \sigma_K^2 M_{t \wedge \tau^{K,R}}^{K,P_{f,n}} = \sum_{i=1}^{3} (\mathbf{A})_i^{K,R} (t \wedge \tau^{K,R})$$

where for all $t \le \tau^{K,R} \wedge T$

$$(\mathbf{A})_1^{K,R}(t) := K^2 \sigma_K^2 (P_{f,n}(\mu_t^{K,R}) - P_{f,n}(\mu_0^{K,R})),$$

$$(\mathbf{A})_2^{K,R}(t) := - \int_0^t \int_{M_1(\mathbb{R})} \theta(z_s^{K,R}) m_2(z_s^{K,R}) \mathcal{L}_{\text{FVc}}^{\lambda(z_s^{K,R})} P_{f,n}(\mu) \Gamma^{K,R}(\text{d}s, \text{d}\mu),$$

$$(\mathbf{A})_3^{K,R}(t) := O\left(\frac{1}{\sqrt{K}} + \sigma_K K^{\frac{3}{2}+\varepsilon} + \int_0^t \frac{M_3(\mu_s^{K,R})}{K} \mathrm{d}s\right),$$

is a martingale. Since $\sigma_K K \to 0$ by Assumption (5), $(\mathbf{A})_1^{K,R}(t \wedge \tau^{K,R}) \to 0$ when $K \to +\infty$. From Corollary 4.4, the sequence $((\mathbf{A})_2^{K,R}(t \wedge \tau^K))_{t \in [0,T], K \in \mathbb{N}^*}$ is uniformly integrable and converges in law, when $K \to +\infty$, to

$$M_t^{P_{f,n}} := - \int_0^t \theta(\zeta_s^R) m_2(\zeta_s^R) \int_{M_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_s^R)} P_{f,n}(\mu) \Gamma^R(\mathrm{d}s, \mathrm{d}\mu).$$

Note that, from CAUCHY-SCHWARZ's inequality and Lemma 4.3, there exists a constant $C > 0$ such that

$$\mathbb{E} \left( \sup_{0 \le t \le T} \int_0^{t \wedge \tau^{K,R}} \frac{M_3(\mu_s^{K,R})}{K} \mathrm{d}s \right)^2 \le \frac{T}{K^2} \mathbb{E} \left( \int_0^{T \wedge \tau^{K,R}} M_6(\mu_s^{K,R}) \mathrm{d}s \right) \le \frac{CT^2}{K^2}.$$

Hence, $\lim_{K \to +\infty} \sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}) = 0$ in $L^2(\mathbb{R})$. In particular, we deduce that a subsequence of $(\sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}))_{K \in \mathbb{N}^*}$ converges almost surely to 0 and the family $(\sup_{0 \le t \le T} (\mathbf{A})_3^{K,R}(t \wedge \tau^{K,R}))_{K \in \mathbb{N}^*}$ is uniformly integrable along this subsequence [84, Theorem 13.7]. Using the same method based on SKOROHOD's representation theorem as in the proof of Theorem 5.1, we deduce that the process $(M_t^{P_{f,n}})_{t \ge 0}$ is a $\mathbb{Q}$-martingale. As it is also a continuous and finite variation process by Lemma 6.1, it must hence be $\mathbb{Q}$-a.s. null [48, Theorem 4.1]. Hence, using Lemma 6.1 again, we have proved that

$$\forall f \in \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R}), \mathbb{Q}-\text{a.s., } dt - \text{a.e., } \int_{M_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0.$$

The space $\mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R})$ equipped with the norm

$$\|f\|_{W_0^{2,\infty}} = \|f\|_{\infty} + \|\nabla f\|_{\infty} + \|\text{Hess}(f)\|_{\infty}$$

is separable. So, we can choose a dense countable family $\mathcal{B} \subset \mathcal{C}_K^3(\mathbb{R}^n, \mathbb{R})$ such that

$$\forall f \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R}), \exists (f_q)_{q \in \mathbb{N}^*} \in \mathcal{B}^{\mathbb{N}^*}, f_q \xrightarrow[q \to +\infty]{\|\cdot\|_{W_0^{2,\infty}}} f.$$
and the announced result follows from the fact that $\vartheta'_{k,m}$ is independent of $(X_t)_{t\ge 0}$, which implies

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) \mathrm{d}u \right) \right) = \mathbb{E}_{\xi_0} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) \mathrm{d}u \right) \right). \quad \square $$

From Lemma 6.5, we deduce that we can choose $m \in \mathbb{N}$ large enough such that $\vartheta'_{1,m} = \tau_1$. In this case, the function inside the brackets in the left-hand side of (50) is defined, for all $t \ge 0$ and for all $x \in \mathbb{R}^{M(0)+1}$, by

$$ \psi_t(x) := \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \tau_1}^{(M(0)+1)}(x) \exp \left( \lambda \int_0^{t \wedge \tau_1} M^2(u) \mathrm{d}u \right) \right). $$

For all $t \ge 0$, $n \in \mathbb{N}^*$, $f \in \mathscr{C}_b^2(\mathbb{R}^n, \mathbb{R})$ and $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$ let us denote

$$ P_{\mathrm{FVc}}(t) P_{f,n}(\mu) := \mathbb{E}_{\mu} (P_{f,n}(X_t)) $$

the semi-group of the centered FLEMING-VIOT process $(X_t)_{t\ge 0}$ with resampling rate $\lambda$.

The following result is the main result of this section. It gives an extension of Lemma 6.5 to a function which appears naturally in the proof of Lemma 6.4.

**Proposition 6.7.** For all $t \ge 0$, for all $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$, for all $\xi_0 \in \mathscr{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$

$$ P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)}(\mu) = P_{V_t, M(0)+1}(\mu) := \langle V_t, \mu^{M(0)+1} \rangle $$

where

$$ V_t := \exp(\alpha_0 t) \psi_t - \alpha_0 \int_0^t \exp(\alpha_0 s) \psi_s ds \quad (52) $$

and $\alpha_0 := \lambda M(0) [2M(0) - 1]$. In addition, for all $t \ge 0$, there exist a constant $C_t > 0$ such that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ |V_t(x)| \le C_t (1 + \|x\|_{\infty}^2). $$

*Proof.* Let $t \ge 0$, $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$, $\xi_0 \in \mathscr{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ be fixed. Note that from Lemma 6.5, there exists a constant $\tilde{C}_t$, depending only on $\xi_0$ and $t$ such that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ |\psi_t(x)| \le C_0(\tau_1, 1) (1 + \|x\|_{\infty}^2) \mathbb{E}_{(\mu, \xi_0)} (\exp(\lambda t M^2(0))) \le \tilde{C}_t (1 + \|x\|_{\infty}^2). \quad (53) $$

Now, from Lemma 6.6 and the fact that $\tau_1$ is independent of $(X_t)_{t\ge 0}$, we obtain that

$$ \langle \psi_t, \mu^{M(0)+1} \rangle = \mathbb{E}_{(\mu, \xi_0)} (P_{\xi_0, M(0)}(X_{t \wedge \tau_1})) = \mathbb{E}_{\xi_0} (P_{\mathrm{FVc}}(t \wedge \tau_1) P_{\xi_0, M(0)}(\mu)). $$

As $\tau_1$ follows an exponential law of parameter $\alpha_0$ we have that

$$ \langle \psi_t, \mu^{M(0)+1} \rangle = P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)}(\mu) \exp(-\alpha_0 t) + \alpha_0 \int_0^t P_{\mathrm{FVc}}(s) P_{\xi_0, M(0)}(\mu) \exp(-\alpha_0 s) \mathrm{d}s. \quad (54) $$
Then,

$$ \mathbb{Q}-\text{a.s., } dt-\text{a.e., } \forall q \in \mathbb{N}^*, \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f_q,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0. $$

From (42) and Lemma 6.1, we have that $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma_t^R(\mathrm{d}\mu) < \infty$. As for all $q \in \mathbb{N}^*$,

$$ \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f_q,n}(\mu) \le C_1 (1 + M_2(\mu)) \|f_q\|_{W_0^{2,\infty}}. $$

for some constant $C_1 > 0$, we obtain by the dominated convergence theorem that

$$ \mathbb{Q}-\text{a.s., } dt-\text{a.e., } \forall f \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0. \quad (48) $$

Let us consider $f \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^n, \mathbb{R})$ and for all $q \in \mathbb{N}^*, x \mapsto \chi_q(x) = \exp\left(-\frac{1}{q^2 - \|x\|_\infty^2}\right) \mathbb{1}_{\|x\|_\infty < q}$ of class $\mathcal{C}^\infty(\mathbb{R}^n, \mathbb{R})$ with compact support. Then, for all $q \in \mathbb{N}^*, f\chi_q \in \mathcal{C}_K^2(\mathbb{R}^n, \mathbb{R})$. Noting that, $\left|\int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f\chi_q,n}(\mu) \gamma_t^R(\mathrm{d}\mu)\right|$ is dominated by $C_2 \int_{\mathcal{M}_1(\mathbb{R})} (1 + M_4(\mu)) \gamma_t^R(\mathrm{d}\mu)$ for some constant $C_2 > 0$, and since $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu) \gamma_t^R(\mathrm{d}\mu) < \infty$, we deduce from (48) applied to $f\chi_q$, by the dominated convergence theorem, that

$$ \mathbb{Q}-\text{a.s., } dt-\text{a.e., } \forall f \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^n, \mathbb{R}), \quad \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda(\zeta_t^R)} P_{f,n}(\mu) \gamma_t^R(\mathrm{d}\mu) = 0. $$

Therefore, it follows from Lemma 6.4 that $\mathbb{Q}-\text{a.s., } dt-\text{a.e., } \gamma_t^R(\mathrm{d}\mu) = \pi^{\lambda(\zeta_t^R)}(\mathrm{d}\mu)$ which concludes the proof.

## 6.2 Proof of Lemma 6.4

In Section 6.2.1 we extend some duality results for the centered FLEMING-VIOT process, obtained in [18], which are be useful to prove Lemma 6.4 in Section 6.2.2.

### 6.2.1 Extension of the duality result for the centered Fleming-Viot process

Let us recall that the dual process $(\xi_t)_{t \ge 0}$ of the centered FLEMING-VIOT process $(X_t)_{t \ge 0}$ with resampling rate $\lambda$, on the state space $\bigcup_{n \in \mathbb{N}^*} \mathcal{C}^2(\mathbb{R}^n, \mathbb{R})$, obtained in [18, Section 3.2] is defined as below.

Let us consider $M := (M(t))_{t \ge 0}$ a MARKOV'S birth and death process in $\mathbb{N}$ whose transition rates $q_{i,j}$ from $i$ to $j$ are given by:

(1) $q_{n,n+1} = \lambda n^2$

(2) $q_{n,n-1} = \lambda n(n-1)$

(3) $q_{i,j} = 0$ otherwise.

For all $M(0) \in \mathbb{N}^*, \xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ and $\lambda > 0$, we define

$$ \xi_t := T_{\lambda}^{(M(\tau_n))} (t - \tau_n) \Lambda_n T_{\lambda}^{(M(\tau_{n-1}))} (\tau_n - \tau_{n-1}) \Lambda_{n-1} \cdots \Lambda_1 T_{\lambda}^{(M(0))} (\tau_1) \xi_0, \quad \tau_n \le t < \tau_{n+1}, n \in \mathbb{N}, \quad (49) $$
Thanks to (53) and since $\mu$ has its moment of order 2 finite, the FUBINI theorem ensures us that

$$ \left\langle \int_{0}^{t} \exp(\alpha_{0}s) \psi_{s} ds, \mu^{M(0)+1} \right\rangle = \int_{0}^{t} \exp(\alpha_{0}s) \left\langle \psi_{s}, \mu^{M(0)+1} \right\rangle ds. $$

From (54) we deduce that

$$ \begin{aligned} & \left\langle \int_{0}^{t} \exp(\alpha_{0}s) \psi_{s} ds, \mu^{M(0)+1} \right\rangle \\ &= \alpha_{0} \int_{0}^{t} \exp(\alpha_{0}s) \int_{0}^{s} \exp(-\alpha_{0}u) P_{FVc}(u) P_{\xi_{0},M(0)}(u) du ds + \int_{0}^{t} P_{FVc}(s) P_{\xi_{0},M(0)}(s) ds \quad (55) \\ &= \exp(\alpha_{0}t) \int_{0}^{t} \exp(-\alpha_{0}u) P_{FVc}(u) P_{\xi_{0},M(0)}(u) du. \end{aligned} $$

Then, the first announced result follows from (54) and (55) and the second one from (53). □

### 6.2.2 Proof of Lemma 6.4

The key point of this proof is to establish that for all $\xi_{0} \in \mathcal{C}_{b}^{4}(\mathbb{R}^{M(0)}, \mathbb{R})$, $t \ge 0$ and $\mu \in \mathcal{M}_{1}^{c,4}(\mathbb{R})$,

$$ \begin{aligned} P_{FVc}(t)P_{\xi_{0},M(0)}(\mu) &= \left\langle \xi_{0}, \mu^{M(0)} \right\rangle + \int_{0}^{t} \mathcal{L}_{FVc}^{\lambda} P_{FVc}(s) P_{\xi_{0},M(0)}(\mu) ds \\ &= P_{\xi_{0},M(0)}(\mu) + \int_{0}^{t} \mathcal{L}_{FVc}^{\lambda} P_{V_{s},M(0)+1}(\mu) ds. \end{aligned} \quad (56) $$

where we use Proposition 6.7 in the second equality. To do this, we will first prove that for all $t \ge 0$, there exists a constant $C_{t} > 0$ such that for all $x \in \mathbb{R}^{M(0)+1}$, $\|\text{Hess}(V_{t})(x)\|_{\infty} \le C_{t}(1 + \|x\|_{\infty}^{2})$, so that $V_{t} \in \mathcal{C}_{\|\cdot\|}^{2}(\mathbb{R}^{M(0)+1}, \mathbb{R})$.

For all $n \in \mathbb{N}^{*}$, for any function $f: \mathbb{R}^{n} \to \mathbb{R}$, $k \in \mathbb{N}$ times differentiable, we denote by $D^{k}$ the differential of order $k$ of $f$. Let $t \ge 0$, $\lambda > 0$, $\mu \in \mathcal{M}_{1}^{c,2}(\mathbb{R})$ and $\xi_{0} \in \mathcal{C}_{b}^{4}(\mathbb{R}^{M(0)}, \mathbb{R})$ be fixed.

**Step 1. Preliminary bounds.** Let $x \in \mathbb{R}^{M(0)}$. It follows from (47) that

$$ \begin{aligned} & \partial_{x_i x_j x_k x_l}^{4} T^{(M(0))}(t) \xi_0(x) \\ &= \sum_{m,p,q,r=1}^{M(0)} \partial_{x_i} (m_{t,x}^{\lambda})_m \partial_{x_j} (m_{t,x}^{\lambda})_p \partial_{x_k} (m_{t,x}^{\lambda})_q \partial_{x_l} (m_{t,x}^{\lambda})_r (g_{t,0}^{\lambda} * \partial_{y_m y_p y_q y_r}^{4} \xi_0) (m_{t,x}^{\lambda}) \end{aligned} $$

where $(m_{t,x}^{\lambda})_p$ designates the $p^{th}$ component of $m_{t,x}^{\lambda}$. Then, for all $k \in \{0, \dots, 4\}$, there exists a constant $C_1 > 0$ independent of $x$ and $t$ such that

$$ \| D^k T^{(M(0))}(t) \xi_0(x) \|_{\infty} \le C_1 \left( 1 + \| D^k \xi_0 \|_{\infty} \right). $$

From (45), (23) and (24), we deduce that there exists a constant $C_2 > 0$ independent of $t$ such
where $(\tau_n)_{n \in \mathbb{N}}$ is the sequence of jump times of $M$ with $\tau_0 = 0$, $(T_\lambda^{(n)}(t))_{t \ge 0}$ is the semi-group of operator associated to the generator $B_\lambda^{(n)}$ given by (23) and where $(\Lambda_n)_{n \in \mathbb{N}}$ is a sequence of random operators. These are conditionally independent given $M$ and satisfy for all $k \in \mathbb{N}$, $n \ge 1$ and $1 \le i \ne j \le n$,

$$ \mathbb{P} \left( \Lambda_k = \Phi_{i,j} \mid \{ M(\tau_k^-) = n, M(\tau_k) = n-1 \} \right) = \frac{1}{n(n-1)} $$

and for all $n \ge 1$ and $1 \le i, j \le n$,

$$ \mathbb{P} \left( \Lambda_k = K_{i,j} \mid \{ M(\tau_k^-) = n, M(\tau_k) = n+1 \} \right) = \frac{1}{n^2}, $$

where $\Phi_{i,j}$ and $K_{i,j}$ are respectively defined in (24) and (25). Note that if $M(0) = 1$, the dual process can only jump from $\xi_0$ to $K_{ij}\xi_0$. As for all $t \ge 0$, $\xi_t \in \mathcal{C}^2(\mathbb{R}^{M(t)}, \mathbb{R})$, $K_{i,j}\xi_t$ is well defined. Moreover, the dual process $(\xi_t)_{t \ge 0}$ with initial condition $\xi_0$ is constructed on the same probability space and independently of the centered FLEMING-VIOT process $(X_t)_{t \ge 0}$ with resampling rate $\lambda$ and initial condition $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$. We shall denote by $\mathbb{P}_{(\mu, \xi_0)}$, the law of the couple $((X_t, \xi_t))_{t \ge 0}$ on this probability space.

We denote by $S_t$ the number of jumps of the process $M$ on $[0, t]$. We start with a result giving bounds on the dual process, which is an extension of similar estimates obtained in [18, Lemma 6.4].

**Lemma 6.5.** For all $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ there exists $C_0 : \bigcup_{k \in \mathbb{N}^*} (0, +\infty)^{k-1} \times \{k\} \to \mathbb{R}_+$, locally bounded, such that

$$ C_0(\tau_1, 1) := \|\xi_0\|_{\infty} + \frac{3}{M(0)} \|\text{Hess}(\xi_0)\|_{\infty} $$

and, for all $(t_j)_{j \in \mathbb{N}} \in (0, +\infty)^{\mathbb{N}}$, $k \mapsto C_0((t_i)_{0 \le i \le k-1}, k)$ is non-decreasing and satisfying

$$ \begin{gathered} \forall k \in \mathbb{N}^*, \forall t \le \tau_k, \forall x \in \mathbb{R}^{M(t)}, \\ |\xi_t(x)| \le C_0((\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k) (1 + \|x\|_{\infty}^{2S_t}). \end{gathered} \quad (50) $$

*Proof.* By induction on $k \in \mathbb{N}^*$, we prove the property

$$ (Q_k) : \quad \forall x \in \mathbb{R}^{M(\tau_k)}, \quad |\xi_{\tau_k}(x)| \le C_0((\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k) (1 + \|x\|_{\infty}^{2k}). $$

**Step 1.** Initial case: computation of $C_0(\tau_1, 1)$. Let us recall some notations of [18, Theorem 6.1]. For all $n \in \mathbb{N}^*$, we denote by $\mathbf{1} \in \mathbb{R}^n$, the vector whose coordinates are all 1. For all $n \in \mathbb{N}^*$, $t \ge 0$, $x \in \mathbb{R}^n$ and $\lambda > 0$, let us consider $g_{t,x}^\lambda$ the density of the Gaussian distribution $\mathcal{N}^{(n)}(m_{t,x}^\lambda, \Sigma_t^\lambda)$ where $m_{t,x}^\lambda := x - \frac{(1-\exp(-2\lambda nt))}{n}(x \cdot \mathbf{1})\mathbf{1}$ and $\Sigma_t^\lambda := P\sigma_t^\lambda P^{-1}$ with

$$ \sigma_t^\lambda := \begin{pmatrix} \frac{1-\exp(-4\lambda nt)}{4\lambda n} & 0 & \dots & 0 \\ 0 & t & 0 & \dots \\ \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & t \end{pmatrix} \quad \text{and} \quad P := \begin{pmatrix} \frac{1}{\sqrt{n}} & \frac{1}{\sqrt{2}} & \cdots & \frac{1}{\sqrt{n(n-1)}} \\ \vdots & -\sqrt{\frac{1}{2}} & \cdots & \frac{1}{\sqrt{n(n-1)}} \\ \vdots & 0 & \ddots & \vdots \\ \vdots & \vdots & \ddots & \frac{1}{\sqrt{n(n-1)}} \\ \frac{1}{\sqrt{n}} & 0 & \cdots & -\sqrt{\frac{n-1}{n}} \end{pmatrix}. $$
that for all $x \in \mathbb{R}^{M(0)+1}$, for all $k \in \{0, 1, 2\}$,

$$ \begin{aligned} \|D^k \xi_{t \wedge \tau_1}^{(M(0)+1)}(x)\|_{\infty} &\le C_1 (1 + \|D^k \xi_0\|_{\infty}) + \sum_{\substack{i,j=1 \\ i \ne j}}^n \|D^k \Phi_{i,j} T_{\lambda}^{(M(0))} (\tau_1) \xi_0 (x_1, \dots, x_{M(0)-1})\|_{\infty} \\ &\quad + \sum_{i,j=1}^n \|D^k K_{i,j} T_{\lambda}^{(M(0))} (\tau_1) \xi_0(x)\|_{\infty} \\ &\le C_2 (1 + \|D^k \xi_0\|_{\infty} + \|D^{k+2} \xi_0\|_{\infty} x_{M(0)+1}^2). \end{aligned} $$

By the theorem of differentiation under the integral sign, for all $k \in \{0, 1, 2\}$ and $x \in \mathbb{R}^{M(0)+1}$, we deduce that

$$ \|D^k \psi_t(x)\|_{\infty} \le \exp(\lambda M(0)^2 t) \mathbb{E}_{(\mu, \xi_0)} \left( \|D^k \xi_{t \wedge \tau_1}^{(M(0)+1)}(x)\|_{\infty} \right) \le C_2 \exp(\lambda M(0)^2 t) (1 + \|x\|_{\infty}^2). $$

Then, from the definition of $V_t$ in (52) we deduce that there exists a constant $C_t$ such that for all $k \in \{0, 1, 2\}$, for all $x \in \mathbb{R}^{M(0)+1}$, $\|D^k V_t(x)\|_{\infty} \le C_t (1 + \|x\|_{\infty}^2)$, so that $V_t \in \mathcal{C}_{\|\cdot\|}^2 (\mathbb{R}^{M(0)+1}, \mathbb{R})$.

**Step 2.** Proof of the key point (56). Thanks to (26), for all $t \ge 0$,

$$ P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)}(\mu) = P_{\xi_0, M(0)}(\mu) + \mathbb{E}_{\mu} \left( \int_{0}^{t} \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_s) \mathrm{d}s \right). $$

From (25), there exists a constant $C_3 > 0$ such that for all $t \ge 0$, $\mathbb{E}_{\mu} (|\mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_t)|) \le C_3 (1 + \mathbb{E}_{\mu} (M_2 (X_t)))$ which is finite from [16, Proposition 2.11] since $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$. Hence, from FUBINI's theorem we obtain that

$$ P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)}(\mu) = P_{\xi_0, M(0)}(\mu) + \int_0^t \mathbb{E}_{\mu} \left( \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_s) \right) \mathrm{d}s. \quad (57) $$

As $\mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)}(\mu)$ is a polynomial in $\mu$ and since $|\mathcal{L}_{\mathrm{FVc}}^{\lambda} (\mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)}(\mu))| \le C_4 (1 + M_4(\mu))$ for some constant $C_4 > 0$, we deduce as in (43) that

$$ \mathbb{E}_{\mu} \left( \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_t) \right) - \mathbb{E}_{\mu} \left( \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_0) \right) = \mathbb{E}_{\mu} \left( \int_{0}^{t} \mathcal{L}_{\mathrm{FVc}}^{\lambda} \left( \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} \right) (X_s) \mathrm{d}s \right). $$

In particular, $t \mapsto \mathbb{E}_{\mu} (\mathcal{L}_{\mathrm{FVc}} P_{\xi_0, M(0)} (X_t))$ is continuous, so

$$ \begin{aligned} \mathbb{E}_{\mu} \left( \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_t) \right) &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} \left( \int_{t}^{t+h} \mathcal{L}_{\mathrm{FVc}} P_{\xi_0, M(0)} (X_s) \mathrm{d}s \right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} \left( P_{\xi_0, M(0)} (X_{t+h}) - P_{\xi_0, M(0)} (X_t) \right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} \left( P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)} (X_h) - P_{\mathrm{FVc}}(t) P_{\xi_0, M(0)}(\mu) \right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} \left( P_{V_t, M(0)+1} (X_h) - P_{V_t, M(0)+1}(\mu) \right) \end{aligned} $$
Note that for all $i, j \in \{1, \dots, n\}$, $\partial_{x_i} m_{t,x}^{\lambda} = \epsilon_i - \frac{1-\exp(-2\lambda nt)}{n} \mathbf{1}$ and $\partial_{x_i x_j}^2 m_{t,x}^{\lambda} = 0$ where $(\epsilon_1, \dots, \epsilon_n)$ is the canonical basis of $\mathbb{R}^n$. For all $\lambda > 0$, the key identity for the sequel is

$$ \forall f \in L^{\infty}(\mathbb{R}^n), \forall t \ge 0, \forall x \in \mathbb{R}^n, \quad T_{\lambda}^{(M(0))}(t)f(x) = (f * g_{t,0}^{\lambda})(m_{t,x}^{\lambda}) \quad (51) $$

where * stands for the convolution product (see (46) in [18, Theorem 6.1]).

On the one hand, note that for all $t < \tau_1$, $M(t) = M(0)$. In this case, for all $x \in \mathbb{R}^{M(0)}$, $\lambda > 0$, $\xi_t(x) = T_{\lambda}^{(M(0))}(t)\xi_0(x)$ and so, from [18, Theorem 6.1], $|\xi_t(x)| \le \|\xi_0\|_{\infty}$. On the other hand, at time $\tau_1$, we make a partition of cases according to whether the dual process loses or gains a variable. Let $i, j \in \{1, \dots, M(0)\}$ be fixed.

* Case $\Lambda_1 = \Phi_{i,j}$. In this case, $M(\tau_1) = M(0) - 1$ and we deduce from (49) that

$$ \forall x \in \mathbb{R}^{M(0)-1}, \quad \xi_{\tau_1}(x) = \Phi_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x). $$

By (24), we deduce that for all $x \in \mathbb{R}^{M(0)-1}$, $|\xi_{\tau_1}(x)| \le \|\xi_0\|_{\infty}$.

* Case $\Lambda_1 = K_{i,j}$. In this case, $M(\tau_1) = M(0) + 1$ and we deduce from (49) and (25) then from [18, Theorem 6.1 (3)] and properties of the convolution product that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ \begin{aligned} \xi_{\tau_1}(x) &= K_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x) \\ &= \partial_{x_i x_j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(\tilde{x}) x_{M(0)+1}^2 \\ &= (\partial_{x_j} m_{\tau_1, \tilde{x}}^{\lambda})^t [(\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda})(m_{\tau_1, \tilde{x}}^{\lambda}) \partial_{x_i} m_{t, \tilde{x}}^{\lambda}] x_{M(0)+1}^2 \end{aligned} $$

where $\tilde{x} = (x_1, \dots, x_{M(0)}) \in \mathbb{R}^{M(0)}$. As,

$$ (\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda})(m_{\tau_1, \tilde{x}}^{\lambda}) = \left( \int_{\mathbb{R}^{M(0)}} \partial_{ij}^2 \xi_0(u) g_{\tau_1, 0}^{\lambda} (m_{\tau_1, \tilde{x}}^{\lambda} - u) \, du \right)_{1 \le i,j \le M(0)} $$

we obtain that $\|(\text{Hess}(\xi_0) * g_{\tau_1, 0}^{\lambda})(m_{\tau_1, \tilde{x}}^{\lambda})\|_{\infty} \le \|\partial_{ij}^2 \xi_0\|_{\infty}$. Noting that

$$ |(\partial_{x_j} m_{\tau_1, \tilde{x}}^{\lambda})^t \partial_{x_i} m_{t, \tilde{x}}^{\lambda}| \le \frac{3}{M(0)}, $$

we deduce that for all $x \in \mathbb{R}^{M(0)+1}$,

$$ |\xi_{\tau_1}(x)| \le \frac{3}{M(0)} \sup_{1 \le i,j \le M(0)} \|\partial_{ij}^2 \xi_0\|_{\infty} x_{M(0)+1}^2 \le \frac{3}{M(0)} \|\text{Hess}(\xi_0)\|_{\infty} \|x\|_{\infty}^2 $$

and $(Q_1)$ follows.

**Step 2. Inductive Step.** We assume that, for $k \in \mathbb{N} \setminus \{0, 1\}$, $(Q_{k-1})$ is satisfied and prove that $(Q_k)$ is also. We make again a partition of cases according to whether the dual process loses or gains a variable. Let $i, j \in \{1, \dots, M(\tau_{k-1})\}$ be fixed.
where we used MARKOV'S property in the third equality and Proposition 6.7 in the last equality. From Step 1, we deduce that there exists a constant $C_5 > 0$ such that

$$ \left| \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(\mu) \right| \le C_5 (1 + M_4(\mu)). \qquad (58) $$

Thus, from the martingale problem (43) and the continuity of $t \mapsto \mathbb{E}_{\mu}(\mathcal{L}_{\text{FVc}}P_{V_t, M(0)}(\mu))$, we have that

$$ \begin{aligned} \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} (P_{V_t, M(0)+1}(X_h) - P_{V_t, M(0)+1}(\mu)) &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu} \left( \int_0^h \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(X_s) \, ds \right) \\ &= \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(\mu). \end{aligned} $$

Therefore, we have proved that for all $t \ge 0$, for all $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$,

$$ \mathbb{E}_{\mu} \left( \mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)} (X_t) \right) = \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(\mu), $$

and so, by (57), we obtain (56).

**Step 3. Conclusion.** From (56) and (58) and since $\int_{\mathcal{M}_1(\mathbb{R})} M_4(\mu)\gamma(d\mu) < \infty$, we deduce that

$$ \begin{aligned} & \int_{\mathcal{M}_1(\mathbb{R})} P_{\text{FVc}}(t) P_{\xi_0, M(0)}(\mu) \gamma(d\mu) \\ &= \int_{\mathcal{M}_1(\mathbb{R})} P_{\xi_0, M(0)}(\mu) \gamma(d\mu) + \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_s, M(0)+1}(\mu) \gamma(d\mu) ds \\ &= \int_{\mathcal{M}_1(\mathbb{R})} P_{\xi_0, M(0)}(\mu) \gamma(d\mu), \end{aligned} $$

where we used the assumption of Lemma 6.4 in the last equality.

As the set of test functions $\{P_{f,n} \mid f \in \mathcal{C}_b^4(\mathbb{R}^n, \mathbb{R}), n \in \mathbb{N}^*\}$ is $\mathcal{M}_1(\mathcal{M}_1(\mathbb{R}))$-convergence determining [22, Lemma 2.1.2], it is $\mathcal{M}_1(\mathcal{M}_1(\mathbb{R}))$-separating [34, Chapter 3, Section 4, p.112], so we have for any bounded continuous function $\phi$ from $\mathcal{M}_1(\mathbb{R})$ to $\mathbb{R}$ that

$$ \int_{\mathcal{M}_1(\mathbb{R})} P_{\text{FVc}}(t) \phi(\mu) \gamma(d\mu) = \int_{\mathcal{M}_1(\mathbb{R})} \phi(\mu) \gamma(d\mu). $$

Hence $\gamma$ is an invariant probability measure for the centered FLEMING-VIOT process with resampling rate $\lambda$. Now, from [16, Theorem 4.1], $\pi^{\lambda}$ is its unique invariant probability measure which ends the proof of Lemma 6.4.

# 7 Characterisation of the limiting values of the slow component

Combining the results of Sections 5 and 6, we have proved that for all $R > 0$, the sequence of laws of $(z^{K,R}, \Gamma^{K,R})_{K \in \mathbb{N}^*}$ is tight in $\mathcal{M}_1(\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R})))$ and for any limiting value $Q$
• Case $\Lambda_k = \Phi_{i,j}$ at the $k^{\text{th}}$ jump. In this case, $M(\tau_k) = M(\tau_{k-1}) - 1$ and we deduce from (49) that for all $x \in \mathbb{R}^{M(\tau_{k-1})-1}$, $\xi_{\tau_k}(x) = \Phi_{i,j} T_{\lambda}^{(M(\tau_{k-1}))}(\tau_k - \tau_{k-1})\xi_{\tau_{k-1}}(x)$. By using (24) and $(\mathcal{Q}_{k-1})$, we deduce from [18, Corollary 6.2 (1)] that for all $x \in \mathbb{R}^{M(\tau_{k-1})-1}$,

$$ \left| \Phi_{i,j} T_{\lambda}^{(M(\tau_{k-1}))} (\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x) \right| \le C_2 (\tau_k - \tau_{k-1}, M(\tau_{k-1})) C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \left( 1 + \|x\|_{\infty}^{2(k-1)} \right), $$

where $C_2C_0$ is locally bounded. As $t \mapsto C_2(t, M(\tau_{k-1}))$ is non-decreasing, we deduce $(\mathcal{Q}_k)$ in that case.

• Case $\Lambda_k = K_{i,j}$ at the $k^{\text{th}}$ jump. In this case, $M(\tau_k) = M(\tau_{k-1}) + 1$. From (49), (25) and [18, Theorem 6.1 (3)], we have for all $x \in \mathbb{R}^{M(\tau_{k-1})+1}$,

$$ \begin{aligned} |\xi_{\tau_k}(x)| &= |K_{i,j} T_{\lambda}^{(M(\tau_{k-1})+1)} (\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x)| \\ &= \left| \left( \partial_{x_j} m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda} \right)^t \left[ \left( \xi_{\tau_{k-1}} * \text{Hess} \left( g_{\tau_k - \tau_{k-1}, 0}^{\lambda} \right) \right) (m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda}) \partial_{x_i} m_{\tau_k - \tau_{k-1}, \tilde{x}}^{\lambda} \right] \right| \\ & \qquad \times x_{M(\tau_{k-1})+1}^2, \end{aligned} $$

where $\tilde{x} = (x_1, \dots, x_{M(\tau_{k-1})})^t \in \mathbb{R}^{M(\tau_{k-1})}$. From $(\mathcal{Q}_{k-1})$ and [18, Corollary 6.2 (2)], we deduce that

$$ \begin{aligned} \left| K_{i,j} T_{\lambda}^{(M(\tau_{k-1})+1)} (\tau_k - \tau_{k-1}) \xi_{\tau_{k-1}}(x) \right| & \le C_3 (\tau_k - \tau_{k-1}, M(\tau_{k-1})) C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \left( 1 + \|x\|_{\infty}^{2k} \right), \end{aligned} $$

where $C_3C_0$ is locally bounded and $(\mathcal{Q}_k)$ follows in that case.

We conclude by the principle of induction.

Step 3. *Proof of (50) for $t < \tau_k$.* Note that from (49), for all $k \in \mathbb{N}$ and $t \in (\tau_k, \tau_{k+1})$,

$$ \forall x \in \mathbb{R}^{M(t)}, \quad \xi_t(x) = T_{\lambda}^{(M(\tau_k))} (t - \tau_k) \xi_{\tau_k}(x), $$

so the announced result follows from [18, Corollary 6.2 (1)]. □

Let us consider for all $k \in \mathbb{N}^*$, $\ell, m \in \mathbb{N}$ the stopping times

$$ \vartheta_{k,\ell} := \inf \left\{ t \ge 0 \mid S_t \ge k \quad \text{or} \quad \exists s \in [0,t], \langle \xi_s, X_{t-s}^{M(s)} \rangle \ge \ell \right\}, $$

$$ \vartheta'_{k,m} := \inf \left\{ t \ge 0 \mid S_t \ge k \quad \text{or} \quad C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \ge m \right\}, $$

$$ \begin{aligned} \vartheta'_{k,\ell,m} := \inf \left\{ t \ge 0 \mid S_t \ge k \quad \text{or} \quad \exists s \in [0,t], \langle \xi_s, X_{t-s}^{M(s)} \rangle \ge \ell \right. \\ \left. \quad \text{or} \quad C_0 \left( (\tau_{i+1} - \tau_i)_{0 \le i \le k-1}, k \right) \ge m \right\}. \end{aligned} $$
of this sequence, the canonical process $(\zeta^R, \Gamma^R)$ on $\mathcal{C}^0([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(\mathcal{M}_1(\mathbb{R}))$ satisfies for all $f \in \mathcal{C}_b^1(\mathbb{T}_R, \mathbb{T}_R)$, for all $t \in [0, T]$, $\mathbb{Q}$-a.s.,

$$f(\zeta_t^R) = f(\zeta_0^R) + \int_0^t \int_{\mathcal{M}_1(\mathbb{R})} \mathcal{L}_{\text{SLOW}} f(\zeta_s^R, \mu) \pi^{\lambda}(\zeta_s^R) (d\mu) \, ds. \quad (59)$$

From now on, we will establish in Section 7.1 with Lemma 7.1 that the sequence of laws of $(z^{K,R})_{K \in \mathbb{N}^*}$ converges weakly in $\mathbb{D}([0, T], \mathbb{T}_R)$ to the solution of an ODE in the torus. Finally, Section 7.2 allows us to get away from the torus and to prove Theorem 2.1.

## 7.1 Convergence of the slow component on the torus

**Lemma 7.1.** The sequence $(z^{K,R})_{K \in \mathbb{N}^*}$ converges in law in $\mathbb{D}([0, T], \mathbb{T}_R)$ to the unique solution of

$$\forall t \in [0, T], \quad z_t = x_0 + \int_0^t \partial_1 \text{Fit}(z_s, z_s) \beta(z_s) m_2(z_s) \, ds \quad (60)$$

on the torus $\mathbb{T}_R$.

*Proof.* For any limiting value $\mathbb{Q}$ of the sequence of laws of $(z^{K,R})_{K \in \mathbb{N}^*}$, from (9) and applying (59) with $f = \text{id} \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, we deduce that $\mathbb{Q}$-a.s.,

$$\forall t \in [0, T], \quad \zeta_t^R = \zeta_0^R + \int_0^t \left( \int_{\mathcal{M}_1(\mathbb{R})} M_2(\mu) \pi^{\lambda}(\zeta_s^R) (d\mu) \right) \partial_1 \text{Fit}(\zeta_s^R, \zeta_s^R) \, ds.$$

From [16, Corollary 4.16], for all $s \in [0, t]$,

$$\int_{\mathcal{M}_1(\mathbb{R})} M_2(\mu) \pi^{\lambda}(\zeta_s^R) (d\mu) = \frac{1}{\lambda(\zeta_s^R)} = \beta(\zeta_s^R) m_2(\zeta_s^R),$$

so $\zeta^R$ is solution of (60) $\mathbb{Q}$-a.s. By uniqueness of the solution of (60), the announced result follows. $\square$

## 7.2 End of the proof of Theorem 2.1

Let $T > 0$ be fixed and recall that and $x_0$ is the mean trait value of $\nu_0^K$. From Lemma 7.1 and since the ODE (60) is non-explosive, we can choose $R > 0$ large enough such that $\lim_{K \to +\infty} \mathbb{P}(z_t^{K,R} \in [x_0 - \frac{R}{2}, x_0 + \frac{R}{2}], \forall t \in [0, T]) = 1$. From Proposition 4.7 and (7), we have that, $\lim_{K \to +\infty} \mathbb{P}(\text{Diam}(\text{Supp } \nu_t^{K,R}) \le \frac{1}{K^{1+\epsilon/2}}, \forall t \in [0, T]) = 1$. Hence,

$$\lim_{K \to +\infty} \mathbb{P}(\text{Supp } \nu_{t/K\sigma^2}^{K,R} \subset [x_0 - R, x_0 + R], \forall t \in [0, T]) = 1.$$

Now, on the event $\{\text{Supp } \nu_{t/K\sigma^2}^{K,R} \subset [x_0 - R, x_0 + R], \forall t \in [0, T]\}$, $\nu_{t/K\sigma^2}^{K,R} = \nu_{t/K\sigma^2}^K$ for all $t \in [0, T]$, identifying $x \in \mathbb{T}_R$ with its unique representant in $[x_0 - 2R, x_0 + 2R]$. In particular,

$$\forall t \in [0, T], \quad z_t^{K,R} = z_t^K.$$

Theorem 2.1 then follows from Lemma 7.1.
As $(M(t), \xi_t)_{t \ge 0}$ is independent of $(X_t)_{t \ge 0}$, note that for all $k \in \mathbb{N}^*$, $m \in \mathbb{N}$, $\vartheta'_{k,m}$ is independent of $(X_t)_{t \ge 0}$. Recall that for all $t \ge 0$, $\xi_t \in \mathcal{C}^2(\mathbb{R}^{M(t)}, \mathbb{R})$. For all $n \ge M(t)$, for all $x \in \mathbb{R}^n$, we denote by $\xi_t^{(n)}(x) = \xi_t^{(n)}(x_1, \dots, x_n) := \xi_t(x_1, \dots, x_{M(t)})$ so that for all $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$,

$$ \langle \xi_t^{(n)}, \mu^n \rangle = \langle \xi_t, \mu^{M(t)} \rangle. $$

The next lemma is an extension of the duality identity proved in [18].

**Lemma 6.6.** Given $(X_t)_{t \ge 0}$, $(\xi_t)_{t \ge 0}$ as above with $X_0 := \mu \in \mathcal{M}_1^{c,2k}(\mathbb{R})$, $k \in \mathbb{N}^*$ and $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$, we have that $\mathbb{P}_{(\mu, \xi_0)}$-a.s., for all $t \ge 0$, for all $m \in \mathbb{N}$,

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k,m}}^{M(0)} \right\rangle \right) = \left\langle \mathbb{E}_{\xi_0} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle. \quad (52) $$

Compared to the duality identity of [18], the key point of this lemma is that the right-hand side of the last equation, is a polynomial in $\mu$.

*Proof.* Let us consider $t \ge 0$, $k \in \mathbb{N}^*$, $l, m \in \mathbb{N}$. As $\vartheta'_{k,l,m} \le \vartheta_{k,l}$, the weak duality identity of [18, Theorem 3.4] implies that:

$$ \begin{aligned} & \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k,l,m}}^{M(0)} \right\rangle \right) \\ &= \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_{t \wedge \vartheta'_{k,l,m}}^{(M(0)+k)}, \mu^{M(0)+k} \right\rangle \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,l,m}} M^2(u) du \right) \right) \\ &= \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_{t \wedge \vartheta'_{k,l,m}}^{(M(0)+k)}, \mu^{M(0)+k} \right\rangle \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,l,m}} M^2(u) du \right) \right). \end{aligned} $$

Now, from Lemma 6.5 and the definition of $\vartheta'_{k,l,m}$, note that for all $x \in \mathbb{R}^{M(0)+k}$,

$$ \begin{aligned} & \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k,l,m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,l,m}} M^2(u) du \right) \right) \\ & \le m (1 + \|x\|_{\infty}^{2k}) \exp \left( \lambda t (M(0) + k)^2 \right). \end{aligned} \quad (53) $$

Since $\mu \in \mathcal{M}_1^{c,2k}(\mathbb{R})$, we deduce from FUBINI's theorem that

$$ \begin{aligned} & \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k,l,m}}^{M(0)} \right\rangle \right) \\ &= \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k,l,m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,l,m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle. \end{aligned} $$

On the one hand, from [18, Section 6.2.1], $\lim_{l \to +\infty} \vartheta'_{k,l,m} = \vartheta'_{k,m} \mathbb{P}_{(\mu, \xi_0)}$-a.s., and since $(X_t)_{t \ge 0}$ has continuous paths for the topology of weak convergence, we have $\mathbb{P}_{(\mu, \xi_0)}$-a.s.

$$ \lim_{l \to +\infty} \langle \xi_0, X_{t \wedge \vartheta'_{k,l,m}}^{M(0)} \rangle = \langle \xi_0, X_{t \wedge \vartheta'_{k,m}}^{M(0)} \rangle. $$
**Funding.** This work was partially funded by the Chair "Modélisation Mathématique et Biodiversité" of VEOLIA-Ecole Polytechnique-MNHN-F.X. N.C. was partially funded by the European Union (ERC, SINGER, 101054787). Views and opinions expressed are however those of the author(s) only and do not necessarily reflect those of the European Union or the European Research Council. Neither the European Union nor the granting authority can be held responsible for them.

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Therefore, we deduce from the LEBESGUE dominated convergence theorem, that for all $\xi_0 \in \mathscr{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$,

$$ \lim_{\ell \to +\infty} \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, \ell, m}}^{M(0)} \right\rangle \right) = \mathbb{E}_{(\mu, \xi_0)} \left( \left\langle \xi_0, X_{t \wedge \vartheta'_{k, m}}^{M(0)} \right\rangle \right). $$

On the other hand, Lemma 6.5 and the dominated convergence theorem imply that for all $x \in \mathbb{R}^{M(0)+k}$,

$$ \lim_{\ell \to +\infty} \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right) \\ = \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, m}} M^2(u) du \right) \right) $$

and the limit satisfies the inequality (53). As $\mu \in \mathcal{M}_1^{c, 2k}(\mathbb{R})$, we deduce again from the dominated convergence theorem that

$$ \lim_{\ell \to +\infty} \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, \ell, m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, \ell, m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle \\ = \left\langle \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k, m}}^{(M(0)+k)} \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k, m}} M^2(u) du \right) \right), \mu^{M(0)+k} \right\rangle $$

and the announced result follows from the fact that $\vartheta'_{k,m}$ is independent of $(X_t)_{t \ge 0}$, which implies

$$ \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) du \right) \right) \\ = \mathbb{E}_{\xi_0} \left( \xi_{t \wedge \vartheta'_{k,m}}^{(M(0)+k)}(x) \exp \left( \lambda \int_0^{t \wedge \vartheta'_{k,m}} M^2(u) du \right) \right). $$

□

From Lemma 6.5, we deduce that we can choose $m \in \mathbb{N}$ large enough such that $\vartheta'_{1,m} = \tau_1$. In this case, the function inside the brackets in the left-hand side of (52) is defined, for all $t \ge 0$ and for all $x \in \mathbb{R}^{M(0)+1}$, by

$$ \psi_t(x) := \mathbb{E}_{(\mu, \xi_0)} \left( \xi_{t \wedge \tau_1}^{(M(0)+1)}(x) \exp \left( \lambda \int_0^{t \wedge \tau_1} M^2(u) du \right) \right). $$

For all $t \ge 0$, $n \in \mathbb{N}^*$, $f \in \mathscr{C}_b^2(\mathbb{R}^n, \mathbb{R})$ and $\mu \in \mathcal{M}_1^{c, 2}(\mathbb{R})$ let us denote

$$ P_{FVc}(t) P_{f,n}(\mu) := \mathbb{E}_{\mu} (P_{f,n}(X_t)) $$

the semi-group of the centered FLEMING-VIOT process $(X_t)_{t \ge 0}$ with resampling rate $\lambda$.

The following result is the main result of this section. It gives an extension of Lemma 6.5 to a function which appears naturally in the proof of Lemma 6.4.
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**Proposition 6.7.** For all $t \ge 0$, for all $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$, for all $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$

$$P_{\text{FVc}}(t)P_{\xi_0,M(0)}(\mu) = P_{V_t,M(0)+1}(\mu) := \langle V_t, \mu^{M(0)+1} \rangle$$

where

$$V_t := \exp(\alpha_0 t) \psi_t - \alpha_0 \int_0^t \exp(\alpha_0 s) \psi_s ds \quad (54)$$

and $\alpha_0 := \lambda M(0) [2M(0) - 1]$. In addition, for all $t \ge 0$, there exist a constant $C_t > 0$ such that for all $x \in \mathbb{R}^{M(0)+1}$,

$$|V_t(x)| \le C_t (1 + \|x\|_{\infty}^2).$$

*Proof.* Let $t \ge 0$, $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$, $\xi_0 \in \mathcal{C}_b^2(\mathbb{R}^{M(0)}, \mathbb{R})$ be fixed. Note that from Lemma 6.5, there exists a constant $\tilde{C}_t$, depending only on $\xi_0$ and $t$ such that for all $x \in \mathbb{R}^{M(0)+1}$,

$$|\psi_t(x)| \le C_0(\tau_1, 1) (1 + \|x\|_{\infty}^2) \mathbb{E}_{(\mu, \xi_0)} (\exp(\lambda t M^2(0))) \le \tilde{C}_t (1 + \|x\|_{\infty}^2). \quad (55)$$

Now, from Lemma 6.6 and the fact that $\tau_1$ is independent of $(X_t)_{t \ge 0}$, we obtain that

$$\langle \psi_t, \mu^{M(0)+1} \rangle = \mathbb{E}_{(\mu, \xi_0)} (P_{\xi_0, M(0)} (X_{t \wedge \tau_1})) = \mathbb{E}_{\xi_0} (P_{\text{FVc}} (t \wedge \tau_1) P_{\xi_0, M(0)} (\mu)).$$

As $\tau_1$ follows an exponential law of parameter $\alpha_0$ we have that

$$\langle \psi_t, \mu^{M(0)+1} \rangle = P_{\text{FVc}}(t) P_{\xi_0, M(0)}(\mu) \exp(-\alpha_0 t) \\ + \alpha_0 \int_0^t P_{\text{FVc}}(s) P_{\xi_0, M(0)}(\mu) \exp(-\alpha_0 s) ds. \quad (56)$$

Thanks to (55) and since $\mu$ has its moment of order 2 finite, the FUBINI theorem ensures us that

$$\left\langle \int_{0}^{t} \exp(\alpha_{0}s) \psi_{s}ds, \mu^{M(0)+1} \right\rangle = \int_{0}^{t} \exp(\alpha_{0}s) \left\langle \psi_{s}, \mu^{M(0)+1} \right\rangle ds.$$

From (56) we deduce that

$$\begin{align}
\left\langle \int_0^t \exp(\alpha_0 s) \psi_s ds, \mu^{M(0)+1} \right\rangle & \nonumber \\
&= \alpha_0 \int_0^t \exp(\alpha_0 s) \int_0^s \exp(-\alpha_0 u) P_{\text{FVc}}(u) P_{\xi_0, M(0)} du ds \nonumber \\
& \qquad + \int_0^t P_{\text{FVc}}(s) P_{\xi_0, M(0)}(\mu) ds \tag{57} \\
&= \exp(\alpha_0 t) \int_0^t \exp(-\alpha_0 u) P_{\text{FVc}}(u) P_{\xi_0, M(0)}(\mu) du. \nonumber
\end{align}$$

Then, the first announced result follows from (56) and (57) and the second one from (55). $\square$
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### 6.2.2 Proof of Lemma 6.4

The key point of this proof is to establish that for all $\xi_0 \in \mathcal{C}_b^4(\mathbb{R}^{M(0)}, \mathbb{R})$, $t \ge 0$ and $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$,

$$ 
\begin{aligned}
P_{\mathrm{FVc}}(t)P_{\xi_0,M(0)}(\mu) &= \langle \xi_0, \mu^{M(0)} \rangle + \int_0^t \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{\mathrm{FVc}}(s)P_{\xi_0,M(0)}(\mu) \mathrm{d}s \\
&= P_{\xi_0,M(0)}(\mu) + \int_0^t \mathcal{L}_{\mathrm{FVc}}^{\lambda} P_{V_s,M(0)+1}(\mu) \mathrm{d}s.
\end{aligned} 
\quad (58) $$

where we use Proposition 6.7 in the second equality. To do this, we will first prove that for all $t \ge 0$, there exists a constant $C_t > 0$ such that for all $x \in \mathbb{R}^{M(0)+1}$, $\|\mathrm{Hess}(V_t)(x)\|_{\infty} \le C_t (1 + \|x\|_{\infty}^2)$, so that $V_t \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^{M(0)+1}, \mathbb{R})$.

For all $n \in \mathbb{N}^*$, for any function $f: \mathbb{R}^n \to \mathbb{R}$, $k \in \mathbb{N}$ times differentiable, we denote by $D^k f$ the differential of order $k$ of $f$. Let $t \ge 0$, $\lambda > 0$, $\mu \in \mathcal{M}_1^{c,2}(\mathbb{R})$ and $\xi_0 \in \mathcal{C}_b^4(\mathbb{R}^{M(0)}, \mathbb{R})$ be fixed.

**Step 1. Preliminary bounds.** Let $x \in \mathbb{R}^{M(0)}$. It follows from (51) that

$$ 
\begin{aligned}
& \partial_{x_i x_j x_k x_l}^4 T^{(M(0))}(t) \xi_0(x) \\
& = \sum_{m,p,q,r=1}^{M(0)} \partial_{x_i} (m_{t,x}^{\lambda})_m \partial_{x_j} (m_{t,x}^{\lambda})_p \partial_{x_k} (m_{t,x}^{\lambda})_q \partial_{x_l} (m_{t,x}^{\lambda})_r (g_{t,0}^{\lambda} * \partial_{y_m y_p y_q y_r}^4 \xi_0) (m_{t,x}^{\lambda})
\end{aligned}
$$

where $(m_{t,x}^{\lambda})_p$ designates the $p^{\mathrm{th}}$ component of $m_{t,x}^{\lambda}$. Then, for all $k \in \{0, \dots, 4\}$, there exists a constant $C_1 > 0$ independent of $x$ and $t$ such that

$$ \|D^k T^{(M(0))}(t) \xi_0(x)\|_{\infty} \le C_1 \left(1 + \|D^k \xi_0\|_{\infty}\right). $$

From (49), (24) and (25), we deduce that there exists a constant $C_2 > 0$ independent of $t$ such that for all $x \in \mathbb{R}^{M(0)+1}$, for all $k \in \{0, 1, 2\}$,

$$ 
\begin{aligned}
\|D^k \xi_{t \wedge \tau_1}^{(M(0)+1)}(x)\|_{\infty} &\le C_1 \left(1 + \|D^k \xi_0\|_{\infty}\right) + \sum_{i,j=1}^n \|D^k K_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x)\|_{\infty} \\
&\quad + \sum_{\substack{i,j=1 \\ i \ne j}}^n \|D^k \Phi_{i,j} T_{\lambda}^{(M(0))}(\tau_1) \xi_0(x_1, \dots, x_{M(0)-1})\|_{\infty} \\
&\le C_2 \left(1 + \|D^k \xi_0\|_{\infty} + \|D^{k+2} \xi_0\|_{\infty} x_{M(0)+1}^2\right).
\end{aligned}
$$

By the theorem of differentiation under the integral sign, for all $k \in \{0, 1, 2\}$ and $x \in \mathbb{R}^{M(0)+1}$, we deduce that

$$ 
\begin{aligned}
\|D^k \psi_t(x)\|_{\infty} &\le \exp(\lambda M(0)^2 t) \mathbb{E}_{(\mu, \xi_0)} \left( \|D^k \xi_{t \wedge \tau_1}^{(M(0)+1)}(x)\|_{\infty} \right) \\
&\le C_2 \exp(\lambda M(0)^2 t) (1 + \|x\|_{\infty}^2).
\end{aligned}
$$

Then, from the definition of $V_t$ in (54) we deduce that there exists a constant $C_t$ such that for all $k \in \{0, 1, 2\}$, for all $x \in \mathbb{R}^{M(0)+1}$, $\|D^k V_t(x)\|_{\infty} \le C_t (1 + \|x\|_{\infty}^2)$, so that $V_t \in \mathcal{C}_{\|\cdot\|}^2(\mathbb{R}^{M(0)+1}, \mathbb{R})$.