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Step 2. *Proof of the key point (58).* Thanks to (27), for all $t \ge 0$,

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

From (26), there exists a constant $C_3 > 0$ such that for all $t \ge 0$, $\mathbb{E}_{\mu}(|\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(X_t)|) \le C_3(1 + \mathbb{E}_{\mu}(M_2(X_t)))$ which is finite from [18, Proposition 2.11] since $\mu \in \mathcal{M}_1^{c,4}(\mathbb{R})$. Hence, from FUBINI'S theorem we obtain that

$$P_{\text{FVc}}(t)P_{\xi_0, M(0)}(\mu) = P_{\xi_0, M(0)}(\mu) + \int_{0}^{t} \mathbb{E}_{\mu}\left(\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(X_s)\right) \mathrm{d}s. \quad (59)$$

As $\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(\mu)$ is a polynomial in $\mu$ and since

$$|\mathcal{L}_{\text{FVc}}^{\lambda}(\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(\mu))| \le C_4 (1 + M_4(\mu))$$

for some constant $C_4 > 0$, we deduce as in (47) that

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

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

$$\begin{aligned} \mathbb{E}_{\mu}(\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(X_t)) &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu}\left(\int_{t}^{t+h} \mathcal{L}_{\text{FVc}} P_{\xi_0, M(0)}(X_s) \mathrm{d}s\right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu}\left(P_{\xi_0, M(0)}(X_{t+h}) - P_{\xi_0, M(0)}(X_t)\right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu}\left(P_{\text{FVc}}(t) P_{\xi_0, M(0)}(X_h) - P_{\text{FVc}}(t) P_{\xi_0, M(0)}(\mu)\right) \\ &= \lim_{h \to 0} \frac{1}{h} \mathbb{E}_{\mu}\left(P_{V_t, M(0)+1}(X_h) - P_{V_t, M(0)+1}(\mu)\right) \end{aligned}$$

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

$$|\mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(\mu)| \le C_5 (1 + M_4(\mu)). \quad (60)$$

Thus, from the martingale problem (47) and the continuity of $t \mapsto \mathbb{E}_{\mu}(\mathcal{L}_{\text{FVc}} P_{V_t, M(0)}(\mu))$, we have that

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

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

$$\mathbb{E}_{\mu}(\mathcal{L}_{\text{FVc}}^{\lambda} P_{\xi_0, M(0)}(X_t)) = \mathcal{L}_{\text{FVc}}^{\lambda} P_{V_t, M(0)+1}(\mu),$$
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and so, by (59), we obtain (58).

Step 3. Conclusion. From (58) and (60) and since $\int_{M_1(\mathbb{R})} M_4(\mu)\gamma(d\mu) < \infty$, we deduce that

$$ \begin{aligned} & \int_{M_1(\mathbb{R})} P_{FVc}(t)P_{\xi_0,M(0)}(\mu)\gamma(d\mu) \\ &= \int_{M_1(\mathbb{R})} P_{\xi_0,M(0)}(\mu)\gamma(d\mu) + \int_0^t \int_{M_1(\mathbb{R})} \mathcal{L}_{FVc}^{\lambda} P_{V_s,M(0)+1}(\mu)\gamma(d\mu)ds \\ &= \int_{M_1(\mathbb{R})} P_{\xi_0,M(0)}(\mu)\gamma(d\mu), \end{aligned} $$

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 $M_1(M_1(\mathbb{R}))$-convergence determining [26, Lemma 2.1.2], it is $M_1(M_1(\mathbb{R}))$-separating [40, Chapter 3, Section 4, p.112], so we have for any bounded continuous function $\phi$ from $M_1(\mathbb{R})$ to $\mathbb{R}$ that

$$ \int_{M_1(\mathbb{R})} P_{FVc}(t)\phi(\mu)\gamma(d\mu) = \int_{M_1(\mathbb{R})} \phi(\mu)\gamma(d\mu). $$

Hence $\gamma$ is an invariant probability measure for the centered FLEMING-VIOT process with resampling rate $\lambda$. Now, from [18, 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 $M_1(\mathbb{D}([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(M_1(\mathbb{R})))$ and for any limiting value $\mathbb{Q}$ of this sequence, the canonical process $(\zeta^R, \Gamma^R)$ on $\mathcal{C}^0([0, T], \mathbb{T}_R) \times \mathcal{M}_m^T(M_1(\mathbb{R}))$ 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.,

$$ f(\zeta_t^R) = f(\zeta_0^R) + \int_0^t \int_{M_1(\mathbb{R})} \mathcal{L}_{SLOW} f(\zeta_s^R, \mu) \pi^{\lambda}(\zeta_s^R) (d\mu) ds. \quad (61) $$

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

$$ \forall t \in [0, T], \quad z_t = x_0 + \frac{1}{2} \int_0^t \partial_1 \text{Fit}(z_s, z_s) \beta(z_s) m_2(z_s) ds \quad (62) $$

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 (61) with $f = \mathrm{id} \in \mathcal{C}_b^2(\mathbb{T}_R, \mathbb{T}_R)$, we deduce that $\mathbb{Q}$-a.s.,

$$ \forall t \in [0, T], \quad \zeta_t^R = \zeta_0^R + \int_0^t \left( \int_{M_1(\mathbb{R})} M_2(\mu) \pi^\lambda(\zeta_s^R) (d\mu) \right) \partial_1 \mathrm{Fit}(\zeta_s^R, \zeta_s^R) \, ds. $$

From [18, Corollary 4.16], for all $s \in [0, t]$,

$$ \int_{M_1(\mathbb{R})} M_2(\mu) \pi^{\lambda}(\zeta_s^R) (d\mu) = \frac{1}{2\lambda(\zeta_s^R)} = \frac{\beta(\zeta_s^R) m_2(\zeta_s^R)}{2}, $$

so $\zeta^R$ is solution of (62) $\mathbb{Q}$-a.s. By uniqueness of the solution of (62), the announced result follows. $\square$

## 7.2 End of the proof of Theorem 2.1

Let $T > 0$ be fixed and recall that $x_0$ is the mean trait value of $\nu_0^K$. From Lemma 7.1 and since the ODE (62) is non-explosive, we can choose $R > 0$ large enough such that

$$ \lim_{K \to +\infty} \mathbb{P} \left( z_t^{K,R} \in \left[ x_0 - \frac{R}{2}, x_0 + \frac{R}{2} \right], \forall t \in [0, T] \right) = 1. $$

From Proposition 4.8 and (7), we have that,

$$ \lim_{K \to +\infty} \mathbb{P} \left( \mathrm{Diam} \left( \mathrm{Supp} \, \nu_t^{K,R} \right) \le \frac{1}{K^{1+\epsilon/2}}, \forall t \in [0, T] \right) = 1. $$

Hence,

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

Now, on the event $\{\mathrm{Supp} \, \nu_{t/K\sigma_K^2}^{K,R} \subset [x_0 - R, x_0 + R], \forall t \in [0, T]\}$, $\nu_{t/K\sigma_K^2}^{K,R} = \nu_{t/K\sigma_K^2}^K$ 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.

**Funding.** This work was partially funded by the Chair “Mod\^elisation Math\^ematique et Biodiversit\^e” 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.

**Acknowledgments.** We thank the two anonymous referees for their useful comments.
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