$\displaystyle 1$
$\displaystyle={\mathbb{E}}_{\mu_{0}^{K}}\left(E_{T_{1}^{K,+}\wedge
t_{0}^{K}}^{K,+}\right)\geqslant{\mathbb{E}}_{\mu_{0}^{K}}\left(E_{T_{1}^{K,+}\wedge
t_{0}^{K}}^{K,+}{\mathds{1}}_{\Gamma}\right)$
$\displaystyle\geqslant\exp\left(-2u_{\ell}-1\right){\mathbb{E}}_{\mu_{0}^{K}}\left(\exp\left(M_{2}^{K,+}\left(T_{1}^{K,+}\wedge
t_{0}^{K}\right)+\frac{\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}T_{1}^{K,+}\wedge
t_{0}^{K}-\frac{1}{2}\left\langle{\rm
Mart}^{K,+}\right\rangle_{T_{1}^{K,+}\wedge
t_{0}^{K}}\right){\mathds{1}}_{\Gamma}\right)$
$\displaystyle\geqslant\exp\left(-2u_{\ell}-1\right){\mathbb{E}}_{\mu_{0}^{K}}\left(\exp\left(M_{2}^{K,+}\left(T_{1}^{K,+}\wedge
t_{0}^{K}\right)\right){\mathds{1}}_{\Gamma}\right)$
$\displaystyle\geqslant\exp\left(-2u_{\ell}-1\right)\left[\exp\left(u_{\ell+1}\right){\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}=\ell+1\right\\}\cap\Gamma\cap\left\\{T_{1}^{K,+}\leqslant
t_{0}^{K}\right\\}\right)\right.$ $\displaystyle\hskip
113.81102pt+\left.{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}\leqslant\ell-1\right\\}\cap\Gamma\cap\left\\{T_{1}^{K,+}\leqslant
t_{0}^{K}\right\\}\right)\right].$

As for all events $A,B,C$, ${\mathbb{P}}\left(A\cap B\cap
C\right)\geqslant{\mathbb{P}}(A)-{\mathbb{P}}\left(A\cap
B^{c}\right)-{\mathbb{P}}\left(A\cap C^{c}\right)$, and denoting
$p:={\mathbb{P}}_{\mu_{0}^{K}}\left(L_{1}^{K,+}=\ell+1\right)$ we have that

$\displaystyle 1$
$\displaystyle\geqslant\exp\left(u_{\ell}-1\right)\left[p-{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}=\ell+1\right\\}\cap\Gamma^{c}\right)-{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}=\ell+1\right\\}\cap\left\\{T_{1}^{K,+}>t_{0}^{K}\right\\}\right)\right]$
$\displaystyle\quad+\exp\left(-2u_{\ell}-1\right)\left[1-p-{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}\leqslant\ell-1\right\\}\cap\Gamma^{c}\right)-{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{L_{1}^{K,+}\leqslant\ell-1\right\\}\cap\left\\{T_{1}^{K,+}>t_{0}^{K}\right\\}\right)\right]$
$\displaystyle\geqslant\left(\exp\left(u_{\ell}-1\right)-\exp\left(-2u_{\ell}-1\right)\right)p+\exp\left(-2u_{\ell}-1\right)$
$\displaystyle\hskip
21.33955pt-\left(\exp\left(u_{\ell}-1\right)+\exp\left(-2u_{\ell}-1\right)\right)\left[{\mathbb{P}}_{\mu_{0}^{K}}\left(\Gamma^{c}\right)+{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}>t_{0}^{K}\right)\right].$

Then, considering
$s_{0}^{K}:=\frac{u_{\ell-1}K^{2}\sigma^{2}}{5\overline{b}u_{\ell+1}+\underline{\theta}\underline{m}_{2}}<t_{0}^{K}$
be fixed, we deduce that

$\displaystyle p$
$\displaystyle\leqslant\frac{1-\exp\left(-2u_{\ell}-1\right)}{\exp\left(u_{\ell}-1\right)-\exp\left(-2u_{\ell}-1\right)}+\frac{\exp\left(u_{\ell}\right)+\exp\left(-2u_{\ell}\right)}{\exp\left(u_{\ell}\right)-\exp\left(-2u_{\ell}\right)}\left[{\mathbb{P}}_{\mu_{0}^{K}}\left(\Gamma^{c}\right)+{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}>t_{0}^{K}\right)\right]$
$\displaystyle\leqslant\exp\left(-u_{\ell}+1\right)\times\frac{1-\exp\left(-2u_{\ell}-1\right)}{1-\exp\left(-u_{\ell+1}\right)}+\frac{\exp\left(u_{\ell}\right)+\exp\left(-2u_{\ell}\right)}{\exp\left(u_{\ell}\right)-\exp\left(-2u_{\ell}\right)}$
$\displaystyle\hskip
85.35826pt\times\left[{\mathbb{P}}_{\mu_{0}^{K}}\left(\Gamma^{c}\cap\left\\{T_{1}^{K,+}\geqslant
s_{0}^{K}\right\\}\right)+{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}<s_{0}^{K}\right)+{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}>t_{0}^{K}\right)\right].$

Denoting $\lceil a\rceil$ the upper integer part of $a\in{\mathbb{R}}$, we
obtain on the one hand that there exists a constant $C_{1}>0$ such that

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(\Gamma^{c}\cap\left\\{T_{1}^{K,+}\geqslant
s_{0}^{K}\right\\}\right)$ $\displaystyle\hskip
14.22636pt={\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{\left\langle{\rm
Mart}^{K,+}\right\rangle_{T_{1}^{K,+}\wedge
t_{0}^{K}}>\frac{2\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}T_{1}^{K,+}\wedge
t_{0}^{K}\right\\}\cap\left\\{T_{1}^{K,+}\geqslant s_{0}^{K}\right\\}\right)$
$\displaystyle\hskip 14.22636pt=\sum\limits_{k\,=\,1}^{\lceil
t_{0}^{K}/s_{0}^{K}\rceil-1}{{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{\left\langle{\rm
Mart}^{K,+}\right\rangle_{T_{1}^{K,+}\wedge
t_{0}^{K}}>\frac{2\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}T_{1}^{K,+}\wedge
t_{0}^{K}\right\\}\cap\left\\{ks_{0}^{K}\leqslant
T_{1}^{K,+}\leqslant(k+1)s_{0}^{K}\right\\}\right)}$ $\displaystyle\hskip
71.13188pt+{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{\left\langle{\rm
Mart}^{K,+}\right\rangle_{T_{1}^{K,+}\wedge
t_{0}^{K}}>\frac{2\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}T_{1}^{K,+}\wedge
t_{0}^{K}\right\\}\cap\left\\{T_{1}^{K,+}\geqslant\lceil
t_{0}^{K}/s_{0}^{K}\rceil s_{0}^{K}\right\\}\right)$ $\displaystyle\hskip
14.22636pt\leqslant\sum\limits_{k\,=\,1}^{\lceil
t_{0}^{K}/s_{0}^{K}\rceil-1}{{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{(k+1)s_{0}^{K}}>\frac{2\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}ks_{0}^{K}\right)+{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{t_{0}^{K}}>\frac{2\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}t_{0}^{K}\right)}$
$\displaystyle\hskip 14.22636pt\leqslant
K^{2}\sigma^{2}\left(\sum\limits_{k\,=\,1}^{\lceil
t_{0}^{K}/s_{0}^{K}\rceil-1}{\frac{{\mathbb{E}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{(k+1)s_{0}^{K}}\right)}{2\underline{b}u_{\ell-1}ks_{0}^{K}}+\frac{{\mathbb{E}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{t_{0}^{K}}\right)}{2\underline{b}u_{\ell-1}t_{0}^{K}}}\right)$
$\displaystyle\hskip
14.22636pt\leqslant\frac{C_{1}}{K^{\frac{\varepsilon}{2}}}$

where we use Markov’s inequality in the second inequality and (34) in the last
inequality. On the other hand, we control
${\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}<s_{0}^{K}\right)$ as follows:

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}<s_{0}^{K}\right)$
$\displaystyle\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\sup_{0\leqslant
t\leqslant s_{0}^{K}}{M_{2}^{K,+}(t)}\geqslant u_{\ell+1}\right\\}\right)$
$\displaystyle\hskip
28.45274pt+{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant s_{0}^{K}}{M_{2}^{K,+}(t)}<u_{\ell-1}\right\\}\right).$

Let us consider $K_{1}\in{\mathbb{N}}^{\star}$ independent of $\ell$
satisfying $K_{1}^{\varepsilon/2}\geqslant 2$. There exists a constant
$C_{1}>0$ such that for all $K\geqslant K_{1}$, we have that

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\sup_{0\leqslant
t\leqslant s_{0}^{K}}{M_{2}^{K,+}(t)}\geqslant u_{\ell+1}\right\\}\right)$
$\displaystyle\hskip
56.9055pt\overset{}{\leqslant}{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\sup_{0\leqslant
t\leqslant
s_{0}^{K}}{\left\\{2u_{\ell}+1-\frac{\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}t+{\rm
Mart}_{t}^{K,+}\right\\}\geqslant u_{\ell+1}}\right\\}\right)$
$\displaystyle\hskip
56.9055pt\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\sup_{0\leqslant
t\leqslant s_{0}^{K}}{{\rm Mart}_{t}^{K,+}}\geqslant
u_{\ell}-1\right\\}\right)$ $\displaystyle\hskip
56.9055pt\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\sup_{0\leqslant t\leqslant
s_{0}^{K}}{\left|{\rm
Mart}_{t}^{K,+}\right|}\geqslant\frac{3^{\ell}K^{\frac{\varepsilon}{2}}}{2}\right)$
$\displaystyle\hskip 56.9055pt\leqslant 4E_{\mu_{0}^{K}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{s_{0}^{K}}\right)\times\frac{4}{3^{2\ell}K^{\varepsilon}}$
$\displaystyle\hskip 56.9055pt\leqslant\frac{C_{1}}{K^{\varepsilon}}$

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
$K_{2}\in{\mathbb{N}}^{\star}$ independent of $\ell$ satisfying
$K_{2}^{\varepsilon/2}\geqslant 4$. 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 $K\geqslant K_{2}$, we have that

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant s_{0}^{K}}{M_{2}^{K,+}(t)}<u_{\ell-1}\right\\}\right)$
$\displaystyle\hskip
28.45274pt\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant
s_{0}^{K}}{\left\\{2u_{\ell-1}-1-\frac{5\overline{b}u_{\ell+1}+\underline{\theta}\underline{m}_{2}}{2K^{2}\sigma^{2}}t+{\rm
Mart}^{K,+}_{t}\right\\}}<u_{\ell-1}\right\\}\right)$ $\displaystyle\hskip
28.45274pt\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<s_{0}^{K}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant s_{0}^{K}}{{\rm
Mart}_{t}^{K,+}}<-\frac{u_{\ell-1}}{2}+1\right\\}\right)$ $\displaystyle\hskip
28.45274pt\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\sup_{0\leqslant t\leqslant
s_{0}^{K}}{\left|{\rm Mart}_{t}^{K,+}\right|}>\frac{u_{\ell-1}}{2}-1\right)$
$\displaystyle\hskip 28.45274pt\leqslant\frac{C_{2}}{K^{\varepsilon}}.$

Hence, for all $K\geqslant\max\left\\{K_{1},K_{2}\right\\}$,
${\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}<s_{0}^{K}\right)\leqslant\frac{C_{1}+C_{2}}{K^{\varepsilon}}$.
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

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(T_{1}^{K,+}>t_{0}^{K}\right)$
$\displaystyle\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{\sup_{0\leqslant
t\leqslant
t_{0}^{K}}{M_{2}^{K,+}(t)}<u_{\ell+1}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant t_{0}^{K}}{M_{2}^{K,+}(t)}\geqslant u_{\ell-1}\right\\}\right)$
$\displaystyle\leqslant{\mathbb{P}}_{\mu_{0}^{K}}\left({\rm
Mart}_{t_{0}^{K}}^{K,+}\geqslant
u_{\ell-1}-2u_{\ell}-1+\frac{\underline{b}u_{\ell-1}}{K^{2}\sigma^{2}}t_{0}^{K}\right)$
$\displaystyle\leqslant\frac{{\mathbb{E}}_{\mu_{0}^{K}}\left(\left\langle{\rm
Mart}^{K,+}\right\rangle_{t_{0}^{K}}\right)}{\left(u_{\ell-1}-1\right)^{2}}$
$\displaystyle\leqslant\frac{C_{3}}{K^{\varepsilon}}.$

Hence, for all $K\geqslant\max{\left\\{K_{1},K_{2}\right\\}}$, we deduce that
$p\leqslant\varepsilon_{K}$ where

$\varepsilon_{K}:=\exp\left(-u_{\ell}+1\right)\times\frac{1-\exp\left(-2u_{\ell}-1\right)}{1-\exp\left(-u_{\ell+1}\right)}+\frac{\exp\left(u_{\ell}\right)+\exp\left(-2u_{\ell}\right)}{\exp\left(u_{\ell}\right)-\exp\left(-2u_{\ell}\right)}\times\frac{C_{1}}{K^{\frac{\varepsilon}{2}}}$

and thus, for all $K\geqslant\max{\left\\{K_{1},K_{2}\right\\}}$,

${\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K}<\check{\tau}^{K}\right\\}\cap\left\\{L_{1}^{K}=\ell+1\right\\}\right)={\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K,+}<\check{\tau}^{K}\right\\}\cap\left\\{L_{1}^{K}=\ell+1\right\\}\right)\leqslant
p\leqslant\varepsilon_{K}.$

Step 2. Control of the second right-hand term of (32). Similarly, let us
consider $K_{3}\in{\mathbb{N}}^{\star}$ independent of $\ell$ satisfying
$K_{3}^{\varepsilon/2}\geqslant 2$. 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 $K\geqslant K_{3}$ we have that

${\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K}<\check{\tau}^{K}\right\\}\cap\left\\{\sup_{0\leqslant
t\leqslant
K\sigma^{2}\wedge\check{\tau}^{K}}{M_{2}\left(\mu_{t}^{K}\right)}\geqslant
u_{\ell+1}\right\\}\right)\leqslant\frac{16{\mathbb{E}}_{\mu_{0}^{K}}\left(\left\langle
M^{K,P_{{\rm{id}}^{2},1}}\right\rangle_{K\sigma^{2}\wedge\check{\tau}^{K}}\right)}{3^{2\ell}K^{\varepsilon}}\leqslant\frac{C_{4}}{K^{1+\varepsilon}}.$

Let us consider $K_{4}\in{\mathbb{N}}^{\star}$ independent of $\ell$
satisfying the relation
$u_{\ell-1}>1+\frac{5\overline{b}u_{\ell+1}+\underline{\theta}\underline{m}_{2}}{2K_{4}}$.
As previously, we establish that there exists a constant $C_{5}>0$ such that
for all $K\geqslant K_{4}$ we have that

$\displaystyle{\mathbb{P}}_{\mu_{0}^{K}}\left(\left\\{T_{1}^{K}<\check{\tau}^{K}\right\\}\cap\left\\{\inf_{0\leqslant
t\leqslant
K\sigma^{2}\wedge\check{\tau}^{K}}{M_{2}\left(\mu_{t}^{K}\right)}<u_{\ell-1}\right\\}\right)\leqslant\frac{4{\mathbb{E}}_{\mu_{0}^{K}}\left(\left\langle
M^{K,P_{{\rm{id}}^{2},1}}\right\rangle_{K\sigma^{2}\wedge\check{\tau}^{K}}\right)}{\left(u_{\ell-1}-1-\frac{5\overline{b}u_{\ell+1}+\underline{\theta}\underline{m}_{2}}{2K}\right)^{2}}\leqslant\frac{C_{5}}{K^{1+\varepsilon}}.$

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

${\mathbb{P}}_{\mu_{0}^{K}}\left(\left[\left\\{L_{1}^{K}=\ell+1\right\\}\cup\left\\{T_{1}^{K}<K\sigma^{2}\right\\}\right]\cap\left\\{T_{1}^{K}<\check{\tau}^{K}\right\\}\right)<\frac{1}{2}-\eta_{K}$

for any
$\eta_{K}\geqslant\frac{1}{2}-\epsilon_{K}-\frac{C_{4}+C_{5}}{K^{1+\varepsilon}}$.
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. ∎

####  4.4.2 Construction of the coupling

The goal of this section is to construct a coupling between
$\left(L_{k}^{K}\right)_{k\in{\mathbb{N}}}$ and a biased random walk on
${\mathbb{N}}^{\star}$ and reflected in $1$. To do this, we will construct a
sequence $\left(\xi_{k}^{K}\right)_{k\in{\mathbb{N}}}$ 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 $\xi_{0}^{K}\in\left\\{-1,1\right\\}$ such that

$\left\\{\xi_{0}^{K}=-1\right\\}\subset\left[\left\\{L_{1}^{K}=L_{0}^{K}-1\right\\}\cap\left\\{K\sigma^{2}\leqslant
T_{1}^{K}<\check{\tau}^{K}\right\\}\right]\cup\left\\{T_{1}^{K}\geqslant\widehat{\tau}^{K}\right\\}$

and ${\mathbb{P}}\left(\xi_{0}^{K}=-1\right)=\frac{1}{2}+\eta_{K}$.

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
$\xi_{1}^{K}\in\left\\{-1,1\right\\}$ such that, conditionally to
${\mathcal{F}}_{T_{1}^{K}}\vee\sigma\left(\xi_{0}^{K}\right)$,

$\left\\{\xi_{1}^{K}=-1\right\\}\subset\left[\left\\{L_{2}^{K}=L_{1}^{K}-1\right\\}\cap\left\\{K\sigma^{2}+T_{1}^{K}\leqslant
T_{2}^{K}<\check{\tau}^{K}\right\\}\right]\cup\left\\{T_{2}^{K}\geqslant\widehat{\tau}^{K}\right\\}$

and
${\mathbb{P}}\left(\xi_{1}^{K}=-1\left|\phantom{1^{1^{1}}}\right.{\mathcal{F}}_{T_{1}^{K}}\vee\sigma\left(\xi_{0}^{K}\right)\right)=\frac{1}{2}+\eta_{K}$.
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
$\left(\xi_{i}^{K}\right)_{0\leqslant i\leqslant k}$ 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
$\xi_{k+1}^{K}\in\left\\{-1,1\right\\}$ such that, conditionally to
${\mathcal{F}}_{T_{k+1}^{K}}\vee\sigma\left(\xi_{0}^{K},\cdots,\xi_{k}^{K}\right)$,

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

and
${\mathbb{P}}\left(\xi_{1}^{K}=-1\left|\phantom{1^{1^{1}}}\right.{\mathcal{F}}_{T_{k+1}^{K}}\vee\sigma\left(\xi_{0}^{K},\cdots,\xi_{k}^{K}\right)\right)=\frac{1}{2}+\eta_{K}$.

To conclude, we set

$\forall K\in{\mathbb{N}}^{\star},\quad\forall k\in{\mathbb{N}},\qquad
Z_{k+1}^{K}-Z_{k}^{K}:=\xi_{k}^{K}\qquad\quad{\rm and}\qquad\quad
Z_{0}^{K}=L_{0}^{K}$ (35)

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

######  Lemma 4.12.

The sequence $\left(Z_{k}^{K}\right)_{k\in{\mathbb{N}}}$, given by (35) is a
biased simple random walk on ${\mathbb{N}}^{\star}$, reflected in $1$ and for
all $k\in{\mathbb{N}}$ as long as $T_{k+1}^{K}<\check{\tau}^{K}$, that

(1) $\displaystyle Z_{k+1}^{K}-Z_{k}^{K}\geqslant L_{k+1}^{K}-L_{k}^{K}$ (2)
$\displaystyle T_{k+1}^{K}-T_{k}^{K}\geqslant K\sigma^{2}\qquad{\rm
when}\qquad Z_{k+1}^{K}-Z_{k}^{K}=-1$

On the one hand, note that for all $k\in{\mathbb{N}}$, $L_{k}^{K}\leqslant
Z_{k}^{K}$. On the other hand, note that Lemma 4.9 implies for all
$k\in{\mathbb{N}}$,
$L_{k+1}^{K}-L_{k}^{K}\in\left\\{\cdots,-3,-2,-1,1\right\\}$ which justifies
(1) of Lemma 4.12. In Figure LABEL:Fig_M2_Marche_aleatoire, we illustrate the
coupling between $\left(L_{k}^{K}\right)_{0\leqslant k\leqslant 7}$ and
$\left(Z_{k}^{K}\right)_{0\leqslant k\leqslant 7}$. In this figure, we observe
that $\xi_{k}^{K}=L_{k+1}^{K}-L_{k}^{K}$ for all
$k\in\left\\{0,1,2,4,5,6\right\\}$ and
$\xi_{3}^{K}=-1\geqslant-2=L_{4}^{K}-L_{3}^{K}$. 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
$\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor$ for the random
walk $\left(Z_{k}^{K}\right)_{k\in{\mathbb{N}}}$.

######  Remark 4.13.

By Lemma 4.12, $N^{K}\leqslant k_{0}$ where $k_{0}$ is the first integer such
that
$L_{k_{0}}^{K}=\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor$,
so $\widehat{\tau}^{K}\geqslant T_{k_{0}}^{K}\geqslant T_{N^{K}}^{K}$.

The following lemma gives an estimate on the problem of exit from a domain for
$Z^{K}$. For all $k\in{\mathbb{N}}^{\star}$, 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,

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

###### Proof.

For all $k\in{\mathbb{N}}^{\star}$, let us consider the stopping time
${\mathcal{T}}_{k}^{K}:=\inf\left\\{n\in{\mathbb{N}}\left|\phantom{1^{1^{1^{1}}}}\right.Z_{n}^{K}=k\right\\}$
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

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

Note that $N^{K}$ is greater than the number of transitions from $2$ to $1$
that occur before
${\mathcal{T}}_{\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor}^{K}$.
Hence, under ${\mathbf{P}}_{1}$, $N^{K}\geqslant X$ where $X$ is random
variable with geometric law of parameter
$p_{K}:=\left(r_{K}-1\right)\exp\left(-\left(\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor-1\right)\log(r_{K})\right)$.
Since $r_{K}\sim_{K\to+\infty}\frac{K^{\varepsilon/2}}{C}$, 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, $\log(1-x)\geqslant-2x$ we deduce that for all $K$ large
enough,

$\displaystyle{\mathbf{P}}_{1}\left(X\geqslant\exp\left(\frac{\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor-1}{2}\log\left(r_{K}\right)\right)\right)$
$\displaystyle\hskip
28.45274pt=\exp\left(\exp\left(\frac{\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor-1}{2}\log\left(r_{K}\right)\right)\log\left(1-p_{K}\right)\right)$
$\displaystyle\hskip
28.45274pt\geqslant\exp\left(-2\left(r_{K}-1\right)\exp\left(-\frac{\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor-1}{2}\log\left(r_{K}\right)\right)\right).$

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

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

and the announced result follows. ∎

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 straightfoward 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

${\mathbb{P}}\left(L_{0}^{K}>1\right)\leqslant{\mathbb{P}}\left(M_{2}\left(\mu_{0}^{K}\right)\geqslant
K^{\frac{\varepsilon}{2}}\right)\leqslant\frac{C_{2}^{\star}}{K^{\varepsilon}}$

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

####  4.4.4 Conclusion

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

$\displaystyle{\mathbb{P}}\left(\widehat{\tau}^{K}\leqslant\check{\tau}^{K}\wedge
T\right)$
$\displaystyle\leqslant{\mathbb{P}}\left(\left\\{\widehat{\tau}^{K}\leqslant\check{\tau}^{K}\wedge
T\right\\}\cap\left\\{\check{\tau}^{K}>T_{N^{K}}^{K}\right\\}\right)$ (36)
$\displaystyle\hskip
28.45274pt+{\mathbb{P}}\left(\left\\{\widehat{\tau}^{K}\leqslant\check{\tau}^{K}\wedge
T\right\\}\cap\left\\{\check{\tau}^{K}\leqslant
T_{N^{K}}^{K}\right\\}\right).$

Step 1. Control of the first right-hand term of (36). As $Z_{0}^{K}=1$ with
probability converging to $1$ and
$Z_{N^{K}}^{K}=\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor$,
we have
$N_{+}^{K}-N_{-}^{K}\leqslant\left\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\right\rfloor$
and $N_{+}^{K}+N_{-}^{K}=N^{K}$. Thus,

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

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

$\widehat{\tau}^{K}\geqslant
T_{N^{K}}^{K}\geqslant\sum\limits_{i\,=\,1}^{N^{K}}{\left(T_{i}^{K}-T_{i-1}^{K}\right)}\geqslant
N_{-}^{K}K\sigma^{2}\geqslant\frac{K\sigma^{2}}{2}\left(N^{K}-\frac{\varepsilon}{2\log(3)}\log(K)\right).$

Therefore,

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

which tends to $0$ when $K\to+\infty$ according to Corollary 4.15.

Step 2. Control of the second right-hand term of (36). If
$\check{\tau}^{K}\leqslant T_{N^{K}}^{K}$, let us consider $\widetilde{N}^{K}$
the last index $k$ such that $T_{k}^{K}<\check{\tau}^{K}$. We have
$\widetilde{N}^{K}<N^{K}$ and so
$\widetilde{N}^{K}+1\leqslant\widetilde{N}^{K}$ which implies, thanks to Lemma
4.9, that for all $t\leqslant T_{\widetilde{N}^{K}+1}$ and $K$ large enough,

$M_{2}\left(\mu_{t}^{K}\right)\leqslant
2u_{\lfloor\frac{\varepsilon}{2\log(3)}\log(K)\rfloor}+1\leqslant
2K^{\frac{\varepsilon}{2}}+1<K^{\varepsilon}.$

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

${\mathbb{P}}\left(\left\\{\widehat{\tau}^{K}\leqslant\check{\tau}^{K}\wedge
T\right\\}\cap\left\\{\check{\tau}^{K}\leqslant
T_{N^{K}}^{K}\right\\}\right)={\mathbb{P}}\left(\varnothing\right)=0.$

Therefore,
$\lim_{K\to+\infty}{{\mathbb{P}}\left(\widehat{\tau}^{K}\leqslant\check{\tau}^{K}\wedge
T\right)}=0$ 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
$\left\\{\left(z^{K},\Gamma^{K}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
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 ${\mathbb{T}}_{R}:=\left[x_{0}-2R,x_{0}+2R\right]$ of
length $4R$ with $R>0$ fixed and $x_{0}$ is the value of the mean trait of
$\nu_{0}^{K}$ as in Theorem 6. We define
${\mathcal{M}}_{1}\left({\mathbb{T}}_{R}\right)$ the set of probability
measure on ${\mathbb{T}}_{R}$. We denote by
${\mathcal{M}}_{m}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$ the
set of measures $\Gamma$ on
${\mathbb{R}}_{+}\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)$ such that
for all $t\geqslant 0$,
$\Gamma\left([0,t]\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)=t$.
For any $t\geqslant 0$, we denote by
${\mathcal{M}}_{m}^{t}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
the set of measures
$\Gamma\in{\mathcal{M}}_{m}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
restricted to $[0,t]\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)$.

Let
$b_{R}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{T}}_{R}^{2},{\mathbb{R}}\right)$
and
$\theta_{R}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{T}}_{R},{\mathbb{R}}\right)$
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
$\nu^{K,R}:=\left(\nu_{t}^{K,R}\right)_{t\in[0,T]}$,
$z^{K,R}:=\left(z_{t}^{K,R}\right)_{t\in[0,T]}$ and
$\mu^{K,R}=\left(\mu_{t}^{K,R}\right)_{t\in[0,T]}$ as follows

$\nu_{t}^{K,R}:=\frac{1}{K}\sum_{i\,=\,1}^{K}{\delta_{x_{i}^{R}(t)}},\qquad
z_{t}^{K,R}:=\left\langle{\rm{id}},\nu_{t/K\sigma^{2}}^{K,R}\right\rangle\qquad{\rm
and}\qquad\mu_{t}^{K,R}:=\left(h_{\frac{1}{\sigma\sqrt{K}}}\circ\tau_{-z_{t}^{K}}\right)\sharp\,\nu_{t/K\sigma^{2}}^{K}$

where for all $i\in\left\\{1,\cdots,K\right\\}$, $\nu_{t}^{K,R}$ is defined on
${\mathcal{M}}_{1}\left({\mathbb{T}}_{R}\right)$ 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
${\mathcal{M}}_{1}\left({\mathbb{T}}_{R}^{K,\sigma}\right)$ where
${\mathbb{T}}_{R}^{K,\sigma}$ is the torus corresponding to the interval
$\left[\frac{1}{\sigma\sqrt{K}}\left(x_{0}-2R-z_{t}^{K}\right),\frac{1}{\sigma\sqrt{K}}\left(x_{0}+2R-z_{t}^{K}\right)\right]$.
However, we will identify in the sequel
$M_{1}\left({\mathbb{T}}_{R}^{K,\sigma}\right)$ as a subset of
${\mathcal{M}}_{1}({\mathbb{R}})$ using the natural embedding of
${\mathbb{T}}_{R}^{K,\sigma}$ in ${\mathbb{R}}$. So,
$\mu_{t}^{K,R}\in{\mathcal{M}}_{1}({\mathbb{R}})$. We define also the stopping
times

$\displaystyle\widehat{\tau}^{K,R}$ $\displaystyle:=\inf\left\\{t\geqslant
0\left|\phantom{1^{1^{1^{1}}}}\right.M_{2}\left(\mu_{t}^{K,R}\right)\geqslant
K^{\varepsilon}\right\\},$ $\displaystyle\check{\tau}^{K,R}$
$\displaystyle:=\inf\left\\{t\geqslant
0\left|\phantom{1^{1^{1^{1}}}}\right.{\rm{Diam}}\left({\rm{Supp\,}}\nu_{t}^{K,R}\right)>\frac{1}{\sigma
K^{\frac{3+\varepsilon}{2}}}\right\\},$

and $\tau^{K,R}:=\check{\tau}^{K,R}\wedge\widehat{\tau}^{K,R}$. We define on
${\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
the sequence of random measures
$\left(\Gamma^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$ as follows

$\Gamma^{K,R}\left(\mathrm{d}t,\mathrm{d}\mu\right)=\delta_{\mu_{t}^{K,R}}\left(\mathrm{d}\mu\right)\mathrm{d}t.$

In the sequel, we study the limit as $K$ tends to $+\infty$ of the sequence of
laws of
$\left(\left(z^{K,R},\Gamma^{K,R}\right)\right)_{K\in{\mathbb{N}}^{\star}}$ in
${\mathcal{M}}_{1}\left({\mathbb{D}}\left(\left[0,T\right],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)\right)$.

######  Theorem 5.1.

Let $T,R>0$. The sequence of laws of
$\left(\left(z^{K,R},\Gamma^{K,R}\right)\right)_{K\in{\mathbb{N}}^{\star}}$ is
tight in the set of probability measures on
${\mathbb{D}}\left(\left[0,T\right],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
and for any limiting value ${\mathbb{Q}}$ of this sequence, the canonical
process $\left(\zeta^{R},\Gamma^{R}\right)$ on
${\mathbb{D}}\left(\left[0,T\right],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
satisfies that for all
$f\in{\mathscr{C}}_{b}^{2}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$,

$N_{t}^{f,R}:=f\left(\zeta_{t}^{R}\right)-f\left(\zeta_{0}^{R}\right)-\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{{\rm
SLOW}}f\left(\zeta_{s}^{R},\mu\right)\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}=0$
(37)

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

${\mathbb{E}}_{{\mathbb{Q}}}\left(\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{M_{5}\left(\mu\right)\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}\right)<\infty,$
(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
$\left(z_{{\tiny\bullet}\wedge\tau^{K,R}}^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$.
In Step 2, we establish a compact containement condition for the stopped
_fast_ component
$\left(\mu_{{\tiny\bullet}\wedge\tau^{K,R}}^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$.
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
$\left(N_{t\wedge\tau^{K,R}}^{f,K,R}\right)_{t\in[0,T],K\in{\mathbb{N}}^{\star}}$
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
$\left(N_{{\tiny\bullet}\wedge\tau^{K,R}}^{f,K,R}\right)_{K\in{\mathbb{N}}^{\star}}$
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
$\left\\{z_{{\tiny\bullet}\wedge\tau^{K,R}}^{K,R}\right\\}_{K\in{\mathbb{N}}^{\star}}$
on ${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)$. Let
$f\in{\mathscr{C}}_{b}^{2}({\mathbb{T}}_{R},{\mathbb{T}}_{R})$. For all
$K\in{\mathbb{N}}^{\star}$ and for all $t\in[0,T]$, let us consider the two
processes $Y_{t}^{K,R}:=f\left(z_{t\wedge\tau^{K,R}}^{K,R}\right)$ and
$Z_{t}^{K,R}$ defined by the relation

$Y_{t}^{K,R}=Y_{0}^{K,R}+\int_{0}^{t}{Z_{s}^{K,R}\mathrm{d}s}+M_{t\wedge\tau^{K,R}}^{f,K,R}$

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 $C>0$
such that

$\displaystyle\sup_{K\in{\mathbb{N}}^{\star}}{{\mathbb{E}}\left[\left(\int_{0}^{T}{\left|Z_{t}^{K,R}\right|^{2}\mathrm{d}t}\right)^{\frac{1}{2}}\right]}$
$\displaystyle\hskip
56.9055pt\leqslant\sup_{K\in{\mathbb{N}}^{\star}}{{\mathbb{E}}\left[1+\int_{0}^{T}{\left|Z_{t}^{K,R}\right|^{2}\mathrm{d}t}\right]}$
$\displaystyle\hskip 56.9055pt\leqslant
C\left\\{1+\sup_{K\in{\mathbb{N}}^{\star}}{{\mathbb{E}}\left[\int_{0}^{T\wedge\tau^{K,R}}{\left(M_{2}^{2}\left(\mu_{t}^{K,R}\right)+\frac{1}{K^{2}}\left[1+M_{2}^{2}\left(\mu_{t}^{K,R}\right)\right]\right)\mathrm{d}t}\right]}\right\\}$
$\displaystyle\hskip 56.9055pt<\infty.$

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

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

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

Step 2. Compact containement condition.

######  Lemma 5.2.

Let $R>0$. For all $T\geqslant 0$, the family of laws of the marginal random
variables of the stopped _fast_ process
$\left\\{\mu_{t\wedge\tau^{K,R}}^{K,R}\right\\}_{t\in[0,T],K\in{\mathbb{N}}^{\star}}$
is tight on ${\mathcal{M}}_{1}\left({\mathbb{R}}\right)$ i.e.

$\forall\eta>0,\
\exists\,D_{\varepsilon,\eta}^{R,T}\subset{\mathcal{M}}_{1}\left({\mathbb{R}}\right)\
{\rm compact},\ \forall t\in\left[0,T\right],\ \forall
K\in{\mathbb{N}}^{\star},\quad{\mathbb{P}}\left(\mu_{t\wedge\tau^{K,R}}^{K,R}\in
D_{\varepsilon,\eta}^{R,T}\right)\geqslant 1-\eta.$

###### Proof.

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

$M_{2}\left(\mu_{t}^{K,R}\right){\mathds{1}}_{t\leqslant\tau^{K,R}}\geqslant{\mathds{1}}_{t\leqslant\tau^{K,R}}\int_{\kappa_{p,\eta}^{c}}{x^{2}\mu_{t}^{K,R}\left(\mathrm{d}x\right)}\geqslant
a_{p,\eta}^{2}\mu_{t\wedge\tau^{K,R}}^{K,R}\left(\kappa_{p,\eta}^{c}\right){\mathds{1}}_{t\leqslant\tau^{K,R}},$

and we deduce that
$\left\\{\mu_{t\wedge\tau^{K,R}}^{K,R}\left(\kappa_{p,\eta}^{c}\right)>\frac{1}{p}\right\\}\cap\left\\{t\leqslant\tau^{K,R}\right\\}\subset\left\\{M_{2}\left(\mu_{t\wedge\tau^{K,R}}^{K,R}\right)>\frac{a_{p,\eta}^{2}}{p}\right\\}\cap\left\\{t\leqslant\tau^{K,R}\right\\}$.
Hence,

$\displaystyle{\mathbb{P}}\left(\left\\{\exists p\in{\mathbb{N}}^{\star},\
\mu_{t\wedge\tau^{K,R}}^{K,R}\left(\kappa_{p,\eta}^{c}\right)>\frac{1}{p}\right\\}\right)$
$\displaystyle\leqslant{\mathbb{P}}\left(\tau^{K,R}<t\right)+\sum_{p\in{\mathbb{N}}^{\star}}{{\mathbb{P}}\left(\left\\{\mu_{t\wedge\tau^{K,R}}^{K,R}\left(\kappa_{p,\eta}^{c}\right)>\frac{1}{p}\right\\}\cap\left\\{t\leqslant\tau^{K,R}\right\\}\right)}$
$\displaystyle\leqslant{\mathbb{P}}\left(\tau^{K,R}<t\right)+\sum_{p\in{\mathbb{N}}^{\star}}{{\mathbb{P}}\left(M_{2}\left(\mu_{t}^{K,R}\right){\mathds{1}}_{t\leqslant\tau^{K,R}}>\frac{a_{p,\eta}^{2}}{p}\right)}$
$\displaystyle\leqslant{\mathbb{P}}\left(\tau^{K,R}<t\right)+\sum_{p\in{\mathbb{N}}^{\star}}{\frac{{\mathbb{E}}\left(M_{2}\left(\mu_{t}^{K,R}\right){\mathds{1}}_{t\leqslant\tau^{K,R}}\right)}{\frac{a_{p,\eta}^{2}}{p}}}$
$\displaystyle\leqslant\eta.$

Therefore, we have proved that

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

where
${\mathcal{K}}_{\eta}:=\left\\{\mu\in{\mathcal{M}}_{1}({\mathbb{R}})\left|\phantom{1^{1^{1^{1}}}}\right.\forall
p\in{\mathbb{N}}^{\star},\mu\left(\kappa_{p,\eta}^{c}\right)\leqslant\frac{1}{p}\right\\}$
which is compact by Prohorov’s theorem.∎

Step 3. Tightness of the family of laws of
$\left\\{\Gamma^{K,R}\right\\}_{K\in{\mathbb{N}}^{\star}}$ on
${\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$.
Let $\eta>0$ be fixed. From Proposition 4.7, there exists
$K_{0}\in{\mathbb{N}}^{\star}$ such that for all $K\geqslant K_{0}$, for all
$t\in[0,T]$,
${\mathbb{P}}\left(t\geqslant\tau^{K,R}\right)\leqslant\frac{\eta}{2}$.
Consider $D_{\varepsilon,\frac{\eta}{2}}^{R,T}$ the compact set in Lemma 5.2.
It follows that for all $t\in[0,T]$, $K\geqslant K_{0}$,

$\displaystyle{\mathbb{E}}\left(\Gamma^{K,R}\left([0,t]\times
D_{\varepsilon,\frac{\eta}{2}}^{R,T}\right)\right)$
$\displaystyle\geqslant{\mathbb{E}}\left(\Gamma^{K,R}\left([0,t]\times
D_{\varepsilon,\frac{\eta}{2}}^{R,T}\right){\mathds{1}}_{t<\tau^{K,R}}\right)$
$\displaystyle=\int_{0}^{t}{{\mathbb{P}}\left(\left\\{\mu_{s\wedge\tau^{K,R}}^{K,R}\in
D_{\varepsilon,\frac{\eta}{2}}^{R,T}\right\\}\cap\\{t<\tau^{K,R}\\}\right)\mathrm{d}s}$
$\displaystyle\geqslant\int_{0}^{t}{{\mathbb{P}}\left(\mu_{s\wedge\tau^{K,R}}^{K,R}\in
D_{\varepsilon,\frac{\eta}{2}}^{R,T}\right)\mathrm{d}s}-\int_{0}^{t}{{\mathbb{P}}\left(t\geqslant\tau^{K,R}\right)\mathrm{d}s}$
$\displaystyle\geqslant t\left(1-\eta\right).$

Therefore, the tightness of the family of laws of
$\left\\{\Gamma^{K,R}\right\\}_{K\in{\mathbb{N}}^{\star}}$ follows from  [49,
Lemma 1.3].

Step 4. Tightness of the family of laws of
$\left\\{\left(z^{K,R},\Gamma^{K,R}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
on
${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$.
From Steps 1 and 3 and Prohorov’s theorem, we deduce that the family of laws
of the couple
$\left\\{\left(z_{{\tiny\bullet}\wedge\tau^{K,R}}^{K,R},\Gamma^{K,R}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
is relatively compact in
${\mathcal{M}}_{1}\left({\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)\right)$.
Hence, there exists a probability measure ${\mathbb{Q}}$ on the canonical
space
${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
and an increasing function $n:{\mathbb{N}}^{\star}\to{\mathbb{N}}^{\star}$
such that the subsequence of laws of
$\left\\{\left(z_{{\tiny\bullet}\wedge\tau^{n\left(K\right),R}}^{n\left(K\right),R},\Gamma^{n\left(K\right),R}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
converges weakly to the limiting value ${\mathbb{Q}}$ when $K\to+\infty$.
Thanks to Proposition 4.7, we deduce that the family of laws of
$\left\\{\left(z^{n(K),R},\Gamma^{n(K),R}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
converges weakly to ${\mathbb{Q}}$ when $K\to+\infty$ and therefore that the
family of laws of
$\left\\{\left(z^{K,R},\Gamma^{K,R}\right)\right\\}_{K\in{\mathbb{N}}^{\star}}$
is relatively compact, thus tight on
${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
by Prohorov’s theorem.

Step 5. Uniform integrability. For all
$f\in{\mathscr{C}}_{b}^{2}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$, let
us consider $\left(N_{t}^{f,K,R}\right)_{t\in[0,T]}$ the stochastic process
defined by

$N_{t}^{f,K,R}:=f\left(z_{t}^{K,R}\right)-f\left(z_{0}^{K,R}\right)-\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{\rm
SLOW}f\left(z_{s}^{K,R},\mu\right)\Gamma^{K,R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}.$

From (9), we have for all
$f\in{\mathscr{C}}_{b}^{2}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$,
there exists a constant $C>0$ such that

$\forall t\in[0,T],\ \forall
K\in{\mathbb{N}}^{\star},\quad\left|N_{t\wedge\tau^{K,R}}^{f,K,R}\right|\leqslant
C\left(1+\int_{0}^{t}{M_{2}\left(\mu_{s}^{K,R}\right){\mathds{1}}_{s\leqslant\tau^{K,R}}\mathrm{d}s}\right),$

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

Step 6. Proof of (38) and almost sure convergence of
$\left(\widetilde{N}_{{\tiny\bullet}\wedge\tau^{K,R}}^{f,K,R}\right)_{K\in{\mathbb{N}}^{\star}}$
to $\widetilde{N}^{f,R}$. From Step 4, Proposition 4.7 and Skorohod’s
representation theorem, there exists an increasing function
$\bar{n}:{\mathbb{N}}^{\star}\to{\mathbb{N}}^{\star}$ and a probability space
on which we define, the random variable
$\widetilde{\tau}^{\bar{n}\left(K\right),R}$, the families
$\left\\{\widetilde{z}^{\bar{n}\left(K\right),R}\right\\}_{K\in{\mathbb{N}}^{\star}}$
and
$\left\\{\widetilde{\Gamma}^{\bar{n}\left(K\right),R}\right\\}_{K\in{\mathbb{N}}^{\star}}$
and $\widetilde{\zeta}^{R},\widetilde{\Gamma}^{R}$ copies of
$\left\\{z^{\bar{n}(K),R}\right\\}_{K\in{\mathbb{N}}^{\star}}$,
$\left\\{\Gamma^{\bar{n}(K),R}\right\\}_{K\in{\mathbb{N}}^{\star}}$,
$\zeta^{R}$, $\Gamma^{R}$ under ${\mathbb{Q}}$ such that for all
$K\in{\mathbb{N}}^{\star}$, the sequence
$\left(\widetilde{z}^{\bar{n}\left(K\right),R},\widetilde{\Gamma}^{\bar{n}\left(K\right),R},\widetilde{\tau}^{\bar{n}\left(K\right),R}\right)_{K\in{\mathbb{N}}^{\star}}$
converges a.s. to
$\left(\widetilde{\zeta}^{R},\widetilde{\Gamma}^{R},+\infty\right)$ when
$K\to+\infty$. Note that for all $t\in[0,T]$,

$\widetilde{\Gamma}^{\bar{n}\left(K\right),R}\left(\left[0,t\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}\right]\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)=t\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}\xrightarrow[K\longrightarrow+\infty]{{\rm
a.s.}}t=\widetilde{\Gamma}\left([0,t]\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right).$

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

$\int_{0}^{t\wedge\widetilde{\tau}^{K,R}}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{5}\left(\mu\right)\widetilde{\Gamma}^{K,R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}\xrightarrow[K\longrightarrow+\infty]{{\rm
a.s.}}\int_{0}^{t}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{5}\left(\mu\right)\widetilde{\Gamma}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}$

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

$\displaystyle\lim_{K\to+\infty}{\int_{0}^{t\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{\rm
SLOW}f\left(\widetilde{z}_{s}^{\bar{n}\left(K\right),R},\mu\right)\widetilde{\Gamma}^{\bar{n}\left(K\right),R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}}$
$\displaystyle\hskip
85.35826pt=\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{\rm
SLOW}f\left(\widetilde{\zeta}^{R},\mu\right)\widetilde{\Gamma}^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}$

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

$\displaystyle\widetilde{N}_{t}^{f,K,R}$
$\displaystyle:=f\left(\widetilde{z}_{t}^{K,R}\right)-f\left(\widetilde{z}_{0}^{K,R}\right)-\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{{\rm
SLOW}}f\left(\widetilde{z}_{t}^{K,R},\mu\right)\widetilde{\Gamma}^{K,R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}},\quad
t\leqslant T\wedge\widetilde{\tau}^{K,R}$
$\displaystyle\widetilde{N}_{t}^{f,R}$
$\displaystyle:=f\left(\widetilde{\zeta}_{t}^{R}\right)-f\left(\widetilde{\zeta}_{0}^{R}\right)-\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{{\mathcal{L}}_{{\rm
SLOW}}f\left(\widetilde{\zeta}_{t}^{R},\mu\right)\widetilde{\Gamma}^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}},\quad
t\leqslant T.$

Step 7. $N^{f,R}$ is a martingale. Let us consider the filtration
$\left(\widetilde{{\mathcal{F}}}_{t}^{K,R}\right)_{t\in[0,T]}$ defined by
$\widetilde{{\mathcal{F}}}_{t}^{K,R}:=\sigma\left(\widetilde{z}_{s}^{K,R},\widetilde{\Gamma}^{K,R}\left([0,s]\times
H\right)\left|\phantom{1^{1^{1^{1}}}}\right.s\leqslant
t,H\in{\mathcal{B}}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)\right)$,
$\left(f_{i}\right)_{1\leqslant i\leqslant p}$, $p\in{\mathbb{N}}^{\star}$
bounded Lipschitz functions from ${\mathbb{T}}_{R}$ to ${\mathbb{T}}_{R}$ and
$0<t_{1}\leqslant\cdots\leqslant t_{p}\leqslant s<t$. Let us denote for all
$K\in{\mathbb{N}}^{\star}$, for all $t\leqslant\widetilde{\tau}^{K,R}\wedge
T$,
$\widetilde{M}_{t}^{f,K,R}:=\widetilde{N}_{t}^{f,K,R}+\widetilde{{\mathcal{E}}}_{t}^{f,K,R}$
where $\widetilde{M}^{f,K,R}$ is constructed from $\widetilde{z}^{K,R}$ and
$\widetilde{\mu}^{K,R}$ as in (18) and $\widetilde{{\mathcal{E}}}_{t}^{f,K,R}$
is an error term. Note that $\widetilde{M}_{t}^{f,K,R}$ is a
$\left(\widetilde{{\mathcal{F}}}_{t}^{K,R}\right)_{t\in[0,T]}-$martingale as
in Proposition 3.1. Hence,

$\displaystyle{\mathbb{E}}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{t\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}\prod_{i\,=\,1}^{p}{f_{i}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{t_{i}\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}\right)}\right)$
$\displaystyle={\mathbb{E}}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{s\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}\prod_{i\,=\,1}^{p}{f_{i}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{t_{i}\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}\right)}\right).$

From Proposition 3.2 and (30),
$\widetilde{{\mathcal{E}}}_{t\wedge\widetilde{\tau}^{K,R}}^{f,K,R}=\frac{1}{K}O\left(\int_{0}^{t}{\left[1+M_{2}\left(\widetilde{\mu}_{s}^{K,R}\right)\right]{\mathds{1}}_{s\leqslant\widetilde{\tau}^{K,R}}\mathrm{d}s}\right)$
and then satisfy the condition
$\lim_{K\to+\infty}{{\mathbb{E}}\left(\sup_{0\leqslant t\leqslant
T}{\left|{\mathcal{E}}_{t\wedge\widetilde{\tau}^{K,R}}^{f,K,R}\right|}\right)}=0$.
As for all $i\in\\{1,\cdots,p\\}$, $f_{i}$ is Lipschitz, there exists a
constant $C>0$ such that

$\displaystyle{\mathbb{E}}\left(\left|f_{i}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{t_{i}\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}-{\mathcal{E}}_{t_{i}\wedge\tau^{\bar{n}\left(K\right),R}}^{f,\bar{n}\left(K\right),R}\right)-f_{i}\left(\widetilde{M}^{f,\bar{n}\left(K\right),R}_{t_{i}\wedge\widetilde{\tau}^{\bar{n}\left(K\right),R}}\right)\right|\right)$
$\displaystyle\hskip
85.35826pt\leqslant\frac{C}{\bar{n}\left(K\right)}{\mathbb{E}}\left(\int_{0}^{t_{i}}{\left[1+M_{2}\left(\mu_{r}^{\bar{n}\left(K\right),R}\right)\right]{\mathds{1}}_{r\leqslant\widetilde{\tau}^{\bar{n}\left(K\right),R}}\mathrm{d}r}\right)$

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
$u\in\\{s,t\\}$,

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

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

${\mathbb{E}}\left(\widetilde{N}^{f,R}_{t}\prod_{i\,=\,1}^{p}{f_{i}\left(\widetilde{N}^{f,R}_{t_{i}}\right)}\right)={\mathbb{E}}\left(\widetilde{N}^{f,R}_{s}\prod_{i\,=\,1}^{p}{f_{i}\left(\widetilde{N}^{f,R}_{t_{i}}\right)}\right).$

Hence,

${\mathbb{E}}\left(N{}^{f,R}_{t}\prod_{i\,=\,1}^{p}{f_{i}\left(N{}^{f,R}_{t_{i}}\right)}\right)={\mathbb{E}}\left(N{}^{f,R}_{s}\prod_{i\,=\,1}^{p}{f_{i}\left(N{}^{f,R}_{t_{i}}\right)}\right).$

Since the last property is true for all $p\in{\mathbb{N}}^{\star}$,
$t_{1}\leqslant\cdots\leqslant t_{p}\leqslant s<t$ and for all bounded
Lipschitz functions $\left(f_{i}\right)_{1\leqslant i\leqslant p}$, the
monotone class theorem ensures us that

${\mathbb{E}}\left(N_{t}^{f,R}\left|\phantom{1^{1^{1^{1}}}}\right.\sigma\left(N_{u}^{f,R}\left|\phantom{1^{1^{1}}}\right.u\leqslant
s\right)\right)=N_{s}^{f,R}.$

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
$f\in{\mathscr{C}}^{2}_{b}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$ and
$t\in[0,T]$

$\displaystyle f^{2}\left(\zeta_{t}^{R}\right)$
$\displaystyle=f^{2}\left(\zeta_{0}^{R}\right)+2\int_{0}^{t}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{f\left(\zeta_{s}^{R}\right){\mathcal{L}}_{\rm
SLOW}f\left(\zeta^{R}_{s},\mu\right)\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}$
(40)
$\displaystyle\quad+2\int_{0}^{t}{f\left(\zeta_{s}^{R}\right)\mathrm{d}N_{s}^{f,R}}+\left\langle
N^{f,R}\right\rangle_{t}+\sum_{0<s\leqslant
t}{\left\\{f\left(\zeta_{s}^{R}\right)-f\left(\zeta_{s^{-}}^{R}\right)\right\\}^{2}}.$

On the other hand, applying (37) with
$f^{2}\in{\mathscr{C}}_{b}^{2}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$,
we obtain that for all $t\in[0,T]$

$f^{2}\left(\zeta_{t}^{R}\right)=f^{2}\left(\zeta_{0}^{R}\right)+\int_{0}^{t}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
SLOW}f^{2}\left(\zeta_{s}^{R},\mu\right)\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}+N_{t}^{f^{2},R}.$
(41)

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

$N_{t}^{f^{2},R}-2\int_{0}^{t}{f\left(\zeta_{s}^{R}\right)\mathrm{d}N_{s}^{f,R}}=\left\langle
N^{f,R}\right\rangle_{t}+\sum_{0<s\leqslant
t}{\left\\{f\left(\zeta_{s}^{R}\right)-f\left(\zeta_{s^{-}}^{R}\right)\right\\}^{2}}$

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

$\left\langle N^{f,R}\right\rangle_{t}=-\sum_{0<s\leqslant
t}{\left\\{f\left(\zeta_{s}^{R}\right)-f\left(\zeta_{s^{-}}^{R}\right)\right\\}^{2}}\leqslant
0$

so that $\left\langle N^{f,R}\right\rangle_{t}=0$. Hence, ${\mathbb{Q}}-$a.s.
$N^{f,R}=0$ which completes the proof. ∎

##  6 Characterisation of the occupation measure limit on the torus

Consider a probability measure ${\mathbb{Q}}$ on
${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}\left({\mathbb{R}}\right)\right)$
and the canonical process $\left(\zeta^{R},\Gamma^{R}\right)$ 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 $T\geqslant 0$ be fixed. With the notations of Theorem 5.1, there exists a
random probability measure-valued process
$\left(\gamma_{t}^{R}\right)_{t\in[0,T]}$ that is predictable in $(\omega,t)$
and such that for all bounded measurable function
$\psi:[0,T]\times{\mathcal{M}}_{1}\left({\mathbb{R}}\right)\to{\mathbb{T}}_{R}$,

$\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{\psi(s,\mu})\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}=\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{\psi(s,\mu)\gamma^{R}_{s}(\mathrm{d}\mu)\mathrm{d}s}}.$
(42)

###### Proof.

The desintegration result of $\Gamma^{R}$ follows directly from  [49,  Lemma
1.4]. ∎

######  Corollary 6.2.

Let $T\geqslant 0$ be fixed. With the notations of Theorem 5.1, we have that
$\zeta^{R}\in{\mathscr{C}}^{0}\left([0,T],{\mathbb{T}}_{R}\right)$ is
differentiable of derivative in $L^{1}({\mathbb{R}})$ ${\mathbb{Q}}-$a.s.

###### Proof.

Applying (37) with
$f={\rm{id}}\in{\mathscr{C}}_{b}^{1}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$,
we deduce from Theorem 5.1, (9) and (42), ${\mathbb{Q}}-$a.s., for all
$t\in[0,T]$

$\displaystyle\zeta_{t}^{R}=\zeta_{0}^{R}+\int_{0}^{t}{\left(\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{2}\left(\mu\right)\gamma_{s}^{R}\left(\mathrm{d}\mu\right)}\right)\partial_{1}{\rm{Fit}}\left(\zeta_{s}^{R},\zeta_{s}^{R}\right)\mathrm{d}s}.$

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. ∎

We now want to characterise the limiting value
$\Gamma^{R}\left(\mathrm{d}t,\mathrm{d}\mu\right)=\gamma_{t}^{R}\left(\mathrm{d}\mu\right)\mathrm{d}t$
under ${\mathbb{Q}}$.

######  Proposition 6.3.

With the notations of Theorem 5.1, for a.e. $t\in[0,T]$, ${\mathbb{Q}}-$a.s.,
$\gamma_{t}^{R}=\pi^{\lambda\left(\zeta^{R}\right)}$ 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 $\mu$:

$P_{f,n}\left(\mu\right):=\left\langle
f,\mu^{n}\right\rangle:=\int_{{\mathbb{R}}}{\cdots\int_{{\mathbb{R}}}{f\left(x_{1},\cdots,x_{n}\right)\mu\left(\mathrm{d}x_{1}\right)\cdots\mu\left(\mathrm{d}x_{n}\right)}}$

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

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

From (25), we can see that if
$f\in{\mathscr{C}}^{4}_{b}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$,
${\mathcal{L}}_{\rm{FVc}}^{\lambda}P_{f,n}\left(\mu\right)$ is a polynomial in
$\mu$ of the form $P_{J,n+1}\left(\mu\right)$ for some function
$J\in{\mathscr{C}}^{2}_{\left\|\cdot\right\|}\left({\mathbb{R}}^{n+1},{\mathbb{R}}\right)$.
We recall from  [16,  Proposition 2.11] that if
$\mu\in{\mathcal{M}}_{1}^{c,5}({\mathbb{R}})$, then $\sup_{0\leqslant
t\leqslant T}{{\mathbb{E}}_{\mu}\left(M_{5}\left(X_{t}\right)\right)}<\infty$
where $\left(X_{t}\right)_{t\geqslant 0}$ denotes the centered Fleming-Viot
process with resampling rate $\lambda$. Since, by (25), for all
$\mu\in{\mathcal{M}}_{1}^{c,4}({\mathbb{R}})$,

$\left|{\mathcal{L}}_{\rm FVc}P_{J,n+1}(\mu)\right|\leqslant
C\left(1+M_{4}\left(\mu\right)\right),$

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 $\left(\widehat{M}_{t}^{P_{J,n+1}}\right)_{t\geqslant 0}$
defined by

$\widehat{M}_{t}^{P_{J,n+1}}:=P_{J,n+1}\left(X_{t}\right)-P_{J,n+1}\left(X_{0}\right)-\int_{0}^{t}{{\mathcal{L}}_{\rm
FVc}^{\lambda}P_{J,n+1}\left(X_{s}\right)\mathrm{d}s}$ (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
$\gamma\in{\mathcal{M}}_{1}\left({\mathcal{M}}_{1}({\mathbb{R}})\right)$
satisfies
$\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{4}\left(\mu\right)\gamma\left(\mathrm{d}\mu\right)}<\infty$
and

$\displaystyle\forall n\in{\mathbb{N}}^{\star},\ \forall
f\in{\mathscr{C}}_{\|\cdot\|}^{2}\left({\mathbb{R}}^{n},{\mathbb{R}}\right),\quad\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda}{P_{f,n}}\left(\mu\right)\gamma\left(\mathrm{d}\mu\right)}=0,$

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

###  6.1 Proof of Proposition 6.3

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

$\displaystyle K^{2}\sigma^{2}M_{t\wedge\tau^{K,R}}^{K,P_{f,n}}$
$\displaystyle=\sum_{i\,=\,1}^{3}{{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{i}\left(t\wedge\tau^{K,R}\right)}$

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

$\displaystyle{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{1}(t)$
$\displaystyle:=K^{2}\sigma^{2}\left(P_{f,n}\left(\mu_{t}^{K,R}\right)-P_{f,n}\left(\mu_{0}^{K,R}\right)\right),$
$\displaystyle{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{2}(t)$
$\displaystyle:=-\int_{0}^{t}{\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{\theta\left(z_{s}^{K,R}\right)m_{2}\left(z_{s}^{K,R}\right){\mathcal{L}}_{\rm
FVc}^{\lambda\left(z_{s}^{K,R}\right)}P_{f,n}\left(\mu\right)\Gamma^{K,R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}},$
$\displaystyle{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{3}(t)$
$\displaystyle:=O\left(\frac{1}{\sqrt{K}}+\sigma
K^{\frac{3}{2}+\varepsilon}+\int_{0}^{t}{\frac{M_{3}\left(\mu_{s}^{K,R}\right)}{K}\,\mathrm{d}s}\right),$

is a martingale. Since $\sigma K\to 0$ by Assumption (5),
${\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{1}\left(t\wedge\tau^{K,R}\right)\to 0$ when
$K\to+\infty$. From Corollary 4.3, the sequence
$\left({\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{2}\left(t\wedge\tau^{K}\right)\right)_{t\in[0,T],K\in{\mathbb{N}}^{\star}}$
is uniformly integrable and converges in law, when $K\to+\infty$, to

$M_{t}^{P_{f,n}}:=-\int_{0}^{t}{\theta\left(\zeta_{s}^{R}\right)m_{2}\left(\zeta_{s}^{R}\right)\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{s}^{R}\right)}P_{f,n}\left(\mu\right)\Gamma^{R}\left(\mathrm{d}s,\mathrm{d}\mu\right)}}.$

Note that, from Cauchy-Schwarz’s inequality and Lemma 4.1, there exists a
constant $C>0$ such that

${\mathbb{E}}\left(\sup_{0\leqslant t\leqslant
T}{\int_{0}^{t\wedge\tau^{K,R}}{\frac{M_{3}\left(\mu_{s}^{K,R}\right)}{K}}}\mathrm{d}s\right)^{2}\leqslant\frac{T}{K^{2}}{\mathbb{E}}\left(\int_{0}^{T\wedge\tau^{K,R}}{M_{6}\left(\mu_{s}^{K,R}\right)\mathrm{d}s}\right)\leqslant\frac{CT}{K^{2}}.$

Hence, $\lim_{K\to+\infty}{\sup_{0\leqslant t\leqslant
T}{{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{3}(t\wedge\tau^{K,R})}}=0$ in
$L^{2}({\mathbb{R}})$. In particular, we deduce that a subsequence of
$\left(\sup_{0\leqslant t\leqslant
T}{{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{3}(t\wedge\tau^{K,R})}\right)_{K\in{\mathbb{N}}^{\star}}$
converges almost surely to $0$ and the family $\left(\sup_{0\leqslant
t\leqslant
T}{{\rm{\textbf{\rm{{(A)}}}}}^{K,R}_{3}(t\wedge\tau^{K,R})}\right)_{K\in{\mathbb{N}}^{\star}}$
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
$\left(M_{t}^{P_{f,n}}\right)_{t\geqslant 0}$ is a ${\mathbb{Q}}-$martingale.
As it is also a continuous and finite variation process by Lemma 42, it must
hence be ${\mathbb{Q}}-$almost surely null  [42,  Theorem 4.1]. Hence, using
Lemma 42 again, we have proved that

$\forall f\in{\mathscr{C}}_{K}^{3}\left({\mathbb{R}}^{n},{\mathbb{R}}\right),\
{\mathbb{Q}}-{\rm a.s.},\ \mathrm{d}t-{\rm
a.e.},\quad\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f,n}\left(\mu\right)\gamma_{t}^{R}\left(\mathrm{d}\mu\right)}=0.$

The space ${\mathscr{C}}^{2}_{K}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$
equipped with the norm
$\left\|f\right\|_{W_{0}^{2,\infty}}=\left\|f\right\|_{\infty}+\left\|\nabla
f\right\|_{\infty}+\left\|{\rm{Hess}}(f)\right\|_{\infty}$ is separable. So,
we can choose a dense countable family
${\mathcal{B}}\subset{\mathscr{C}}^{3}_{K}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$
such that

$\forall f\in{\mathscr{C}}^{2}_{K}\left({\mathbb{R}}^{n},{\mathbb{R}}\right),\
\exists\,\left(f_{p}\right)_{p\in{\mathbb{N}}^{\star}}\in{\mathcal{B}}^{{\mathbb{N}}^{\star}},\quad
f_{p}\xrightarrow[p\longrightarrow+\infty]{\|\cdot\|_{W_{0}^{2,\infty}}}f.$

Then,

${\mathbb{Q}}-{\rm a.s.},\ \mathrm{d}t-{\rm a.e.},\ \forall
p\in{\mathbb{N}}^{\star},\quad\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f_{p},n}\left(\mu\right)\gamma_{t}^{R}\left(\mathrm{d}\mu\right)}=0.$

From (38) and Lemma 42, we have that
$\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{4}\left(\mu\right)\gamma_{t}^{R}(\mathrm{d}\mu)}<\infty$.
As for all $p\in{\mathbb{N}}^{\star}$,

${\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f_{p},n}\left(\mu\right)\leqslant
C_{1}\left(1+M_{2}\left(\mu\right)\right)\left\|f_{p}\right\|_{W_{0}^{2,\infty}}.$

for some constant $C_{1}>0$, we obtain by the dominated convergence theorem
that

${\mathbb{Q}}-{\rm a.s.},\ \mathrm{d}t-{\rm a.e.},\ \forall
f\in{\mathscr{C}}_{K}^{2}\left({\mathbb{R}}^{n},{\mathbb{R}}\right),\quad\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f,n}\left(\mu\right)\gamma_{t}^{R}\left(\mathrm{d}\mu\right)}=0.$
(44)

Let us consider
$f\in{\mathscr{C}}^{2}_{\|\cdot\|}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$
and for all $q\in{\mathbb{N}}^{\star}$,
$x\mapsto\chi_{q}(x)=\exp\left(-\frac{1}{q^{2}-\left\|x\right\|_{\infty}^{2}}\right){\mathds{1}}_{\left\|x\right\|_{\infty}<q}$
of class ${\mathscr{C}}^{\infty}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$
with compact support. Then, for all $q\in{\mathbb{N}}^{\star}$,
$f\chi_{q}\in{\mathscr{C}}_{K}^{2}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$.
Noting that, $\left|\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f\chi_{q},n}\left(\mu\right)\gamma_{t}^{R}\left(\mathrm{d}\mu\right)}\right|$
is dominated by
$C_{2}\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{\left(1+M_{4}\left(\mu\right)\right)\gamma_{t}^{R}(\mathrm{d}\mu)}$
for some constant $C_{2}>0$, and since
$\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{4}\left(\mu\right)\gamma_{t}^{R}(\mathrm{d}\mu)}<\infty$,
we deduce from (44) applied to $f\chi_{q}$, by the dominated convergence
theorem, that

${\mathbb{Q}}-{\rm a.s.},\ \mathrm{d}t-{\rm a.e.},\ \forall
f\in{\mathscr{C}}^{2}_{\|\cdot\|}\left({\mathbb{R}}^{n},{\mathbb{R}}\right),\quad\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda\left(\zeta_{t}^{R}\right)}P_{f,n}\left(\mu\right)\gamma_{t}^{R}\left(\mathrm{d}\mu\right)}=0.$

Therefore, it follows from Lemma 6.4 that ${\mathbb{Q}}-$a.s,
$\mathrm{d}t-$a.e.,
$\gamma_{t}^{R}\left(\mathrm{d}\mu\right)=\pi^{\lambda\left(\zeta_{t}^{R}\right)}\left(\mathrm{d}\mu\right)$
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 $\left(\xi_{t}\right)_{t\geqslant 0}$ of
the centered Fleming-Viot process $\left(X_{t}\right)_{t\geqslant 0}$ with
resampling rate $\lambda$, on the state space
$\bigcup_{n\in{\mathbb{N}}^{\star}}{{\mathscr{C}}^{2}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)}$,
obtained in  [16,  Section 3.2] is defined as below.

Let us consider $M:=\left(M(t)\right)_{t\geqslant 0}$ a Markov’s birth and
death process in ${\mathbb{N}}$ whose transition rates $q_{i,j}$ from $i$ to
$j$ are given by:

${\rm{\textbf{\rm{{(1)}}}}}\ \ q_{n,n+1}=\lambda
n^{2}\qquad\quad{\rm{\textbf{\rm{{(2)}}}}}\ \ q_{n,n-1}=\lambda
n(n-1)\qquad\quad{\rm{\textbf{\rm{{(3)}}}}}\ \ q_{i,j}=0\ {\rm{otherwise}}.$

For all $M(0)\in{\mathbb{N}}^{\star}$,
$\xi_{0}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$
and $\lambda>0$, we define

$\xi_{t}:=T_{\lambda}^{\left(M\left(\tau_{n}\right)\right)}\left(t-\tau_{n}\right)\Lambda_{n}T_{\lambda}^{\left(M\left(\tau_{n-1}\right)\right)}\left(\tau_{n}-\tau_{n-1}\right)\Lambda_{n-1}\cdots\Lambda_{1}T_{\lambda}^{\left(M\left(0\right)\right)}\left(\tau_{1}\right)\xi_{0},\\\
\tau_{n}\leqslant t<\tau_{n+1},\ n\in{\mathbb{N}},$ (45)

where $\left(\tau_{n}\right)_{n\in{\mathbb{N}}}$ is the sequence of jump times
of $M$ with $\tau_{0}=0$, $\left(T_{\lambda}^{(n)}(t)\right)_{t\geqslant 0}$
is the semi-group of operator associated to the generator $B^{(n)}_{\lambda}$
given by (22) and where $\left(\Lambda_{n}\right)_{n\in{\mathbb{N}}}$ is a
sequence of random operators. These are conditionally independent given $M$
and satisfy for all $k\in{\mathbb{N}}$, $n\geqslant 1$ and $1\leqslant i\neq
j\leqslant n$,

${\mathbb{P}}\left(\Lambda_{k}=\Phi_{i,j}\left|\phantom{1^{1^{1^{1}}}}\right.\left\\{M\left(\tau_{k}^{-}\right)=n,M\left(\tau_{k}\right)=n-1\right\\}\right)=\frac{1}{n(n-1)}$

and for all $n\geqslant 1$ and $1\leqslant i,j\leqslant n$,

${\mathbb{P}}\left(\Lambda_{k}=K_{i,j}\left|\phantom{1^{1^{1^{1}}}}\right.\left\\{M\left(\tau_{k}^{-}\right)=n,M\left(\tau_{k}\right)=n+1\right\\}\right)=\frac{1}{n^{2}},$

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 $t\geqslant 0$,
$\xi_{t}\in{\mathscr{C}}^{2}\left({\mathbb{R}}^{M(t)},{\mathbb{R}}\right)$,
$K_{i,j}\xi_{t}$ is well defined. Moreover, the dual process
$\left(\xi_{t}\right)_{t\geqslant 0}$ with initial condition $\xi_{0}$ is
constructed on the same probability space and independently of the centered
Fleming-Viot process $\left(X_{t}\right)_{t\geqslant 0}$ with resampling rate
$\lambda$ and initial condition $\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$.
We shall denote by ${\mathbb{P}}_{\left(\mu,\xi_{0}\right)}$, the law of the
couple $\left(X_{t},\xi_{t}\right)_{t\geqslant 0}$ 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
$\xi_{0}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$
there exists a function $C_{0}$ from
$\bigcup_{k\in{\mathbb{N}}^{\star}}{(0,+\infty)^{k-1}\times\\{k\\}}$ to
${\mathbb{R}}_{+}$, locally bounded, such that

$C_{0}\left(\tau_{1},1\right):=\left\|\xi_{0}\right\|_{\infty}+\frac{3}{M(0)}\left\|{\rm{Hess}}\left(\xi_{0}\right)\right\|_{\infty}$

and, for all
$\left(t_{j}\right)_{j\in{\mathbb{N}}}\in(0,+\infty)^{{\mathbb{N}}}$,
$k\mapsto C_{0}\left(\left(t_{i}\right)_{0\leqslant i\leqslant k-1},k\right)$
is non-decreasing and satisfying

$\forall k\in{\mathbb{N}}^{\star},\ \forall t\leqslant\tau_{k},\ \forall
x\in{\mathbb{R}}^{M(t)},\qquad\left|\xi_{t}(x)\right|\leqslant
C_{0}\left(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant i\leqslant
k-1},k\right)\left(1+\left\|x\right\|_{\infty}^{2S_{t}}\right).$ (46)

###### Proof.

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

$\displaystyle({\mathcal{Q}}_{k}):\quad$ $\displaystyle\forall
x\in{\mathbb{R}}^{M\left(\tau_{k}\right)},\quad\left|\xi_{\tau_{k}}(x)\right|\leqslant
C_{0}(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant i\leqslant
k-1},k)\left(1+\left\|x\right\|_{\infty}^{2k}\right).$

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

$\sigma_{t}^{\lambda}:=\begin{pmatrix}\frac{1-\exp\left(-4\lambda
nt\right)}{4\lambda n}&0&\dots&\dots&0\vspace{0.1cm}\\\ 0&t&0&\dots&0\\\
\vdots&0&\ddots&\ddots&\vdots\\\ \vdots&\vdots&\ddots&\ddots&0\\\
0&0&\dots&0&t\end{pmatrix}\quad{\rm and}\quad
P:=\begin{pmatrix}\frac{1}{\sqrt{n}}&\frac{1}{\sqrt{2}}&\cdots&\cdots&\frac{1}{\sqrt{n(n-1)}}\\\
\vdots&-\sqrt{\frac{1}{2}}&\ddots&\cdots&\frac{1}{\sqrt{n(n-1)}}\\\
\vdots&0&\ddots&\ddots&\vdots\\\
\vdots&\vdots&\ddots&\ddots&\frac{1}{\sqrt{n(n-1)}}\\\
\frac{1}{\sqrt{n}}&0&\cdots&0&-\sqrt{\frac{n-1}{n}}\end{pmatrix}.$

Note that for all $i,j\in\\{1,\cdots,n\\}$,
$\partial_{x_{i}}m_{t,x}^{\lambda}=\epsilon_{i}-\frac{1-\exp\left(-2\gamma
nt\right)}{n}\bm{1}$ and $\partial_{x_{i}x_{j}}^{2}m_{t,x}^{\lambda}=0$ where
$\left(\epsilon_{1},\cdots,\epsilon_{n}\right)$ is the canonical basis of
${\mathbb{R}}^{n}$. For all $\lambda>0$, the key identity for the sequel is

$\forall f\in L^{\infty}\left({\mathbb{R}}^{n}\right),\ \forall t\geqslant 0,\
\forall x\in{\mathbb{R}}^{n},\quad T_{\lambda}^{(M(0))}(t)f(x)=\left(f\ast
g_{t,0}^{\lambda}\right)\left(m_{t,x}^{\lambda}\right)$ (47)

where $\ast$ 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 $x\in{\mathbb{R}}^{M(0)}$, $\lambda>0$,
$\xi_{t}(x)=T_{\lambda}^{\left(M(0)\right)}(t)\xi_{0}(x)$ and so, from  [16,
Theomem 6.1],
$\left|\xi_{t}(x)\right|\leqslant\left\|\xi_{0}\right\|_{\infty}$. 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
$i,j\in\left\\{1,\cdots,M(0)\right\\}$ be fixed.

  * $\bullet$

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

$\forall
x\in{\mathbb{R}}^{M(0)-1},\qquad\xi_{\tau_{1}}(x)=\Phi_{i,j}T_{\lambda}^{\left(M(0)\right)}\left(\tau_{1}\right)\xi_{0}(x).$

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

  * $\bullet$

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 $x\in{\mathbb{R}}^{M(0)+1}$,

$\displaystyle\xi_{\tau_{1}}(x)=K_{i,j}T_{\lambda}^{\left(M(0)\right)}\left(\tau_{1}\right)\xi_{0}(x)$
$\displaystyle=\partial_{x_{i}x_{j}}T_{\lambda}^{\left(M(0)\right)}\left(\tau_{1}\right)\xi_{0}\left(\widetilde{x}\right)x_{M(0)+1}^{2}$
$\displaystyle=\left(\partial_{x_{j}}m_{\tau_{1},\widetilde{x}}^{\lambda}\right)^{t}\left[\left({\rm{Hess}}\left(\xi_{0}\right)\ast
g_{\tau_{1},0}^{\lambda}\right)\left(m_{\tau_{1},\widetilde{x}}^{\lambda}\right)\partial_{x_{i}}m_{t,\widetilde{x}}^{\lambda}\right]x_{M(0)+1}^{2}$

where
$\widetilde{x}=\left(x_{1},\cdots,x_{M(0)}\right)\in{\mathbb{R}}^{M(0)}$. As,

$\left({\rm{Hess}}\left(\xi_{0}\right)\ast
g_{\tau_{1},0}^{\lambda}\right)\left(m_{\tau_{1},\widetilde{x}}^{\lambda}\right)=\left(\int_{{\mathbb{R}}^{M(0)}}{\partial_{ij}^{2}\xi_{0}(u)g_{\tau_{1},0}^{\lambda}\left(m_{\tau_{1},\widetilde{x}}^{\lambda}-u\right)\mathrm{d}u}\right)_{1\leqslant
i,j\leqslant M(0)}$

we obtain that $\left\|\left({\rm{Hess}}\left(\xi_{0}\right)\ast
g_{\tau_{1},0}^{\lambda}\right)\left(m_{\tau_{1},\widetilde{x}}^{\lambda}\right)\right\|_{\infty}\leqslant\left\|\partial_{ij}^{2}\xi_{0}\right\|_{\infty}$.
Noting that
$\left|\left(\partial_{x_{j}}m_{\tau_{1},\widetilde{x}}^{\lambda}\right)^{t}\partial_{x_{i}}m_{t,\widetilde{x}}^{\lambda}\right|\leqslant\frac{3}{M(0)}$,
we deduce that for all $x\in{\mathbb{R}}^{M(0)+1}$,

$\left|\xi_{\tau_{1}}(x)\right|\leqslant\frac{3}{M(0)}\sup_{1\leqslant
i,j\leqslant
M(0)}{\left\|\partial_{ij}^{2}\xi_{0}\right\|_{\infty}}x_{M(0)+1}^{2}\leqslant\frac{3}{M(0)}\left\|{\rm{Hess}}(\xi_{0})\right\|_{\infty}\left\|x\right\|_{\infty}^{2}$

and $({\mathcal{Q}}_{1})$ follows.

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

  * $\bullet$

Case $\Lambda_{k}=\Phi_{i,j}$ at the $k^{\rm{th}}$ jump. In this case,
$M(\tau_{k})=M(\tau_{k-1})-1$ and we deduce from (45) that for all
$x\in{\mathbb{R}}^{M\left(\tau_{k-1}\right)-1}$,
$\xi_{\tau_{k}}(x)=\Phi_{i,j}T_{\lambda}^{\left(M(\tau_{k-1})\right)}(\tau_{k}-\tau_{k-1})\xi_{\tau_{k-1}}(x)$.
By using (23) and $\left({\mathcal{Q}}_{k-1}\right)$, we deduce from  [16,
Corollary 6.2 (1)] that for all
$x\in{\mathbb{R}}^{M\left(\tau_{k-1}\right)-1}$,

$\displaystyle\left|\Phi_{i,j}T_{\lambda}^{\left(M\left(\tau_{k-1}\right)\right)}\left(\tau_{k}-\tau_{k-1}\right)\xi_{\tau_{k-1}}(x)\right|$
(48) $\displaystyle\hskip 56.9055pt\leqslant
C_{2}\left(\tau_{k}-\tau_{k-1},M\left(\tau_{k-1}\right)\right){C}_{0}\left(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant
i\leqslant k-1},k\right)\left(1+\left\|x\right\|_{\infty}^{2(k-1)}\right),$

where $C_{2}{C}_{0}$ is locally bounded. As $t\mapsto
C_{2}\left(t,M\left(\tau_{k-1}\right)\right)$ is non-decreasing, we deduce
$\left({\mathcal{Q}}_{k}\right)$ in that case.

  * $\bullet$

Case $\Lambda_{k}=K_{i,j}$ at the $k^{\rm{th}}$ 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 $x\in{\mathbb{R}}^{M\left(\tau_{k-1}\right)+1}$,

$\displaystyle\left|\xi_{\tau_{k}}(x)\right|$
$\displaystyle=\left|K_{i,j}T_{\lambda}^{\left(M\left(\tau_{k-1}\right)+1\right)}\left(\tau_{k}-\tau_{k-1}\right)\xi_{\tau_{k-1}}(x)\right|$
$\displaystyle=\left|\left(\partial_{x_{j}}m^{\lambda}_{\tau_{k}-\tau_{k-1},\widetilde{x}}\right)^{t}\left[\left(\xi_{\tau_{k-1}}\ast{\rm{Hess}}\left(g_{\tau_{k}-\tau_{k-1},0}^{\lambda}\right)\right)(m^{\lambda}_{\tau_{k}-\tau_{k-1},\widetilde{x}})\partial_{x_{i}}m^{\lambda}_{\tau_{k}-\tau_{k-1},\widetilde{x}}\right]\right|x_{M\left(\tau_{k-1}\right)+1}^{2},$

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

$\displaystyle\left|K_{i,j}T_{\lambda}^{\left(M\left(\tau_{k-1}\right)+1\right)}\left(\tau_{k}-\tau_{k-1}\right)\xi_{\tau_{k-1}}(x)\right|$
(49) $\displaystyle\hskip 56.9055pt\leqslant
C_{3}\left(\tau_{k}-\tau_{k-1},M\left(\tau_{k-1}\right)\right)C_{0}\left(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant
i\leqslant k-1},k\right)\left(1+\left\|x\right\|_{\infty}^{2k}\right),$

where $C_{3}{C}_{0}$ is locally bounded and $\left({\mathcal{Q}}_{k}\right)$
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 $t\in\,\left]\tau_{k},\tau_{k+1}\right[$,

$\forall
x\in{\mathbb{R}}^{M\left(t\right)},\qquad\xi_{t}(x)=T_{\lambda}^{\left(M\left(\tau_{k}\right)\right)}\left(t-\tau_{k}\right)\xi_{\tau_{k}}(x),$

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

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

$\displaystyle\vartheta_{k,\ell}$ $\displaystyle:=\inf{\left\\{t\geqslant
0\left.\phantom{1^{1^{1^{1}}}}\right|S_{t}\geqslant
k\quad{\rm{or}}\quad\exists s\in[0,t],\
\left\langle\xi_{s},X_{t-s}^{M(s)}\right\rangle\geqslant\ell\right\\}},$
$\displaystyle\vartheta_{k,m}^{\prime}$
$\displaystyle:=\inf{\left\\{t\geqslant
0\left.\phantom{1^{1^{1^{1}}}}\right|S_{t}\geqslant k\quad{\rm{or}}\quad
C_{0}\left(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant i\leqslant
k-1},k\right)\geqslant m\right\\}},$
$\displaystyle\vartheta_{k,\ell,m}^{\prime}$
$\displaystyle:=\inf{\left\\{t\geqslant
0\left.\phantom{1^{1^{1^{1}}}}\right|S_{t}\geqslant
k\quad{\rm{or}}\quad\exists s\in[0,t],\
\left\langle\xi_{s},X_{t-s}^{M(s)}\right\rangle\geqslant\ell\right.}$
$\displaystyle\hskip
83.93553pt\left.\phantom{1^{1^{1^{1}}}}\quad{\rm{or}}\quad
C_{0}\left(\left(\tau_{i+1}-\tau_{i}\right)_{0\leqslant i\leqslant
k-1},k\right)\geqslant m\right\\}.$

As $\left(M(t),\xi_{t}\right)_{t\geqslant 0}$ is independent of
$\left(X_{t}\right)_{t\geqslant 0}$, note that for all
$k\in{\mathbb{N}}^{\star}$, $m\in{\mathbb{N}}$, $\vartheta_{k,m}^{\prime}$ is
independent of $\left(X_{t}\right)_{t\geqslant 0}$. Recall that for all
$t\geqslant 0$,
$\xi_{t}\in{\mathscr{C}}^{2}\left({\mathbb{R}}^{M(t)},{\mathbb{R}}\right)$.
For all $n\geqslant M(t)$, for all $x\in{\mathbb{R}}^{n}$, we denote by
$\xi_{t}^{(n)}(x)=\xi_{t}^{(n)}\left(x_{1},\cdots,x_{n}\right):=\xi_{t}\left(x_{1},\cdots,x_{M(t)}\right)$
so that for all $\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$,

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

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

######  Lemma 6.6.

Given $\left(X_{t}\right)_{t\geqslant 0}$, $\left(\xi_{t}\right)_{t\geqslant
0}$ as above with $X_{0}:=\mu\in{\mathcal{M}}_{1}^{c,2k}({\mathbb{R}})$,
$k\in{\mathbb{N}}^{\star}$ and
$\xi_{0}\in{\mathscr{C}}_{b}^{2}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$,
we have that ${\mathbb{P}}_{\left(\mu,\xi_{0}\right)}-$a.s., for all
$t\geqslant 0$, for all $m\in{\mathbb{N}}$,

$\displaystyle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{0},X_{t\wedge\vartheta_{k,m}^{\prime}}^{M(0)}\right\rangle\right)=\left\langle{\mathbb{E}}_{\xi_{0}}\left(\xi_{t\wedge\vartheta_{k,m}^{\prime}}^{\left(M(0)+k\right)}\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right),\mu^{M(0)+k}\right\rangle.$
(50)

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 $t\geqslant 0$, $k\in{\mathbb{N}}^{\star}$,
$\ell,m\in{\mathbb{N}}$. As
$\vartheta_{k,\ell,m}^{\prime}\leqslant\vartheta_{k,\ell}$, the weak duality
identity of  [16,  Theorem 3.4] implies that:

$\displaystyle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{0},X_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{M(0)}\right\rangle\right)$
$\displaystyle={\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}},\mu^{M\left(t\wedge\vartheta_{k,\ell,m}^{\prime}\right)}\right\rangle\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right)$
$\displaystyle={\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{\left(M(0)+k\right)},\mu^{M(0)+k}\right\rangle\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right).$

Now, from Lemma 46 and the definition of $\vartheta_{k,\ell,m}^{\prime}$, note
that for all $x\in{\mathbb{R}}^{M(0)+k}$,

${\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{\left(M(0)+k\right)}(x)\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right)\leqslant
m\left(1+\left\|x\right\|_{\infty}^{2k}\right)\exp\left(\lambda
t\left(M(0)+k\right)^{2}\right).$ (51)

Since $\mu\in{\mathcal{M}}_{1}^{c,2k}({\mathbb{R}})$, we deduce from Fubini’s
theorem that

${\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{0},X_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{M(0)}\right\rangle\right)=\left\langle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{\left(M(0)+k\right)}\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right),\mu^{M(0)+k}\right\rangle.$

On the one hand, from  [16,  Section 6.2.1],
$\lim_{\ell\to+\infty}{\vartheta_{k,\ell,m}^{\prime}}=\vartheta_{k,m}^{\prime}$
${\mathbb{P}}_{\left(\mu,\xi_{0}\right)}-$a.s., and since
$\left(X_{t}\right)_{t\geqslant 0}$ has continous paths for the topology of
weak convergence, we have ${\mathbb{P}}_{\left(\mu,\xi_{0}\right)}-$a.s.

$\lim_{\ell\to+\infty}{\left\langle\xi_{0},X_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{M(0)}\right\rangle}=\left\langle\xi_{0},X_{t\wedge\vartheta_{k,m}^{\prime}}^{M(0)}\right\rangle.$

Therefore, we deduce from the dominated convergence theorem, that for all
$\xi_{0}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$,

$\lim_{\ell\to+\infty}{{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{0},X_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{M(0)}\right\rangle\right)}={\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\langle\xi_{0},X_{t\wedge\vartheta_{k,m}^{\prime}}^{M(0)}\right\rangle\right).$

On the other hand, Lemma 46 and the dominated convergence theorem imply that
for all $x\in{\mathbb{R}}^{M(0)+k}$,

$\displaystyle\lim_{\ell\to+\infty}{{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{\left(M(0)+k\right)}(x)\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right)}$
$\displaystyle\hskip
56.9055pt={\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,m}^{\prime}}^{\left(M(0)+k\right)}(x)\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right)$

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

$\displaystyle\lim_{\ell\to+\infty}{\left\langle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,\ell,m}^{\prime}}^{\left(M(0)+k\right)}\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,\ell,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right),\mu^{M(0)+k}\right\rangle}$
$\displaystyle\hskip
56.9055pt=\left\langle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,m}^{\prime}}^{\left(M(0)+k\right)}\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right),\mu^{M(0)+k}\right\rangle$

and the announced result follows from the fact that $\vartheta_{k,m}^{\prime}$
is independent of $\left(X_{t}\right)_{t\geqslant 0}$, which implies

$\displaystyle{\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\xi_{t\wedge\vartheta_{k,m}^{\prime}}^{\left(M(0)+k\right)}(x)\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right)$
$\displaystyle\hskip
85.35826pt={\mathbb{E}}_{\xi_{0}}\left(\xi_{t\wedge\vartheta_{k,m}^{\prime}}^{\left(M(0)+k\right)}(x)\exp\left(\lambda\int_{0}^{t\wedge\vartheta_{k,m}^{\prime}}{M^{2}(u)\mathrm{d}u}\right)\right).\qed$

From Lemma 46, we deduce that we can choose $m\in{\mathbb{N}}$ large enough
such that $\vartheta^{\prime}_{1,m}=\tau_{1}$. In this case, the function
inside the brackets in the left-hand side of (50) is defined, for all
$t\geqslant 0$ and for all $x\in{\mathbb{R}}^{M(0)+1}$, by

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

For all $t\geqslant 0$, $n\in{\mathbb{N}}^{\star}$,
$f\in{\mathscr{C}}^{2}_{b}\left({\mathbb{R}}^{n},{\mathbb{R}}\right)$ and
$\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$ let us denote

$P_{\rm{FVc}}(t)P_{f,n}\left(\mu\right):={\mathbb{E}}_{\mu}\left(P_{f,n}\left(X_{t}\right)\right)$

the semi-group of the centered Fleming-Viot process $(X_{t})_{t\geqslant 0}$
with resampling rate $\lambda$.

The following result is the main result of this section. It gives an extension
of Lemma 46 to a function which appears naturally in the proof of Lemma 6.4.

######  Proposition 6.7.

For all $t\geqslant 0$, for all $\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$,
for all
$\xi_{0}\in{\mathscr{C}}^{2}_{b}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$

$P_{\rm{FVc}}(t)P_{\xi_{0},M(0)}(\mu)=P_{V_{t},M(0)+1}\left(\mu\right):=\left\langle
V_{t},\mu^{M(0)+1}\right\rangle$

where

$V_{t}:=\exp\left(\alpha_{0}t\right)\psi_{t}-\alpha_{0}\int_{0}^{t}{\exp\left(\alpha_{0}s\right)\psi_{s}\mathrm{d}s}$
(52)

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

$\left|V_{t}(x)\right|\leqslant C_{t}\left(1+\|x\|^{2}_{\infty}\right).$

###### Proof.

Let $t\geqslant 0$, $\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$,
$\xi_{0}\in{\mathscr{C}}_{b}^{2}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$
be fixed. Note that from Lemma 46, there exists a constant
$\widetilde{C}_{t}$, depending only on $\xi_{0}$ and $t$ such that for all
$x\in{\mathbb{R}}^{M(0)+1}$,

$\left|\psi_{t}(x)\right|\leqslant
C_{0}\left(\tau_{1},1\right)\left(1+\left\|x\right\|_{\infty}^{2}\right){\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\exp\left(\lambda
tM^{2}(0)\right)\right)\leqslant\widetilde{C}_{t}\left(1+\left\|x\right\|_{\infty}^{2}\right).$
(53)

Now, from Lemma 50 and the fact that $\tau_{1}$ is independent of
$\left(X_{t}\right)_{t\geqslant 0}$, we obtain that

$\displaystyle\left\langle\psi_{t},\mu^{M(0)+1}\right\rangle={\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(P_{\xi_{0},M(0)}\left(X_{t\wedge\tau_{1}}\right)\right)$
$\displaystyle={\mathbb{E}}_{\xi_{0}}\left(P_{\rm{FVc}}\left(t\wedge\tau_{1}\right)P_{\xi_{0},M(0)}\left(\mu\right)\right).$

As $\tau_{1}$ follows an exponential law of parameter $\alpha_{0}$ we have
that

$\left\langle\psi_{t},\mu^{M(0)+1}\right\rangle=P_{\rm{FVc}}\left(t\right)P_{\xi_{0},M(0)}\left(\mu\right)\exp\left(-\alpha_{0}t\right)+\alpha_{0}\int_{0}^{t}{P_{\rm{FVc}}\left(s\right)P_{\xi_{0},M(0)}\left(\mu\right)\exp\left(-\alpha_{0}s\right)\mathrm{d}s}.$
(54)

Thanks to (53) and since $\mu$ has its moment of order 2 finite, the Fubini
theorem ensures us that

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

From (54) we deduce that

$\displaystyle\left\langle\int_{0}^{t}{\exp\left(\alpha_{0}s\right)\psi_{s}\mathrm{d}s},\mu^{M(0)+1}\right\rangle$
(55) $\displaystyle\hskip
21.33955pt=\alpha_{0}\int_{0}^{t}{\exp\left(\alpha_{0}s\right)\int_{0}^{s}{\exp\left(-\alpha_{0}u\right)P_{\rm{FVc}}(u)P_{\xi_{0},M(0)}\mathrm{d}u}\mathrm{d}s}+\int_{0}^{t}{P_{\rm{FVc}}(s)P_{\xi_{0},M(0)}(\mu)\mathrm{d}s}$
$\displaystyle\hskip
21.33955pt=\exp\left(\alpha_{0}t\right)\int_{0}^{t}{\exp\left(-\alpha_{0}u\right)P_{\rm{FVc}}(u)P_{\xi_{0},M(0)}(\mu)\mathrm{d}u}.$

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
$\xi_{0}\in{\mathscr{C}}_{b}^{4}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$,
$t\geqslant 0$ and $\mu\in{\mathcal{M}}_{1}^{c,4}({\mathbb{R}})$,

$\displaystyle P_{\rm FVc}(t)P_{\xi_{0},M(0)}(\mu)$
$\displaystyle=\left\langle\xi_{0},\mu^{M(0)}\right\rangle+\int_{0}^{t}{{\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\rm FVc}(s)P_{\xi_{0},M(0)}(\mu)\mathrm{d}s}$ (56)
$\displaystyle=P_{\xi_{0},M(0)}(\mu)+\int_{0}^{t}{{\mathcal{L}}_{\rm
FVc}^{\lambda}P_{V_{s},M(0)+1}(\mu)\mathrm{d}s}.$

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

For all $n\in{\mathbb{N}}^{\star}$, 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 $t\geqslant 0$,
$\lambda>0$, $\mu\in{\mathcal{M}}_{1}^{c,2}({\mathbb{R}})$ and
$\xi_{0}\in{\mathscr{C}}^{4}_{b}\left({\mathbb{R}}^{M(0)},{\mathbb{R}}\right)$
be fixed.

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

$\displaystyle\partial_{x_{i}x_{j}x_{k}x_{\ell}}^{4}T^{\left(M(0)\right)}(t)\xi_{0}(x)$
$\displaystyle\hskip
28.45274pt=\sum_{m,p,q,r\,=\,1}^{M(0)}\partial_{x_{i}}\left(m_{t,x}^{\lambda}\right)_{m}\partial_{x_{j}}\left(m_{t,x}^{\lambda}\right)_{p}\partial_{x_{k}}\left(m_{t,x}^{\lambda}\right)_{q}\partial_{x_{\ell}}\left(m_{t,x}^{\lambda}\right)_{r}\left(g_{t,0}^{\lambda}\ast\partial_{y_{m}y_{p}y_{q}y_{r}}^{4}{\xi_{0}}\right)\left(m_{t,x}^{\lambda}\right)$

where $\left(m_{t,x}^{\lambda}\right)_{p}$ designates the $p^{\rm th}$
component of $m_{t,x}^{\lambda}$. Then, for all
$k\in\left\\{0,\cdots,4\right\\}$, there exists a constant $C_{1}>0$
independent of $x$ and $t$ such that

$\left\|D^{k}T^{\left(M(0)\right)}(t)\xi_{0}(x)\right\|_{\infty}\leqslant
C_{1}\left(1+\left\|D^{k}\xi_{0}\right\|_{\infty}\right).$

From (45), (23) and (24), 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
$k\in\\{0,1,2\\}$,

$\displaystyle\left\|D^{k}\xi_{t\wedge\tau_{1}}^{\left(M(0)+1\right)}(x)\right\|_{\infty}$
$\displaystyle\leqslant
C_{1}\left(1+\left\|D^{k}\xi_{0}\right\|_{\infty}\right)+\sum_{\begin{subarray}{c}i,j\,=\,1\\\
i\,\neq\,j\end{subarray}}^{n}{\left\|D^{k}\Phi_{i,j}T_{\lambda}^{\left(M(0)\right)}\left(\tau_{1}\right)\xi_{0}\left(x_{1},\cdots,x_{M(0)-1}\right)\right\|_{\infty}}$
$\displaystyle\hskip
56.9055pt+\sum_{i,j\,=\,1}^{n}{\left\|D^{k}K_{i,j}T_{\lambda}^{\left(M(0)\right)}\left(\tau_{1}\right)\xi_{0}(x)\right\|_{\infty}}$
$\displaystyle\leqslant
C_{2}\left(1+\left\|D^{k}\xi_{0}\right\|_{\infty}+\left\|D^{k+2}\xi_{0}\right\|_{\infty}x_{M(0)+1}^{2}\right).$

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

$\displaystyle\left\|D^{k}\psi_{t}(x)\right\|_{\infty}\leqslant\exp\left(\lambda
M(0)^{2}t\right){\mathbb{E}}_{\left(\mu,\xi_{0}\right)}\left(\left\|D^{k}\xi_{t\wedge\tau_{1}}^{\left(M(0)+1\right)}(x)\right\|_{\infty}\right)\leqslant
C_{2}\exp\left(\lambda
M(0)^{2}t\right)\left(1+\left\|x\right\|_{\infty}^{2}\right).$

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
$x\in{\mathbb{R}}^{M(0)+1}$, $\left\|D^{k}V_{t}(x)\right\|_{\infty}\leqslant
C_{t}\left(1+\left\|x\right\|_{\infty}^{2}\right)$, so that
$V_{t}\in{\mathscr{C}}^{2}_{\|\cdot\|}\left({\mathbb{R}}^{M(0)+1},{\mathbb{R}}\right)$.

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

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

From (25), there exists a constant $C_{3}>0$ such that for all $t\geqslant 0$,
${\mathbb{E}}_{\mu}\left(\left|{\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\xi_{0},M(0)}\left(X_{t}\right)\right|\right)\leqslant
C_{3}\left(1+{\mathbb{E}}_{\mu}\left(M_{2}\left(X_{t}\right)\right)\right)$
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

$P_{\rm
FVc}(t)P_{\xi_{0},M(0)}(\mu)=P_{\xi_{0},M(0)}(\mu)+\int_{0}^{t}{{\mathbb{E}}_{\mu}\left({\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\xi_{0},M(0)}\left(X_{s}\right)\right)\mathrm{d}s}.$ (57)

As ${\mathcal{L}}_{\rm FVc}^{\lambda}P_{\xi_{0},M(0)}(\mu)$ is a polynomial in
$\mu$ and since $\left|{\mathcal{L}}_{\rm
FVc}^{\lambda}\left({\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\xi_{0},M(0)}(\mu)\right)\right|\leqslant
C_{4}\left(1+M_{4}\left(\mu\right)\right)$ for some constant $C_{4}>0$, we
deduce as in (43) that

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

In particular, $t\mapsto{\mathbb{E}}_{\mu}\left({\mathcal{L}}_{\rm
FVc}P_{\xi_{0},M(0)}\left(X_{t}\right)\right)$ is continuous, so

$\displaystyle{\mathbb{E}}_{\mu}\left({\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\xi_{0},M(0)}\left(X_{t}\right)\right)$
$\displaystyle=\lim_{h\to
0}{\frac{1}{h}{\mathbb{E}}_{\mu}\left(\int_{t}^{t+h}{{\mathcal{L}}_{\rm
FVc}P_{\xi_{0},M(0)}\left(X_{s}\right)\mathrm{d}s}\right)}$
$\displaystyle=\lim_{h\to
0}\frac{1}{h}{\mathbb{E}}_{\mu}\left(P_{\xi_{0},M(0)}\left(X_{t+h}\right)-P_{\xi_{0},M(0)}\left(X_{t}\right)\right)$
$\displaystyle=\lim_{h\to
0}{\frac{1}{h}}{\mathbb{E}}_{\mu}\left(P_{\rm{FVc}}\left(t\right)P_{\xi_{0},M(0)}\left(X_{h}\right)\right)-P_{\rm{FVc}}\left(t\right)P_{\xi_{0},M(0)}(\mu)$
$\displaystyle=\lim_{h\to
0}\frac{1}{h}{\mathbb{E}}_{\mu}\left(P_{V_{t},M(0)+1}\left(X_{h}\right)-P_{V_{t},M(0)+1}\left(\mu\right)\right)$

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

$\left|{\mathcal{L}}_{\rm FVc}^{\lambda}P_{V_{t},M(0)+1}(\mu)\right|\leqslant
C_{5}\left(1+M_{4}\left(\mu\right)\right).$ (58)

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

$\displaystyle\lim_{h\to
0}\frac{1}{h}{\mathbb{E}}_{\mu}\left(P_{V_{t},M(0)+1}\left(X_{h}\right)-P_{V_{t},M(0)+1}\left(\mu\right)\right)$
$\displaystyle=\lim_{h\to
0}{\frac{1}{h}{\mathbb{E}}_{\mu}\left(\int_{0}^{h}{{\mathcal{L}}_{\rm
FVc}^{\lambda}P_{V_{t},M(0)+1}\left(X_{s}\right)\mathrm{d}s}\right)}$
$\displaystyle={\mathcal{L}}_{\rm
FVc}^{\lambda}{P_{V_{t},M(0)+1}\left(\mu\right)}.$

Therefore, we have proved that for all $t\geqslant 0$, for all
$\mu\in{\mathcal{M}}_{1}^{c,4}({\mathbb{R}})$,

$\displaystyle{\mathbb{E}}_{\mu}\left({\mathcal{L}}_{\rm
FVc}^{\lambda}P_{\xi_{0},M(0)}\left(X_{t}\right)\right)={\mathcal{L}}_{\rm
FVc}^{\lambda}P_{V_{t},M(0)+1}(\mu),$

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

Step 3. Conclusion. From (56) and (58) and since
$\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{4}\left(\mu\right)\gamma(\mathrm{d}\mu)}<\infty$,
we deduce that

$\displaystyle\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{P_{\rm
FVc}(t)P_{\xi_{0},M(0)}(\mu)\gamma(\mathrm{d}\mu)}$ $\displaystyle\hskip
42.67912pt=\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{P_{\xi_{0},M(0)}(\mu)\gamma(\mathrm{d}\mu)}+\int_{0}^{t}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
FVc}^{\lambda}{P_{V_{s},M(0)+1}\left(\mu\right)}\gamma(\mathrm{d}\mu)\mathrm{d}s}}$
$\displaystyle\hskip
42.67912pt=\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{P_{\xi_{0},M(0)}(\mu)\gamma(\mathrm{d}\mu)},$

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

As the set of test functions
$\left\\{P_{f,n}\left|\phantom{1^{1^{1}}}\right.f\in{\mathscr{C}}_{b}^{4}({\mathbb{R}}^{n},{\mathbb{R}}),n\in{\mathbb{N}}^{\star}\right\\}$
is ${\mathcal{M}}_{1}\left({\mathcal{M}}_{1}({\mathbb{R}})\right)-$convergence
determining  [22,  Lemma 2.1.2], it is
${\mathcal{M}}_{1}\left({\mathcal{M}}_{1}({\mathbb{R}})\right)-$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

$\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{P_{\rm
FVc}(t)\phi(\mu)\gamma(\mathrm{d}\mu)}=\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{\phi(\mu)\gamma(\mathrm{d}\mu)}.$

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
$\left(z^{K,R},\Gamma^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$ is tight in
${\mathcal{M}}_{1}\left({\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}({\mathbb{R}})\right)\right)$
and for any limiting value ${\mathbb{Q}}$ of this sequence, the canonical
process $\left(\zeta^{R},\Gamma^{R}\right)$ on
${\mathscr{C}}^{0}\left([0,T],{\mathbb{T}}_{R}\right)\times{\mathcal{M}}_{m}^{T}\left({\mathcal{M}}_{1}({\mathbb{R}})\right)$
satisfies for all
$f\in{\mathscr{C}}^{1}_{b}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$, for
all $t\in[0,T]$, ${\mathbb{Q}}-$a.s.,

$f\left(\zeta_{t}^{R}\right)=f\left(\zeta_{0}^{R}\right)+\int_{0}^{t}{\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{{\mathcal{L}}_{\rm
SLOW}f\left(\zeta_{s}^{R},\mu\right)\pi^{\lambda\left(\zeta_{s}^{R}\right)}\left(\mathrm{d}\mu\right)\mathrm{d}s}}.$
(59)

From now on, we will establish in Section 7.1 with Lemma 7.1 that the sequence
of laws of $\left(z^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$ converges weakly
in ${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)$ 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 6.

###  7.1 Convergence of the slow component on the torus

######  Lemma 7.1.

The sequence $\left(z^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$ converges in
law in ${\mathbb{D}}\left([0,T],{\mathbb{T}}_{R}\right)$ to the unique
solution of

$\forall t\in[0,T],\quad
z_{t}=x_{0}+\int_{0}^{t}{\partial_{1}{\rm{Fit}}\left(z_{s},z_{s}\right)\beta\left(z_{s}\right)m_{2}\left(z_{s}\right)\mathrm{d}s}$
(60)

on the torus ${\mathbb{T}}_{R}$.

###### Proof.

For any limiting value ${\mathbb{Q}}$ of the sequence of laws of
$\left(z^{K,R}\right)_{K\in{\mathbb{N}}^{\star}}$, from (9) and applying (59)
with
$f={\rm{id}}\in{\mathscr{C}}_{b}^{2}\left({\mathbb{T}}_{R},{\mathbb{T}}_{R}\right)$,
we deduce that ${\mathbb{Q}}-$a.s.,

$\forall
t\in[0,T],\quad\zeta_{t}^{R}=\zeta_{0}^{R}+\int_{0}^{t}{\left(\int_{{\mathcal{M}}_{1}({\mathbb{R}})}{M_{2}\left(\mu\right)\pi^{\lambda\left(\zeta_{s}^{R}\right)}(\mathrm{d}\mu)}\right)\partial_{1}{\rm{Fit}}\left(\zeta_{s}^{R},\zeta_{s}^{R}\right)\mathrm{d}s}.$

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

$\int_{{\mathcal{M}}_{1}\left({\mathbb{R}}\right)}{M_{2}\left(\mu\right)\pi^{\lambda\left(\zeta_{s}^{R}\right)}\left(\mathrm{d}\mu\right)}=\frac{1}{\lambda\left(\zeta_{s}^{R}\right)}=\beta\left(\zeta_{s}^{R}\right)m_{2}\left(\zeta_{s}^{R}\right),$

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

###  7.2 End of the proof of Theorem 6

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
$\lim_{K\to+\infty}{{\mathbb{P}}\left(z_{t}^{K,R}\in\left[x_{0}-\frac{R}{2},x_{0}+\frac{R}{2}\right],\forall
t\in[0,T]\right)}=1$. From Proposition 4.7 and (7), we have that,
$\lim_{K\to+\infty}{{\mathbb{P}}\left({\rm{Diam}}\left({\rm{Supp\,}}\nu_{t}^{K,R}\right)\leqslant\frac{1}{K^{1+\varepsilon/2}},\forall
t\in[0,T]\right)}=1$. Hence,

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

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

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

Theorem 6 then follows from Lemma 7.1.

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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