数理解析研究所講究録
1193 巻 2001 年 20-29

20

Ergodic properties of Fleming-Viot processes with selection andrecombination
板津誠– Seiichi Itatsu
静岡大学理学部

Department of $Mathematics_{f}$ Faculty of Science, Shizuoka University

1

Introduction

Let $E$ be a locally compact separable metric space and $P(E)$ be the spaceof all probability measures on $E$ . For.
let us denote$f_{m}\in D(A)$ and $F\in C^{2}(R^{m})$ let
For any.
$\langle f, \mu\rangle=\int_{E}fd\mu$$\mu\in \mathcal{P}(E)$$f_{1},$$\varphi(\mu)=F(\langle f1, \mu\rangle, \cdots , \langle f_{m},\mu\rangle)$$\cdots,$$=F(\langle \mathrm{f}, \mu\rangle)$$=$$\frac{1}{2}\sum_{i,j=1}^{m}(\langle fifj, \mu\rangle-\langle fi, \mu\rangle\langle fj, \mu\rangle)F\mathcal{Z}iz_{j}(\langle \mathrm{f}, \mu\rangle)$$+$$\sum_{i=1}^{m}(\langle Afi, \mu\rangle+\langle Bfi, \mu\rangle 2)F\mathcal{Z}_{i}(\langle \mathrm{f}, \mu\rangle)$$+$$\sum_{i=1}\{\langle(mfi\otimes 1)\sigma,\mu\rangle 2-\langle fi, \mu\rangle\langle\sigma, \mu^{2}\rangle\}Fzi(\langle \mathrm{f}, \mu\rangle)$$\mathcal{L}\varphi(\mu)$(1).

Here $E$ is the space of genetic types andis a mutation operator inspace of bounded continuous functions on $E$ ) which is the gen(erator for a Feller semigrouponthe space of continuousfunctions vanishing at infinity). Here $\sigma=\sigma(x, y)$ is a bounded symmetricfunction on $E\cross E$ which is selection parameters for types $y\in EB$ is arecombination operator defined by$A$$\overline{C}(E)$$\equiv \mathrm{t}\mathrm{h}\mathrm{e}$$\{\tau(t)\}$$\hat{C}(E)(\equiv$$x,$$Bf(x,y)= \alpha\int_{E}(f(x)-f(X))R’((x, y),$ $dX’)$

where
and $R((x, y),$is a one step transition function on $E^{2}\cross B(E)$ ,and we denotethe -fold product of . According to [3], this operator
defines a generator corresponding to a Markov process on $P(E)$ in the sensethat themartingale problem for is well posed. This process$\alpha\geq 0$$d_{X’)}$$\mu^{n}$$c_{P(E)}[\mathrm{o}, \infty)$

$n$

$\mu$

$\mathcal{L}$21

is called the Fleming-Viot process. The aim of this paper is to consider
ergodicity for this process by using the duality in the form.
$E_{\mu}[ \langle f,\mu_{t}^{n}\rangle]=\sum^{\infty}k=1\langle f_{k}(t), \mu^{k}\rangle$

for any $t\geq 0,$
andwith $\sup$-norm. Hereand satisfy
for some $\gamma>1$ and $f_{n}(0)=f$ and $f_{k}(0)=0$
for $k\neq n$ , and we consider a semigroup for this process.
$n\in \mathrm{N}$$f\in\overline{C}(E^{n})$$f_{k}(t)\in\overline{C}(E^{k})$$||\cdot||$$\Sigma_{k=1\gamma}^{\infty k}||fk(t)||<\infty$Construction of a semigroup2

We consider that is a locally compact separable metric space, and treat
the case of the formula (1) and assume $\{T(t)\}$ is a Feller semigroup on$\mathrm{E}$$\hat{C}(E)$

$k$

times

with the generator . Denote the semigroup $T_{k}(t)=\sim T(t)\otimes\cdots\otimes\tau(t)$ on
and its generator.
We now consider duality under general condition for the diffusion. In thissection we consider the operator of the form
$A$$A^{(k)}$$\overline{C}(E^{k})$(2)$\mathcal{L}\varphi(\mu)$

Here
:

$=$

$\frac{1}{2}\sum_{i,j=1}^{m}(\langle fif_{j}, \mu\rangle-\langle f_{i}, \mu\rangle\langle fj, \mu\rangle)F(z_{i}zj\langle \mathrm{f}, \mu\rangle)$$+$$\sum_{i=1}^{m}(\langle Afi, \mu\rangle+\langle\tilde{B}fi, \mu\rangle\infty)Fzi(\langle \mathrm{f}, \mu\rangle)$

is an operator from

.

to
with
and
a bounded operator andfor some$\gamma>1$ and. In the formula (1) we consider$\tilde{B}f(x)=Bf(x_{1}, x_{2})+\sigma(x_{1,2}X)f(X1)-\sigma(X_{2}, X_{3})f(x_{1})$and in this case iswell defined. Let us define the space $S_{1}=\{f=(f_{1}, f_{2}, \cdots)\in\Sigma_{k=1}^{\infty}\hat{c}(Ek)$ :. Denote
for $f=f=$. Let,$f=(f_{1},f_{2},\cdots)\in$
and
. For
and $\mu\in P(E)$ define
We will construct a semigroup $\{U(t)\}$ corresponding toon Banachspacewith the norm.
$B_{l}$$\tilde{B}$$\hat{C}(E)$$\overline{C}(E^{\infty})$$\hat{c}(E)arrow\hat{C}(E^{l})$$\tilde{B}f=\sum_{l=1}^{\infty}B_{\iota}f$$\Sigma_{l=1}^{\infty}||B_{l}||\gamma^{\iota_{-1}}<\infty$$\langle\tilde{B}f_{i}, \mu^{\infty}\rangle=\sum_{k=1}^{\infty}\langle B_{k}fi, \mu^{k}\rangle$$\mathcal{L}$$||f||_{\gamma} \equiv\sup_{k\geq 1}\gamma^{k}||f_{k}||<\infty\}$$(f_{1}, f_{2}, \cdots)\in S_{1}$$\varphi_{f}(\mu)=\Sigma_{k=1}^{\infty}\langle fk, \mu^{k}\rangle$$C=\{\varphi_{f}(\mu)=\Sigma_{k=1}^{\infty}\langle f_{k}, \mu^{k}\rangle :f_{k}\in\hat{C}(E^{k}), ||f||_{\gamma}<\infty\}$$D=\{\varphi_{f}(\mu)=\Sigma_{k=1}^{\infty}\langle fk, \mu^{k}\rangle\in C:f_{k}\in D(A^{(k}))\}$$S_{1}$$\langle f, \mu^{\infty}\rangle=\sum_{k=1}^{\infty}\langle fk, \mu^{k}\rangle$$\hat{\mathcal{L}}$$S_{1}$$||\cdot||_{\gamma}$

22

Theorem 1. Assume $E$ is a locally compact and assume above andof(2) defined on $D$ is well defined, closable, and dissipative, and conservative,corresponding to a Markov process
and generates a semigroupand constants and , and it holds
then there exists a semigroup $U(t)$ onthat$\mathcal{L}$$(P_{\mu}, \mu_{t})$$\{\mathcal{T}(t)\}$$S_{1}$(3)$\rho$$c_{0}$$\mathcal{T}(t)\varphi f(\mu)=E_{\mu}[\langle f, \mu t\infty\rangle]=\langle U(t)f, \mu^{\infty}\rangle$

for any $t\geq 0$ and $f\in S_{1}$ and$||U(t)||\leq(1-\rho)-1ec0t$ .

Proof. For

equationandfollows from the formula$\varphi_{f}(\mu)=\sum_{k=1}^{\infty}\langle f_{k},\mu^{k}\rangle\in D$$c_{\varphi_{f}}(\mu)=\varphi_{g}(\mu)$$\varphi_{g}(\mu)=\sum_{k=1}^{\infty}\langle g_{k}, \mu^{k}\rangle\in C$, the$\hat{\mathcal{L}}f=g$

where$( \hat{\mathcal{L}}f)_{k}\equiv\sum_{i1\leq<j\leq k+1}\Phi fk+1+(Ak)-i(k+1j)()f_{k}+\sum_{1l=}^{k}B_{l}^{(})k-l+1fk-l+1$

for $k\geq 1$ , and$B_{l}^{(k)}$:$\hat{C}(E^{k})arrow\hat{C}(E^{k+\iota_{-}}1)$

defined by$B_{l}^{(k}f)(_{X_{1}}, \cdots, Xk+l-1)=\sum_{i=1}kB\iota f(X_{1}, \cdots, xi-1, \cdot, Xi, \cdots, xk-1)(Xk, \cdots, Xk+l-1)$

for$f\in\overline{C}(E^{k})$, and for $i<j$$\Phi_{ij}^{(k)}fk(x_{1}, \cdots, X_{k}-1)=f_{k}(X_{1}, \cdots, Xj-1, xi, x_{j}, \cdots, x_{k}-1)$for$f_{k}\in\overline{C}(E^{k})$.

Because $||B_{l}(k)||\leq k||B_{l}||$ , for any $\delta>0$ let a positive constant be $L=$
. Thensuch that $k\leq L+\delta$ and let$\lambda\geq 0$$L( \delta)=\frac{9\delta^{2}-10\delta+4}{8\delta}$$\frac{(\begin{array}{l}k2\end{array})\gamma^{k-1}}{(\lambda+(\begin{array}{l}k-12\end{array}))\gamma^{k}}+\sum_{l=1}^{\infty}\frac{||B_{\iota^{k}}^{()}||\gamma k+\iota-1}{(\lambda+(\begin{array}{l}k+l-12\end{array}))\gamma^{k}}$for any

$k$

where

$\leq$$\frac{(\begin{array}{l}k2\end{array})/\gamma+kd(\gamma)}{\lambda+(\begin{array}{l}k-12\end{array})}$23$d( \gamma)=\sum_{\iota=1}\infty||B_{\iota}||\gamma^{\iota_{-1}}$,

and put
so that $\rho=(1+\delta)/\gamma+\delta d(\gamma)<1$ .
For given
we consider $f(t)=(f_{1}(t), f2(t),$andsuch that$\delta>0$$h\in S_{1}$with$\ldots)$$f_{k}(t)\in\overline{C}(E^{k})$$f(\mathrm{O})=h$(4)$\frac{d}{dt}f_{k}(t)$$(\hat{\mathcal{L}}f(t))_{k}$

$=$

$=$

$1 \leq i<j\leq+1\sum_{k}\Phi^{(+1}ijf_{k}+1()t)k$$+(A^{(k)}-)f_{k}(t)+ \sum B-l+1)f_{k}l-\mathrm{t}+1((kt)\iota=k1$

for $k\geq 1$ and $t>0$ . This is equivalent to(5)$f_{k}(t)$$=$$e^{-(\begin{array}{l}k2\end{array})(t-}u)Tk(t-u)fk(u)$$+ \int_{u}^{t}e^{-(\begin{array}{l}k2\end{array})(t}-S)T_{k}(t-S)\{\sum_{i<1}\Phi_{i}+1)fjk+1(_{S}(k)1\leq j\leq k+$$+ \sum_{1l=}^{k}B_{l}^{(+1}-)fk-l+1(_{S})\}k\iota d_{S}$for $k\geq 1$ and $t>u$ , and we have that$||f_{k}(t)||$$\leq$$||f_{k}(u)||$$+ \int_{u}^{t}e^{-(\begin{array}{l}k2\end{array}))}-((ts||f_{k+1}(_{S})||+\sum_{=\iota 1}||B_{l}^{(}-\iota+1)||||f_{k\iota+}-kk(1S)||)ds$

Let
then

$m(t)= \sup_{k\geq 1,s\leq t}\gamma^{k-\lambda S}e||f_{k}(s)||$$||f_{k}(s)||\leq\gamma^{-k}e^{\lambda}m(SS)$$e^{-\lambda t}\gamma^{k}||fk(t)||$and,$\Sigma_{l=1}^{k}||B_{l}^{(-}kl+1$) $||\gamma^{\iota-1}\leq kd(\gamma)$, and we have$\leq$$e^{-\lambda t}\gamma^{k}||f_{k}(u)||$$+$$\int_{u}^{t}e^{-}-(\{+\lambda\}(ts)/\gamma+kd(\gamma))m(S)d_{S}$$\leq$$m(u)+ \frac{((\begin{array}{l}k+12\end{array})/\gamma+kd(\gamma))}{(\begin{array}{l}k2\end{array})+\lambda}m(t)$

.

.

24

Let

$\lambda\geq c_{0}\equiv L(\gamma^{-1}+d(\gamma))/\rho$$\rho<1$, then $m(t)\leq m(u)+\rho m(t)$ . Therefore by, we have$m(t)\leq(1-\rho)^{-1}m(u)$.

Therefore(6)$t>0$ .

for$\gamma^{k}||f_{k}(t)||\leq(1-\rho)^{-1Ct}e\sup\gamma 0kk||f_{k}(0)||$

By this inequality $f(\mathrm{O})=0$ implies $f(t)–0$ . So the equation (4) has aunique solution for $f(\mathrm{O})=h\in S_{1}$ and implies$\frac{d}{dt}\varphi_{f}(t)(\mu)=c\varphi_{f}(t)(\mu)$.

Therefore $f(t)$ satisfies$\mathcal{T}(t)\varphi_{h}(\mu)=\langle f(t),\mu\rangle\infty$.

So we have
$E_{\mu}[ \langle h,\mu_{t}^{\infty}\rangle]=\sum^{\infty}k=1\langle f_{k}(t), \mu^{k}\rangle$.

By the inequality (6) there exists a semigroup $\{U(t)\}$ on
to such that$||U(t)||\leq(1-\rho)^{-}1e^{ct}0$ .

$S_{1}$

corresponding$\hat{L}$Q.E.D.

Let us denote the semigroup $\{U(t)\}$ by$\{U_{0}(t)\}$when$\tilde{B}=0$. Then we have

Lemma 1. Assume the assumption of Theorem 1 , then $\{U_{0}(t)\}$ and $\{U(t)\}$on$S_{1}$

satisfies.

$||U(t)-U0(t)||\leq(1-\rho_{0})^{-1}(1-\rho)^{-1}\beta d(\gamma)e^{c}\mathrm{o}t$, and are constants depends only on.
$h\in S_{1}$ we consider $f^{0}(t)=(f_{1}^{0}(t), f^{0}2(t),$
For
given
Proof.
and $f(\mathrm{O})=h$ such that

where$\rho,$$\rho_{0},$$\beta$$c_{0}$$\gamma,$$d(\gamma)$$\ldots)$with$\overline{C}(E^{k})$(7)$\frac{d}{dt}f_{k}^{0}(t)$$=$$(\hat{\mathcal{L}}_{\mathrm{o}f^{0}}(t))_{k}$$=$$1 \leq i<j\leq+\sum_{k1}\Phi i(kj+1)f^{0}k+1(t)+(A^{(}k)-)f_{k}^{0}(t)$$f_{k}^{0}(t)\in$25

for $k\geq 1$ and $t>0$ . This is equivalent to(8)$f_{k}^{0}(t)$–) $Tk(t-u)f^{0}k(u)$$e^{-(\begin{array}{l}k2\end{array})(-u}t$$+ \int_{u}^{t}e-(t-s)T_{k}(t-s)\{\sum_{1\leq j\leq k+1}\Phi^{(}+1f_{k1}ij+()k0i<S)\}dS$

for $k\geq 1$ and $t>u$ , and we have that$||f_{k}(t)-f_{k}^{0}(t)||$$||f_{k}(u)-f_{k}^{0}(u)||$$\leq$$+ \int_{u}^{t}e^{-(\begin{array}{l}k2\end{array})t)}\{(-S||f_{k+}1(S)-f^{0}k+1(s)||+$$+ \sum_{l=1}^{k}||B(k-\iota+1)|l|||f_{k-l+1}(S)||\}ds$

Let $l(t)= \sup_{k>1,S}\leq t\gamma^{k-\lambda S}e||f_{k}(S)-fk(0s)||$ , thenand
, and we have.

$||f_{k}(s)-fk0(S)||\leq\gamma^{-k}e^{\lambda s}\iota(s)$$\Sigma_{l=1}^{k}||B_{l}^{(}+1\overline{)}|k-\iota|\gamma^{\iota_{-1}}\leq kd(\gamma)$$e^{-\lambda t}\gamma^{k}||f_{k}(t)-f_{k}^{0}(t)||$Let$\lambda\geq c_{0}$$\leq$$\int_{u}^{t}e^{-\{}+\lambda\}(t-s)((1/\gamma)l(_{S)kd()}+\gamma m(s))d_{S}$$\leq$$m(u)+ \frac{((\begin{array}{l}k+12\end{array})(1/\gamma)l(t)+kd(\gamma)m(t))}{(\begin{array}{l}k2\end{array})+\lambda}$, and put$\rho_{0}l(t)+\beta d(\gamma)m(t)$$\rho_{0}=\sup\frac{(\begin{array}{l}k+12\end{array})(1/\gamma)}{(\begin{array}{l}k2\end{array})+\lambda},$$\beta=\sup_{k}\frac{k}{(\begin{array}{l}k2\end{array})+\lambda}$.

, then $l(t)\leq$. Therefore by $\rho_{0}<1$ , we have$l(t)\leq(1-\rho 0)-1\beta d(\gamma)m(t)$.

Therefore(9)$\gamma^{k}||f_{k}(t)-f_{k}^{0}(t)||\leq(1-\rho_{0)^{-}(-}11\rho)-1\beta d(\gamma)e^{c_{0}}t\sup\gamma^{k}k||f_{k}(0)||$

for $t>0$ .
By the inequality (9) semigroups$\{U_{0}(t)\}$and $\{U(t)\}$ on$||U(t)-U0(t)||\leq(1-\rho_{0})^{-1}(1-\rho)-1\beta d(\gamma)ec_{0}t$$S_{1}$

satisfies.

Q.E.D.

26

3

Ergodicity of semigroups

We define $\{T(t)\}$ is uniforniy ergodic if there exist a stationary distribution.
such that$||T(i)-\langle\cdot, \pi_{0}\rangle 1||arrow 0(tarrow\infty)$$\pi_{0}$is uniformly ergodic and that for some
Theorem 2. Assume and that$M$
and a stationary distribution
andpositive constants$\{\tau(t)\}$$\pi_{0}$$\lambda_{0}$$||T(t)f-\langle f, \pi 0\rangle 1||\leq Me^{-\lambda_{1}t}||f||$Let.

. Then there exists a stationary distribution such thatsatisfying that
there exist constants$\Pi$$\lambda_{1}=\min(\lambda_{0},1)$

for any$\epsilon>0$$M_{1}=M_{1}(\epsilon),$ $\delta=\delta(\epsilon)>0$$||T(t)\varphi f(\mu)-\langle\varphi f(\mu), \square \rangle 1||\leq M_{1}e-(\lambda_{1}-\epsilon)t||f||_{\gamma}$for$f\in S_{1}$if$||\sigma||+\alpha<\delta$.

We denote $h_{0}=(1,0,0, \cdots)\in S_{1}$Theorem 3. Under the assumption of Theorem 2 it holds thatis ergodic in the sense that for a positive constant $M_{2}>0$
corresponding tosuch that
and
and$\{U_{0}(t)\}$$\hat{\mathcal{L}}_{0}$$m\in S_{1}^{*}$$h_{0}\in S_{1}$$||U_{0}(t)f-\langle f, m\rangle h_{0}||_{\gamma}\leq M_{2}e-\lambda_{1}t||f||\gamma$.

$m_{k}\in P(E^{k})$ .
where $m=(m_{1}, m_{2}, \cdots),$
Proof. Let $N(t)$ be a death process with rate
the hitting time of . Put an operator’ then by (5)$\langle f,m\rangle=\Sigma_{k}\langle f_{k}, m_{k}\rangle,$$j$$\Phi_{k}=\varpi_{2}^{1}\sum_{i<j}\Phi^{(k)}ij$$(U_{0}(t)f)j= \sum_{jk\geq}Ek[T_{j}(t-\mathcal{T}j)\Phi_{j+1}\cdots\tau k(\tau_{k}-1)f_{k};\tau j\leq t<\tau_{j+1}]$Let$Y_{k}=\Phi_{j+1}\cdots T_{k}(\mathcal{T}k-1)f_{k}$$||U_{0}(t)f-\langle f, m\rangle h_{0}||_{\gamma}$

on$\tau_{j}\leq t<\tau_{j+1}$.

, then$\leq$$\sum_{k}|E[T(t-\mathcal{T}_{1})Y_{k^{-}}\langle Y_{k}, \pi_{0}\rangle;t>\tau_{1}]|$$+$$2(\gamma-1)-1P(\tau_{1}\geq t)||f||_{\gamma}$$\leq$$\gamma(\gamma-1)-1(||T(t-\tau_{1})-\langle\cdot, \pi 0\rangle 1||+2P(\tau 1\geq t))||f||_{\gamma}$

foris defined bywhere $m=(m_{1}, m_{2}, \cdots)$ and$f\in C(E^{k}),$ $m_{1}=\pi_{0}$ , and $m_{k}=R_{k}()^{*}(\Sigma i<j)\Phi_{ij}(k)^{*}m_{k}-1(k\geq 2)$ . Here$m_{k}$$\langle f, m_{k}\rangle=\int\langle f, \mu^{k}\rangle\Pi_{0}(d\mu)$27$R_{k}(\lambda)$

is the resolvent of$T_{k}(t)$. By [3] $P(\tau_{1}\geq t)\leq 3e^{-t}$ , so the Theorem

holds.
Q.E.D.
Lemma 2. Let be a Banach space and $h\in L$ and $m\in L^{*}$ with $||h||=a$
is a bounded operator on with uniform norm
and $||m||=b$ . Assume. Let
and $U=P_{0}+B$ , then$||B||<1/(2+4ab)$ andwe have, $\zeta-U$ is invertible in L. Put(a) For
$L$

$L$

$B$

$P_{0}=\langle\cdot, m\rangle h$$\langle h, m\rangle=1$$\zeta\in\Gamma\equiv\{\zeta\in \mathrm{C}:|\zeta-1|=\frac{1}{2}\}$$P_{1}= \frac{1}{2\pi i}\oint_{\Gamma}(\zeta-U)^{-1}d\zeta$,.
is thethen $\dim P_{1}L=\dim P_{1^{*}}L^{*}=1,$ $P_{1}U=UP_{1}$ , and:eigenspace of corresponding to the eigenvalue , contained in$|(-1|<1/2$ }. It becomes that the eigenvalue in $D$ is unique with multiplicity
1. Similar results hold as:and(b) Assume
has an eigenvalue with $eigenvector\varphi_{0}$ and $|(_{0}-1|<1/2$ ,then we have that $\varphi_{0}=c(\zeta_{0}-B)^{-}1h$ and$P_{1}^{2}=P_{1}$$U$$P_{1}L$$D\equiv\{(\in \mathrm{C}$$U^{*}$$P_{1}^{*}$$U$$\zeta_{0}$$\langle\varphi_{0}, m\rangle=c$

and
(10)

$P_{1}=\langle(\zeta_{0}-B)^{-2}h, m\rangle^{-1}\langle\cdot, (\overline{\zeta}_{0}-B^{*})^{-}1m\rangle((0-B)^{-}1h$$UP_{1}=P_{1}U=P_{1}$(c) Under the assumption of$(b)_{f}$,,the next relation holds.

$||U-\zeta_{0}P1||\leq 8||B||$

if $||B||<1/(4+8ab)$ .
Lemma 3. Under the assumption of theorem 1 for any, then there exists
such that if
and $M_{1}>0$ such that$d(\gamma)<\delta$$\delta=\delta(\epsilon)>0$$h_{1}\in S_{1}$$||U(t)f-\langle f, m_{1}\rangle h_{1}||\gamma\leq M_{1}e^{-(\lambda_{1})t}-\epsilon||f||_{\gamma}$

and$\langle h_{0}, m_{1}\rangle\langle h_{1\mu^{\infty}},\rangle=1$.

there exists
and$\epsilon>0$,$m_{1}\in S_{1}^{*}$

28

Proof. By Theorem 3 we have that for any$t_{0}$$0<\epsilon<\lambda_{1}$there exist$h_{0},m$, andsuch that.

$||U(t_{0})f- \langle f, m\rangle h0||_{\gamma}\leq\frac{1}{16}e-(\lambda_{1}-\epsilon)t\mathrm{o}||f||_{\gamma}$

By Lemma 1 we have that there exists $\delta>0$ such that for$||U(t_{0})f-U \mathrm{o}(t0)f||\gamma\leq\frac{1}{16}e-(\lambda 1-\epsilon)t\mathrm{o}||f||_{\gamma}$

According to Lemma 3 we have that there exist$m_{1}$,$h_{1}$$d(\gamma)<\delta$

.

, and

$||U(t_{0})f-\zeta_{0}\langle f, m1\rangle h_{1}||\gamma\leq e^{-(\lambda_{1\epsilon})t}-0||f||\gamma$$\zeta_{0}$such that.

So we have for any $n>0$$||U(nt_{0})f-\zeta^{n}0\langle f, m1\rangle h_{1}||\gamma\leq e^{-(\lambda_{1})t0}-\epsilon n||f||_{\gamma}$.

By Theorem 1 there exists $M’>0$ such that $||U(s)||\leq M’$ for $0\leq s\leq t_{0}$ .
We have that$||U(nt_{0}+s)f-\zeta_{0}n\langle U(S)f, m_{1}\rangle h_{1}||_{\gamma}\leq M’e^{-(\lambda_{1}}-\epsilon)t||f||_{\gamma}$

and$||U(nt_{0}+s)f-(_{0}^{n}\langle f, m_{1}\rangle U(s)h1||_{\gamma}\leq M^{;_{e^{-(\lambda}}}1-\epsilon)t||f||\gamma$

for $0\leq s\leq t_{0}$ . Then$|\zeta_{0}|\leq 1$

and if$|\zeta_{0}|=1$, then$\langle U(s)f, m_{1}\rangle h_{1}=\langle f,m_{1}\rangle U(s)h1=c(s)\langle f, m_{1}\rangle h1$

with some constant $c(s)$ . Because
we have$\mathcal{T}(t)1=1$, by the above equations and (3)$1=( \mathcal{T}(nt_{0}+s)1)(\mu)=\langle U(nt_{0}+s)h_{0,\mu^{\infty}\rangle}=c(s)\langle h0, m1\rangle\langle h_{1}, \mu^{\infty}\rangle narrow\infty\lim\zeta^{n}0$

Therefore$\zeta_{0}=1$$M_{1}=M’e^{(\lambda}1-\epsilon)t0$.

. Because $U(\mathrm{O})=I,$ $c(s)=c(0)=1$ holds. Therefore let, then the inequality of the Theorem holds.
Q.E.D.

Proof of Theorem 2. Because$\mathcal{T}(t)1=1$, by Lemma 3$1=(\mathcal{T}(t)1)(\mu)=\langle U(t)h_{0}, \mu\rangle\infty=\langle h_{0}, m_{1}\rangle\langle h1, \mu\rangle\infty$

.

29

Let

$m_{2}=(m_{2}^{(1)}, m_{2}^{(}, \cdot)2)..=\frac{1}{\langle h0,m_{1}\rangle}m_{1}$$P(E^{k})$and$\langle h_{2,\mu^{\infty}}\rangle=1$. Becauseand$\varphi_{f}(\mu)=\langle f, \mu^{\infty}\rangle$$|T(t)\varphi_{f(}\mu)-\langle f, m_{2}\rangle\langle.h2, \mu^{\infty}\rangle|$, then, Lemma 3 implies that$h_{2}=\langle h_{0}, m_{1}\rangle h_{1}$$=$$|\langle U(t)f-\langle f, m_{2}\rangle h_{2},\mu\rangle\infty|$$\leq$$M_{1}\gamma(\gamma-1)-1e-(\lambda 1-\epsilon)t||f||_{\gamma}$$m_{2}^{(k)}\in$so Theorem 2 holds.
Q.E.D.

References
Proba[1] Dawson, D. A. Measure-valued Markov Processes. Ecolebilit\’es de Saint-Flour XXI-1991 Springer-Verlag LNM 1541(1993), 1-260.
$d^{f}Et\acute{e}de$[2] Ethier, S. N. and Griffiths, R. C. The transition function of a FlemingViot process. Ann. Prob. 21(1993), 1571-1590.
[3] Ethier, S. N. and Kurtz, T. G. Fleming-Viot processes in population
genetics. SIAM J. Control and Optim. 31(1993) 345-386.

[4] Ethier, S. N. and Kurtz, T. G. Convergence to Fleming-Viot processes in
the weak atomic topology. Stochastic Processes Appl. 54(1994) 1-27.
[5] Ethier, S. N. and Kurtz, T. G. Coupling and ergodic theorems for
Fleming-Viot processes. Ann. Prob. 26(1998) 533-561.
[6] Itatsu, S., Particle systems corresponding to Fleming Viot processes withselection.,RIMS KOKYUROKU Kyoto Univ.,1157(2000)68-73.