Examples

2.2.1 Branching Brownian motions

Example 2.16. Suppose that d = 1. Let M^ν, ν ∈ S1, be the killed Brownian motion with respect to exp (−A^ν_t) and let M^ν be the branching Brownian motion with motion component M^ν and branching rate µ∈ K^R_∞. First take µ=δ₀ and ν =δ_−a+δ_a fora >0. Since

inf

½1

2D(u, u) +u(−a)²+u(a)² :u∈C₀^∞(R), u(0)² = 1

¾

= 2

1 + 2a by Example 6.2 below, the branching processM^ν extincts if and only if

Q(0)≤1 + 2 1 + 2a.

Next takeµ=δ_−b+δ_b forb >0 and ν=δ₀. Suppose thatQ(b) =Q(−b) =Q. Since inf

½1

2D(u, u) +u(0)² :u∈C₀^∞(R), u(−b)²+u(b)² = 1

¾

= 1

2(1 +b) by Example 6.2, the branching process M^ν extincts if and only if

Q≤1 + 1 2(1 +b).

Example 2.17. LetM be a spherically symmetric Riemannian manifold with a poleoand con- sider the Brownian motion on M. Denote by (E,F) the associated Dirichlet form on L²(M;V):

E(u, u) =1 2

Z

M

|∇u|²dV,

F = the closure of C₀^∞(M) with respect toE(·,·) +∥ · ∥²_L²_(M;V),

where V is the Riemannian volume of M. Let B(r) = {x∈M :d(x, o)< r} and ∂B(r) = {x∈M :d(x, o) =r}, where d is the Riemannian distance of M. Denote by δr the surface measure of∂B(r) and let

S(r) =δr(∂B(r)) and G(r) = Z _∞

r

1 S(ρ)dρ.

We now set

λ(δˇ _R;M\B(r)) = inf (

E(u, u) :u∈C₀^∞(M \B(r)), Z

∂B(R)

u²dδ_R= 1 )

for r and R with R > r > 0. Then the following results are shown by Takeda [56]; if (Q− 1)S(R)G(R)>1/2, then

λ(δˇ R;M\B(r))≥Q−1 ⇐⇒ r0≤r < R, where the positive constantr₀ is a unique root of

G(r) = 2(Q−1)S(R)G(R)² 2(Q−1)S(R)G(R)−1. On the other hand, if (Q−1)S(R)G(R)≤1/2, then

ˇλ(δR;M \B(r))> Q−1 for anyr < R.

Let us denote byMthe Brownian motion onM and byM^rthe absorbing Brownian motion onM\B(r). LetM^rbe the branching Brownian motion with motion componentM^r, branching rate δ_R and branching mechanism {p_n(x)}^∞_n=0 such that Q(x) = P_∞

n=0np_n(x) ≡ Q. Theorem 2.11 and Remark 6.6 then imply the following; if (Q−1)S(R)G(R) > 1/2, then M^r extincts locally if and only if r0 ≤ r < R. On the other hand, if (Q−1)S(R)G(R) ≤ 1/2, then M^r extincts locally for anyr < R. For instance, take the d-dimensional hyperbolic space H^d as M (see Example 3.3 of [32] for definition).

(i) Ford= 2,S(R)G(R) is strictly increasing, limR↓0S(R)G(R) = 0 and lim

R→∞S(R)G(R) = 1

([52, Example 2.6]). Hence if Q > 3/2, then there exists a unique root R0 such that (Q− 1)S(R₀)G(R₀) = 1/2. Moreover, if R > R₀, then M^r extincts locally if and only if r₀ ≤r < R.

If R ≤ R0, then M^r extincts locally for any r < R. On the other hand, if Q ≤ 3/2, then (Q−1)S(R)G(R)<1/2 for all R >0, and consequently M^r extincts locally for anyr < R.

(ii) For d= 3,

(Q−1)S(R)G(R)>1/2 ⇐⇒ Q >2 + 1

e^2R−1 (2.13)

by Example 2.6 of [52]. Hence, if Q satisfies the right hand side of (2.13), then M^r extincts locally if and only if r₀ ≤r < R. Otherwise, M^r extincts locally for anyr < R.

(iii) For d≥4, S(R)G(R)<1/(d−1) by Example 2.6 of [52]. Therefore, if Q≤(d+ 1)/2, thenM^r extincts locally for any r < R.

More detailed properties are studied for branching Brownian motions on H² in [39], and for branching Markov processes onH^d in [36].

2.2.2 Branching symmetric α-stable processes

LetM^α = (X_t, P_x) be the symmetricα-stable process onR^d. LetM^D be the absorbingα-stable process on an open set DinR^d and (E^D,F^D) the associated Dirichlet form on L²(D). Define

ˇλ(µ;D) = inf

½

E^D(u, u) :u∈C₀^∞(D), Z

D

u²dµ= 1

¾

forµ∈ K_∞^D.

Example 2.18. Suppose that d = 1 and 1 < α ≤ 2. Then M^α is recurrent and one point has positive capacity. Let D = R\ {0}, µ = δ_a, a > 0, and p₀(x) +p₂(x) ≡ 1 on D. Then Q(x) = 2p₂(x). Since

λ(δ_a;R^d\ {0}) =−Γ(α) cos

³πα 2

´

2a^α−1

by Example 6.6 below, we see from Theorem 2.4 and Theorem 2.8 that Q(a)<1−Γ(α) cos

³πα 2

´

2a^α−1 ⇒Px(e0<∞)≡1, sup

x∈R\{0}Ex

£N_{0}¤

<∞

Q(a) = 1−Γ(α) cos

³πα 2

´

2a^α−1 ⇒P_x(e₀<∞)≡1, sup

x∈R\{0}E_x£ N_{0}¤

=∞

Q(a)>1−Γ(α) cos

³πα 2

´

2a^α−1 ⇒P_x(e₀<∞)<1, sup

x∈R\{0}E_x£ N_{0}¤

=∞. In particular, if

0< a≤



−Γ(α) cos

³πα 2

´

2





1/(α−1)

,

then this branching symmetric α-stable process extincts even if p₂(a) = 1.

Example 2.19. Suppose that 1< α≤2 andd > α. ThenM^α is transient and the surface of a sphere has positive capacity. Let δ_r be the surface measure of a sphere∂B(r) ={x∈R^d:|x|= r}. Take µ=δr and assume thatQ(x) ≡Q. Using Example 4.1 of [58], we see from Theorem 2.11 and Remark 6.6 that, ifQ >1, then M^α extincts locally if and only if

0< r≤









√πΓ

µd+α 2 −1

¶ Γ

³α 2

´

(Q−1)Γ

µα−1 2

¶ Γ

µd−α 2

¶









1/(α−1)

. (2.14)

On the other hand, if Q≤1, then M^α extincts locally for anyr >0.

Let δ_r be the normalized surface measure of ∂B(r), δ_r = δ_r/δ_r(∂B(r)). Take µ = δ_r and assume that Q(x) ≡ Q. Noting that λ¡

δ_r;R^d¢

= δ_r(∂B(r))λ(δ_r;R^d) and δ_r(∂B(r)) = 2π^d/2r^d−1/Γ (d/2). we see that ifQ >1, thenM^α extincts locally if and only if

r≥









(Q−1)Γ µd

2

¶ Γ

µα−1 2

¶ Γ

µd−α 2

¶

2π^(d+1)/2Γ

³α 2

´ Γ

µd+α 2 −1

¶









1/(d−α)

. (2.15)

On the other hand, if Q≤1, then M^α extincts locally for anyr >0.

Example 2.20. Suppose that 0< α≤2 andd > α. Let1_B(r)dxbe thed-dimensional Lebesgue measure restricted on a ballB(r) ={x∈R^d:|x|< r}. Take µ(dx) =1_B(r)dxand assume that Q(x)≡Q. IfQ >1 and

0< r≤







 2^α−1Γ

µd 2

¶ Γ

³α 2 + 1

´

(Q−1)Γ

µd−α 2

¶









1/α

, (2.16)

thenM^αextincts locally by Example 6.4 below. On the other hand, ifQ≤1, then M^αextincts locally for any r >0.

Chapter 3

Exponential growth of the numbers of particles for branching symmetric Markov processes

We study the exponential growth of the numbers of particles for branching symmetric Hunt processes by the principal eigenvalues of associated Schr¨odinger operators under the assumption that the Schr¨odinger operators have spectral gaps.

3.1 Exponential growth of the numbers of particles

LetX be a locally compact separable metric space and ma positive smooth Radon measure on X with full support. Let M = (X_t, P_x) be an m-symmetric Hunt process on X. Throughout this section, we assume thatMis transient and satisfies Assumption 1.3. LetM= (X_t,P_x,Gt) be the branching symmetric Hunt process with motion component M, branching rate µ∈ K_∞ and branching mechanism {pn(x)}n≥0. Recall that Q(x) = P_∞

n=0npn(x) and suppose that sup_x∈XQ(x)<∞.

We proved in Lemma 2.9 that, if Px(ζ <∞) = 1 for any x∈X, then {e₀ =∞}=

n

tlim→∞Z_t=∞o

P_x-a.s.

for anyx∈X. This result says that, if the branching process Mdoes not extinct, then P_x

µ

tlim→∞Z_t=∞¯¯

¯¯e₀ =∞

¶

= 1.

We first study the exponential growth of Z_t in terms of the principal eigenvalue λ₁((Q−1)µ) = inf

½

E(u, u)− Z

X

u²(Q−1)dµ:u∈ F, Z

X

u²dm= 1

¾

. (3.1)

From now on, we suppose that λ1 := λ1((Q−1)µ) < 0. We denote by h the ground state corresponding toλ₁. Then

h(x) =e^λ1tE_x h

exp

³

A^(Q_t ^−1)µ

´

h(X_t);t < ζ i

. (3.2)

Define

M_t=e^λt

Zt

X

i=1

h(Xⁱ_t), t≥0. (3.3)

ThenMtis aPx-martingale by (1.19) and (3.2), and thus there exists a limitM_∞:= limt→∞Mt∈ [0,∞) Px-a.s. Furthermore, it follows from (1.20) and (3.2) that

E_x£ M_t²¤

=e^2λ1tE_x h

exp

³

A^(Q_t ^−1)µ

´

h(X_t)²;t < ζ i

+Ex

·Z _t∧ζ

0

exp

³

2λ1s+A^(Q_s ^−1)µ

´

h(Xs)²dA^Rµ_s

¸ ,

(3.4)

whereR(x) =P_∞

n=0n(n−1)p_n(x).

Lemma 3.1. Assume that sup_x∈XR(x)<∞. Then M_t is square integrable.

Proof. Since e^2λ1tEx

h exp

³

A^(Q_t ^−1)µ

´

h(Xt)²;t < ζ

i≤e^λ1t∥h∥∞e^λ1tEx

h exp

³

A^(Q_t ^−1)µ

´

h(Xt);t < ζ i

=e^λ1t∥h∥_∞h(x)

by (3.2), the last term above converges to 0 ast→ ∞. Hence it follows from (3.4) that

tlim→∞Ex

£M_t²¤

=Ex

·Z _ζ

0

exp

³

2λ1s+A^(Q_s ^−1)µ

´

h(Xs)²dA^Rµ_s

¸

≤ ∥h∥²_∞∥R∥∞sup

x∈X

Ex

·Z ζ 0

exp

³

2λ1s+A^(Q_s ^−1)µ

´ dA^µ_s

¸ .

(3.5)

Since inf

½

E(u, u)− Z

X

u²(Q−1)dµ−2λ1

Z

X

u²dm:u∈ F, Z

X

u²dm= 1

¾

=−λ1>0 by the definition ofλ₁, Lemma 3.5 of [52] shows that

inf

½

E(u, u) + Z

X

u²dµ−2λ₁ Z

X

u²dm:u∈ F, Z

X

u²Qdµ= 1

¾

>1.

Then the last term of (3.5) is finite by Theorem 1.2, whenceM_t is square integrable.

Lemma 3.1 tells us that Ex[M_∞] =h(x) >0, which yields that Px(M_∞∈(0,∞))> 0 for any x∈X. It also holds that

Ex

£M_∞² ¤

=Ex

·Z _ζ

0

exp

³

2λ1s+A^(Q_s ^−1)µ

´

h(Xs)²dA^Rµ_s

¸ .

Recall that the extinction probability ue is a minimal solution to (2.1) by Proposition 2.1.

We then obtain

Lemma 3.2. Suppose that P_x(ζ <∞) = 1 for any x∈X. IfRR

X×XG^µ(x, y)µ(dx)µ(dy)<∞, then the equation (2.1)has just two solutions, u≡1 andue.

Proof. Letube a solution to (2.1) such thatu(x₀)<1 forx₀∈X. Sinceuis finely continuous by Lemma 2.2, it follows from (2.1) that Px0(σO∩F^µ <∞)>0, where O ={x∈X :u(x)<1} and F^µ is the fine support of the measureµdefined in (1.8). Moreover, by the irreducibility of M, it holds that P_x(σ_O∩F^µ <∞)>0 for any x∈X, which implies that u <1 on X.

As a direct calculation yields that Ex

h exp

³−A^µ_ζ´i

= 1−Ex

·Z _ζ

0

exp (−A^µ_t)dA^µ_t

¸ ,

the equation (2.1) is equivalent to that

v=G^µ((F(1)−F(1−v))µ) on X,

wherev= 1−u >0. Since the functionve= 1−ue >0 is a solution to the equation above, we see that

Z

X

v(F(1)−F(1−v_e)))dµ= Z

X

G^µ((F(1)−F(1−v))µ) (F(1)−F(1−v_e))dµ

= Z

X

G^µ((F(1)−F(1−ve))µ) (F(1)−F(1−v))dµ

= Z

X

ve(F(1)−F(1−v))dµ.

Here the integrability of the terms above follows by the assumption onµand the second equality holds by Theorem 3.2 (iv) of [2]. Since F(·) is strictly convex andv_e≥v >0, it holds that

F(1)−F(1−v)

1−(1−v) = F(1)−F(1−v_e)

1−(1−ve) µ-a.e., which shows thatu=ue µ-a.e. Using (2.1), we have u=ue onX.

Proposition 3.3. Suppose that P_x(ζ <∞) = 1 for anyx∈X. If sup_x∈XR(x)<∞ and RR

X×XG^µ(x, y)µ(dx)µ(dy)<∞, then

{e₀=∞}={M_∞>0} P_x-a.s.

for anyx∈X.

Proof. Since λ1<0 and

M_t≤e^λ1tZ_t∥h∥∞, (3.6)

it holds that

{M_∞>0} ⊂ {e₀ =∞}. By the assumption on the lifetime,

P_x(T =∞, e₀ =∞) =E_x h

exp

³−A^µ_ζ

´

;ζ =∞i

= 0.

Noting that

{T =∞} ⊂ {e0 <∞} ⊂ {M_∞= 0},

we see that

P_x(M_∞= 0) =P_x(M_∞= 0, T =∞) +P_x(M_∞= 0, T <∞)

=Px(T =∞) +Px(M_∞= 0, T <∞)

=Ex

h exp

³−A^µ_ζ

´

;ζ <∞i +Ex

·Z ζ 0

exp (−A^µ_t)F(P_·(M_∞= 0))(Xt)dA^µ_t

¸ , that is, the functionP_x(M_∞= 0) is a solution to (2.1). SinceP_x(M_∞= 0)<1, it follows from Lemma 3.2 thatPx(M_∞= 0) =ue(x) for any x∈X. Namely,Px(M_∞>0) =Px(e0=∞) for any x∈X, which completes the proof.

Theorem 3.4. Suppose thatPx(ζ <∞) = 1 for anyx∈X.

(i) If sup_x∈XR(x)<∞ and RR

X×XG^µ(x, y)µ(dx)µ(dy)<∞, then

Px(M_∞∈(0,∞)|e0=∞) = 1, x∈X. (3.7) As a consequence,

P_x µ

lim inf

t→∞ e^λ1tZ_t>0¯¯

¯¯e₀=∞

¶

= 1, x∈X. (3.8)

(ii) If sup_x∈XR(x)<∞ and RR

X×XG^µ(x, y)µ(dx)µ(dy)<∞, then for anyκ > λ1, Px

µ

tlim→∞e^κtZt=∞¯¯

¯¯e0 =∞

¶

= 1, x∈X. (3.9)

(iii) For any κ < λ₁,

P_x Ã

tlim→∞e^κt

Zt

X

i=1

h(Xⁱ_t) = 0

!

= 1, x∈X (3.10)

and

Px

³ lim inf

t→∞ e^κtZt= 0

´

= 1, x∈X. (3.11)

Furthermore, if X is Green bounded forM, that is, ifsup_x∈XE_x[ζ]<∞, then, for anyκ < λ₁, P_x

³

tlim→∞e^κtZ_t= 0

´

= 1, x∈X. (3.12)

Proof. The equation (3.7) follows from Proposition 3.3. Since {M_∞>0} ⊂n

lim inf

t→∞ e^λ1tZ_t>0 o⊂n

tlim→∞e^κtZ_t=∞o forκ > λby (3.6), we have (3.8) and (3.9).

Suppose that κ < λ. Then the equation (3.10) holds by Lemma 3.1. By (1.19), e^κtEx[Zt] =Ex

h exp

³

κt+A^(Q_t ^−1)µ

´

;t < ζ i

=e^κtE_x

·

exp (−A^µ_t) Z _t

0

exp¡ A^Qµ_s ¢

dA^Qµ_s ;t < ζ

¸

+e^κtE_x[exp (−A^µ_t) ;t < ζ].

(3.13)

Choose a positive constant εsuch that 0< ε < λ1−κ. Then the last term above is not greater than

e^{(κ−λ1+ε)t}E_x

·Z _ζ

0

exp

³

(λ₁−ε)s+A^(Q_s ^−1)µ

´ dA^Qµ_s

¸

+e^κtE_x[exp (−A^µ_t) ;t < ζ]. (3.14)

By the same argument as in Lemma 3.1, it follows that sup

x∈X

E_x

·Z _ζ

0

exp

³

(λ₁−ε)s+A^(Q_s ^−1)µ

´ dA^Qµ_s

¸

<∞, and thus the term of (3.14) converges to 0 ast→ ∞. Hence by Fatou’s lemma,

E_x h

lim inf

t→∞ e^κtZ_t

i≤ lim

t→∞e^κtE_x[Z_t] = 0, which implies (3.11).

From now on, we assume that X is Green bounded for M. Let u_κ(x) =E_x

h exp

³

κζ+A^(Q_ζ ^−1)µ

´i .

Then sup_x∈Xuκ(x) < ∞ by Theorem 5.2 of [14] or Theorem 2.4 of [52]. Moreover, Jensen’s inequality yields that

xinf∈Xuκ(x)≥exp µ

κsup

x∈X

Ex[ζ]−sup

x∈X

Ex

h A^µ_ζ

i¶

>0, where we note that sup_x∈XE_x

h A^µ_ζ

i

<∞ by (1.10). By the definition ofu_κ and (1.19),

e^κtEx

"_Z Xt

i=1

uκ(Xⁱ_t)

#

=e^κtEx

h exp

³

A^(Q_t ^−1)µ

´

uκ(Xt);t < ζ i

=e^κtEx

h exp

³

A^(Q_t ^−1)µ

´ EXt

h exp

³

κζ+A^(Q_ζ ^−1)µ

´i

;t < ζ i

. Then the last term above is equal to

E_x h

exp

³

κζ+A^(Q_ζ ^−1)µ

´

;t < ζ

i≤u_κ(x)

by the Markov property. Sincee^κtP_Zt

i=1uκ(Xⁱ_t) is a nonnegative Px-supermartingale such that sup

(x,t)∈X×[0,∞)

e^κtE_x

"_Z Xt

i=1

u_κ(Xⁱ_t)

#

≤ sup

x∈X

u_κ(x)<∞, there exists a limit lim_t→∞e^κtP_Zt

i=1u_κ(Xⁱ_t)<∞ P_x-a.s. for any x∈X. Moreover, we see that lim sup_t→∞e^κtZt<∞ Px-a.s. because infx∈Xuκ(x)>0 and

µ

xinf∈Xuκ(x)

¶

e^κtZt≤e^κt

Zt

X

i=1

uκ(Xⁱ_t).

Noting that κ < λ₁ is arbitrary, we have (3.12).

We next study the exponential growth of the number of particles in every open set. In the sequel, we assume that λ1 := λ1((Q−1)µ) < 0. Let A be an open set in X. Note that, if P_x(L_A=∞)>0 for somex∈X, then P_x(L_A=∞)>0 for any x∈X by the irreducibility of M.

Lemma 3.5. Assume that the support of the branching rate µis compact. Then, for any non- empty open setA in X,

Px

µ lim sup

t→∞ Zt(A) =∞

¶

>0, x∈X. (3.15)

Moreover, if P_x(ρ_A<∞) = 1 for anyx∈X, then P_x

µ lim sup

t→∞ Z_t(A) = 0 or ∞

¶

= 1, x∈X. (3.16)

Namely,

{L_A=∞}=

½ lim sup

t→∞ Z_t(A) =∞

¾

P_x-a.s., x∈X.

To prove Lemma 3.5, we consider the following equation:

u(x) =E_x h

exp

³−A^µ_ζ

´i +E_x

·Z _ζ

0

exp (−A^µ_t)F(u)(X_t)dA^µ_t

¸

, 0≤u≤1. (3.17) We can then prove the following by the same way as in Lemma 3.2.

Lemma 3.6. Assume that RR

X×XG^µ(x, y)µ(dx)µ(dy) < ∞. If the functions u₁ and u₂ are solutions to (3.17) respectively, and u₁≤u₂ <1 onX, then u₁=u₂ onX.

Proof of Lemma 3.5. LetO be a relatively compact open set in X such thatO includes the support ofµ andλ1 < λ1(µ, Q;O)<0, where

λ₁(µ, Q;O) = inf

½

E^O(u, u)− Z

O

u²(Q−1)dµ:u∈ F^O, Z

O

u²dm= 1

¾ .

Since the measure µ|O belongs to S_∞^O, the branching process M^O = (P^O_x) does not extinct by Theorem 2.4, and thus,

P_x

³

tlim→∞Z_t(O) =∞´

≥P^O_x

³

tlim→∞Z_t=∞´

>0, x∈O.

Furthermore, the left hand side above is positive for anyx∈X by the irreducibility ofM.

Let us denote by p^(Q_t ^−1)µ(x, y) the integral kernel of the Feynman-Kac semigroupp^(Q_t ^−1)µ as defined in (1.14). Then p^(Q_t ^−1)µ(x, A) := R

Ap^(Q_t ^−1)µ(x, y)m(dy) is bounded and continuous on X by Theorem 1.4 (i) and

p:= inf

x∈Op^(Q₁ ^−1)µ(x, A)>0 by the irreducibility ofM. Since

Ex[Zt(A)] =Ex

h exp

³

A^(Q_t ^−1)µ

´

;t < ζ, Xt∈A i

by (1.19), it holds that

xinf∈OEx[Z1(A)] =p >0, and thus

x∈Oinf Px(Z1(A)≥1)>0. (3.18)

Let q be a nonnegative constant such that e^−q= sup

x∈O

E_x[exp (−Z₁(A))].

Then it holds that 0< q≤p because the right hand side above is less than one by (3.18) and sup

x∈O

E_x[exp (−Z₁(A))] ≥exp µ

−inf

x∈OE_x[Z₁(A)]

¶

by Jensen’s inequality. Choose a positive constant q such that 0 < q < q. Then for any xⁿ= (x¹, x², x³,· · ·xⁿ)∈O⁽ⁿ⁾,

Pxⁿ(Z1(A)< qZ0(O)) =Pxⁿ(exp (−Z1(A))>exp (−qZ0(O)))

≤e^nq Yn i=1

E_xⁱ[exp (−Z1(A))]

by Chebyshev’s inequality. Since the last term above is not greater thane^(q−q)<1 for anyn≥1 by the definition ofq, it holds that

sup

n≥1,xⁿ∈O⁽ⁿ⁾

Pxⁿ(Z1(A)< qZ0(O))<1.

Namely,

inf

n≥1,xⁿ∈O⁽ⁿ⁾

Pxⁿ(Z1(A)≥qZ0(O))>0.

Let us define

A_m={Z_m(A)≥qZ_m−1(O)} for any positive integer m≥1 and

Ω₀= n

tlim→∞Z_t(O) =∞o

. (3.19)

Then by the Markov property,

P_x(A_m+1| Gm)(ω) =P_Xm(ω)(Z₁(A)≥qZ₀(O))

≥ inf

n≥1,xⁿ∈O⁽ⁿ⁾

P_xⁿ(Z₁(A)≥qZ₀(O))>0 for anyx∈X andω ∈Ω₀, and hence

X∞ m=0

P_x(A_m+1| Gm)(ω) =∞.

Noting that ( _∞

X

m=0

P_x(A_m+1| Gm) =∞ )

=

\∞ k=1

[∞ m=k

A_m

by [25, p.237, Corollary 3.2], we obtain (3.15).

From now on, we assume that A is an open set in X such that P_x(ρ_A < ∞) = 1 for any x∈X. Set

u₁(x) =P_x

³

tlim→∞Z_t(A) = 0

´

and

u2(x) =Px

µ lim sup

t→∞ Zt(A)<∞

¶ .

We then see in a similar way to Proposition 3.3 that the functions u1 and u2 are solutions to (3.17) respectively, by the assumption on A. Since it holds that u1 ≤ u2 < 1 by definition, Lemma 3.6 implies that u₁ =u₂ on X, which leads us to (3.16).

Proposition 3.7. Assume that the support of the branching rate µ is compact. Then, for any non-empty open set A in X and κ > λ₁,

P_x µ

lim sup

t→∞ e^κtZ_t(A) =∞

¶

>0, x∈X. (3.20)

Moreover, if P_x(ρ_A<∞) = 1 for anyx∈X, then {L_A=∞}=

½ lim sup

t→∞ e^κtZt(A) =∞

¾

Px-a.s., x∈X, and

{LA<∞}= n

tlim→∞e^κtZt(A) = 0 o

Px-a.s., x∈X.

Proof. For any κ > λ1, there exists a relatively compact open set O in X such that O includes the support ofµ andλ₁ < λ₁(µ, Q;O)< κ. Then, by Theorem 3.4 (ii),

Px

³

tlim→∞e^κtZt(O) =∞´

≥P^O_x

³

tlim→∞e^κtZt=∞´

>0, x∈O.

Moreover, the left hand side above is positive for any x ∈X by the irreducibility of M. If we replace Ω0 defined in (3.19) with

n

tlim→∞e^κtZt(O) =∞o , then (3.20) follows by the same way as in Lemma 3.5.

From now on, letA be an open set inX such thatP_x(ρ_A<∞) = 1 for anyx∈X. Set u₁(x) =P_x

³

tlim→∞e^κtZ_t(A) = 0

´

and

u₂(x) =P_x µ

lim sup

t→∞ e^κtZ_t(A)<∞

¶ .

Then it follows from (3.20) that uA ≤ u1 ≤ u2 < 1 on X, where uA(x) = Px(LA < ∞).

Furthermore, by noting that u_A, u₁ and u₂ are solutions to (3.17) respectively, Lemma 3.6 implies thatuA=u1=u2 onX, which completes the proof.

Theorem 3.8. (i) For any relatively compact open setA in X, Px

µ lim sup

t→∞ e^λ1tZt(A)<∞

¶

= 1, x∈X. (3.21)

As a consequence, for any κ < λ₁, P_x

³

tlim→∞e^κtZ_t(A) = 0

´

= 1, x∈X.

(ii)Assume that the support of the branching rateµ is compact. Then, for any non-empty open set A in X such that Px(ρA<∞) = 1 for anyx∈X andκ > λ1,

P_x µ

lim sup

t→∞ e^κtZ_t(A) =∞¯¯¯L_A=∞

¶

= 1, x∈X. (3.22)

Proof. Let A be a relatively compact open set in X. Then e^λ1tZ_t(A)≤ 1

inf_x∈Ah(x)M_t. Since

lim sup

t→∞ e^λ1tZt(A)≤ 1

infx∈Ah(x)M_∞<∞ Px-a.s., (3.21) holds. The equation (3.22) follows from Proposition 3.7.

Remark 3.9. Engl¨ander and Kyprianou [26] studied the exponential growth of the numbers of particles in every relatively compact open set for branching diffusion processes such that the branching rates are nonnegative, bounded and continuous functions. On the other hand, we can take unbounded functions as branching rates in (3.20) of Proposition 3.7 and Theorem 3.8 (i). For instance, let us consider a branching Brownian motion onR³. Then, since the measure µ(dx) = 1/|x|χ_|x|≤1dxbelongs toK_∞^R3, we can take the measureµas branching rate. Moreover, the ground state ofλ₁(µ;R³) satisfies (1.24) because the support of µis compact.

Assume that Mis Harris recurrent. Let us consider the branching symmetric Hunt process M= (Px) onX such that the branching rateµbelongs toK∞. Denote by T the first splitting time of M. Since P_x(A^µ_∞=∞) = 1 for any x∈X (see [46, p.426, Proposition 3.11]), it follows that

Px(T =∞) =Ex[exp (−A^µ_∞)] = 0

for anyx∈X. Using this fact, we can show Theorem 3.4 (i), (ii) and Theorem 3.8 by the same argument. Here the condition RR

X×XG^µ(x, y)µ(dx)µ(dy)<∞is replaced with µ(X)<∞ and the condition on the lifetime or the last exit times is not imposed.

Remark 3.10. Let M^D be an absorbing symmetric α-stable process on an open set D in R^d and assume that M^D is transient. As we mentioned in Remark 2.12, any measure µ ∈ S_∞^D satisfies µ|O ∈ S_∞^O for every bounded C^1,1 domain O inD. Hence, if we take a branching rate µ∈ S_∞^D such that RR

D×DG^µ,D(x, y)µ(dx)µ(dy) <∞, then the arguments from Lemma 3.5 to Theorem 3.8 work, whereG^µ,D(x, y) is the Green function of the exp (−A^µ_t)-subprocess ofM^D,

that is, Z

D

G^µ,D(x, y)f(y)dy=E_x

·Z _τD

0

exp (−A^µ_t)f(X_t)dt

¸ .

ドキュメント内 PDF Asymptotic Properties of Branching Symmetric Markov Processes - Osaka U (ページ 35-47)