The proof of Theorem 1.6 mimic the proof of Theorem 2.4 in [8]. It relies on the next two lemmas. We only give the proof of Lemma 8.2 because it differs from its analogue in [8].
Lemma 8.1. We consider the product measureNx1⊗Nx2on the spaceC(R+,W)2. The canonical process on this space is denoted by(W1, W2). Assumed >2dc−1.
Then for every(x1, x2)∈D2, we haveNx1⊗Nx2-a.e.
suppYD(W1)∩suppYD(W2)= ∅.
Lemma 8.2. Forε >0,δ >0, set gε(δ)=supNy
suppYD∩∂D\B∂D(z, ε)= ∅ ,
where the supremum is taken over(y, z)∈D×∂D, such thatd(y, ∂D)= |y−z|<
δ. Then for everyε >0, limδ↓0gε(δ)=0.
Proof. Since the boundary ofDisC2, we have the uniform exterior sphere condi-tion. There existsδ0∈(0, ε/3), for everyz∈ ∂D, we can findz0∈Dc(unique)
such thatB(z0, δ0)⊂Dcand∂B(z0, δ0)∩∂D= {z}. We defineBr =B(z0, rδ0). We have fory∈B2\B1,Ny-a.e.
{suppYD∩∂D\B∂D(z, ε)= ∅}
⊂
∃s∈(0, σ);ζs =τD(Ws) and Wˆs ∈∂D\B∂D(z, ε)
⊂
∃s∈(0, σ );τB¯c
3(Ws) <∞, τB¯c
3(Ws) < τB1(Ws) .
The first inclusion is a consequence of the definition ofLR+×R+×Dand the second is a consequence of the snake property. By the special Markov property (cf [4]
proposition 7), ifN is the number of excursions of the Brownian snake outside R+×R+×B2\B1that reachR+×R+×B3cbeforeR+×R+×B1, then we have
Ny
∃s∈(0, σ );τB¯c
3(Ws) <∞, τB¯c
3(Ws) < τB1(Ws)
=Ny[N >0]
≤Ny[N]
=Ny
YB2\B1(dy)Ny[τB¯c
3(Ws) <∞, τB¯c 3 < τB1]
≤Ny
∂B2
YB2\B1(dy)Ny[τB¯3c<+∞] .
We used the fact that ify ∈∂B1, then from the snake property, we haveNy-a.e.
for alls ∈ (0, σ),τB1(Ws) =0. By symmetry, we get thatNy[τB¯c
3 <+∞] =c0
is independent ofy∈∂B2. It is also finite since(Wˆs, s ≥0)is continuous under E(0,0,y). We then deduce from (6) that
Ny
suppYD∩∂D\B∂D(z, ε)= ∅
≤c0Ey[κB2 < κB1]. Thus we get that forδ∈(0, δ0),
gε(δ)≤c0Ey[κB(0,2δ0)< κB(0,δ0)],
where|y| = δ0+δ. The lemma is then a consequence of classical results on
Brownian motion.
Proof of Theorem 1.6. Let(Dk, k≥0)be an increasing sequence of open subsets ofDsuch thatD¯k ⊂Dk+1andd(y, ∂D)≤1/kfor ally ∈∂Dk. From the special Markov property (see [4] proposition 7) and proposition 2.4, we get that the law XDunderPXν is the same as the law of
i∈IYD(Wi), where conditionally onXDk, the random measure
i∈IδWiis a Poisson measure onC(R+,W)with intensity XDk(dy)Ny[·]. With a slight abuse of notation, we may assume that the point measure
i∈IYD(Wi)is also defined underPXν. It follows from lemma 8.1 and properties of Poisson measures that a.s. for everyi=j,
suppYD(Wi)∩suppYD(Wj)= ∅.
Forε >0, letUε denote the event “suppXD is contained in a finite union of disjoint compact sets of∂Dwith diameter less thanε”. It is easy to check thatUε
is measurable. Letkbe large enough. Furthermore, by the previous observations, and denoting byyi ∈Dkthe common starting point of the pathsWsi, and byzithe only point in∂Dsuch that|yi −zi| =d(yi, ∂D), we have
PXν[Uε]≥PνX
∀i∈I,diam(suppYD(Wi))≤ε
≥PνX
∀i∈I,suppYD(Wi)⊂B∂D(zi, ε/2)
=EνX
exp−
XDk(dy)Ny[suppYD∩∂D\B∂D(z, ε/2)= ∅]
≥EXν
exp−gε/2(1/k)(XDk,1) ,
where forB ∈ B(Rd), diam(B)=sup{x−x;(x, x)∈B×B}. We can now letkgo to+∞, using lemma 8.2, to conclude thatPXν[Uε]=1. Since this holds for everyε >0, we conclude that suppXDis totally disconnectedPXν-a.s.
9. Appendix
Lemma 9.1. Let(St, t≥0)be a stable subordinator. Forr >0, letLr =inf{u >
0, Su > r}. Then(St, t ∈[0, Lr))and(SLr−−S(Lr−t)−, t ∈[0, Lr))are identi-cally distributed.
We write P for the law of the subordinatorS=(St, t≥0)started at 0. We recall that the Laplace transform ofSis given byη(λ)=cρ∗λρ, wherec∗ρ =2−ρ/ 1(1+ρ). Its L´evy measure is given byP(ds) = 1(0,∞)(s)[2ρ1(ρ)1(1−ρ)]−1s−1−ρds. Notice thatLr is the last exit time of [0, r] forS. LetQ = (Qt, t ≥ 0)be the transition kernel of S andU = ∞
0 Qt dt its potential. The transition kernels and the potential are absolutely continuous with respect to the Lebesgue measure l onR. And we have Qt(x, dy) = qt(y−x)dy andU(x, dy) = u(y −x)dy, whereu(y) = ρ2ρyρ−11y≥0. LetQˆ = (Qˆt, t ≥ 0)be the transition kernel of (−St, t≥0). This is the dual kernel ofQwith respect tol. We consider the process V defined by
Vt =
S(Lr−t)− if 0≤t < L,
if t≥L,
whereis a cemetery point added toR. Notice the law ofS0isδ0, the Dirac mass at 0, and thus, the density ofδ0Uw.r.t. the reference measurel is justu. Thanks to XVIII 45 and 51 of [7], the processV is under P a Markov process with kernel (Q˜t, t≥0)defined as theu-transform ofQˆ, that is
Q˜t(x, dy)= 1
u(x)u(y)qt(x−y)dy.
We define the processY by Yt =
V0−Vt if 0≤t < L,
if t ≥L.
Notice thatY0=0 P-a.s. and the processY is right continuous and nondecreasing up to its lifetime. We want to prove that Y and the process S killed at timeLr have the same law. It will be enough to check that for every integern≥1, every sequencetn >· · · > t1 >0, andf1, . . . , fn, measurable nonnegative functions onR,
E
f1(Yt1) . . . fn(Ytn)
=E
f1(St1) . . . fn(Stn)1Stn<r . Using the transition kernel ofV, we get
I =E
f1(Yt1) . . . fn(Ytn)
=E
f1(V0−Vt1) . . . fn(V0−Vtn)
=
Rν(dv0)
RQ˜t1(v0, dv1)f1(v0−v1). . .
RQ˜tn−tn1(vn−1, dvn)fn(v0−vn), whereνis the law ofV0=SLr−. Thanks to [3] proposition 2 p.76, we have that
ν(dv0)=u(v0)1v0<rdv0
∞
r−v0
P(ds)=cρu(v0)(r−v0)−ρ1v0<rdv0.
Thus we have I=cρ
Rdv0u(v0)(r−v0)−ρ1v0<r
Rndv1. . . dvn u(v1)
u(v0)qt1(v0−v1)f1(v0−v1). . .
× u(vn)
u(vn−1)qtn−tn−1(vn−1−vn)fn(v0−vn)
=cρ
Rdv0(r−v0)−ρ1v0<r
Rndv1. . . dvnu(vn) qt1(v0−v1)f1(v0−v1) . . .
×qtn−tn−1(vn−1−vn)fn(v0−vn).
We use the change of variablez=v0,y1=v0−v1,· · · , yn=v0−vn, and the definition ofuto get
I =cρ
Rndy1. . . dynqt1(y1)f1(y1) . . . qtn−tn−1(yn−yn−1)fn(yn)
×
Rdz (r−z)−ρρ2ρ(z−yn)ρ−11r>z>yn
=E
f1(St1) . . . fn(Stn)1Stn<r , becausecρ
Rdz (r−z)−ρρ2ρ(z−yn)ρ−11r>z>yn =1r>yn.
Notations
dc =(α+1)/(α−1) critical dimension.
θ(dy) Lebesgue measure on∂D. Bε =B∂D(y0, ε) ball on∂D. PD Poisson kernel ofD. GD Green function ofD. γt Brownian motion inRd. Ex law ofγ started atx. ρ=(α−1) .
St ρ-stable subordinator.
ξt residual life time ofS. Lt time change, inverse ofS. 1t =γLt “freezed” Brownian motion.
E=R+×R+×Rd: state space ofξ =(ξ, L, 1). Pz law ofξt started atz∈E.
Px law ofξstarted at(0,0, x).
PDx law ofξ killed out ofR+×R+×D. Prx law ofξkilled at timer.
κB exit time ofBforγ. τB exit time ofBfor1=γL.
w = (w, ζ) E-valued path with life timeζ; fort ∈ [0, ζ), we writew(t) = (ξt(w), Lt(w), 1t(w)).
τB(w) exit time ofBfor1(w). κB(w) exit time ofBfor1L−1(w).
ˆ
w=1ζ(w) spatial end point.
Notations for the snake
ζs life time of the snake at times.
Ws snake at times; fort∈[0, ζs),Ws(t)=(ξt(Ws), Lt(Ws), 1t(Ws)). St(Ws) inverse of the time changeLt(Ws).
γt(Ws)=1St(Ws)(Ws) spatial motion of the snake pathWs.
Wˆs =1ζs(Ws) end point of the spatial motion of the snake pathWs. Lˆs =Lζs(Ws) end point of the time change of the snake pathWs. Ew law ofWs started at pathw.
E∗w law ofWs started at pathwand killed when its lifeζstime reaches 0.
E∗(r) =
Prx(dw)E∗w law ofWs killed when its life time reaches 0 and started with a typical (random) path of life timer.
Nz excursion measure of the snake started at the trivial pathz∈E. Nx excursion measure of the snake started at the trivial path(0,0, x). σ duration of the snake excursion.
LD exit local time ofD. YD exit measure of the snake.
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