Williams Decomposition for Superprocesses

Home , Jump process

arXiv:1609.03265v1 [math.PR] 12 Sep 2016 n,say one, H yuigtotasomtost hnetebacigmechan branching the change d to spatially transformations two with using superprocesses by for established was sition eedo h rgnlmto.Teeoe nti etn,t setting, this in Therefore, lineag ancestral motion. the original of the law on the consequence, depend a mo As spatial the structure. an mechanism, Delmas branching of independent terminology spatially the with analogy in decomposition, fteacsrllnaeo h atidvda lv should alive individual last the time of nonhomoge extinction lineage with ancestral superprocess the a for of contrary, the On account. time extinction lv,cniindon conditioned alive, ne oecniin on conditions some under ptal eedn udai rnhn ehns given mechanism branching quadratic dependent spatially a of decomposition spinal a give we precisely, Let Introduction 1 genealogy. mechanism; Phrases and Keywords Classification: Subject Mathematics 2010 AMS ∗ † ‡ = h eerho hsato sspotdb SC(rn o.1 Nos. (Grant NSFC by supported is author this of research The eerhspotdi atb rn rmteSmn Foundat Simons the from grant a by part in supported Research h eerho hsato sspotdb SC(rn o 11 No. (Grant NSFC by supported is author this of research The X oapitms tisetnto ie hsgnrlzsarsl fTribe of result a generalizes This general. time. more is extinction mechanism its branching at no our mass the point superprocesses, a some an to for super As that, only prove where considered. we Hénard were [4] decomposition, and mechanism Delmas branching decompositio of quadratic (Williams result dependent alive main individual the last of the alization to respect with mechanism h easprrcs ihasailydpnetbacigmech branching dependent spatially a with superprocess a be a eddcdfo baa n ems[]ween pta mo spatial no where [1] Delmas and Abraham from deduced be may edcmoetegnaoyo eea uepoeswt spat with superprocess general a of genealogy the decompose We ψ 0 n hnuigtegnaoyo uepoesswt branch with superprocesses of genealogy the using then and , h H emsadHead[]gv ilasdcmoiinfrsu for decomposition Williams a gave Hénard [4] and Delmas . ilasdcmoiinfrsuperprocesses for decomposition Williams a-i Ren Yan-Xia of H X = sfiie nti ae esuytegnaoia tutr o structure genealogical the study we paper this In finite. is uepoess ilasdcmoiin ptal depe spatially decomposition; Williams superprocesses; : β h ( x with and ) > h ∗ α Ψ( emn Song Renming ( x en osat hsdcmoiini aldaWilliams a called is decomposition This constant. a being 0 ,z x, se( (see ) = ) X Abstract β H novn h neta ieg ftels individual last the of lineage ancestral the involving ( )ad( and 2) x 1 ) z + α † 02;6G5 60J80. 60G55; 60J25; ( H pnetqartcbacigmechanism branching quadratic ependent x n u Zhang Rui and o #233 emn Song). Renming (#429343, ion ) )i 4) n[] h ilasdecompo- Williams the [4], In [4]). in 3) 213 n 11671017) and 1271030 601354) z eedo h pta oinadthe and motion spatial the on depend ini needn ftegenealogical the of independent is tion 2 ftels niiulaiede not does alive individual last the of e ´ nr 4.Frasprrcs with superprocess a For Hénard [4]. d by s Ψ( ism , eu rnhn ehns,telaw the mechanism, branching neous edsrpinof description he mlzdttlms ilconverge will mass total rmalized plcto fteWilliams the of application ,z x, al eedn branching dependent ially ns.W sueta the that assume We anism. rcse ihspatially with processes oasailyindependent spatially a to ) 1]i h es that sense the in [15] n mechanism ing ) hsi gener- a is This n). ‡ X ini ae into taken is tion epoesswith perprocesses dn branching ndent odtoe on conditioned f X ψ More . 0 given by the Brownian snake. As mentioned in [4], the drawback of the approach in [4] is that one has to restrict to quadratic branching mechanisms with bounded and smooth parameters. The goal of this paper is to establish a Williams decomposition for more general superprocesses. Our superprocesses are more general in two aspects: first the spatial motion can be a general Markov process and secondly the branching mechanism is general and spatially dependent (see (2.1) below). We will give conditions that guarantee our general superprocesses admit a Williams decomposition. The conditions should be satisfied by a lot of superprocesses. We obtain a Williams decomposition by direct construction. For any fixed constant h> 0, we first describe the motion of a spine up to time h and then construct three kinds of immigrations (continuous immigration, jump immigration and immigration at time 0) alone the spine. We prove that, conditioned on H = h, the sum of the contributions of the three types of immigrations has the same distribution as X before time h, see Theorem 3.5 below. Note that for quadratic branching mechanisms, there is no jump immigration. As an application of the Williams decomposition, we prove that, for some superprocesses, the normalized total mass will converge to a point mass at its extinction time, see Theorem 3.7 below. This generalizes a result of Tribe [15] in the sense that our branching mechanism is more general.

2 Preliminary

2.1 Superprocesses and assumptions In this subsection, we describe the superprocesses we are going to work with and formulate our assumptions. Suppose that E is a locally compact separable metric space. Let E∂ := E ∪ {∂} be the one- point compactification of E. ∂ will be interpreted as the cemetery point. Any function f on E is automatically extended to E∂ by setting f(∂) = 0. Let DE be the set of all the càdlàg functions from [0, ∞) into E∂ having ∂ as a trap. The 0 0 filtration is defined by Ft = Ft+, where Ft is the natural canonical filtration, and F = t≥0 Ft. Consider the canonical process ξ on (D , {F } ). We will assume that ξ = {ξ , Π } is a Hunt t E t t≥0 t x W process on E and ζ := inf{t > 0 : ξt = ∂} is the lifetime of ξ. We will use {Pt : t ≥ 0} to denote + the semigroup of ξ. We will use Bb(E) (Bb (E)) to denote the set of (non-negative) bounded Borel 0 functions on E. We will use MF (E) to denote the family of finite measures on E and MF (E) to denote the family of non-trivial finite measures on E. Suppose that the branching mechanism is given by

Ψ(x, z)= −α(x)z + b(x)z2 + (e−zy − 1+ zy)n(x,dy), x ∈ E, z> 0, (2.1) Z(0,+∞) + where α ∈Bb(E), b ∈Bb (E) and n is a kernel from E to (0, ∞) satisfying

sup (y ∧ y2)n(x,dy) < ∞. (2.2) x∈E Z(0,+∞) Then there exists a constant K > 0, such that

|α(x)| + b(x)+ (y ∧ y2)n(x,dy) ≤ K. Z(0,+∞)

Let MF (E) be the space of ﬁnite measures on E, equipped with the topology of weak convergence. As usual, hf,µi := E f(x)µ(dx) and kµk := h1,µi. According to [13, Theorem 5.12], there R 2 + is a Hunt process X = {Ω, G, Gt, Xt, Pµ} taking values in MF (E), such that, for every f ∈Bb (E) and µ ∈ MF (E), −hf,Xti − log Pµ e = huf (t, ·),µi, (2.3) where uf (t,x) is the unique positive solution to the equation

t uf (t,x) + Πx Ψ(ξs, uf (t − s,ξs))ds = Πxf(ξt), (2.4) Z0

where Ψ(∂, z) = 0, z > 0. X = {Xt : t ≥ 0} is called a superprocess with spatial motion ξ = {ξt, Πx} and branching mechanism Ψ, or sometimes a (Ψ,ξ)-superprocess. In this paper, the superprocess we deal with is always this Hunt realization. For the existence of X, see also [3] and [5].

Deﬁne v(t,x) := − log Pδx (kXtk = 0), and H := inf{t ≥ 0 : kXtk = 0}. It is obvious that v(0,x) = ∞. In this paper, we will consider the critical and subcritical case. More precisely, throughout this paper, we assume that X satisfy the following uniform global extinction property.

(H1) For any t> 0, sup v(t,x) < ∞ and lim v(t,x) = 0. (2.5) x∈E t→∞

Remark 2.1 Note that Assumption (H1) is equivalent to

inf Pδx (kXtk = 0) > 0 for all t> 0 and Pδx (H < ∞) = lim Pδx (kXtk = 0) = 1. (2.6) x∈E t→∞

Remark 2.2 If ∞ Ψ(x, z) ≥ Ψ(z) := bz2 + e−yz − 1+ yz n(dy), (2.7) Z0 ∞ 2 e where b ≥ 0, 0 (y ∧ y )n(dy) < ∞ and Ψ satisﬁes the Grey condition: R ∞ 1 e dz < ∞, Z Ψ(z) then Assumption (H1) holds. e We also assume that

(H2) For any x ∈ E and t> 0, ∂v w(t,x) := − (t,x) (2.8) ∂t exists. Moreover, for any 0

sup sup w(s,x) < ∞. (2.9) r≤s≤t x∈E

Note that, since t → v(t,x) is decreasing, we have w(t,x) ≥ 0. We also use vt and wt to denote the function x → v(t,x) and x → w(t,x) respectively.

3 Example 1 Assume that the spatial motion ξ is conservative, that is Pt(1) ≡ 1, and the branching mechanism is spatially independent, that is

∞ Ψ(x, z)=Ψ(z)= az + bz2 + e−yz − 1+ yz n(dy), Z0 ∞ 2 where a ≥ 0, b ≥ 0 and 0 (y ∧ y )n(dy) < ∞. We also assume that Ψ satisﬁes the Grey condition: R ∞ 1 dz < ∞. Ψ(z) Z

Then {kXtk,t ≥ 0} is a continuous state branching process with branching mechanism Ψ(z). So v(t,x) = v(t) < ∞ does not depend on x, and limt→∞ v(t) = 0, thus Assumption (H1) holds immediately. Moreover, for t> 0, we have that d w(t) := − v(t)=Ψ(v(t)). dt Thus Assumption (H2) is satisﬁed. See [10, Theorem 10.1] for more details.

In Section 5, we will give more examples, including some class of superdiﬀusions, that satisfy Assumptions (H1)-(H2).

2.2 Excursion law of {Xt, t ≥ 0}

We use D to denote the space of MF (E)-valued c`adl`ag functions t 7→ ωt on (0, ∞) having zero as a trap. We use (A, At) to denote the natural σ-algebras on D generated by the coordinate process. Let {Qt(µ, ·) := Pµ (Xt ∈ ·) : t ≥ 0, µ ∈ MF (E)} be the transition semigroup of X. Then by (2.3), we have

−hf,νi e Qt(µ,dν) = exp{−hVtf,µi} for µ ∈ MF (E) and t ≥ 0, ZMF (E) where Vtf(x) := uf (t,x),x ∈ E. This implies that Qt(µ1 + µ2, ·) = Qt(µ1, ·) ∗ Qt(µ2, ·) for any µ1,µ2 ∈ MF (E), and hence Qt(µ, ·) is an inﬁnitely divisible probability measure on MF (E). By the semigroup property of Qt, Vt satisﬁes that

VsVt = Vt+s for all s,t ≥ 0.

Moreover, by the inﬁnite divisibility of Qt, each operator Vt has the representation

−hf,νi + Vtf(x)= λt(x,f)+ 1 − e Lt(x,dν) for t> 0, f ∈Bb (E), (2.10) M (E)0 Z F where λt(x,dy) is a bounded kernel on E and (1 ∧ ν(1))Lt(x,dν) is a bounded kernel from E to 0 0 0 MF (E) . Let Qt be the restriction of Qt to MF (E) . Let E0 := {x ∈ E : λt(x, E) = 0 for all t> 0}. For λ> 0, we use Vtλ to denote Vtf when the function f ≡ λ. It then follows from (2.10) that for every x ∈ E and t> 0,

−λh1,νi Vtλ(x)= λt(x, E)λ + 1 − e Lt(x,dν). M (E)0 Z F 4 The left hand side tends to − log Pδx (Xt = 0) as λ → +∞. Therefore, Assumption (H1) implies that λt(x, E) = 0 for all t> 0 and hence x ∈ E0, which says that E = E0. For x ∈ E, we get from (2.10) that

−hf,νi + Vtf(x)= 1 − e Lt(x,dν) for t> 0, f ∈Bb (E). M (E)0 Z F It then follows from [13, Proposition 2.8 and Theorem A.40] that for every x ∈ E, the family of 0 measures {Lt(x, ·) : t > 0} on MF (E) constitutes an entrance law for the restricted semigroup 0 {Qt : t ≥ 0}. It is known (see [13, Section 8.4]) that one can associate with {Pδx : x ∈ E} a family of σ-ﬁnite measures {Nx : x ∈ E} deﬁned on (D, A) such that Nx({0}) = 0,

−hf,ωti −hf,Xti + (1 − e )Nx(dω)= − log Pδx (e ), f ∈Bb (E), t> 0, (2.11) ZD and, for every 0

Nx(ωt1 ∈ dµ1, · · · ,ωtn ∈ dµn)

= Nx(ωt1 ∈ dµ1)Pµ1 (Xt2−t1 ∈ dµ2) · · · Pµn−1 (Xtn−tn−1 ∈ dµn). (2.12)

This measure Nx is called the Kuznetsov measure corresponding to the entrance law {Lt(x, ·) : t> 0} or the excursion law for superprocess X. For earlier work on excursion law of superprocesses, see [6, 8, 12]. It follows from (2.11) that for any t> 0,

Nx(kωtk=0)= 6 − log Pδx (kXtk = 0) < ∞. (2.13)

3 Main results

In this and the next section we will always assume that Assumptions (H1)-(H2) hold. Recall that H := inf{t ≥ 0 : kXtk = 0}. Note that

−hvt,µi FH (t) := Pµ(H ≤ t)= Pµ(kXtk =0)= e . (3.1)

By the continuity of v(t,x) with respect to t ∈ (0, ∞), we get that for any t> 0,

−hvt−ǫ,µi −hvt,µi Pµ(H

For h> 0, deﬁne hw , X ie−hvh−t,Xti M h := h−t t , 0 ≤ t < h. (3.3) t −hv ,X0i hwh, X0ie h h Then, under Pµ, {Mt , 0 ≤ t < h} is a nonnegative martingale with mean one (see Lemma 4.2 below).

Theorem 3.1 For any h> 0 and t < h,

1 h lim Pµ(A|h ≤ H < h + ǫ)= Pµ( AMt ), ∀A ∈ Gt. ǫ↓0

5 We deﬁne, for each h> 0,

Pµ(·|H = h) := lim Pµ(·|h ≤ H < h + ǫ). ǫ↓0

h h Then, by Theorem 3.1, {Xt,t

Corollary 3.2 For any A ∈ Gt, we have ∞ h Pµ(A ∩ {H>t})= Pµ(A)FH (dh). Zt Proof: It follows from Fubini’s theorem that ∞ ∞ h h Pµ(A)FH (dh)= Pµ(1AMt )FH (dh) t t Z Z ∞ −hvh−t,Xti = Pµ(1Ahwh−t, Xtie ) dh t Z ∞ −hvh−t,Xti = Pµ 1A hwh−t, Xtie dh t Z ∞ −hvh,Xti = Pµ 1A hwh, Xtie dh Z0 = Pµ (A ∩ {Xt 6= 0})= Pµ (A ∩ {H>t}) , where in the ﬁfth equality we use the fact that

∞ −hvh,Xti −hvh,Xti −hvh,Xti hwh, Xtie dh = lim e − lim e = 1{X 6=0}. h→∞ h→0 t Z0 ✷ For any h> 0 and t ∈ [0, h), we deﬁne

w(h − t,ξ ) t ′ h t − R0 Ψz(ξu,v(h−u,ξu)) du Yt := e , w(h, ξ0)

′ ∂Ψ(x,z) where Ψz(x, z)= ∂z . Then we have the following result whose proof will be given in Section 4.

h h Lemma 3.3 Under Πx, {Yt ,t

Remark 3.4 In Example 1, w(t,x) and v(t,x) do not depend on x, and for any h > 0 and h 0 ≤ t < h, Yt ≡ 1.

Now we state our main result: the Williams decomposition. We will construct a new process h {Λt ,t

6 h Under Πx, (ξt)0≤t

h h Πν := Πx ν(dx). ZE h Then, under Πν , (ξt)0≤t 0 : kωtk = 0}, ω ∈ D. h h w(h,x) h Let ξ := {(ξt)0≤t

Continuous immigration Suppose that N 1,h(ds, dω) is a Poisson random measure on [0, h) × D

with intensity measure 21[0,h)(s)1H(ω)

1,h,N 1,h Xt := ωt−sN (ds, dω). (3.4) Z[0,t) ZD

Jump immigration Suppose that N 2,h(ds, dω) is a Poisson random measure on [0, h) × D with ∞ P intensity measure 1[0,h)(s)1H(ω)

R t 2,h,P 2,h Xt := ωt−sN (ds, dω). (3.5) Z0 ZD

0,h Immigration at time 0 Let {Xt , 0 ≤ t < h} be a process distributed according to the law Pµ(X ∈ ·|H < h). We assume that the three processes X0,h, X1,h,N and X2,h,P are independent given the trajectory of ξh. Deﬁne h 0,h 1,h,N 2,h,P Λt := Xt + Xt + Xt . (3.6) h (h) We write the law of Λ as Pµ .

P(h) h Theorem 3.5 Under µ , the process {Λt ,t

h If we deﬁne Λt = 0, for any t ≥ h, then we get the following result.

Corollary 3.6 {Xt; Pµ} has the same ﬁnite dimensional distribution as

∞ (h) h Pµ (Λ ∈ ·)FH (dh). Z0 + Proof: Let fk ∈ Bb (E), k = 1, 2, · · · ,n and 0 = t0 < t1 < t2 < · · · < tn. We put tn+1 = ∞ and deﬁne (tn,tn+1] := (tn, ∞). We will show that

n n P (h) h µ exp − hfj, Xtj i = Pµ exp − hfj, Λtj i FH (dh).    (0,∞)     Xj=1  Z  Xj=1         

7 h Since Λt = 0, for t ≥ h, we get that

n (h) h Pµ exp − hfj, Λtj i FH (dh) (0,∞)    Z  Xj=1  n  r  (h)   h = Pµ exp − hfj, Λtj i FH (dh) (tr ,tr+1]    Xr=0 Z  Xj=1  n  r    = Ph exp − hf , X i F (dh) µ   j tj  H r=0 (tr ,tr+1] X Z  Xj=1  n  r    = P exp − hf , X i ; t

(H3) For any bounded open set B ⊂ E and any t> 0, the function

t c x →− log Pδx Xs(B ) ds = 0 0 Z is ﬁnite for x ∈ B and locally bounded.

Theorem 3.7 Assume that (H1)-(H3) hold and that for any µ ∈ MF (E),

h lim ξt = ξh−, Πν -a.s., (3.7) t↑h

w(h,x) where ν(dx)= hw(h,·),µi µ(dx). Then there exists an E-valued random variable Z such that

Xt lim = δZ , Pµ-a.s., t↑H kXtk where the limit above is in the sense of weak convergence. Moreover, conditioned on {H = h}, Z h + has the same law as {ξh−, Πν }, that is, for any f ∈Cb (E), ∞ h Pµf(Z)= Πν (f(ξh−))FH (dh). (3.8) Z0 h Note that, if the martingale {Yt , 0 ≤ t < h} is uniformly integrable, then condition (3.7) holds. Now we give an example that satisﬁes Assumption (H3).

8 Example 2 Assume that ξ is a diffusion on Rd with infinitesimal generator ∂ ∂ ∂ L = aij(x) + bj(x) , ∂xi ∂xj ∂xj X X which satisfies the following two conditions: (A) (Uniform ellipticity) There exists a constant γ > 0 such that

2 d ai,j(x)uiuj ≥ γ uj , x ∈ R . X X (B) aij and bj are bounded H¨older continuous functions. Suppose that the branching mechanism Ψ(x, z) satisﬁes that, for some α ∈ (1, 2] and c > 0, Ψ(x, z) ≥ czα for all x ∈ Rd. α Let {X, Pµ} and {X, Pµ} be a (ξ, Ψ)-superprocess and a (ξ, z )-superprocess respectively. Then, for any open set B ⊂ Rd,

e e t t c c − log Pδx exp −λ Xs(B ) ds ≤− log Pδx exp −λ Xs(B ) ds Z0 Z0 ≤− log Peδx (R⊂ B), e (3.9) where R is the range of X, which is the minimal closed subsete of Rd which supports all the measures Xt, t ≥ 0. Thus, we have that

t e t e c c − log Pδx Xs(B ) ds = 0 = lim − log Pδx exp −λ Xs(B ) ds ≤− log Pδx (R⊂ B). 0 λ→∞ 0 Z Z e By [5, Theorem 8.1], x →− log Pδx (R⊂ B) is continuous in x ∈ B. Therefore the superprocess X satisfies Assumption (H3). e Remark 3.8 Now we consider the superprocess in Example 1. We assume that ξ is a diffusion in Rd satisfying the conditions in Example 2, and the branching mechanism Ψ(z) satisfies that, α h for some α ∈ (1, 2] and c > 0, Ψ(z) ≥ cz . Thus, Assumption (H3) holds. Since Yt = 1 and h Πx = Πx, condition (3.7) holds automatically. Therefore, Theorem 3.7 holds and Z has the same law as ξH , where ξ0 ∼ ν(dx)= µ(dx)/kµk. Moreover, ξ and H are independent. Compared with [15], the example above assumes that the spatial motion ξ is a diffusion, while in [15], the spatial motion is a Feller process. However, in [15], the branching mechanism is binary (Ψ(z)= z2), while in the example above, the branching mechanisms is more general.

4 Proofs of Main Results

We will use Pr,δx to denote the law of X starting from the unit mass δx at time r> 0. Similarly, we will use Πr,x to denote the law of ξ starting from x at time r> 0. First, we give an useful lemma.

+ + Lemma 4.1 Suppose that f ∈ Bb (E) and gi ∈ Bb (E), i = 1, 2, · · · ,n. For any 0 < t1 ≤ t2 ≤ ···≤ tn and 0 ≤ r ≤ tn, we have

P hf, X i exp − hg , X i r,µ  tn j tj  n j:Xtj ≥r o   9 tn ′ −hUg (r,·),µi = Πr,x exp − Ψz(ξu,Ug(u, ξu)) du f(ξtn ) µ(dx)e , (4.1) ZE Zr where

Ug(r, x) := − log Pr,δx exp − hgj, Xtj i . n j:Xtj ≥r o + + In particular, for any f ∈Bb (E) and g ∈Bb (E), we have

t −hg,Xti ′ −ug(t,x) Pδx hf, Xtie = Πx exp − Ψz(ξu, ug(t − u, ξu)) du f(ξt) e . (4.2) 0 Z Proof: By [13, Propersition 5.14], we have that, for 0 ≤ r ≤ tn,

− log P exp − hg , X i− θhf, X i = hF (r, ·),µi, r,µ  j tj tn  θ n j:Xtj ≥r o   where Fθ(r, x) is the unique bounded positive solution on [0,tn] × E of

Fθ(r, x) + Πr,x Ψ(ξu, Fθ(u, ξu)) du = Πr,xgj(ξtj )+ θΠr,xf(ξtn ). (4.3) r Z j:Xtj ≥r ′ ∂ Let Fθ(r, x) := ∂θ Fθ(r, x). Then,

∂ P hf, X i exp − hg , X i = − e−hFθ(r,·),µi = hF ′ (r, ·),µie−hUg (r,·), µi. r,µ  tn j tj  ∂θ θ=0+ 0 j:t ≥r n Xj o   Diﬀerentiating both sides of (4.3) with respect to θ and then letting θ → 0, we get that

tn ′ ′ ′ F0(r, x) + Πr,x Ψz(ξu,Ug(u, ξu))F0(u, ξu) du = Πr,xf(ξtn ), Zr which implies that tn ′ ′ − Rr Ψz(ξu,Ug(u,ξu)) du F0(r, x) = Πr,x e f(ξtn ) . Therefore (4.1) holds. h i ✷

∂v Recall that v(t,x) := − log Pδx (kXtk = 0) and w(t,x) := − ∂t (t,x) ≥ 0. Recall the deﬁnition of h Mt in (3.3).

h h Lemma 4.2 Under Pµ, {Mt ,t

Proof: For any h > 0 and 0 ≤ t < h, by Assumption (H2) and the dominated convergence theorem, we get that ∂ P hw , X ie−hvh−t,Xti = P e−hvh−t,Xti µ h−t t ∂h µ h i ∂ = e−hvh,µi = hw ,µie−hvh,µi, (4.4) ∂h h h where in the second equality, we used the Markov property of X. Thus, it follows that Pµ(Mt ) = 1.

10 By the Markov property of X, we obtain that, for s

−hvh−t,Xti −hvh−t,Xt−si −hvh−s,Xsi Pµ hwh−t, Xtie Gs]= PXs hwh−t, Xt−sie = hwh−s, Xsie , h h i h ✷ which implies that, under Pµ, {Mt ,t

Proof of Theorem 3.1: For any A ∈ Gt, by the Markov property of X,

Pµ(A ∩ {h ≤ H < h + ǫ}) Pµ(A|h ≤ H < h + ǫ) = Pµ(h ≤ H < h + ǫ) P (1 P (h − t ≤ H < h − t + ǫ)) = µ A Xt e−hvh+ǫ,µi − e−hvh,µi

−hv + ,Xti −hv ,Xti Pµ 1A e h−t ǫ − e h−t = . e−hvh+ǫ,µi − e−hvh,µi By Assumption (H2), we get that

1 −hvh+ǫ,µi −hvh,µi −hvh,µi lim e − e = hwh,µie (4.5) ǫ↓0 ǫ and 1 −hvh−t+ǫ,Xti −hvh−t,Xti −hvh−t,Xti lim e − e = hwh−t, Xtie . (4.6) ǫ↓0 ǫ Note that, for 0 <ǫ< 1, 1 1 e−hvh−t+ǫ,Xti − e−hvh−t,Xti ≤ 1 − exp{−hv − v , X i} ǫ ǫ h−t h−t+ǫ t hvh−t − vh−t+ǫ, Xti ≤ ≤ sup sup w(s,x)h1, Xti. ǫ h−t≤s≤h−t+1 x∈E

Thus, it follows from the dominated convergence theorem that 1 −hvh−t+ǫ,Xti −hvh−t,Xti −hvh−t,Xti lim Pµ 1A e − e = Pµ 1Ahwh−t, Xtie . (4.7) ǫ↓0 ǫ Thus, by (4.5) and (4.7), we have that

h lim Pµ(A|h ≤ H < h + ǫ)= Pµ(1AMt ). ǫ↓0

The proof is now complete. ✷

Proof of Lemma 3.3: By the Markov property of X, we get that,

−v(t+s,x) −hvs,Xti e = Pδx (Xt+s =0)= Pδx (PXt (Xs = 0)) = Pδx (e ), (4.8)

which implies that uvs (t,x)= v(t + s,x). By (4.4) with h = t + s and µ = δx, we get that

−v(t+s,x) −hvs,Xti w(t + s,x)e = Pδx (hws, Xtie ) t ′ −v(t+s,x) = Πx exp − Ψz(ξu, v(t + s − u, ξu)) du w(s,ξt) e , Z0

11 where in the last equality we used Lemma 4.1 and the fact that uvs (t,x) = v(t + s,x). Thus, it follows immediately that

t ′ w(t + s,x) = Πx exp − Ψz(ξu, v(t + s − u, ξu)) du w(s,ξt) . (4.9) Z0 For 0

t ′ − R Ψ (ξu,v(h−u,ξu)) du Πx w(h − t,ξt)e 0 z |Fs

s ′ t ′ − R Ψ (ξu,v(h−u,ξu)) du − R Ψ (ξu,v(h−u,ξu)) du = e 0 z Πx w(h − t,ξt)e s z |Fs

s ′ t−s ′ − R0 Ψz(ξu,v(h−u,ξu)) du − R0 Ψz(ξu,v(h−s−u,ξu)) du = e Πξs w(h − t,ξt−s)e

s ′ − R Ψ (ξu,v(h−u,ξu)) du = e 0 z w(h − s,ξs), where the last equality above follows from (4.9). The proof is now complete. ✷

4.1 Williams decomposition + Proof of Theorem 3.5: Let fk ∈Bb (E), k = 1, 2, · · · ,n and 0 = t0

n n Ph exp − hf , X i = P(h) exp − hf , Λh i . µ   j tj  µ   j tj   Xj=1   Xj=1      h By the deﬁnition of Λt , we have   

n (h) h Pµ exp − hfj, Λtj i n Xj=1 o n w(h, x) h (h) h h = µ(dx)Πx Pµ exp − hfj, Λtj i |ξ . (4.10) E hw(h, ·),µi Z h n Xj=1 o i h By the construction of Λt , we have

n (h) h h Pµ exp − hfj, Λtj i |ξ n Xj=1 o n n P (h) 1,h,N h = µ exp − hfj, Xtj i |H < h × Pµ exp − hfj, Xtj i |ξ n Xj=1 o n Xj=1 o n (h) 2,h,P h × Pµ exp − hfj, Xtj i |ξ n Xj=1 o =:(I) × (II) × (III). (4.11)

Deﬁne, for s < h,

n 1 − Pj=1hfj ,Xtj −si s≤tj Js(h, x) := − log Pδx e ; kXh−sk = 0 . (4.12) h i

12 We ﬁrst deal with part (I). By (4.12), we have

J0(h, x)= − log Pδx exp − hfj, Xtj i ; kXhk = 0 . (4.13) n Xj=1 o −hv , µi By (3.2), Pµ(H < h)= Pµ(H ≤ h)= e h . Thus we have

(I)= ehv(h,·), µie−hJ0(h,·), µi. (4.14)

Next we deal with part (II). By the deﬁnition of X1,h,N and Fubini’s theorem, we have

n n t 1,h,N 1,h hfj, Xtj i = hfj,ωtj −si1s

t n (h) 1,h h (II) = Pµ exp − hfj,ωtj −si1s

n 1 − Pj=1hfj ,ωtj−si s

Hence, t (II) = exp − 2b(ξs) Js(h, ξs) − v(h − s,ξs) ds . (4.16) 0 Z Now we deal with (III). Using arguments similar to those leading to (4.15), we get that

n t n 2,h,P 2,h hfj, Xtj i = hfj,ωtj −si1s≤tj N (ds, dω). 0 D Xj=1 Z Z Xj=1

13 Thus,

t n (h) 2,h h (III)= Pµ exp − hfj,ωtj −si1s≤tj N (ds, dω) |ξ   0 D    Z Z Xj=1  t ∞  − n hf ,X i1   P Pj=1 j tj −s s≤tj = exp − ds yn(ξs, dy) yδξs 1 − e 1H

(II) × (III) t ∞ −yJs(h,ξs) =exp − 2b(ξs)Js(h, ξs)+ y 1 − e n(ξs, dy) ds 0 0 Z t Z ∞ −yv(h−s,ξs) × exp 2b(ξs)v(h − s,ξs) − y 1 − e n(ξs, dy) ds 0 0 tZ Z t ′ ′ =exp − Ψz(ξs, Js(h, ξs)) ds × exp Ψz(ξs, v(h − s,ξs)) ds . (4.18) Z0 Z0 By (4.11), (4.14) and (4.18), we get that, for h>t,

n Πh P(h) exp − hf , Λh i |ξh x µ   j tj   h  Xj=1  i  t  t hv(h,·), µi −hJ0(h,·), µi h ′ ′ =e e  Πx exp − Ψz(ξs, Js(h, ξs)) ds × exp Ψz(ξs, v(h − s,ξs)) ds 0 0 h Z t Z i hv(h,·),µi −hJ0(h,·), µi w(h − t,ξt) ′ =e e Πx exp − Ψz(ξs, Js(h, ξs)) ds . w(h, x) 0 h Z i So, by (4.10), we obtain that

n P(h) exp − hf , Λh i µ   j tj   Xj=1  hv(h,·),µi  t e −hJ0(h,·), µi ′ = e  Πxw(h − t,ξt)exp − Ψz(ξs, Js(h, ξs)) ds µ(dx). (4.19) hwh,µi E 0 Z h Z i Now we calculate Js(h, x) deﬁned in (4.12). For 0 ≤ s

t −hJ0(h,·), µi ′ e Πx w(h − t,ξt)exp − Ψz(ξs, Js(h, ξs)) ds µ(dx) E 0 Z h n Z i

=Pµ hw(h − t, ·), Xti exp − hfj, Xtj i − hv(h − t, ·), Xti . h n Xj=1 oi Thus, by (4.19), we get that

n n (h) h P h Pµ exp − hfj, Λtj i = µ exp − hfj, Xtj i Mt . n Xj=1 o h n Xj=1 o i Now, the proof is complete. ✷

4.2 The behavior of Xt near extinction h h w(h,x) Recall that, for any µ ∈ MF (E), ξ = {(ξt)0≤t

Lemma 4.3 Suppose that Assumptions (H1)-(H3)) hold and that for any µ ∈ MF (E),

h lim ξt = ξh−, Πν -a.s., t↑h

w(h,x) where ν(dx)= hw(h,·),µi µ(dx). Then, for any h> 0,

h Λt P(h) lim h = δξh− , µ -a.s. t↑h kΛt k

Proof: By the decomposition (3.6), we have

h 0,h 1,h,N 2,h,P Λt := Xt + Xt + Xt .

Deﬁne 0,h h h H0 := inf{t ≥ 0 : Xt = 0} and H(Λ ) := inf{t ≥ 0 : Λt = 0}. 0,h h Then by the deﬁnition of X , we have H0 < h. By Theorem 3.5, H(Λ )= h. It follows that

0,h Xt (h) lim h = 0, Pµ -a.s. (4.21) t↑h kΛt k

Note that E∂ is a compact separable metric space. According to [14, Exercise 9.1.16 (iii)], Cb(E∂; R), + the space of bounded continuous R-valued functions f on E∂, is separable. Therefore, Cb (E), the space of nonnegative bounded continuous R-valued functions f on E, is also a separable space. It + suﬃces to prove that, for any f ∈ Cb (E),

1,h,N 2,h,P (h) hfh, Xt i + hfh, Xt i Pµ lim h = 0 = 1, (4.22) t↑h kΛt k !

15 where fh(x)= f(x) − f(ξh−). Note that

1,h,N 2,h,P 1,h,N 2,h,P (h) hfh, Xt i + hfh, Xt i (h) (h) hfh, Xt i + hfh, Xt i h Pµ lim h = 0 = Pµ Pµ lim h = 0 ξ . t↑h kΛt k ! " t↑h kΛt k !#

+ Therefore, it suﬃces to prove that, for any f ∈ Cb (E),

1,h,N 2,h,P (h) hfh, Xt i + hfh, Xt i h (h) Pµ lim h = 0 ξ = 1, Pµ -a.s. (4.23) t↑h kΛt k !

Step 1 We ﬁrst prove that given ξh,

1,h,N hfh, Xt i (h) lim h = 0, Pµ -a.s. (4.24) t↑h kΛt k

Note that given ξh, t 1,h,N 1,h hfh, Xt i := hfh,ωt−siN (ds, dω), Z0 ZD where N 1,h(ds, dω) is a Poisson random measure on [0, h) × D with intensity measure

21[0,h)(s)1H(ω)

1,h Let I1 be the support of the measure N . Note that I1 is a random subset of [0, h) × D. h + In the remainder of this proof, we always assume that ξ is given. Since f ∈ Cb (E), for any ǫ > 0, there exists δ1 > 0, depending on ξh−, such that |f(x) − f(ξh−)| ≤ ǫ for all |x − ξh−| ≤ δ1. It follows from the fact that ξh− = lims↑h ξs there exists δ2 ∈ (0, h), depending on ξh−, such that |ξs − ξh−| < δ1/2 for all s ∈ (h − δ2, h). Let B := B(ξh−, δ1)= {x ∈ E : |x − ξh−| < δ1}. Then, for any t ∈ (h − δ2/2, h), we have

1,h,N 1,h,N 1,h,N |hfh, Xt i| =|hfh1B¯ , Xt i + hfh1B¯c , Xt i| 1,h,N 1,h,N ≤ǫh1, Xt i + 2kfk∞h1B¯c , Xt i h−δ2 h 1,h ≤ǫh1, Λt i + 2kfk∞ h1,ωt−siN (ds, dω) 0 D t Z Z 1,h + 2kfk∞ h1B¯c ,ωt−siN (ds, dω) Zh−δ2 ZD h =:ǫh1, Λt i + 2kfk∞J1(t) + 2kfk∞J2(t). (4.25)

It follows that

1,h,N |hfh, Xt i| J1(t) J2(t) h ≤ ǫ + 2kfk∞ h + 2kfk∞ h . (4.26) kΛt k kΛt k kΛt k

First we deal with J1. For s ∈ (0, h − δ2) and t ∈ (h − δ2/2, h), we have t − s > δ2/2. Thus, for t ∈ (h − δ2/2, h), we have

h−δ2 1,h J1(t)= h1,ωt−siN (ds, dω)= h1,ωt−si, 0 ω(δ2/2)6=0,H(ω)

16 where S1 := {(s,ω) : s ∈ [0, h − δ2), w(δ2/2) 6=0 and H(ω) < h − s}. (4.27) Notice that

21[0,h)(s)1H(ω)

≤2K Nξs (w(δ2/2) 6= 0)ds Z0 h−δ2

=2K v(δ2/2,ξs)ds ≤ 2Khkvδ2/2k∞ < ∞, (4.28) Z0 which implies that given ξh, 1,h (h) N (S1) < ∞, Pµ -a.s.

h (h) That is, given ξ , ♯{I1 ∩ S1} < ∞, Pµ -a.s. For any (s,ω) ∈ (I1 ∩ S1), we have s + H(ω) < h, which implies that H1 := max(s,ω)∈(I1∩S1)(s + H(ω)) < h. Thus, for any t ∈ (H1, h), J1(t) = 0, which implies that given ξh, J1(t) (h) lim h = 0, Pµ -a.s. (4.29) t↑h kΛt k

To deal with J2, we deﬁne

D1 := {ω : ∃u ∈ (0, δ2), such that h1B¯c ,ωui > 0}, and S2 = [h − δ2, h) × D1. (4.30)

Then,

J2(t)= h1B¯c ,ωt−si1s

21[0,h)(s)1H(ω)

D1 ={ω ∈ D : ∃u ∈ (0, δ2), such that h1B¯c ,ωui > 0} δ2 = ω ∈ D : h1B¯c ,ωui du > 0 Z0 δ2 ⊂ ω ∈ D : h1Bc ,ωui du > 0 . Z0

17 Thus,

δ2 Nx(D1) ≤Nx h1Bc ,ωui du > 0 Z0 δ2 = lim Nx 1 − exp −λ h1Bc ,ωui du λ→∞ Z0 δ2 c = lim − log Pδx exp −λ h1B , Xui du λ→∞ Z0 δ2 c = − log Pδx h1B ,ωui du = 0 . (4.33) Z0 Combining (4.33) and Assumption (H3), we get

21[0,h)(s)1H(ω)

2,h,P hfh, Xt i (h) lim h = 0, Pµ -a.s. (4.34) t↑h kΛt k

Note that given ξh, t 2,h,P 2,h hfh, Xt i := hfh,ωt−siN (ds, dω), Z0 ZD where N 2,h(ds, dω) is a Poisson random measure on [0, h) × D with intensity measure

∞ P 1[0,h)(s)1H(ω)

h−δ2 2,h,P h 2,h hfh, Xt i≤ǫh1, Λt i + 2kfk∞ h1,ωt−siN (ds, dω) 0 D t Z Z 2,h + 2kfk∞ h1B¯c ,ωt−siN (ds, dω) Zh−δ2 ZD h =ǫh1, Λt i + 2kfk∞ h1,ωt−si + 2kfk∞ h1B¯c ,ωt−si (s,ω)∈X(I2∩S1) (s,ω)∈X(I2∩S2) h =:ǫh1, Λt i + 2kfk∞J3(t) + 2kfk∞J4(t), where S1 and S2 are the set deﬁned in (4.27) and (4.30). It follows that

2,h,P |hfh, Xt i| J3(t) J4(t) h ≤ ǫ + 2kfk∞ h + 2kfk∞ h . (4.35) kΛt k kΛt k kΛt k

18 So, to prove (4.34), we only need to prove that

J3(t) (h) lim h = 0, Pµ -a.s., (4.36) t↑h kΛt k and J4(t) (h) lim h = 0, Pµ -a.s. (4.37) t↑h kΛt k Note that ∞ P 1[0,h)(s)1H(ω)

P P −yv(δ2/2,ξs) yδξs (Xδ2/2 6= 0) = 1 − yδξs (Xδ2/2 = 0) = 1 − e ≤ yv(δ2/2,ξs).

2,h Thus, N (S1) < ∞, a.s., which implies that (4.36). To prove (4.37) we only need to show that, given ξh,

∞ P yn(ξs, dy) yδξs (X ∈ dω)ds < ∞. (4.39) ZS2 Z0 In fact,

∞ P yn(ξs, dy) yδξs (X ∈ dω)ds ZS2 Z0 h ∞ δ2 P c ≤ yn(ξs, dy) yδξs h1B , Xui du > 0 ds Zh−δ2 Z0 Z0 h ∞ h δ2 1 2 P c ≤ yn(ξs, dy)ds + − log δξs h1B , Xui du = 0 y n(ξs, dy)ds Zh−δ2 Z1 Zh−δ2 Z0 Z0 δ2 c ≤Kh + Kh sup − log Pδx h1B , Xui du = 0 < ∞, x∈B(ξ ,δ1/2) 0 h− h Z i where in the second inequality, we used the fact that

δ2 δ2 P c P c yδξs h1B , Xui du > 0 =1 − exp y log δξs h1B ,ωui du = 0 Z0 Z0 n δ2 o P c ≤− y log δξs h1B , Xui du = 0 . 0 Z The proof is now complete. ✷

19 Proof of Theorem 3.7: Since {Xt,t ≥ 0} is a Hunt process, t → Xt is right continuous, which implies that X X lim t exists = lim t exists , t↑H kX k t∈Q↑H kX k t t where Q is the set of all rational numbers in [0, ∞). And, note that

H = inf{t ∈ Q : kXtk = 0}.

Thus, by Corollary 3.6 and Lemma 4.3, we get that

∞ h Xt (h) Λt Pµ lim exists = Pµ lim exists FH (dh) = 1. Q Q h t∈ ↑H kXtk 0 t∈ ↑h kΛt k h i Z h i Let V := lim Xt . Then, for any f ∈B+(E), by Lemma 4.3, t↑H kXtk b

hf, Xti Pµ[exp{−hf, V i}]=Pµ lim exp − t∈Q↑H kXtk h∞ n oi h (h) hf, Λt i = lim Pµ exp − FH (dh) Q h 0 t∈ ↑h kΛt k Z ∞ h = Πν [exp(−f(ξh−))] FH (dh). Z0

Thus, V is a Dirac measure of the form V = δZ and the law of Z satisﬁes (3.8). The proof is now complete. ✷

5 Examples

In this section, we will list some examples that satisfy Assumptions (H1) and (H2). The purpose of these examples is to show that Assumptions (H1) and (H2) are satisﬁed in a lot of cases. We will not try to give the most general examples possible.

Example 3 Suppose that Pt is conservative and preserves Cb(E). Let A be the inﬁnitesimal generator of Pt in Cb(E) and D(A) be the domain of A. Also assume that

Ψ(x, z)= −α(x)z + b(x)z2, where supx∈E α(x) ≤ 0 and infx∈E b(x) > 0 and 1/b ∈ D(A). Then by Remark 2.2, we know that Auumption (H1) is satisﬁed. One can check that

−1 b (ξt) t e− R0 (b(ξs)A(1/b)(ξs )) ds,t ≥ 0 b−1(x) 1/b is a positive martingale under Πx. Thus we deﬁne another probability measure Πx by

1/b −1 Πx b (ξ ) t t − R0 (b(ξs)A(1/b)(ξs )) ds = −1 e , t ≥ 0. Πx Ft b (x)

20 Let A1/b be the inﬁnitesimal generator of ξ under Π1/b. If −α(x) − b(x)A(1/b)(x) ∈ D(A1/b), then it follows from [4, (3.10) and Lemma 4.9] that w(t,x) exists and satisﬁes

1 β2eβ0t w(t,x) ≤ ect 0 , β0t 2 infx∈E b(x) (e − 1) where c, β0 are positive constants. Using this, one can check that Assumption (H2) is satisﬁed. This example shows that our result covers Delmas and H´enard [4, Corollary 4.14].

Now we give some examples of superprocesses, with general branching mechanisms, satisfying Assumptions (H1) and (H2). Recall that the general form of branching mechanism is given by ∞ Ψ(x, z)= −α(x)z + b(x)z2 + (e−yz − 1+ yz)n(x,dy). Z0 By (2.2), there exists K > 0, such that

∞ |α(x)| + b(x)+ (y ∧ y2)n(x,dy) ≤ K. Z0 Thus we have |Ψ(x, z)|≤ 3K(z + z2), x ∈ Rd. (5.1) In the next two examples, we always assume that E = Rd and that Ψ satisﬁes (2.7) and the following condition: for any M > 0, there exist c> 0 and γ0 ∈ (0, 1] such that

|Ψ(x, z) − Ψ(y, z)|≤ c|x − y|γ0 , x,y ∈ Rd, z ∈ [0, M]. (5.2)

By Remark 2.2, condition (2.7) implies that Assumption (H1) is satisﬁed. Therefore, in the following examples, we only need to check that Assumption (H2) is satisﬁed.

Example 4 Assume that the spatial motion ξ is a diffusion on Rd satisfying the conditions in Example 2. The branching mechanism Ψ is of the form in (2.1) and satisfies (2.7) and (5.2). Then the (ξ, Ψ)-superprocess X satisfies Assumptions (H1) and (H2).

We now proceed to prove the second assertion of the example above.

d d Lemma 5.1 For f ∈ Bb(R ) and x ∈ R , the function t → Ptf(x) is diﬀerentiable on (0, ∞). d d Furthermore, there exists a constant c such that for any t ∈ (0, 1], x ∈ R and f ∈Bb(R ), ∂ P f(x) ≤ ckfk t−1. (5.3) ∂t t ∞

Proof: For t ∈ (n,n + 1], Ptf(x)= Pt−n(Pn f)(x). Thus, we only need to prove the diﬀerentia- bility for t ∈ (0, 1]. It follows from [11, IV.(13.1)] that

c x y 2 ∂ − d −1 − 2| − | p(t,x,y) ≤ c t 2 e t . (5.4) ∂t 1

d Thus by the dominated convergence theorem we have that for all t ∈ (0, 1] and x ∈ R , ∂ ∂ Ptf(x)= p(t,x,y)f(y) dy, ∂t d ∂t ZR 21 and that for all t ∈ (0, 1], x ∈ Rd and bounded Borel function f on Rd,

∂ P f(x) ≤ c kfk t−1. ∂t t 3 ∞

✷ The proof is now complete. d Lemma 5.2 Assume that fs(x) is uniformly bounded in (s,x) ∈ [0, 1] × R , that is, there is a d constant L> 0 so that, for all s ∈ [0, 1] and x ∈ R , |fs(x)|≤ L. Then there is a constant c such that for any t ∈ (0, 1] and x,x′ ∈ Rd,

t t ′ ′ Pt−sfs(x) ds − Pt−sfs(x ) ds ≤ cL(|x − x |∧ 1). Z0 Z0

Proof: It follows from [11, IV.(13.1)] that there exist constants c1, c2 > 0 such that for all t ∈ (0, 1] and x,x′ ∈ Rd, c x y 2 − d+1 − 2| − | |∇xp(t,x,y)|≤ c1t 2 e t . (5.5) Thus c x y 2 c x′ y 2 ′ −1/2 ′ −d/2 − 4| − | − 4| − | p(t,x,y) − p(t,x ,y) ≤ c3((t |x − x |) ∧ 1)t e t + e t . (5.6)

Hence for any t ∈ (0, 1] and x,x ′ ∈ Rd,

t t 1 ′ −1/2 ′ ′ Pt−sfs(x) ds − Pt−sfs(x ) ds ≤ c5L s |x − x | ds ≤ c6L|x − x |. (5.7) Z0 Z0 Z0

✷

Lemma 5.3 Assume that fs(x) satisﬁes the following conditions:

d (i) There is a constant L so that, for all (s,x) ∈ [0, 1] × R , |fs(x)|≤ L.

(ii) For any t0 ∈ [0, 1], lims→t0 supx∈Rd |fs(x) − ft0 (x)| = 0.

(iii) There exist constants s0 ∈ (0, 1), C > 0 and γ ∈ (0, 1] such that for all s ∈ [0,s0] and x,x′ ∈ Rd with |x − x′|≤ 1,

′ ′ γ |fs(x) − fs(x )|≤ C|x − x | . (5.8)

t Then, t → 0 Pt−sfs(x) ds is diﬀerentiable on (0,s0), and for t ∈ [0,s0), R ∂ t t ∂ P f (x) ds = P f (x) ds + f (x). (5.9) ∂t t−s s ∂t t−s s t Z0 Z0 t d Proof: Let G(t,x) := 0 Pt−sfs(x) ds. First, we will show that for any x ∈ R , R t −1 lim t Pt−sfs(x) ds = f0(x). (5.10) t↓0 Z0 d Since f0 ∈ Cb(R ), we have lims→0 Psf0(x)= f0(x), which implies that

t t −1 −1 lim t Pt−sf0(x) ds = lim t Psf0(x) ds = f0(x). t→0 t→0 Z0 Z0 22 Thus, it suﬃces to prove that

t −1 lim t Pt−s(fs − f0)(x) ds = 0. (5.11) t→0 Z0 Notice that t −1 t |Pt−s(fs − f0)(x)| ds ≤ sup kfs − f0k∞ → 0, Z0 s≤t as t → 0. Thus, (5.10) is valid. For any 0

By (5.10), we have lim(II)= ft(x). (5.12) r↓0

Now we deal with part (I). For 0

Pt+r−sfs(x) − Pt−sfs(x) r

p(t + r − s,x,y) − p( t − s,x,y) = (fs(y) − fs(x)) dy Rd r Z 2 d c4|x−y| − 2 −1 − t s ≤ c 3 |fs(y) − fs(x)|(t − s) e − dy Rd Z 2 c |x−y| γ − d −1 − 4 ≤ c5 |x − y| (t − s) 2 e t−s dy d ZR γ/2−1 ≤ c6(t − s) . (5.13)

Thus, using the dominated convergence theorem, we get that, for any 0 ≤ t

t t Pt+r−sfs(x) − Pt−sfs(x) ∂ lim(I)= lim ds = Pt−sfs(x) ds. (5.14) r↓0 r↓0 r ∂t Z0 Z0 Combining (5.12) and (5.14), we get that

G(t + r, x) − G(t,x) t ∂ lim = Pt−sfs(x) ds + ft(x). r↓0 r ∂t Z0 Using similar arguments, we can also show that

G(t,x) − G(t − r, x) t ∂ lim = Pt−sfs(x) ds + ft(x). r↓0 r ∂t Z0 Thus, (5.9) follows immediately. The proof is now complete.

23 ✷ Recall that v(s, ·) is a bounded function and

t v(t + s,x)+ Pt−u(Ψs+u)(x) du = Pt(vs)(x), Z0 where Ψu(x)=Ψ(x, v(u, x)). (5.15)

Lemma 5.4 For any s> 0, there is a constant c(s) such that for t ∈ [0, 1/2) and x,y ∈ Rd,

|vt+s(x) − vt+s(y)|≤ c(s)|x − y|.

Moreover, c(s) is decreasing in s> 0.

1∧s Proof: Let e(s) := 2 . Note that t + e(s) ∈ (e(s), 1). Thus

t+e(s) v(t + s,x)+ Pt+e(s)−u(Ψ(·, vs−e(s)+u(·)))(x) du = Pt+e(s)(vs−e(s))(x). Z0 d It follows from (5.6) that there exists a constant c1 such that for all x,y ∈ R ,

|Pt+e(s)(vs−e(s))(x) − Pt+e(s)(vs−e(s))(y)| −1/2 ≤ckvs−e(s)k∞((t + e(s)) |x − y|) ∧ 1) −1/2 ≤ckvs−e(s)k∞(t + e(s)) |x − y| −1/2 ≤ckvs/2k∞(e(s)) |x − y|. (5.16)

Since v(s − e(s)+ u, x) ≤ v(s − e(s),x) ≤ v(s/2,x), we have for u> 0,

2 kΨ(·, vs−e(s)+u(·))k∞ ≤ 3K(kvs/2k∞ + kvs/2k∞).

d Applying Lemma 5.2, we get that there is a constant c2 > 0 such that for t ∈ [0, 1/2) and x,y ∈ R ,

t+e(s) t+e(s) Pt+e(s)−u(Ψ(·, vs−e(s)+u(·)))(x) du − Pt+e(s)−u(Ψ(·, vs−e(s)+u(·)))(y) du 0 0 Z Z 2 ≤c 23K(kvs/2k∞ + kvs/2k∞)(|x − y|∧ 1). (5.17)

The conclusions of the lemma now follow immediately from (5.16) and (5.17). ✷

Lemma 5.5 The function Ψu(x) given by (5.15) satisﬁes the following two properties:

(1) For any u0 > 0,

lim sup |Ψu(x) − Ψu0 (x)| = 0; u→u0 x∈Rd

′ (2) For t0 ∈ (0, 1), there exists a constant c > 0 such that for any |x − x | ≤ 1, s>t0 and t ∈ [0, 1/2], ′ ′ γ0 |Ψs+t(x) − Ψs+t(x )|≤ c|x − x | .

24 Proof: (1) For z1 < z2 ∈ [0, a], we can easily check that

|Ψ(x, z1) − Ψ(x, z2)| ∞ 2 2 −yz1 −yz2 ≤|α(x)||z1 − z2| + b(x)|z1 − z2 | + e + yz1 − e − yz2 n(x,dy) Z0 ∞ ≤K(1 + 2a)|z1 − z2| + (2 ∧ (ya))y| z1 − z2|n(x,dy) ≤ K(3 + 3 a)|z1 − z2|, (5.18) Z0 where in the second inequality above we use the fact that d | (e−x + x)| = 1 − e−x ≤ 2 ∧ x. dx

Thus, for |u − u0|≤ u0/2, we have that

|Ψu(x) − Ψu0 (x)|≤ 3K(1 + kvu0/2k∞)|vu(x) − vu0 (x)|. (5.19)

Thus, it suﬃces to show that t 7→ vt(x) is continuous on (0, ∞) uniformly in x. It follows from Lemma 5.4 that, for any t> 0, x 7→ vt(x) is uniformly continuous, thus

lim kPrvt − vtk∞ = 0. r↓0

For r> 0 and t> 0, we have that

r |vt(x) − vt+r(x)| ≤ |Prvt(x) − vt(x)| + | Pr−u(Ψt+u)(x) du| 0 Z 2 ≤ kPrvt − vtk∞ + 3K(kvtk∞ + kvtk∞)r → 0, r ↓ 0, where in the last inequality we used (5.1) and the fact that vt+u(x) ≤ vt(x). The proof of limr↓0 kvt − vt−rk∞ = 0 is similar and omitted. The proof of part (1) is now complete.

(2) For any s>t0, and t ∈ [0, 1/2], v(t + s,x) ≤ kvt0 k∞. By our assumption on Ψ, there exist c1 > 0 and γ0 ∈ (0, 1] such that for |x − y|≤ 1, s>t0 and t ∈ [0, 1/2],

γ0 |Ψ(x, vs+t(x)) − Ψ(y, vs+t(x))|≤ c1|x − y| .

By Lemma 5.4, there exists c2 = c2(t0) such that for s>t0 and t ∈ [0, 1/2],

|vs+t(x) − vs+t(y)|≤ c2|x − y|.

Thus, for |x − y|≤ 1, s>t0, and t ∈ [0, 1/2],

≤ |Ψ(x, vs+t(x)) − Ψ(y, vs+t(x))| + 3K(1 + kvt0 k∞)|vs+t(x) − vs+t(y)| γ0 ≤ c1|x − y| + 3K(1 + kvt0 k∞)c2|x − y| γ0 ≤ c3|x − y| . (5.20)

The proof of (2) is now complete. ✷

25 Lemma 5.6 The function t → vt(x) is diﬀerentiable in (0, ∞), and for any s> 0 and t ∈ [0, 1/2), ∂ w(t + s,x)= − ∂t vt+s(x) satisﬁes that ∂ t ∂ w(t + s,x)= − P (v )(x)+ P (Ψ )(x) du +Ψ (x). (5.21) ∂t t s ∂t t−u s+u t+s Z0

Moreover, t → w(t,x) is continuous and for any s0 > 0, sups>s0 supx∈Rd w(s,x) < ∞.

Proof: For any t,s > 0,

t v(t + s,x)+ Pt−u(Ψs+u)(x) du = Pt(vs)(x). Z0 Thus, combining Lemmas 5.1, 5.3 and 5.5, (5.21) follows immediately. For ﬁxed t ∈ (0, 1/2), we deal with the three parts on right hand side of (5.21) separately. Since t → v(t,x) is continuous, the function s → Ψt+s(x)=Ψ(x, v(t + s,x)) is continuous and, by (5.1), 2 sup |Ψt+s(x)|≤ 3K(kvt0 k∞ + kvt0 k∞) < ∞. (5.22) s>t0 By (5.26), ∂ −1 sup | Pt(vs)(x)|≤ c4kvt0 k∞t < ∞. (5.23) s>t0 ∂t

By (5.13) and Lemma 5.5 (2), we get that, for any s>t0,

t ∂ sup sup | Pt−u(Ψs+u)(x) du| < ∞. (5.24) d ∂t s>t0 x∈R Z0

Combining (5.22) -(5.24), we get that, for t0 > 0,

sup sup w(t + s,x) < ∞, s>t0 x∈Rd ✷ which implies that, for any s0 > 0, sups>s0 supx∈Rd w(s,x) < ∞.

Now we give an example of a superprocess with discontinuous spatial motion and general branching mechanism such that Assumptions (H1) and (H2) are satisﬁed.

d Example 5 Suppose that B = {Bt} is a Brownian motion in R and S = {St} is an independent subordinator with Laplace exponent ϕ, that is

Ee−λSt = e−tϕ(λ), t> 0,λ> 0.

d The process ξt = BSt is called a subordinate Brownian motion in R . Subordinate Brownian motions form a large class of L´evy processes. When S is an (α/2)-stable subordinator, that is, ϕ(λ)= λα/2, ξ is a symmetric α-stable process in Rd. Suppose that Ψ is of the form in (2.1) satisfying (2.7) and (5.2). Suppose further that ϕ satisﬁes the following conditions:

1 ϕ(r2) 1. 0 r dr < ∞.

2.R There exist constants δ ∈ (0, 2] and a1 ∈ (0, 1) such that

δ/2 a1λ ϕ(r) ≤ ϕ(λr), λ ≥ 1, r ≥ 1. (5.25)

26 then X satisﬁes Assumptions (H1) and (H2).

Now we proceed to prove the second assertion of the example above. The arguments are similar to that for the second assertion of Example 4. Without loss of generality, we will assume that ϕ(1) = 1. First we introduce some notation. Put Φ(r)= ϕ(r2) and let Φ−1 be the inverse function of Φ. For t> 0 and x ∈ Rd, we deﬁne

1 −1 1 −d ρ(t,x):=Φ + |x| + |x| . Φ−1(t−1) Φ−1(t−1) ! For t> 0, x ∈ Rd and β, γ ∈ R, we deﬁne

β −1 −1 −γ β d ργ (t,x):=Φ (t ) (|x| ∧ 1)ρ(t,x) , t> 0,x ∈ R .

Let p(t,x,y) = p(t,x − y) be the transition density of ξ and let {Pt : t ≥ 0} be the transition semigroup of ξ. It is well known that {Pt : t ≥ 0} satisﬁes the strong Feller property, that is, for d d any t> 0, Pt maps bounded Borel functions on R to bounded continuous functions on R . Now we list some other properties of the semigroup {Pt : t ≥ 0} which will be used later.

d d Lemma 5.7 For f ∈ Bb(R ) and x ∈ R , the function t → Ptf(x) is diﬀerentiable on (0, ∞). d d Furthermore, there exists a constant c such that for any t ∈ (0, 1], x ∈ R and f ∈Bb(R ), ∂ P f(x) ≤ ckfk t−1. (5.26) ∂t t ∞

Proof: For t ∈ (n,n + 1], Ptf(x)= Pt−n(Pn f)(x). Thus, we only need to prove the diﬀerentia- bility for t ∈ (0, 1]. It follows from [9, Lemma 3.1(a) and Theorem 3.4] that ∂ p(t,x) ≤ c ρ(t,x). (5.27) ∂t 1

By [9, Lemma 2.6(a)], we have

−1 ρ(t,x)dx < c2t , t ∈ (0, 1]. (5.28) d ZR Thus by the dominated convergence theorem we have that for all t ∈ (0, 1] and x ∈ Rd, ∂ ∂ Ptf(x)= p(t,x,y)f(y) dy, ∂t d ∂t ZR and that for all t ∈ (0, 1], x ∈ Rd and bounded Borel function f on Rd, ∂ P f(x) ≤ c kfk t−1. ∂t t 3 ∞

The proof is now complete. ✷

d Lemma 5.8 Assume that fs(x) is uniformly bounded in (s,x) ∈ [0, 1] × R , that is, there is a d constant L> 0 so that, for all s ∈ [0, 1] and x ∈ R , |fs(x)|≤ L. Then there is a constant c such that for any t ∈ (0, 1] and x,x′ ∈ Rd,

t t ′ ′ δ/2 Pt−sfs(x) ds − Pt−sfs(x ) ds ≤ cL(|x − x | ∧ 1). Z0 Z0

27 Proof: It follows from [9, Proposition 3.3] that there exists a constant c1 > 0 such that for all t ∈ (0, 1] and x,x′ ∈ Rd,

′ −1 −1 ′ ′ p(t,x) − p(t,x ) ≤ c1 (Φ (t )|x − x |) ∧ 1 t ρ(t,x)+ ρ(t,x ) . (5.29)

′ d Thus using (5.28) we get that for any t ∈ (0, 1] and x,x ∈ R , t t t ′ −1 −1 ′ Pt−sfs(x) ds − Pt−sfs(x ) ds ≤ c2L (Φ (s )|x − x |) ∧ 1 ds. (5.30) Z0 Z0 Z0 ′ ′ −1 When |x − x | < 1, Φ(|x − x | ) ≥ Φ(1) = 1. Thus, t 1 (Φ−1(s−1)|x − x′|) ∧ 1 ds ≤ |x − x′| Φ−1(s−1)ds + Φ(|x − x′|−1) −1. ′ −1 −1 Z0 Z(Φ(|x−x | )) It is well known that ϕ, the Laplace exponent of a subordinator, satisﬁes

ϕ(λr) ≤ λϕ(r), λ ≥ 1,r > 0.

Using this, we immediately get that

Φ−1(λr) ≥ λ1/2Φ−1(r), λ ≥ 1,r > 0.

For s ∈ [(Φ(|x − x′|−1)−1, 1], by taking r = s−1 and λ = sΦ(|x − x′|−1) in the display above, we get

Φ−1(s−1) ≤ |x − x′|−1s−1/2(Φ(|x − x′|−1))−1/2.

Therefore 1 |x − x′| Φ−1(s−1)ds ′ −1 −1 Z(Φ(|x−x | )) 1 ′ −1 −1/2 −1/2 ′ −1 −1/2 ≤ (Φ(|x − x | )) s ds ≤ c3(Φ(|x − x | )) . ′ −1 −1 Z(Φ(|x−x | )) Consequently for all t ∈ (0, 1] and x,x′ ∈ Rd with |x − x′| < 1, we have

t −1 −1 ′ ′ −1 −1/2 ((Φ (t )|x − x |) ∧ 1) ds ≤ c4(Φ(|x − x | )) . Z0 By taking r = 1 and λ = |x − x′|−1 in (5.25), we get

′ −δ ′ −1 a1|x − x | ≤ Φ(|x − x | ).

Thus for all t ∈ (0, 1] and x,x′ ∈ Rd with |x − x′| < 1, we have

t −1 −1 ′ −1/2 ′ δ/2 ((Φ (s )|x − x |) ∧ 1) ds ≤ c4a1 |x − x | . (5.31) Z0 Combining (5.30) and (5.31), we immediately get the desired conclusion. ✷

Lemma 5.9 Assume that fs(x) satisﬁes the following conditions:

d (i) There is a constant L so that, for all (s,x) ∈ [0, 1] × R , |fs(x)|≤ L.

28 (ii) For any t0 ∈ [0, 1], lims→t0 supx∈Rd |fs(x) − ft0 (x)| = 0.

(iii) There exist constants s0 ∈ (0, 1), γ ∈ (0, δ/2] and C > 0 such that for all s ∈ [0,s0] and x,x′ ∈ Rd with |x − x′|≤ 1,

′ ′ γ |fs(x) − fs(x )|≤ C|x − x | . (5.32)

t Then, t → 0 Pt−sfs(x) ds is diﬀerentiable on (0,s0), and for 0 ≤ t

Using the same arguments as those leading to (5.10), we get

t −1 lim t Pt−sfs(x) ds = f0(x), t↓0 Z0 which implies that lim(II)= ft(x). (5.34) r↓0

Now we deal with part (I). For 0

Pt+r−sfs(x) − Pt−sfs(x) r

p(t + r − s,x,y) − p( t − s,x,y) = (fs(y) − fs(x)) dy d r ZR

≤ c 3 |fs(y) − fs(x)|ρ(t − s,x − y) dy d ZR γ ≤ c4 ρ0 (t − s,x − y) dy Rd Z −1 −1 −1 −γ ≤ c5(t − s) Φ ((t − s) ) , (5.35) where in the last inequality we used [9, Lemma 2.6(a)]. It follows from [9, Lemma 2.3] that

t −1 −1 −1 −γ −1 −1 −γ (t − s) Φ ((t − s) ) ds ≤ c6Φ (t ) . (5.36) Z0 Thus, using the dominated convergence theorem, we get that, for any 0 ≤ t

t t Pt+r−sfs(x) − Pt−sfs(x) ∂ lim(I)= lim ds = Pt−sfs(x) ds. (5.37) r↓0 r↓0 r ∂t Z0 Z0 29 Combining (5.34) and (5.37), we get that

G(t + r, x) − G(t,x) t ∂ lim = Pt−sfs(x) ds + ft(x). r↓0 r ∂t Z0 Using similar arguments, we can also show that

G(t,x) − G(t − r, x) t ∂ lim = Pt−sfs(x) ds + ft(x). r↓0 r ∂t Z0 Thus, (5.33) follows immediately. The proof is now complete. ✷

Lemma 5.10 For any s> 0, there is a constant c(s) such that for t ∈ [0, 1/2) and x,y ∈ Rd,

δ/2 |vt+s(x) − vt+s(y)|≤ c(s)|x − y| .

Moreover, c(s) is decreasing in s> 0.

Proof: The proof of this lemma is similar as that of Lemma 5.4. We use Lemma 5.8 instead of Lemma 5.2. Here we omit the details. ✷

Lemma 5.11 The function Ψu(x) satisﬁes the following two properties:

(1) For any u0 > 0,

lim sup |Ψu(x) − Ψu0 (x)| = 0; u→u0 x∈Rd

′ (2) For t0 ∈ (0, 1), there exists a constant c> 0 and γ1 ∈ (0, δ/2] such that for any |x − x |≤ 1, s>t0 and t ∈ [0, 1/2], ′ ′ γ1 |Ψs+t(x) − Ψs+t(x )|≤ c|x − x | .

Proof: The proof of part (1) is exactly the same as that of part (1) of Lemma 5.5. Using arguments similar to that in the proof of part (2) of Lemma 5.5 and using Lemma 5.10 instead of Lemma 5.4, we can get the result in part (2). Here we omit the details. ✷

Lemma 5.12 The functiom t → vt(x) is diﬀerentiable in (0, ∞), and for any s> 0 and t ∈ [0, 1/2), ∂ w(t + s,x)= − ∂t vt+s(x) satisﬁes that

∂ t ∂ w(t + s,x)= − P (v )(x)+ P (Ψ )(x) du +Ψ (x). (5.38) ∂t t s ∂t t−u s+u t+s Z0

Moreover, t → w(t,x) is continuous and for any s0 > 0, sups>s0 supx∈Rd w(s,x) < ∞.

Proof: Combining Lemmas 5.7, 5.9 and 5.11, and using arguments similar to that in the proof of Lemma 5.6, Lemma 5.12 follows immediately. ✷

Remark 5.13 Actually, by the same arguments and the results from [9], one check that in the example above, we could have replaced the subordinate Brownian motion by the non-symmetric jump process considered there, which contains the non-symmetric stable-like process discussed in [2].

30 Let L be as in Example 2. Let E be a bounded smooth domain in Rd and let p(t,x,y) be the Dirichlet heat kernel of L in E. It follows from [7, Theorem 2.1, p. 247] that there exist ci > 0, i = 1, 2, 3, 4, such that for all t ∈ (0, 1],

c x y 2 ∂ − d −1 − 2| − | p(t,x,y) ≤ c t 2 e t , and ∂t 1 2 d+1 c4|x−y| − 2 − t |∇ xp(t,x,y)|≤ c3t e . Using these instead of (5.4) and (5.5), and repeating the arguments for Example 4, we can get the following example.

Example 6 Assume that E be is bounded smooth domain in Rd and that the spatial motion is ξE, which is the diffusion ξ of Example 2 killed upon exiting E. The branching mechanism Ψ is of the form in (2.1) and satisfies (2.7) and (5.2) on E. Then the (ξE, Ψ)-superprocess X satisfies Assumptions (H1) and (H2).

References

[1] Abraham, R. and Delmas, J.-F.: Williams decomposition of the Lévy continuum random tree and simultaneous extinction probability for populations with neutral mutations. Stochastic Process. Appl. 119 (2009), 1124–1143. [2] Chen, Z.-Q. and Zhang, X.: Heat kernels and analyticity of non-symmetric jump diffusion semigroups. Probab. Theory Relat. Fields, DOI 10.1007/s00440-015-0631-y. [3] Dawson, D. A. : Measure-Valued Markov Processes. Springer-Verlag, 1993. [4] Delmas, J. F. and Hénard, O.: A Williams decomposition for spatially dependent super-processes. Electron. J. Probab. 18 (2013), 1–43. [5] Dynkin, E. B.: Superprocesses and partial differential equations. Ann. Probab. 21 (1993), 1185–1262. [6] Dynkin, E. B. and Kuznetsov, S. E.: N-measure for branching exit Markov system and their applications to differential equations. Probab. Theory Rel. Fields 130 (2004), 135–150. [7] Garroni, M. G. and Menaldi, J.-L.: Green functions for second order parabolic integro-differential prob- lems. Longman, Harlow, 1992. [8] El Karoui, N. and Roelly, S.: Propriétés de martingales, explosion et représentation de Lévy-Khintchine d’une classe de processus de branchment àvaleurs mesures. Stoch. Proc. Appl. 38 (1991), 239–266. [9] Kim, K., Song, R. and Vondracek, Z.: Heat kernels of non-symmetric jump processes: beyond the stable case. arXiv:1606.02005 [10] Kyprianou, A. E. : Introductory Lectures on Fluctruations of Le´vy Processes with Applications. Springer. [11] Ladyzenskaja, O. A., Solonnikov, V. A. and Ural’ceva, N. N.: Linear and Quasi-linear Equations of Parabolic Type. American Math. Soc., Providence, Rhode Island, 1968. [12] Li, Z.: Skew convolution semigroups and related immigration processes. Theory Probab. Appl. 46 (2003), 274–296. [13] Li, Z.: Measure-Valued Branching Markov Processes. Springer, Heidelberg, 2011. [14] Stroock, D. W.: Probability Theory. An Analytic View. 2nd ed. Cambridge University Press, Cambridge, 2011. [15] Tribe, R.: The behavior of superprocesses near extinction. Ann. Probab. 20 (1992), 286-311.

31 Yan-Xia Ren: LMAM School of Mathematical Sciences & Center for Statistical Science, Peking University, Beijing, 100871, P.R. China. Email: [email protected] Renming Song: Department of Mathematics, University of Illinois, Urbana, IL 61801, U.S.A. Email: [email protected] Rui Zhang: School of Mathematical Sciences, Capital Normal University, Beijing, 100048, P.R. China. Email: [email protected]