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So the un- derline topology is a very important factor to consider when establishing large deviations. In the new preprint [6], the authors applied the criteria in [5] to obtain a large deviation principle for stochastic partial differential equations driven by Poisson random measures on some nuclear spaces where tightness of measures are relatively easy to establish. Often the real state space of the solution of a stochastic partial differential equation is a smaller Hilbert space contained in the nuclear space. This makes it interesting to directly consider large deviations on the actual state space. The aim of this paper is to establish a large deviation principle for a fully non-linear stochastic evolution equation driven by both Brownian motions and Poisson random mea- sures like (1.1) on a given Hilbert space H. We will apply the criteria in [5]. Among other things , we need to prove the tightness of the solutions of random perturbations of the equation (1.1) on the space D([0, T ]; H). To this end, we split the time interval [0, T ] into [0, t0] and [t0, T ] for a given arbitrarily small positive constant t0 > 0 because two different treatments are needed for these two intervals. This also make the proofs involved. Finally we mention that large deviations for L´evy processes on Banach spaces and large deviations for solutions of stochastic differential equations driven by Poisson measures in finite dimensions were studied in [1], [2]. The rest of the paper is organized as follows. In Section 2, we recall the general criteria of large deviations obtained in [5] and formulate precisely the stochastic evolution equations we are going to study. Section 3 is devoted to the proof of the large deviation principle. A number of preparing propositions and lemmas will be proved in this section. We end this section with some notations. For a topological space , denote the corre- E sponding Borel σ-field by ( ). We will use the symbol ” = ” to denote convergence in B E d ⇒ distribution. Let N, N0, R, R+, R denote the set of positive integers, non-negative inte- gers, real numbers, positive real numbers, and d-dimensional real vectors respectively. For a Polish space X, denote by C([0, T ], X), D([0, T ], X) the space of continuous functions and right continuous functions with left limits from [0,T] to X respectively. For a metric space , E denote by Mb( ), Cb( ) the space of real valued bounded ( )/ (R)-measurable maps and real valued boundedE continuousE functions respectively. ForB Ep >B0, a measure ν on , and E a Hilbert space H, denote by Lp( , ν; H) the space of measurable functions f from to H p E E such that E f(v) ν(dv) < , where H is the norm on H. For a function x : [0, T ] , we use the notationk k x and x∞(t) interchangeablyk·k for the evaluation of x at t [0, T ]. Similar→E R t ∈ convention will be followed for stochastic processes. We say a collection Xǫ of -valued random variables is tight if the probability distributions of Xǫ are tight in{ (} ) (theE space of probability measures on ). P E E 2 Preliminaries
In the first part of this section, we will recall the general criteria for a large deviation principle given in [5]. To this send, we closely follow the framework and the notations in [6] and [5]. In the second part, we will precisely formulate the stochastic evolution equations we will study.
2 2.1 Large Deviation Principle Let Xǫ,ǫ > 0 Xǫ be a family of random variables defined on a probability space { }≡{ } (Ω, , P) and taking values in a Polish space (i.e., a complete separable metric space) . DenoteF expectation with respect to P by E. The theory of large deviations is concernedE with events A for which probability P(Xǫ A) converge to zero exponentially fast as ǫ 0. ∈ → The exponential decay rate of such probabilities is typically expressed in terms of a “rate function” I mapping into [0, ]. E ∞ Definition 2.1 (Rate function) A function I : [0, ] is called a rate function on , E → ∞ E if for each M < the level set x : I(x) M is a compact subset of . For A ( ), .∞ { ∈E ≤ } E ∈ B E we define I(A) = infx∈A I(x).
Definition 2.2 (Large deviation principle) Let I be a rate function on . The sequence Xǫ is said to satisfy the large deviation principle on with rate function IEif the following { } E two conditions hold. a. Large deviation upper bound. For each closed subset F of , E lim sup ǫ log P(Xǫ F ) I(F ). ǫ→0 ∈ ≤ − b. Large deviation lower bound. For each open subset G of , E lim sup ǫ log P(Xǫ G) I(G). ǫ→0 ∈ ≥ − If a sequence of random variables satisfies a large deviation principle with some rate function, then the rate function is unique.
2.2 Poisson Random Measure and Brownian Motion 2.2.1 Poisson Random Measure Let X be a locally compact Polish space. Let (X) be the space of all measures ν MF C on (X, (X)) such that ν(K) < for every compact K in X. Endow F C(X) with the weakestB topology such that for every∞ f C (X) (the space of continuousM functions with ∈ c compact supports), the function ν f, ν = X f(u)dν(u), ν F C (X) is continuous. This topology can be metrized such→ that h i (X) is a Polish∈ space M (see e.g. [5]). Fix F CR T (0, ) and let X = [0, T ] X. Fix a measureM ν (X), and let ν = λ ν, where ∈ ∞ T × ∈MF C T T ⊗ λT is Lebesgue measure on [0, T ]. We recall that a Poisson random measure n on XT with mean measure (or intensity measure) ν is a (X ) valued random variable such that for each B (X ) with T MF C T ∈ B T νT (B) < , n(B) is Poisson distributed with mean νT (B) and for disjoint B1, , Bk (X ), n(∞B ), , n(B ) are mutually independent random variables (cf. [13]). Denote··· by∈ B T 1 ··· k P the measure induced by n on ( F C(XT ), ( F C(XT ))). Then letting M = F C(XT ), P is the unique probability measureM on (M,B (MM)) under which the canonicalM map, N : . M M, N(m) = m, is a Poisson randomB measure with intensity measure ν . With → T applications to large deviations in mind, we also consider, for θ > 0, probability measures P on (M, (M)) under which N is a Poissson random measure with intensity θν . The θ B T corresponding expectation operators will be denoted by E and Eθ, respectively. Let Y = X [0, ) and Y = [0, T ] Y. Let M¯ = (Y ) and let P¯ be the unique T F C T . probability measure× ∞ on (M¯ , (M¯ )) under× which the canonicalM map, N¯ : M¯ M¯ , N¯(m) = m, B → 3 is a Poisson random measure with intensity measureν ¯T = λT ν λ∞, with λ∞ being Lebesgue measure on [0, ). The corresponding expectation operator⊗ ⊗ will be denoted by E¯. . Let = σ N¯((0,s] A):0∞ s t, A (Y) , and let ¯ denote the completion under P¯. Ft { × ≤ ≤ ∈ B } Ft We denote by ¯ the predictable σ-field on [0, T ] M¯ with the filtration ¯ :0 t T on P × {Ft ≤ ≤ } (M¯ , (M¯ )). Let ¯ be the class of all ( ¯ (X))/ [0, )-measurable maps ϕ : XT M¯ [0, B). For ϕ A¯, define a counting processP ⊗ B N ϕ onB X ∞by × → ∞ ∈ A T N ϕ((0, t] U)= 1 (r)N¯(dsdxdr), t [0, T ], U (X). (2.2) × [0,ϕ(s,x)] ∈ ∈ B Z(0,t]×U Z(0,∞) N ϕ is the controlled random measure, with ϕ selecting the intensity for the points at location x and time s, in a possibly random but non-anticipating way. When ϕ(s, x, m¯ ) θ (0, ), we write N ϕ = N θ. Note that N θ has the same distribution with respect to ≡P¯ ∈ ∞ as N has with respect to Pθ.
2.2.2 PRM and BM Denote C([0, T ], R∞) by W, where R∞ is the infinite product space of the real line R and endowed with the product topology. Let V = W M . Then let the mapping N : V M be × ∞ → defined by N(w, m)= m for (w, m) V, and let β =(βi)i=1 be defined by βi(w, m)= wi for (w, m) V. Define the σ filtration∈ := σ N((0,s] A), β (s):0 s t, A (Y), i ∈ − Gt { × i ≤ ≤ ∈ B ≥ 1 . For every θ> 0, Pθ denotes the unique probability measure on (V, (V)) such that : } ∞ B (a) (βi)i=1 is an i.i.d. family of standard Brownian motions. (b) N is a PRM with intensity measure θνT .
If controlled Poisson random measure is also considered, we set V¯ := W M¯ , and let the × mapping N¯ : V¯ M¯ be defined by N¯(¯w, m¯ )=m ¯ for (¯w, m¯ ) V¯ accordingly. Analogously, → ∈ we define (P¯θ, ¯t). We denote by ¯t the P¯ completion of ¯t and ¯ the predictable σ filed on [0,G T ] V¯ with the filtration{F } ¯ − on (V¯, B(V¯)).{G Let} ¯ beP the class of all − × {Ft} A ( ¯ (X))/ [0, )-measurable maps ϕ : X V¯ [0, ). Define l : [0, ) [0, ) by P ⊗ B B ∞ T × → ∞ ∞ → ∞ l(r)= r log r r +1, r [0, ). − ∈ ∞ For any ϕ ¯ the quantity ∈ A
LT (ϕ)= l(ϕ(t, x, ω))νT (dtdx) (2.3) ZXT is well defined as a [0, ]-valued random variable. Let H be a separable∞ Hilbert space. Define function space
T := ψ : ψ is ¯ (H) measurable and ψ(s) 2 ds < , a.s. P¯ . (2.4) L2 { P \ B k kH ∞ − } Z0 Set = ¯. Define L˜ (ψ) := 1 T ψ(s) 2 ds for ψ , and L¯ (u) := L˜ (ψ)+ U L2 × A T 2 0 k kH ∈ L2 T T L (ϕ) for u =(ψ,ϕ). T R
4 2.3 A General Criteria In this section, we recall a general criteria for a large deviation principle established in [5]. ǫ Let ǫ>0 be a family of measurable maps from V¯ to U, where V¯ is introduced in Section 2.2.1{G and} U is some Polish space. We present below a sufficient condition for large deviation −1 principle (LDP in abbreviation) to hold for the family Zǫ = ǫ(√ǫβ, ǫN ǫ ), as ǫ 0. G → Define
SN = g : X [0, ): L (g) N , (2.5) { T → ∞ T ≤ } and
S˜N = f : L2([0, T ]: H): L˜ (f) N . (2.6) { T ≤ } A function g SN can be identified with a measure νg M, defined by ∈ T ∈ νg (A)= g(s, x)ν (dsdx), A (X ). T T ∈ B T ZA This identification induces a topology on SN under which SN is a compact space, see the Appendix of [6]. Throughout we use this topology on SN . Set S¯N = S˜N SN . Define ¯N × S = N≥1 S , and let
S N = u =(ψ,ϕ) : u(ω) S¯N , P¯ a.e. ω , U { ∈ U ∈ } where is introduced in Section 2.2.2. U ǫ The following condition will be sufficient to establish a LDP for a family Z ǫ>0 defined −1 by Zǫ = ǫ(√ǫβ, ǫN ǫ ). { } G Condition 2.1 There exists a measurable map 0 : V U such that the following hold. a. For N N, let (f ,g ), (f,g) S¯N be suchG that→(f ,g ) (f,g) as n . Then ∈ n n ∈ n n → →∞ · · 0( f (s)ds,νgn ) 0( f(s)ds,νg ) in U. G n T → G T Z0 Z0 N b. For N N, let uǫ =(ψǫ,ϕǫ), u =(ψ,ϕ) be such that uǫ converges in distribution to u as ǫ 0∈. Then ∈ U → · · −1 ǫ(√ǫβ + ψ (s)ds, ǫN ǫ ϕǫ ) 0( ψ(s)ds,νϕ). G ǫ ⇒ G T Z0 Z0 U S S 0 · g U For φ , define φ = (f,g) : φ = ( 0 f(s)ds,νT ) . Let I : [0, ] be defined by∈ { ∈ G } → ∞ R
I(φ) = inf L¯T (q) , φ U. (2.7) q=(f,g)∈Sφ{ } ∈
By convention, I(φ)= if Sφ = . The following criteria∞ was established∅ in [5].
−1 Theorem 2.3 For ǫ > 0, let Zǫ be defined by Zǫ = ǫ(√ǫβ, ǫN ǫ ), and suppose that Condition 2.1 holds. Then I defined as in (2.7) is a rateG function on U and the family Zǫ satisfies a large deviation principle with rate function I. { }ǫ>0 5 For applications, the following strengthened form of Theorem 2.3 is useful. Let Kn ∞ { ⊂ X, n =1, 2, be an increasing sequence of compact sets such that n=1Kn = X. For each n let ···} ∪ . ¯ = ϕ ¯ : for all (t, ω) [0, T ] M¯ , n ϕ(t, x, ω) 1/n if x K Ab,n { ∈ A ∈ × ≥ ≥ ∈ n and ϕ(t, x, ω)=1 if x Kc , ∈ n} and let ¯ = ∞ ¯ . Define ˜N = N (ψ,φ): φ ¯ . Ab ∪n=1Ab,n U U ∩{ ∈ Ab} Theorem 2.4 Suppose Condition 2.1 holds with N replaced by ˜N . Then the conclusions U U of Theorem 2.3 continue to hold .
2.4 SPDEs In this section we introduce the stochastic partial differential equations (SPDEs in addrevi- ation) that will be studied in this paper. Let H, V be two separable Hilbert spaces such that V is continuously, densely imbedded in H. Identifying H with its dual we have V H = H′ V ′, ⊂ ∼ ⊂ where V ′ stands for the topological dual of V . Let be a bounded linear operator from V ′ A to V satisfying the following coercivity hypothesis: There exist constants α> 0 and λ0 0 such that ≥ 2 u,u + λ u 2 α u 2 , for all u V. (2.8) hA i 0k kH ≥ k kV ∈ Example 2.5 Let H = L2(D), where D Rd is a bounded domain, and set ⊂ k·k 1,2 ∞ V = H0 (D)= C0 (D) , ∞ where C0 (D) is the space of infinite differentiable functions with compact supports and the norm is defined as follows 2 2 2 f := f 2 + f 2 . k k k kL k∇ kL Denote by a(x)=(aij(x)) a matrix-valued function on D satisfying the uniform ellipticity condition: 1 I a(x) cI for some constant c (0, ). c d ≤ ≤ d ∈ ∞ Let b(x) be a vector field on D with b Lp(D) for some p>d. Define ∈ u = div(a(x) u(x)) + b(x) u(x). A − ∇ ·∇ Then (2.8) is fulfilled for (H,V, ). A Example 2.6 Stochastic evolution equations associated with fractional Laplacian:
dYt = ∆αYtdt + dLt, (2.9) Y = h H, (2.10) 0 ∈ where ∆ denotes the generator of the symmetric α-stable process in Rd, 0 <α 2. ∆ is α ≤ α called the fractional Laplace operator. Lt stands for a L´evy process. It is well known that the Dirichlet form associated with ∆α is given by (u(x) u(y))(v(x) v(y)) (u, v)= K(d,α) − d+α − dxdy, E d d x y Z ZR ×R | − | 6 2 2 d u(x) u(y) D( )= u L (R ): | − d+α | dxdy < , E { ∈ d d x y ∞} Z ZR ×R | − | d+2 απ d+α α where K(d,α) = α2α−3π− 2 sin( )Γ( )Γ( ). We choose H = L2(Rd), and V = D( ) 2 2 2 E with the inner product
βt = βk(t)ek Xk=1 where β (t),k 1 are independent standard Brownian motions. k ≥ Denote by L2(H) the space of all Hilbert-Schmidt operators from H to H. Let σ : [0, T ] H L (H), G : [0, T ] H X H be maps satisfying the following conditions: × → 2 × × → Condition 2.2 There exists K( ) L1([0, T ], R+) such that (1)(Growth) For· all∈ t [0, T ], and u H, ∈ ∈ σ(t, u) 2 + G(t, u, v) 2 ν(dv) K(t)(1 + u 2 ); k kL2(H) k kH ≤ k kH ZX (2) (Lipschitz) For all t [0, T ], and u , u H, ∈ 1 2 ∈ σ(t, u ) σ(t, u ) 2 + G(t, u , v) G(t, u , v) 2 ν(dv) K(t) u u 2 . k 1 − 2 kL2(H) k 1 − 2 kH ≤ k 1 − 2kH ZX Consider the following stochastic evolution equation:
t t t −1 Xǫ = Xǫ Xǫds + √ǫ σ(s,Xǫ)dβ(s)+ ǫ G(s,Xǫ , v)N ǫ (dsdv). (2.11) t 0 − A s s s− Z0 Z0 Z0 ZX Here the precise definition of the solution to (2.11) is as follows. e
Definition 2.7 Let (V¯, (V¯), P¯, ¯t ) be the filtered probability space described in Section B ¯ {F } ¯ 2 2.2. Suppose that X0 is a 0-measurable H-valued random variable such that E X0 H < . A stochastic process Xǫ F defined on V¯ is said to be a H-valued solution tok (2.11)k with∞ { t }t∈[0,T ] initial value X0, if ǫ ¯ a) Xt is a H-valued t-measurable random variable, for all t [0, T ]; b) Xǫ D([0, T ],H) F L2([0, T ],V ) a.s.; ∈ ∈ ∩
7 c) For all t [0, T ], every φ V , ∈ ∈ t t ǫ ǫ ǫ Xt ,φ = X0,φ Xs,φ ds + √ǫ σ(s,Xs)dβ(s),φ h i h i − 0 hA i h 0 i t Z Z −1 +ǫ G(s,Xǫ , v),φ N ǫ (ds,dv), a.s.. (2.12) h s− i Z0 ZX Definition 2.8 (Pathwise uniqueness) We say thate the H-valued solution for the stochas- tic evolution equation (2.11) has the pathwise uniqueness if any two H-valued solutions X and X′ defined on the same filtered probability space with respect to the same Poisson random measure and Brownian motion starting from the same initial condition X0 coincide almost surely.
3 Large Deviation Principle
ǫ Assume X0 is deterministic. Let X be the H-valued solution to (2.11) with initial value X . In this section, we establish an LDP for Xǫ under suitable assumptions. 0 { } We begin by introducing the map 0 that will be used to define the rate function and also G ¯N ¯N used for verification of Condition 2.1. Recall that S = N≥1 S , where S is defined in last section. As a first step we show that under the conditions below, for every q = (f,g) S, ∈ the deterministic integral equation S
t t t Xq = X Xqds + σ(s, Xq)f(s)ds + G(s, Xq, v)(g(s, v) 1)ν(dv)ds (3.13) t 0 − A s s s − Z0 Z0 Z0 ZX hase a unique continuouse solution. Heree q =(f,g) plays the rolee of a control. Let G(t, u, v) G(t, v) = sup k kH , (t, v) [0, T ] X. k k0,H 1+ u ∈ × u∈H k kH G(t, u , v) G(t, u , v) G(t, v) = sup k 1 − 2 kH , (t, v) [0, T ] X. k k1,H u u ∈ × u1,u2∈H,u16=u2 k 1 − 2kH i Condition 3.1 (Exponential Integrability) For i = 0, 1, there exists δ1 > 0 such that for all E ([0, T ] X) satisfying ν (E) < , the following holds ∈ B × T ∞
i 2 eδ1kG(s,v)ki,H ν(dv)ds < . ∞ ZE
Remark 1 Condition 3.1 implies that, for every δ2 > 0 and for all E ([0, T ] X) satisfying ν (E) < , ∈ B × T ∞ eδ2kG(s,v)k0,H ν(dv)ds < . ∞ ZE
Now recall the following inequalities from [6], which will be used later. a) For a, b, σ (0, ), there exists C(σ) only depending on σ, such that ∈ ∞ 1 1 ab C(σ)eσa + (b log b b +1)= C(σ)eσa + l(b); (3.14) ≤ σ − σ
8 b) For each β > 0 there exists c (β) > 0, such that c (β) 0 as β and 1 1 → →∞ x 1 c (β)l(x) whenever x 1 β; | − | ≤ 1 | − | ≥ c) For each β > 0 there exists c (β) < , such that 2 ∞ x 1 2 c (β)l(x) whenever x 1 β. | − | ≤ 2 | − | ≤ The following lemma was proved in [6].
Lemma 3.1 Under Condition 2.2 and Condition 3.1, for i =0, 1 and every N N, ∈
N 2 Ci,2 := sup G(s, v) i,H(g(s, v)+1)ν(dv)ds < , (3.15) N k k ∞ g∈S ZXT
N Ci,1 := sup G(s, v) i,H g(s, v) 1 ν(dv)ds < . (3.16) N k k | − | ∞ g∈S ZXT We also need the following lemma whose proof can be found in Chapter III of [17].
Lemma 3.2 Assume that
2 ′ 1 f = f1 + f2, f1 L ([0, T ],V ), f2 L ([0, T ],H), u H, ∈ ∈ 0 ∈ then there exists a unique function u (denote by u′ its derivative) which satisfies
u L2([0, T ],V ) C([0, T ],H),u′ L2([0, T ],V ′)+ L1([0, T ],H), ∈ ∩ t ∈t u ,φ = u ,φ u ,φ ds + f ,φ ds, φ V, h t i h 0 i − 0 hA s i 0 h s i ∀ ∈ d u,u =2 u′,u . dt h i h i R R Lemma 3.3a). If Y C([0, T ],H), for any q =(f,g) S, then ∈ ∈ σ( ,Y ( ))f( ) L1([0, T ],H), G( ,Y ( ), v)(g( , v) 1)ν(dv) L1([0, T ],H); · · · ∈ · · · − ∈ ZX b). If Y C([0, T ],H), n 1 with C = sup sup Y (s) < , then n ∈ ≥ n s∈[0,T ] k n kH ∞ T T CN := sup sup G(s,Yn(s), v)(g(s, v) 1)ν(dv) H ds + σ(s,Yn(s))f(s) Hds ¯N n k − k k k q=(f,q)∈S Z0 ZX Z0 < . h i e ∞ Proof: Since
t G(s,Y (s), v)(g(s, v) 1)ν(dv) H ds 0 k X − k Z t Z G(s,Y (s), v)(g(s, v) 1) ν(dv)ds ≤ k − kH Z0 ZX t G(s,Y (s), v) = k kH g(s, v) 1 (1 + Y (s) )ν(dv)ds 1+ Y (s) | − | k kH Z0 ZX k kH 9 t (1 + sup Y (s) ) G(s, v) g(s, v) 1 ν(dv)ds, ≤ k kH k k0,H| − | s∈[0,T ] Z0 ZX and by Condition 2.2,
t t
σ(s,Y (s))f(s) H ds σ(s,Y (s)) L2(H) f(s) Hds 0 k k ≤ 0 k k k k Z Z t t 2 2 σ(s,Y (s)) L2(H)ds + f(s) H ds ≤ 0 k k 0 k k Z t t Z 2 2 K(s) Y (s) H ds + f(s) H ds ≤ 0 k k 0 k k Z T Z T 2 2 sup Y (s) H K(s)ds + f(s) Hds, ≤ s∈[0,T ] k k 0 0 k k Z Z the lemma follows from Lemma 3.1.
Theorem 3.1 Fix q = (f,g) S. Suppose Condition 2.2 and Condition 3.1 hold. Then ∈ there exists a unique Xq C([0, T ],H) such that for every φ V , ∈ ∈ t t q e q ∗ q Xt ,φ = X0,φ Xs , φ ds + σ(s, Xs )f(s),φ ds h i h i − 0 h A i 0 h i Z t Z e + e G(s, Xq, v),φ (g(s, ve) 1)ν(dv)ds. (3.17) h s i − Z0 ZX Moreover, for fixed N N, there exists C > 0 suche that ∈ N T q 2 q 2 sup sup Xs H + Xs V ds CN . (3.18) q∈SN s∈[0,T ] k k 0 k k ≤ Z e e Proof: Existence of solution: Let Y0(t) := X0, t 0. Suppose Yn−1 has been defined, by ≥ 2 Lemma 3.2 and Lemma 3.3, there exists a unique function Yn L ([0, T ],V ) C([0, T ],H) such that ∈ ∩
t t Yn(t),φ = X0,φ Yn(s),φ ds + σ(s,Yn−1(s))f(s),φ ds h i h i − 0 hA i 0 h i t Z Z + G(s,Y (s), v)(g(s, v) 1),φ ν(dv)ds, φ V ; (3.19) h n−1 − i ∈ Z0 ZX and
2 Yn+1(t) Yn(t) H k t − k = 2 (Yn+1(s) Yn(s)),Yn+1(s) Yn(s) ds − 0 hA − − i Z t +2 (σ(s,Yn(s)) σ(s,Yn−1(s)))f(s),Yn+1(s) Yn(s) ds 0 h − − i Z t +2 G(s,Y (s), v) G(s,Y (s), v),Y (s) Y (s) (g(s, v) 1)ν(dv)ds. h n − n−1 n+1 − n i − Z0 ZX
10 In view of (2.8), t 2 2 Yn+1(t) Yn(t) H + α Yn+1(s) Yn(s) V ds k − k 0 k − k t Z 2 (σ(s,Yn(s)) σ(s,Yn−1(s)))f(s) H Yn+1(s) Yn(s) H ds ≤ 0 k − k k − k Z t +2 G(s,Yn(s), v) G(s,Yn−1(s), v) H Yn+1(s) Yn(s) H g(s, v) 1 ν(dv)ds 0 X k − k k − k | − | Z Zt 2 +λ0 Yn+1(s) Yn(s) H ds 0 k − k tZ 2 K(s) Y (s) Y (s) Y (s) Y (s) f(s) ds ≤ k n − n−1 kH k n+1 − n kHk kH Z0 pt +λ Y (s) Y (s) 2 ds 0 k n+1 − n kH Z0 t G(s,Y (s), v) G(s,Y (s), v) + k n − n−1 kH Y (s) Y (s) Y (s) Y (s) k n − n−1 kH Z0 ZX k n − n−1 kH Yn+1(s) Yn(s) H g(s, v) 1 ν(dv)ds t·k − k | − | 2 C K(s) Yn+1(s) Yn(s) H ds ≤ 0 k − k Z t 2 2 +C Yn(s) Yn−1(s) H f(s) H ds 0 k − k k k Z t 2 +λ0 Yn+1(s) Yn(s) H ds 0 k − k Zt + G(s, v) 1,H g(s, v) 1 ν(dv) Yn(s) Yn−1(s) H Yn+1(s) Yn(s) Hds 0 X k k | − | k − k k − k Z t Z 2 C K(s) Yn+1(s) Yn(s) H ds ≤ 0 k − k Z t 2 2 +C Yn(s) Yn−1(s) H f(s) H ds 0 k − k k k Z t 2 +λ0 Yn+1(s) Yn(s) H ds 0 k − k Zt 2 + G(s, v) 1,H g(s, v) 1 ν(dv) Yn(s) Yn−1(s) H ds 0 X k k | − | k − k Z t Z + G(s, v) g(s, v) 1 ν(dv) Y (s) Y (s) 2 ds k k1,H| − | k n+1 − n kH Z0 ZX We denote J(s) = K(s) + f(s) 2 + G(s, v) g(s, v) 1 ν(dv) + λ , and set k kH X k k1,H| − | 0 a (t)= Y (t) Y (t) 2 . n n n−1 H R Thek above inequality− k yields that t t a (t) C a (s)J(s) ds + C a (s)J(s) ds, (3.20) n+1 ≤ n n+1 Z0 Z0 and furthermore, we have t t t a (t)e−C R0 J(s) dsJ(t) Ce−C R0 J(s) dsJ(t) a (s)J(s) ds n+1 ≤ n Z0 11 t t −C R0 J(s) ds +Ce J(t) an+1(s)J(s) ds Z0 t Set An+1(t)= 0 an+1(s)J(s) ds. It follows that,
Rd t t A (t)e−C R0 J(s) ds CJ(t)e−C R0 J(s) dsA (t). dt n+1 ≤ n So that,
t t s A (t)e−C R0 J(s) ds C J(s)e−C R0 J(u) duA (s) ds. n+1 ≤ n Z0 Thus,
t t C R0 J(s) ds An+1(t) Ce J(s)An(s) ds ≤ 0 Z t t CeC R0 J(s) dsA (t) J(s) ds ≤ n Z0 C A (t). ≤ T n It follows from (3.20) that
t a (t) (C + C ) a (s)J(s) ds. (3.21) n+1 ≤ T n Z0 Iterating the above inequality, we get
n T n (C + CT ) ( 0 J(s) ds) an+1(t) sup a1(s). (3.22) ≤ nR! × s∈[0,T ] Therefore, we have
∞ a (t) < . n+1 ∞ n=0 X Hence there exists Y C([0, T ],H) such that lim sup Y (s) Y (s) 2 = 0. ∈ n→∞ s∈[0,T ] k − n kH On the other hand, by Lemma 3.2 and Lemma 3.3, there exists a unique function Y ′ L2([0, T ],V ) C([0, T ],H) such that ∈ ∩ t t ′ ′ Y (t),φ = X0,φ Y (s),φ ds + σ(s,Y (s))f(s),φ ds h i h i − 0 hA i 0 h i t Z Z + G(s,Y (s), v)(g(s, v) 1),φ ν(dv)ds, φ V. (3.23) h − i ∈ Z0 ZX Using the same argument leading to (3.22), we have
′ 2 lim sup Y (s) Yn(s) H =0. (3.24) n→∞ s∈[0,T ] k − k
Hence Y ′ = Y, t [0, T ] is a solution to (3.17). We have proved the∈ existence of the solution.
12 Uniqueness: Assume X and X′ are two solutions of equation (3.17). Then, as the proof of (3.21), we have,
T sup X′(s) X(s) 2 (C + C ) X′(s) X(s) 2 J(s) ds k − kH ≤ T k − kH s∈[0,T ] Z0 By Gronwall’s inequality, we conclude X′ = X. Finally we prove the estimate (3.18). By Lemma 3.2, we have
t q 2 q 2 Xt H + α Xs V ds k k 0 k k Z t e 2 e q q X0 H +2 σ(s, Xs ) L2(H) f(s) H Xs H ds ≤ k k 0 k k k k k k t Z t q e q e q 2 +2 G(s, Xs , v) H Xs H g(s, v) 1 ν(dv)ds + λ0 Xs H ds 0 X k k k k | − | 0 k k Z Z t Z = X 2 +2 eK(s) f(se) Xq 2 ds e k 0kH k kHk s kH Z0 t G(s,pXq, v) +2 k s kH (1 + Xeq ) Xq g(s, v) 1 ν(dv)ds q k s kH k s kH | − | Z0 ZX 1+ Xs H t ke k q 2 e e +λ0 Xs H dse (3.25) 0 k k Z t 2 e X0 H + C G(s, v) 0,H g(s, v) 1 ν(dv)ds ≤ k k 0 X k k | − | t Z Z q 2 2 +2 Xs H λ0 + K(s)+ f(s) H + G(s, v) 0,H g(s, v) 1 ν(dv) ds(3.26) 0 k k k k X k k | − | Z h Z i By Gronwall’s inequality,e
q 2 sup Xs H s∈[0,t] k k t e2 C X0 H + G(s, v) 0,H g(s, v) 1 ν(dv)ds (3.27) ≤ k k 0 X k k | − | h Z Z t t t i exp C λ + K(s)ds + f(s) 2 ds + G(s, v) g(s, v) 1 ν(dv)ds . × 0 k kH k k0,H| − | Z0 Z0 Z0 ZX h i By Lemma 3.1, (3.25) and (3.27), we obtain (3.18).
We can now present the main large deviations result. Recall that for q = (f,g) S, g ∈ νT (dsdv)= g(s, v)ν(dv)ds. Define · 0( f(s)ds,νg )= Xq for q =(f,g) S as given in Theorem 3.1. (3.28) G T ∈ Z0 Let I : D([0, T ],H) [0, ] bee defined as in (2.7). → ∞ Theorem 3.2 Suppose that Condition 2.2 and Condition 3.1 hold. Then I is a rate function ǫ on D([0, T ],H), and the family X ǫ>0 satisfies a large deviation principle on D([0, T ],H) with rate function I. { }
13 The rest of the paper is devoted to the proof of this theorem. According to Theorem 2.4, we need to prove that Condition 2.1 is fulfilled. The verification of Condition 2.1 a) will be given by Proposition 3.3. Condition 2.1 b) will be established in Theorem 3.5 and a number of preparing lemmas. Let Tt, t 0 denote the semigroup generated by . It is easy to see that Tt, t 0 are compact operators.≥ For f L1([0, T ],H), denote the−A operator ≥ ∈ t Rf(t)= T f(s)ds, t 0, t−s ≥ Z0 which is the mild solution of the equation:
t t Z(t)= Z(s)ds + f(s)ds. − A Z0 Z0 The proof of the following lemma was given in [18].
Lemma 3.4 If L1([0, T ],H) is uniformly integrable, then = R( ) is relatively com- pact in C([0, T ],HD) ⊂. Y D We also need the following lemma, the proof of which can be found in [6].
Lemma 3.5 Let h : [0, T ] X R be a measurable function such that × → h(s, v) 2ν(dv)ds < , | | ∞ ZXT and for all δ (0, ) ∈ ∞ exp(δ h(s, v) )ν(dv)ds < , | | ∞ ZE for all E ([0, T ] X) satisfying νT (E) < . a). Fix∈N B N, and× let g ,g SN be such∞ that g g as n . Then ∈ n ∈ n → →∞
lim h(s, v)(gn(s, v) 1)ν(dv)ds = h(s, v)(g(s, v) 1)ν(dv)ds; n→∞ − − ZXT ZXT b). Fix N N. Given ǫ> 0, there exists a compact set K X, such that ∈ ǫ ⊂ sup h(s, v) g(s, v) 1 ν(dv)ds ǫ. N c | || − | ≤ g∈S Z[0,T ] ZKǫ
We now proceed to verify the first part of Condition 2.1. Recall the map 0 defined by (3.28). G
Proposition 3.3 Fix N N, and let q =(f ,g ), q =(f,g) S¯N be such that q q as ∈ n n n ∈ n → n . Then →∞ · · 0( f (s)ds,νgn ) 0( f(s)ds,νg ) in C([0, T ],H). G n T → G T Z0 Z0
14 Proof: 0 · gn Firstly, we prove that ( 0 fn(s)ds,νT ) n∈N is relatively compact in C([0, T ],H). By Theorem 3.1 and the{G relation between} mild solution and weak solution, R · 0 gn ( fn(s)ds,νT )(t) (3.29) G 0 Z t qn qn = T (t)X0 + T (t s) σ(s, X (s))fn(s)+ G(s, X (s), v)(gn(s, v) 1)ν(dv) ds. 0 − X − Z h Z i By Lemma 3.4, it is sufficient toe prove that e
qn qn 1 = σ( , X ( ))fn( )+ G( , X ( ), v)(gn( , v) 1)ν(dv) L ([0, T ],H) D · · · X · · · − ⊂ n Z o is uniformly integrable.e e We know that is uniformly integrable in L1([0, T ],H) iff D T (I) There exists a finite constant K such that, 0 h(s) H ds K, for every h ,; (II) For every η > 0 there exists δ > 0 such that, fork everyk measurable≤ subset A∈D[0, T ] R ⊂ with λ (A) δ and every h , bh(s) ds η. b T ≤ ∈D A k kH ≤ In fact (I) follows from Lemma 3.3 and Theorem 3.1. We need to check (II). For any R n N and any measurable subset A [0, T ], ∈ ⊂
G(s, Xqn (s), v)(g (s, v) 1)ν(dv) ds k n − kH ZA ZX G(s, Xeqn (s), v) g (s, v) 1 ν(dv)ds ≤ k kH| n − | ZA ZX qn G(s, Xe (s), v) H qn k k (1 + X (s) H ) gn(s, v) 1 ν(dv)ds qn ≤ A X 1+ X (s) H k k | − | Z Z ke k (1 + sup Xqn (s) ) G(es, v) g (s, v) 1 ν(dv)ds. ≤ k e kH k k0,H| n − | s∈[0,T ] ZA ZX Given ǫ> 0, by Lemma 3.5 wee can find a compact subset K X such that ǫ ⊂
G(s, v) 0,H gn(s, v) 1 ν(dv)ds ǫ, n 1. (3.30) c k k | − | ≤ ∀ ≥ ZA ZKǫ On the other hand, by (3.14) for any L> 0, we have
G(s, v) g (s, v) 1 ν(dv)ds k k0,H| n − | ZA ZKǫ 1 T C(L) exp(L G(s, v) )ν(dv)ds + l(g (s, v))ν(dv)ds ≤ k k0,H L n ZA ZKǫ Z0 ZX + exp( G(s, v) )ν(dv)ds. (3.31) k k0,H ZA ZKǫ We also have
σ(s, Xqn (s))f (s) ds k n kH ZA e 15 σ(s, Xqn (s)) f (s) ds ≤ k kL2(H)k n kH ZA K(se) Xqn (s) 2 +1 f (s) ds ≤ k kH k n kH ZA q p T 1/2 qn e 2 2 sup X (s) H +1 ( K(s)ds)( fn(s) Hds) ≤ s∈[0,T ] k k × A 0 k k r h Z Z i [ K(s)dse]1/2 (1 + C )1/2N 1/2, (3.32) ≤ × N ZA
qn 2 T 2 where sups∈[0,T ] X (s) H CN and 0 fn(s) H ds N. Now for anykη > 0,k first≤ choose ǫ > 0k such thatk (1≤ + C )ǫ η/5, then select L > 0 so R N ≤ that (1 + C )N/Le η/5. Finally since ν ([0, T ] K ) = T ν(K ) < , there exists δ > 0 N ≤ T × ǫ ǫ ∞ such that for every measurable subset A [0, T ] satisfying λ (A) δ, we have ⊂ T ≤ (1 + C )C(L) exp(L G(s, v) )ν(dv)ds η/5, N k k0,H ≤ ZA ZKǫ
(1 + C ) exp( G(s, v) )ν(dv)ds η/5, N k k0,H ≤ ZA ZKǫ and [ K(s)ds]1/2 C1/2N 1/2 η/5. × N ≤ ZA Combining all these inequalities gives (II).
Let X be any limit point of the sequence Xqn , n 1 . Now we will prove that X = Xq. { ≥ } Without loss of generality, we assume the whole sequence Xqn converges. { } An applicatione of dominated convergence theoreme shows that, along the convergente sub-e sequence, e
t t q ∗ ∗ X n (s), φ ds X(s), φ ds, φ H ∗ , (3.33) h A i → h A i ∀ ∈ A Z0 Z0 as n . Furthermore,e using the convergencee of Xqn to X, and Lemma 3.1, we have that →∞ t qn G(s, X (s), v),φ (gn(s, v)e 1)ν(edv)ds 0 Xh i − Z t Z G(s, Xe(s), v),φ (g (s, v) 1)ν(dv)ds 0, φ V. (3.34) − h i n − → ∀ ∈ Z0 ZX Also, since X C([0, T ],H),e we have for some κ (0, ) ∈ ∈ ∞ G(s, X(s), v),φ κ G(s, v),φ , φ V, (s, v) X . |he i| ≤ kh ik0,H ∀ ∈ ∀ ∈ T Combining this with Condition 2.2 and Lemma 3.5, we have, as n , e →∞ t G(s, X(s), v),φ (gn(s, v) 1)ν(dv)ds 0 Xh i − Z t Z G(s, Xe(s), v),φ (g(s, v) 1)ν(dv)ds, φ V. (3.35) → h i − ∀ ∈ Z0 ZX e 16 We also have, for every φ V , ∈ t t qn σ(s, X (s))fn(s),φ ds σ(s, X(s))f(s),φ ds | 0 h i − 0 h i | Z t Z e qn e φ H σ(s, X (s)) σ(s, X(s)) L2(H) fn(s) H ds ≤ k k 0 k − k k k th Z e e + σ(s, X(s)) L2(H) fn(s) f(s) H ds 0 k k k − k Z t i φ eK(s) Xqn (s) X(s) f (s) ds ≤ k kH k − kH k n kH Z0 h pt + K(s)e X(s) 2e +1 f (s) f(s) ds k kH k n − kH Z0 p q T Ti φ sup Xqn (s) eX(s) K(s)ds + f (s) 2 ds ≤ k kH k − kH × k n kH s∈[0,T ] Z0 Z0 h T T 1 e 2e 2 2 1/2 + sup X(s) H +1 ( K(s)ds) ( fn(s) f(s) H ds) s∈[0,T ] k k × 0 0 k − k r Z Z i 0, as n . → →∞ e (3.36) Combining (3.33), (3.34), (3.35) and (3.36), we see that X must satisfy t t Xt,φ = X0,φ Xs,φ ds + σ(s, Xe(s))f(s),φ ds h i h i − 0 hA i 0 h i t Z Z e + G(s, X e, v),φ (g(s, v) 1)νe(dv)ds, φ H ∗ . (3.37) h s i − ∀ ∈ A Z0 ZX q Since HA∗ is dense in V , we have Xe= X , completing the proof.
Remark 2 Fix N N. By the proof of (3.30), it is easy to see that, for ever η > 0, there ∈ e e exists δ > 0 such that for any A [0, T ] satisfying λ (A) < δ ⊂ T
sup G(s, v) 0,H g(s, v) 1 ν(dv)ds η. (3.38) N k k | − | ≤ g∈S ZA ZX We now verify the second part of Condition 2.1. The next theorem can be proved similarly as in Section 3 in [18].
Theorem 3.4 Under Condition 2.2, if X0 H, there exists a unique H-valued progressively measurable process (Xǫ) such that, Xǫ L2∈(0, T ; V ) D([0, T ]; H) for any T > 0 and t ∈ ∩ t t t −1 Xǫ = Xǫ Xǫds + √ǫ σ(s,Xǫ)dβ(s)+ ǫ G(s,Xǫ, v)N˜ ǫ (dsdv), a.s.. (3.39) t 0 − A s s s Z0 Z0 Z0 ZX It follows from this theorem that, for every ǫ > 0, there exists a measurable map ǫ: −1 V D([0, T ]; H) such that, for any Poisson random measure nǫ on [0, T ] X with meanG → −1 ǫ ǫ−1 × measure ǫ λT v given on some probability space, (√ǫβ, ǫn ) is the unique solution of −×1 −1 G (3.39) with N˜ ǫ replaced byn ˜ǫ . N 1 N Let φǫ = (ψǫ,ϕǫ) ˜ and φǫ = ˜ . The following lemma follows from Lemma ∈ U ϕǫ ∈ U 2.3 in [5].
17 Lemma 3.6
ǫ (φ ) := exp −1 log(φ (s, x))N¯( ds dx dr) Et ǫ [0,t]×X×[0,ǫ ] ǫ R + ( φǫ(s, x)+1)¯νT ( ds dx dr) −1 − Z[0,t]×X×[0,ǫ ] and 1 t 1 t ˜ǫ(ψ ) := exp ψ (s) dβ(s) ψ (s) 2 ds Et ǫ {√ǫ ǫ − 2ǫ k ǫ k } Z0 Z0 are ¯t martingales. Set {F }− ¯ǫ(ψ ,φ ) := ˜ǫ(ψ ) ǫ(φ ) Et ǫ ǫ Et ǫ Et ǫ then Qǫ(G)= ¯ǫ(ψ ,φ ) dP¯, for G (V¯) t Et ǫ ǫ ∈ B ZG defines a probability measure on V¯.
−1 −1 ǫ ϕǫ ǫ ǫ ¯ By the fact that ǫN under QT has the same law as that of ǫN under P, we know · −1 that X˜ ǫ = ǫ(√ǫβ + ψ (s) ds, ǫN ǫ ϕǫ ) is the unique solution of the following controlled G 0 ǫ stochastic differential equation: R t t t ˜ ǫ ˜ ǫ ˜ ǫ ǫ Xt = X0 Xsds + σ(s, Xs)ϕǫ(s)ds + √ǫ σ(s,Xs)dβ(s) − 0 A 0 0 t Z Z Z −1 ˜ ǫ ǫ ϕǫ + G(s, Xs−, v) ǫN (dsdv) ν(dv)ds 0 X − Z Z t t t ˜ ǫ ˜ ǫ ǫ = X0 Xsds + σ(s, Xs)ϕǫ(s)ds + √ǫ σ(s,Xs)dβ(s) − 0 A 0 0 t Z Z Z ˜ ǫ + G(s, Xs, v)(ϕǫ(s, v) 1)ν(dv)ds 0 X − Z t Z −1 ˜ ǫ ǫ ϕǫ −1 + ǫG(s, Xs−, v) N (dsdv) ǫ ϕǫ(s, v)ν(dv)ds (3.40) 0 X − Z Z The following estimates will be used later.
Lemma 3.7 There exists ǫ0 > 0, such that,
˜ ǫ 2 sup E sup Xt H < . (3.41) 0<ǫ<ǫ0 0≤t≤T k k ∞
Proof: By Ito’s formula,
˜ ǫ 2 Xt H k k t t t 2 ˜ ǫ ∗ ˜ ǫ ˜ ǫ ˜ ǫ ˜ ǫ ˜ ǫ = X0 H 2 Xs, Xt ds +2 Xs, σ(s, Xt )ψǫ(s) ds +2√ǫ σ(s, Xs)Xs dβs k k − 0 h A i 0 h i 0 t Z Z Z ˜ ǫ ˜ ǫ + 2 < Xs,G(s, Xs, v) > (ϕǫ(s, v) 1)ν(dv)ds 0 X − Zt Z −1 ˜ ǫ ˜ ǫ ˜ ǫ 2 ǫ ϕǫ −1 + 2 < ǫG(s, Xs−, v), Xs > + ǫG(s, Xs−, v) H N (dsdv) ǫ ϕǫ(s, v)ν(dv)ds 0 X k k − Z Z
18 t t + ǫ G(s, X˜ ǫ, v) 2 ϕ (s, v)ν(dv)ds + ǫ σ(s, X˜ ǫ) 2 ds. (3.42) k s kH ǫ k s kL2(H) Z0 ZX Z0 The third term on right hand side of last equation is estimated as follows.
t t ˜ ǫ ˜ ǫ ˜ ǫ ˜ ǫ 2 Xs, σ(s, Xt )ψǫ(s) ds 2 Xs H σ(s, Xt ) L2(H) ψǫ(s) H ds 0 h i ≤ 0 k k k k k k Z Zt t ˜ ǫ 2 2 ˜ ǫ 2 Xs H ψǫ(s) H ds + K(s)(1 + Xs H ) ds ≤ 0 k k k k 0 k k Z t Z t X˜ ǫ 2 ψ (s) 2 + K(s) ds + K(s) ds. (3.43) ≤ k skH k ǫ kH Z0 Z0 Denote the forth term on the right hand side of equation (3.42) by Wt. Then we have
E[ sup Ws ] 4E [W ]T 0≤s≤T | | ≤ p T √ ˜ ǫ 2 ˜ ǫ 2 8 ǫE[ σ(s, Xs) L2(H) Xs H ds] ≤ s 0 k k k k Z T √ ˜ ǫ ˜ ǫ 2 8 ǫE[ sup Xs H σ(s, Xs) L2(H) ds] ≤ 0≤s≤T k k · s 0 k k Z T 4√ǫE[ sup X˜ ǫ 2 ]+4√ǫE[ σ(s, X˜ ǫ) 2 ds] ≤ k skH k s kL2(H) 0≤s≤T Z0 T 4√ǫE[ sup X˜ ǫ 2 ]+4√ǫE[ K(s)(1 + X˜ ǫ 2 ) ds] ≤ k skH k skH 0≤s≤T Z0 T T 4√ǫ(1 + K(s) ds)E[ sup X˜ ǫ 2 ]+4√ǫE[ K(s) ds]. (3.44) ≤ k skH Z0 0≤s≤T Z0 The fifth term on the right hand side of equation (3.42) has the following bound.
t ˜ ǫ ˜ ǫ 2 < Xs,G(s, Xs, v) > (ϕǫ(s, v) 1)ν(dv)ds 0 X − Z t Z ˜ ǫ ˜ ǫ 2 Xs H G(s, Xs, v) H ϕǫ(s, v) 1 ν(dv)ds ≤ 0 X k k k k | − | Z t Z ˜ ǫ ˜ ǫ 2 G(s, v) 0,H(1 + Xs H ) Xs H ϕǫ(s, v) 1 ν(dv)ds ≤ 0 X k k k k k k | − | Z t Z ˜ ǫ 2 2 G(s, v) 0,H(1+2 Xs H ) ϕǫ(s, v) 1 ν(dv)ds ≤ 0 X k k k k | − | Z Z t 2CN +4 G(s, v) X˜ ǫ 2 ϕ (s, v) 1 ν(dv)ds, (3.45) ≤ 0,1 k k0,Hk skH | ǫ − | Z0 ZX N where C0,1 is the constant from Lemma 3.1. The seventh term in (3.42) is bounded by,
t ˜ ǫ 2 ǫ G(s, Xs, v) Hϕǫ(s, v)ν(dv)ds 0 X k k Z Z t 2ǫ G(s, v) 2 (1 + X˜ ǫ 2 )ϕ (s, v)ν(dv)ds ≤ k k0,H k skH ǫ Z0 ZX 19 t 2ǫCN +2ǫ G(s, v) 2 X˜ ǫ 2 ϕ (s, v)ν(dv)ds, (3.46) ≤ 0,2 k k0,Hk skH ǫ Z0 ZX N where constant C0,2 was defined in Lemma 3.1. The last term in (3.42) is bounded by,
t t ˜ ǫ 2 ˜ ǫ 2 ǫ σ(s, Xs) L2(H) ds ǫ K(s)(1 + Xs H ) ds 0 k k ≤ 0 k k Z Z t t ǫ X˜ ǫ 2 K(s) ds + ǫ K(s) ds. (3.47) ≤ k skH Z0 Z0 Set t −1 ˜ ǫ ˜ ǫ ǫ ϕǫ −1 Mt = < 2ǫG(s, Xs−, v), Xs > N (dsdv) ǫ ϕǫ(s, v)ν(dv)ds 0 X − Z Z and t −1 ˜ ǫ 2 ǫ ϕǫ −1 Kt = ǫG(s, Xs−, v) H N (dsdv) ǫ ϕǫ(s, v)ν(dv)ds . 0 X k k − Z Z We have t −1 E[ sup K ] E[ sup ǫG(s, X˜ ǫ , v) 2 N ǫ ϕǫ (dsdv)] | t| ≤ k s− kH 0≤t≤T 0≤t≤T Z0 ZX T ˜ ǫ 2 +ǫE[ G(s, Xs, v) Hϕǫ(s, v)ν(dv)ds] 0 X k k Z ZT ˜ ǫ 2 2ǫE[ G(s, Xs, v) Hϕǫ(s, v)ν(dv)ds] ≤ 0 X k k Z T Z 4ǫE[ G(s, v) 2 (1 + X˜ ǫ 2 )ϕ (s, v)ν(dv)ds] ≤ k k0,H k skH ǫ Z0 ZX N ˜ ǫ 2 4ǫC0,2(1 + E[ sup Xs H ]). (3.48) ≤ 0≤t≤T k k
For the martingale Mt, we have
E[ sup Mt ] 4E [M]T 0≤t≤T | | ≤ 1 p T 2 −1 ˜ ǫ ˜ ǫ 2 ǫ ϕǫ 4E < 2ǫG(s, Xs−, v), Xs > N (dsdv) ≤ 0 X | | Z Z 1 T 2 −1 ˜ ǫ 2 ˜ ǫ 2 ǫ ϕǫ 8ǫE G(s, Xs−, v) H Xs H N (dsdv) ≤ 0 X k k k k Z Z 1 T 2 −1 8ǫE( sup X˜ ǫ ) G(s, X˜ ǫ , v) 2 N ǫ ϕǫ (dsdv) ≤ k skH k s− kH 0≤t≤T Z0 ZX T 1 −1 E[ sup X˜ ǫ 2 ]+16ǫ2CE[ G(s, X˜ ǫ , v) 2 N ǫ ϕǫ (dsdv)] ≤ C k skH k s− kH 0≤t≤T Z0 ZX 1 T = E[ sup X˜ ǫ 2 ]+16ǫCE[ G(s, X˜ ǫ , v) 2 ϕ (s, v)ν(dv)ds] C k skH k s− kH ǫ 0≤t≤T Z0 ZX 1 ˜ ǫ 2 N ˜ ǫ 2 E[ sup Xs H ]+32ǫCC0,2(1 + E[ sup Xs H ]) ≤ C 0≤t≤T k k 0≤t≤T k k
20 N 1 N ˜ ǫ 2 32ǫCC0,2 +( + 32ǫCC0,2)E[ sup Xs H ], (3.49) ≤ C 0≤t≤T k k where C is any positive number. T Set K = 0 K(s)ds. Combining the estimates (3.43), (3.45), (3.46) and (3.47), we have RX˜ ǫ 2 k t kH ˜ 2 N N ( X0 H +(1+ ǫ)K + sup Wt +2C0,1 +2ǫC0,2 + sup Kt + sup Mt ) ≤ k k 0≤t≤T | | 0≤t≤T | | 0≤t≤T | | t + X˜ ǫ 2 ( ψ (s) 2 +(1+ ǫ)K(s)+ (4 G(s, v) ϕ (s, v) 1 k skH · k ǫ kH k k0,H| ǫ − | Z0 ZX +2ǫ G(s, v) 2 ϕ (s, v))ν(dv)) ds. k k0,H ǫ By Grownwall’s inequality and Lemma 3.1, we get ˜ ǫ 2 ˜ 2 N N Xt H ( X0 H +(1+ ǫ)K + sup Wt +2C0,1 +2ǫC0,2 + sup Kt + sup Mt ) k k ≤ k k 0≤t≤T | | 0≤t≤T | | 0≤t≤T | | N N e(N+(1+ǫ)K+4C0,1+2ǫC0,2). · N N (N+(1+ǫ)K+4C0,1+2ǫC0,2) Set C0 = e . By (3.44), (3.48) and (3.49), we have
˜ ǫ 2 ˜ 2 N N N N E[ sup Xt H ] C0 E X0 H +2C0,1 +2ǫC0,2 +4ǫC0,2 + 32ǫCC0,2 +(1+ ǫ +4√ǫ)K 0≤t≤T k k ≤ k k 1 +C 4ǫCN + + 32ǫCCN +4√ǫ(1 + K) E[ sup X˜ ǫ 2 ]. 0 0,2 C 0,2 k skH 0≤t≤T Since constant C can be arbitrarily large, we can select C and ǫ0 small enough, such that 1 1 C 4ǫ CN + + 32ǫ CCN +4√ǫ (1 + K) < 0 0 0,2 C 0 0,2 0 2 Therefore, we have ˜ ǫ 2 sup E sup Xt H 0<ǫ<ǫ0 0≤t≤T k k 2C E X˜ 2 +2CN +2ǫ CN +4ǫ CN + 32ǫ CCN +(1+ ǫ +4√ǫ )K ≤ 0 k 0kH 0,1 0 0,2 0 0,2 0 0,2 0 0 < . ∞ The following proof of estimates (3.50) and (3.60) was inspired by the method in [18].
Lemma 3.8 There exists ǫ > 0, such that, for any t [0, T ], 0 <ǫ<ǫ , 0 0 ∈ 0 ∞ 1 1/2 ˜ ǫ 2 ( 4ǫ )E[ sup Xt , ei ] (3.50) 2 − 0≤t≤t0 |h i| Xi=k ∞ t0 ˜ ǫ 2 ˜ ǫ 2 X0, ei + (8N + 129ǫ)( K(s)ds)E[(1 + sup Xt H ) ] ≤ h i 0 0≤t≤t0 k k Xi=k Z t0 2 ˜ ǫ 2 +4E[(1 + sup Xt H ) ] sup G(s, v) 0,H g(s, v) 1 ν(dv)ds k k · N k k | − | 0≤t≤t0 g∈S Z0 ZX t0 1/2 ˜ ǫ 2 2 +(4ǫ +2ǫ)E[(1 + sup Xt H ) ] sup G(s, v) 0,Hg(s, v)ν(dv)ds. k k · N k k 0≤t≤t0 g∈S Z0 ZX
21 Proof: Recall that e , k 1 is an orthonormal basis of the Hilbert space H. For { k ≥ } convenience, let Y ǫ = Xǫ, e . Then i h ii Y ǫ(t) = < X˜ ǫ, e > i t i e t t t ǫ ˜ ǫ ˜ ǫ =
t t ǫ 2 2 ǫ 2 ǫ ˜ ǫ Yi (t) =
t0 ∞ ˜ ǫ ǫ 2E[ σ(s, Xs) L2(H) ψǫ(s) H Yi (s)ei H ds] 0 k k k k k k Z Xi=k t0 ∞ t0 1 2 ǫ 2 ˜ ǫ 2 E[ ψǫ(s) H Yi (s)ei H ds]+8NE[ K(s)(1 + Xs H ) ds] ≤ 8N 0 k k k k 0 k k Z Xi=k Z 22 ∞ t0 1 ǫ 2 ˜ ǫ 2 E[ sup Yi (t) ]+8N( K(s)ds)E[(1 + sup Xt H ) ]. (3.53) ≤ 8 0≤t≤t0 | | 0 0≤t≤t0 k k Xi=k Z The third term is bounded by,
∞ t ǫ ˜ ǫ 2√ǫE[ sup Yi (s)σ(s, Xs)dβ(s), ei ] 0≤t≤t0 h 0 i Xi=k Z t0 ∞ 8√ǫE[ σ(s, X˜ ǫ) 2 Y ǫ(t)e 2 ds] ≤ v k s kL2(H)k | i ikH u 0 i=k uZ X t ∞ t0 √ ǫ ˜ ǫ 2 8 ǫE[( sup Yi (t)ei H ) σ(s, Xs) L2(H) ds] ≤ 0≤t≤t0 k k s 0 k k Xi=k Z ∞ t0 1 ǫ 2 ˜ ǫ 2 E[ sup Yi (t) ]+128ǫE[ σ(s, Xs) L2(H) ds] ≤ 8 0≤t≤t0 | | 0 k k Xi=k Z ∞ t0 1 ǫ 2 ˜ ǫ 2 E[ sup Yi (t) ]+128ǫ( K(s)ds)E[(1 + sup Xt H ) ]. (3.54) ≤ 8 0≤t≤t0 | | 0 0≤t≤t0 k k Xi=k Z The forth term is estimated as follows,
t0 ∞ ǫE[ σ∗(s, X˜ ǫ)e 2 ds] k s ikH Z0 i=k t0 X ˜ ǫ 2 ǫE[ σ(s, Xs) L2(H) ds] ≤ 0 k k Zt0 ǫ( K(s)ds)E[(1 + sup X˜ ǫ )2]. (3.55) ≤ k t kH Z0 0≤t≤t0 The fifth term is bounded by,
t0 ∞ ˜ ǫ ǫ E[
E[ sup Mt ] 4E [M]t0 0≤t≤t0 | | ≤
p t0 ∞ −1 1 ˜ ǫ ǫ 2 ǫ ϕǫ 2 4E < ǫG(s, Xs−, v), Yi (s )ei > N (dsdv) ≤ { 0 X | − | } Z Z Xi=k t0 ∞ −1 1 ˜ ǫ 2 ǫ 2 ǫ ϕǫ 2 4E ǫG(s, Xs−, v) ( Yi (s ) )N (dsdv) ≤ { 0 X k k | − | } Z Z Xi=k ∞ t0 1 1 3 −1 1 2 ǫ 2 2 2 ˜ ǫ 2 ǫ ϕǫ 2 4E[ ǫ sup Yi (t) ǫ G(s, Xs−, v) N (dsdv) ] ≤ { 0≤t≤t0 | | } ·{ 0 X k k } Xi=k Z Z ∞ t0 1 3 −1 2 ǫ 2 2 ˜ ǫ 2 ǫ ϕǫ 2E[ǫ sup Yi (t) ]+2E[ ǫ G(s, Xs−, v) N (dsdv)] ≤ 0≤t≤t0 | | 0 X k k Xi=k Z Z ∞ t0 1 1 2 ǫ 2 2 ˜ ǫ 2 2ǫ E[ sup Yi (t) ]+2ǫ E[ G(s, Xs−, v) ϕǫ(s, v)ν(v)ds] ≤ 0≤t≤t0 | | k k i=k Z0 ZX X∞ 1 2 ǫ 2 2ǫ E[ sup Yi (t) ] ≤ 0≤t≤t0 | | i=k X t0 1 2 ˜ ǫ 2 2 +2ǫ E[(1 + sup Xt ) ]( G(s, v) 0,Hϕǫ(s, v)ν(v)ds) 0≤t≤t0 k k 0 X k k ∞ Z Z 1 2 ǫ 2 2ǫ E[ sup Yi (t) ] ≤ 0≤t≤t0 | | i=k X t0 1 2 ˜ ǫ 2 2 +2ǫ E[(1 + sup Xt ) ] sup ( G(s, v) 0,Hg(s, v)ν(v)ds) k k · N k k 0≤t≤t0 g∈S Z0 ZX (3.57)
The upper bound of the last term in (3.52) is given by,
t0 −1 ˜ ǫ 2 ǫ ϕǫ E[ ǫG(s, Xs, v) HN (dsdv)] 0 X k k Z t0Z ˜ ǫ 2 ǫE[ G(s, Xs, v) Hϕǫ(s, v)ν(dv)ds] ≤ 0 X k k Z Z t0 ˜ ǫ 2 2 ǫE[(1 + sup Xt ) ] sup ( G(s, v) 0,Hg(s, v)ν(v)ds). (3.58) ≤ k k · N k k 0≤t≤t0 g∈S Z0 ZX Therefore, combining the above inequalities, we get
∞ ǫ 2 E[ sup Yi (t) ] 0≤t≤t0 | | i=k ∞ X ∞ 1 1 2 2 ǫ 2
24 t0 ˜ ǫ 2 2 +4E[(1 + sup Xt ) ] sup ( G(s, v) 0,H g(s, v) 1 ν(dv)ds) k k · N k k | − | 0≤t≤t0 g∈S Z0 ZX t0 1 2 ˜ ǫ 2 2 +(4ǫ +2ǫ)E[(1 + sup Xt ) ] sup ( G(s, v) 0,Hg(s, v)ν(v)ds) (3.59) k k · N k k 0≤t≤t0 g∈S Z0 ZX
Lemma 3.9 There exists ǫ > 0, such that, for any t [0, T ], 0 0 ∈ ∞ ˜ ǫ 2 −2ζkt0 sup sup E sup Xt , ei e C. (3.60) N 0<ǫ<ǫ0 φǫ=(ψǫ,ϕǫ)∈U t0≤t≤T |h i| ≤ e Xi=k Proof: By (3.51), we have
∞ ∞ t ∞ Y ǫ(t) 2 X , e 2 2ζ Y ǫ(s) 2ds | i | ≤ h 0 ii − k | i | i=k i=k Z0 i=k X X t X∞ +2 G(s, Xǫ, v), Y ǫ(s)e ) (ϕ (s, v) 1)ν(dv)ds h s i i i ǫ − Z0 ZX i=k t X∞ e −1 +2 ǫG(s, Xǫ, v), Y ǫ(s)e ) N ǫ ϕǫ (dv,ds) h s i i i Z0 ZX i=k t ∞ X e −1e ǫ 2 ǫ ϕǫ + ǫG(s, Xs, v), ei) N (dv,ds) 0 X |h i| Z Z Xi=k t e ∞ +2 √ǫσ(s, Xǫ)dβ , Y ǫ(s)e h s s i ii Z0 i=k t ∞X ǫ e ǫ +2 σ(s, Xs)ψǫ(s), Yi (s)ei ds 0 h i Z Xi=k t e +ǫ σ(s, Xǫ) 2 ds k s kL2(H) Z0 ∞ t ∞ X , e 2 e 2ζ Y ǫ(s) 2ds ≤ h 0 ii − k | i | i=k Z0 i=k X T X∞ +2 G(s, Xǫ, v), Y ǫ(s)e ) (ϕ (s, v) 1) ν(dv)ds |h s i i i ǫ − | Z0 ZX i=k t′ X ∞ e −1 ǫ ǫ ǫ ϕǫ +2 sup ǫG(s, Xs, v), Yi (s)ei) N (dv,ds) 0≤t′≤T | h i | Z0 ZX i=k T ∞ X e −1 e + ǫG(s, Xǫ, v), e ) 2N ǫ ϕǫ (dv,ds) |h s i i| Z0 ZX i=k X t′ ∞ e ǫ ǫ +2 sup √ǫσ(s, Xs)dβs, Yi (s)ei ′ | h i| t ∈[0,T ] Z0 i=k T ∞ X ǫ e ǫ +2 σ(s, Xs)ψǫ(s) H Yi (s)ei H ds 0 k k k k Z Xi=k e 25 T +ǫ σ(s, Xǫ) 2 ds k s kL2(H) Z0 = I I +2I +2I + I +2I +2I + I , 1 − 2 3 e 4 5 6 7 8 by the Gronwall’s inequality this implies that
∞ Y ǫ(t) 2 e−2ζkt[I +2I +2I + I +2I +2I + I ]. | i | ≤ 1 3 4 5 6 7 8 Xi=k Hence, for any t0 > 0, we have
∞ ǫ 2 −2ζkt0 sup Yi (t) e [I1 +2I3 +2I4 + I5 +2I6 +2I7 + I8]. (3.61) t0≤t≤T | | ≤ Xi=k
T ǫ ǫ EI3 E G(s, Xs, v) H Xs H ϕǫ(s, v) 1 ν(dv)ds ≤ 0 X k k k k | − | Z Z T E sup ( Xǫ e+ Xǫ 2 e) G(s, v) ϕ (s, v) 1 ν(dv)ds (3.62) ≤ k skH k skH · k k0,H| ǫ − | 0≤s≤T Z0 ZX h T i e ǫ 2e E (1 + 2 sup Xs H ) sup G(s, v) 0,H g(s, v) 1 ν(dv)ds, ≤ 0≤s≤T k k · g∈SN 0 X k k | − | h i Z Z e T ∞ −1 1/2 ǫ ǫ 2 ǫ ϕǫ EI4 E ǫG(s, Xs, v), Yi (s)ei) N (dv,ds) ≤ 0 X |h i| h Z Z i=k i T X −1 1/2 2 e ǫ 2 ǫ 2 ǫ ϕǫ E ǫ G(s, Xs, v) H Xs H N (dv,ds) ≤ 0 X k k k k h Z Z T i 1/2 −1 1/2 E ǫ1/2 sup Xǫ 2e e ǫ3/2 G(s, Xǫ, v) 2 N ǫ ϕǫ (dv,ds) ≤ k skH k s kH 0≤s≤T Z0 ZX h T i ǫ1/2E sup Xeǫ 2 + ǫ1/2E G(s,eXǫ, v) 2 ϕ (s, v)ν(dv)ds ≤ k skH k s kH ǫ 0≤s≤T Z0 ZX 2 1/2 e ǫ 2 1/2 eǫ ǫ E sup Xs H + ǫ E 1 + sup Xs H ≤ 0≤s≤T k k 0≤s≤T k k h T e e 2 G(s, v) 0,Hϕǫ(s, v)ν(dv)ds · 0 X k k Z Z 2 i 1/2 ǫ 2 1/2 ǫ ǫ E sup Xs H + ǫ E 1 + sup Xs H ≤ 0≤s≤T k k 0≤s≤T k k T e e 2 sup G(s, v) 0,Hg(s, v)ν(dv)ds, (3.63) · N k k g∈S Z0 ZX
T EI ǫE G(s, Xǫ, v) 2 ϕ (s, v)ν(dv)ds 5 ≤ k s kH ǫ Z0 ZX 2 T ǫ e 2 ǫE 1 + sup Xs H sup G(s, v) 0,Hg(s, v)ν(dv)ds, (3.64) ≤ 0≤s≤T k k · g∈SN 0 X k k Z Z e 26 T ǫ 2 ǫ 2 1/2 EI6 √ǫE [ σ(s, Xs) L2(H) Xs H ds] ≤ 0 k k k k n Z T o √ǫE[ sup Xǫ 2e+ 1] [ e K(s)ds]1/2, (3.65) ≤ k skH × s∈[0,T ] Z0 e T ǫ 2 ǫ sup EI7 E K(s) Xs H +1 Xs H ψǫ(s) Hds ψ ∈SN ≤ 0 k k k k k k ǫ e Z q p T ǫ 2 e e E[ sup Xs H + 1] K(s)ds + N , (3.66) ≤ s∈[0,T ] k k × 0 Z e T EI ǫE[ sup Xǫ 2 + 1] K(s)ds, (3.67) 8 ≤ k skH × s∈[0,T ] Z0 e By (3.61)-(3.67), Lemma 3.1 and (3.41), we have
∞ ǫ 2 −2ζkt0 sup sup E sup Yi (t) e CN . N 0<ǫ≤ǫ0 φǫ=(ψǫ,ϕǫ)∈U t0≤t≤T | | ≤ e Xi=k
Remark 3 By Lemma 3.1, Remark 2, (3.41), (3.50) and the integrability of K(s), for every 0 <ǫ<ǫ and η > 0, there exists t [0, T ] such that 0 0 ∈ ∞ ∞ 1/2 ˜ ǫ 2 ˜ ǫ 2 1/2 (1/2 4ǫ )E[ sup Xt , ei ] X0, ei + Cη + C(ǫ + ǫ) (3.68) − 0≤t≤t0 |h i| ≤ h i Xi=k Xi=k In the sequel, the next two tightness results in D([0, T ],H) and D([0, T ], R) will be used.
Lemma 3.10 ([14]) Let H be a separable Hilbert space with the inner product , . For an 2 + h· ·i orthonormal basis e N in H, define the function r : H R by { k}k∈ N → r2 (x)= x, e 2, N N. N h ki ∈ k≥XN+1 Let D be closed under addition which is a total subset of H. n Then the sequence X N of stochastic processes with trajectories in D([0, T ],H) is { }n∈ tight iff it is D-weakly tight and for every ǫ> 0
2 n lim lim sup P rN (X (s)) >ǫ for some s, 0 s T 0. (3.69) N→∞ n→∞ ≤ ≤ → n n Here we say a H-valued sequence X N is“ D-weakly tight” (in ([14])) if X ,φ as a { }n∈ h i R-valued sequence is tight, for every φ D. In order to prove “D-weakly tight”∈ in Lemma 3.10, we need the tightness result in D([0, T ], R); and one can refer to [3].
27 ǫ Let Y ǫ∈(0,ǫ0] be a sequence of random elements of D([0, T ], R), and τǫ, δǫ be such that: { } { } (a) for each ǫ, τǫ is a stopping time with respect to the natural σ-fields, and takes only finitely many values. (b) for each ǫ, the constant δǫ [0, T ] and δǫ 0 as ǫ 0. We introduce the following condition∈ on Y ǫ →: for each→ sequence τ , δ satisfying (a)(b), { } { ǫ ǫ} Condition (A) Y ǫ(τ + δ ) Y ǫ(τ ) 0, as ǫ 0, in probability. ǫ ǫ − ǫ → → For f D([0, T ], R), let J(f) denote the maximum of the jump f(t) f(t ) . ∈ | − − | ǫ ǫ Lemma 3.11 ([14]) Suppose that Y ǫ∈N satisfies Condition (A), and either Y (0) and J(Y ǫ) are tight on the line; or {Y ǫ(}t) is tight on the line for each t [0, T ]{, then } Y ǫ {is tight} in D([0, T ], R). { } ∈ { }
Theorem 3.5 Fix M N, and let φ =(ψ ,ϕ ),φ =(ψ,ϕ) M be such that φ converges ∈ ǫ ǫ ǫ ∈ U ǫ in distribution to u as ǫ 0. Then → · · e −1 ǫ(√ǫβ + ψ (s)ds, ǫN ǫ ϕǫ ) 0( ψ(s)ds,νϕ). G ǫ ⇒ G Z0 Z0 −1 ǫ · ǫ ϕǫ Proof: First, we prove that (√ǫβ + 0 ψǫ(s)ds, ǫN ) is tight in D([0, T ],H). We will use Lemma 3.10 and LemmaG 3.11 to prove this result. R By (3.68) and (3.60), it follows that for any δ > 0 and t T , 0 ≤ ∞ ˜ ǫ 2 sup sup P ( sup Xt , ei > δ) 0≤ǫ≤ǫ0 ˜N 0≤t≤T h i φǫ=(ψǫ,ϕǫ)∈U i=k X ∞ −1 ˜ ǫ 2 δ sup sup E( sup Xt , ei ) ≤ 0≤ǫ≤ǫ0 ˜N 0≤t≤t0 h i φǫ=(ψǫ,ϕǫ)∈U i=k X∞ −1 ˜ ǫ 2 +δ sup sup E( sup Xt , ei ) N 0≤ǫ≤ǫ0 φǫ=(ψǫ,ϕǫ)∈U˜ t0≤t≤T h i Xi=k =: (I)+(II).
By estimates (3.50), for anyǫ ˜ > 0, there exists K1 > 0 and t˜0 > 0, such that for any ˜ ˜ǫ k>K1 and t0 < t0, we have (I) 2 . Fixing a constant 0 < t t˜≤, by estimates (3.60), we know that there exists constant 0 ≤ 0 K2 > 0, such that for any k>K2, ǫ˜ (II) e−2ζkt0 C . (3.70) ≤ ≤ 2 Sinceǫ ˜, we have
∞ ˜ ǫ 2 lim sup sup P ( sup Xt , ei > δ)=0. k→∞ N 0≤ǫ≤ǫ0 qǫ=(ψǫ,ϕǫ)∈U˜ 0≤t≤T h i Xi=k ǫ Hence by Lemma 3.10, we only need to prove that, for every φ HA∗ , (√ǫβ + · −1 ∈ hG ψ (s)ds, ǫN ǫ ϕǫ ),φ is tight in D([0, T ], R). 0 ǫ i R 28 −1 ǫ · ǫ ϕǫ ǫ ǫ For convenience, denote (√ǫβ + 0 ψǫ(s)ds, ǫN ),φ by Y . We will check that Y satisfies the condition of LemmahG 3.11. i R By (3.41), sup P( Y ǫ(t) > L) C/L2, 0<ǫ<ǫ0 | | ≤ hence Y ǫ(t) is tight on the line for each t [0, T ]. { } ǫ ∈ Hence it remains to prove Y satisfy Condition (A). For each sequence τǫ, δǫ satisfying (a)(b), { }
τǫ+δǫ Y ǫ(τ + δ ) Y ǫ(τ ) = Xǫ(s), ∗φ ds ǫ ǫ − ǫ − h A i Zτǫ τǫ+δǫ −1 + e G(s, Xǫ(s), v),φ ǫN ǫ ϕǫ (dv,ds) ν(dv)ds h i − Zτǫ ZX τǫ+δǫ + √ǫσ(s, Xeǫ(s))dβ ,φ h s i Zτǫ τǫ+δǫ + σ(s, Xǫ(es))ψ (s),φ ds h ǫ i Zτǫ τǫ+δǫ = Xǫ(s)e, ∗φ ds − h A i Zτǫ τǫ+δǫ −1 + e G(s, Xǫ(s), v),φ ǫN ǫ ϕǫ (dv,ds) h i Zτǫ ZX τǫ+δǫ + G(s, Xe ǫ(s), v),φ (ϕe 1)ν(dv)ds h i ǫ − Zτǫ ZX τǫ+δǫ + √ǫσ(s, Xeǫ(s))dβ ,φ h s i Zτǫ τǫ+δǫ + σ(s, Xǫ(es))ψ (s),φ ds h ǫ i Zτǫ = I1 + I2 + I3 + Ie4 + I5, (3.71)
τǫ+δǫ ∗ 2 ǫ 2 ∗ 2 ǫ 2 E( I1 ) δǫ φ H + E X (s) H ds δǫ φ H + E( sup X (s) H ) , (3.72) | | ≤ kA k τǫ k k ≤ kA k s∈[0,T ] k k Z e e
τǫ+δǫ E( I 2) E ǫG(s, Xǫ(s), v) 2 φ 2 ǫ−1ϕ (v,s)ν(dv)ds (3.73) | 2| ≤ k kH k kH ǫ Zτǫ ZX T 2 ǫ e 2 2 ǫ φ H E sup X (s) H +1 sup G(s, v) 0,Hg(v,s)ν(dv)ds, ≤ k k s∈[0,T ] k k g∈SM 0 X k k Z Z e
τǫ+δǫ E( I ) E G(s, Xǫ(s), v) φ ϕ 1 ν(dv)ds | 3| ≤ k kHk kH| ǫ − | Zτǫ ZX τǫ+δǫ e ǫ φ H E (1 + sup X (s) H ) G(s, v) 0,H ϕǫ 1 ν(dv)ds , ≤ k k s∈[0,T ] k k τǫ X k k | − | Z Z e 29 By the same argument leading to (3.30), we can show that, for every η > 0, there exists δ > 0, such that
sup G(s, v) 0,H g(v,s) 1 ν(dv)ds η, A [0, T ], λT (A) δ. M k k | − | ≤ ∀ ∈ ≤ g∈S ZA ZX Hence, if δ δ, we deduce that ǫ ≤ ǫ E( I3 ) η φ HE(1 + sup X (s) H ). (3.74) | | ≤ k k s∈[0,T ] k k e For the terms in I4 and I5, we have
T 2 2 ǫ 2 E( I4 ) ǫ φ H E σ(s, X (s)) L2(H)ds | | ≤ k k 0 k k Z T ǫ φ 2 (E[ sup Xǫe(s) 2 ]+1) K(s)ds; (3.75) ≤ k kH k kH × s∈[0,T ] Z0 e
τǫ+δǫ E( I ) φ E σ(s, Xǫ(s)) ψ (s) ds | 5| ≤ k kH k kL2(H)k ǫ kH Zτǫ τǫ+δǫ φ E K(se) Xǫ(s) 2 +1 ψ (s) ds ≤ k kH k kH k ǫ kH Zτǫ q p τǫ+δǫ 1/2 ǫe 2 1/2 M φ H E sup X (s) H +1 [ K(s)ds] , ≤ k k s∈[0,T ] k k × τǫ n q Z o e where M = T ψ (s) 2 ds. 0 k ǫ kH Since K( ) L1([0, T ], R), there exists δ > 0 such that, if δ δ, R· ∈ ǫ ≤
1/2 ǫ 2 E( I5 ) ηM φeH E[ sup X (s) H +e 1]. (3.76) | | ≤ k k s∈[0,T ] k k r e By (3.72)-(3.76) and Chebyshev inequality, we see that Condition (A) holds. Thus we · −1 have proved that ǫ(√ǫβ + ψ (s)ds, ǫN ǫ ϕǫ ) is tight in D([0, T ],H). G 0 ǫ · · −1 Finally, we prove that 0(R ψ(s)ds,νϕ) is the unique limit point of ǫ(√ǫβ+ ψ (s)ds, ǫN ǫ ϕǫ ). G 0 G 0 ǫ Note that Xǫ satisfies R R t ǫ ǫ ∗ e X (t),φ = X0,φ X (s), φ ds h i h i − 0 h A i t Z −1 e + G(s, Xe ǫ(s ), v),φ ǫN ǫ ϕǫ (dv,ds) 0 Xh − i Z t Z e ǫ e + G(s, X (s), v),φ (ϕǫ 1)ν(dv)ds (3.77) 0 Xh i − Z t Z eǫ + √ǫσ(s, X (s))dβs,φ 0 h i Z t + σ(s, Xǫ(es))ψ (s),φ ds, φ V. h ǫ i ∀ ∈ Z0 e 30 ǫ −1 t ǫ ǫ t ǫ ǫ ϕǫ Denote M (t) = 0 √ǫσ(s, X (s))dβs and M (t) = 0 X G(s, X (s ), v)ǫN (dv,ds). Since − R R R e T e e ǫ E( sup M (s) 2 ) ǫE σ(s, Xǫ(s)) 2 ds k kH ≤ k kL2(H) s∈[0,T ] Z0 T e ǫ K(s)ds(E sup Xǫ(s) 2 + 1), ≤ k kH Z0 s∈[0,T ] and e
T ǫ 2 ǫ 2 −1 E( sup M (s) H ) E G(s, X (s ), v)ǫ H ǫ ϕǫν(dv)ds s∈[0,T ] k k ≤ 0 X k − k Z Z T ǫ 2 G(s,eX (s ), v) H ǫ 2 = ǫE k − k (1 + X (s ) H ) ϕǫν(dv)ds ǫ 2 k − k Z0 ZX (1 + X (s ) H ) ke − k T ǫ 2 e 2 ǫE (1 + sup X (es) H ) G(s, v) 0,Hϕǫν(dv)ds , ≤ s∈[0,T ] k k 0 X k k Z Z ǫ e by Lemma 3.1 and (3.41),M 0 and M ǫ 0, as ǫ 0. ⇒ ⇒ǫ ǫ →ǫ Choose a subsequence along which (X ,uǫ, M , M ) converges to (X, u, 0, 0) in distri- ǫ bution. By the Skorokhod representation theorem, we may assume (Xǫ,u , M , M ǫ) ǫ → (X, u, 0, 0) almost surely e e Note that convergence in Skorokhod topology to a continuous limit iseequivalent to the uniforme convergence, and C([0, T ],H) is a closed subset of D([0, T ],H). Hence
ǫ 2 lim sup M (s) H =0, P a.s.. ǫ→0 s∈[0,T ] k k −
Since Xǫ M ǫ C([0, T ],H) and Xǫ M ǫ X almost surely in D([0, T ],H), we have − ∈ − → X C([0, T ],H), and ∈ e e e ǫ 2 e lim sup X (s) X(s) H =0, P a.s.. ǫ→0 s∈[0,T ] k − k −
Along the same lines of the proofe of Theoreme 3.1 and Proposition 3.3, letting, ǫ 0, we see that X must solve →
t t ∗ e X(t),φ = X0,φ X(s), φ ds + σ(s, X(s))ψ(s),φ ds h i h i − 0 h A i 0 h i t Z Z e + G(s, Xe(s), v),φ (ϕ 1)ν(dv)ds,e φ V. (3.78) h i − ∀ ∈ Z0 ZX e · By the uniqueness, this gives that X = 0( ψ(s)ds,νϕ). Proof is completed. G 0 R e
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