Existence, Uniqueness and Regularity of Parabolic SPDEs Driven by Poisson Random Measure

In this paper we investigate SPDEs in certain Banach spaces driven by a Poisson random measure. We show existence and uniqueness of the solution, investigate certain integrability properties and verify the cadlag property.


Introduction
The classical Itô stochastic integral with respect to Brownian motion has been generalized in several directions. One of these directions is to define the stochastic integral in the L p -mean, 1 < p ≤ 2 (see Bichteler [6,7]). This can be done easily if the integrator is of integrable p-variation. The works of Dettweiler [19], Giné and Marcus [24], Dang Hung Thang [44] and Neidhardt [36] have extended this procedure to certain types of Banach spaces in which most of the probabilistic theorems necessary for defining a stochastic integral are valid.
If the integrator is a Poisson random measure it is quite natural to define the stochastic integral in L p -mean. In the case of SPDEs, examples can be covered which are not easily treated using the Hilbert space or M type 2 Banach space theory (see Example 2.2 and Example 2.3).
Let E be a separable Banach space of M type p, 1 < p ≤ 2, with Borel σ-algebra E. Let A be an infinitesimal generator of an analytic semigroup (S t ) t≥0 on E. Let Z be a Banach space and Z be the Borel σ-algebra on Z. Let η be a Poisson random measure defined on (Z, Z) with symmetric Lévy measure ν : Z → R + satisfying certain moment conditions. Assume that f : E → E, and g : E × Z → E are Lipschitz continuous and measurable functions. We consider the following SPDE written in the Itô-form du(t) = (Au(t−) + f (u(t−))) dt + Z g(u(t−); z)η(dz; dt), u(0) = u 0 .
Under a mild solution of equation (1) we understand a predictable càdlàg process u taking values in a certain Banach space and satisfying the integral equation S t−s f (u(s−)) ds + t 0 + Z S t−s g(u(s−); z) η(ds, dz), a.s. , t ≥ 0.
In Theorem 2.1 we find conditions which guarantee existence and uniqueness of the solution u of problem (1).
In contrast to the Wiener process, the Lévy process itself is a.s. discontinuous, however L = {L(t), t ≥ 0} is càdlàg. Parabolic SPDEs driven by the Gaussian white noise were initially introduced and discussed by Walsh [46,47], where he also mentioned as an example the cable equation driven by a Poisson random measure. Kallianpur and Xiong [28,29] showed existence and uniqueness for equation (1) in the space of distributions, while Albeverio, Wu and Zhang [1] investigated equation (1) and showed existence and uniqueness in Hilbert spaces under the L 2 -integrability condition of the Poisson random measure under suitable hypothesis. Existence and uniqueness of SPDEs driven by time and space Poisson random measure were considered by, among others, Applebaum and Wu [2], Bié [8], Knoche [30], Mueller [34], and Mytnik [35]. Stochastic integration in Banach spaces has been discussed e.g. by Brooks and Dinculeanu [9], Dettweiler [19], Kussmaul [31], Neidhardt [36], Rüdiger [42], in which article Banach spaces of M type 1 and M type 2 were considered. SPDEs in M type 2 Banach spaces driven by Wiener noise have been investigated by Brzeźniak [10,11], Elworthy and Brzeźniak [12].
We extend the Itô stochastic integral in the L p -mean, 1 ≤ p ≤ 2 to a wide class of Banach spaces. This allows us to investigate the existence and uniqueness of the solution to (1) a large class of Banach spaces (see e.g. Example 2.2).
This article is organized as follows: In section two we state our main theorem and some examples. In section three we recall some results about Poisson random measures, Lévy processes, and stochastic integration. We then prove our results in sections four and five.

The main result
There exist many connections between the validity of certain theorems for Banach space valued processes and the geometric structure of the underlying Banach space. We have omitted a detailed introduction to this topic as this would exceed the scope of the paper. A short summary of stochastic integration in Banach spaces is given in Chapter 3.2. Remark 2.1. This definition is equivalent to uniform p-smoothability. For literature on this subject see e.g. Brzeźniak [10], Burkholder [13], Dettweiler [18,19], Pisier [39] and Woyczyński [48,49]. Moreover, for δ ∈ R we denote by V δ the domain of the fractional power of −A (for the exact definition we refer to Appendix B). Now we can formulate our main result. Theorem 2.1. Let 1 < p ≤ 2 and E and Z be separable Banach spaces, where E is of M type p and let E and Z be theirs Borel σ-algebras. Let A be an infinitesimal generator of a compact, analytic semigroup on E. Let (Ω, F, P ) be a probability space with given right-continuous filtration (F t ) t≥0 and η : Z×B(R + ) → R be a Poisson random measure with characteristic measure ν ∈ L sym (Z) over (Ω, F, P). Let p, q ∈ [1, ∞) be two constants such that 1 < p ≤ 2 and p < q. Let δ f and δ g be two constants and f : E → V −δ f and g : E → L(Z, V −δg ) two mappings satisfying the following global Lipschitz conditions x, y ∈ E, l = 1, . . . , n, where C 1 and C 2 are some constants. Let γ ≥ 0 be fixed. Then the following holds: a.) Let q < ∞. Assume that the constants δ f , δ g , δ and γ satisfy the following conditions (i) δ g q < 1, and δ f < 1, (ii) (δ g − γ)p < 1 − 1 q , and δ f − γ < 1 − 1 q , (iii) γ < 1 q , (iv) δ > max( 1 q + δ g , δ f − 1 + 1 q , 1 q ). If the initial condition u 0 satisfies E|u 0 | q −γ < ∞, then there exists a unique mild solution to Problem (1) such that for any T > 0 b.) Let q < ∞. Assume that the constants δ f , δ g , δ and γ satisfy the following conditions (i) δ g p < 1, and δ f < 1, (ii) (δ g − γ)p < 1 − 1 q , and . If the initial condition u 0 satisfies E|u 0 | p −γ < ∞, then there exists a unique mild solution to Problem (1), such that for any T > 0 c.) Let q = ∞ and γ = 0 and δ > max(δ f , δ g ). Assume that δ g p < 1 and δ f < 1. If initial condition u 0 satisfies E|u 0 | p < ∞, then there exists a unique mild solution to Problem (1), such that for any T > 0 sup 0≤s≤T E|u(s)| p < ∞, and for all δ > 0 u ∈ L 0 (Ω; ID([0, T ]; V −δ ))).
where C 1 and C 2 are some generic constants.
Let ν : B(Z) → R + be a not necessarily symmetric Lévy measure and η : B(E)×B(R + ) → R + be a Poisson random measure with characteristic measure ν. Let γ(t) : B(Z)×B(R + ) → R + be the compensator of η, i.e. the unique predictable random measure such that is a martingale for each A ∈ B(Z). Letη := η − γ be the compensated Poisson random measure. Then by small modification (see Remark 3.3 and Remark 3.6) it can be shown, that the Theorem 2.1 holds also for the following SPDE Example 2.2. Let 1 < α < 2. Take any p ∈ (α, ∞) and put E = W p Then the formula defines an Z-valued, α-stable symmetric process. Moreover, the process is of integrable p-variation (see 3.3). Our interest is the existence and uniqueness of the solutions (u(t)) t≥0 to the following parabolic equation First, note that by Examples 3.9 and 3.10 the spaces W ϑ,p (O), ϑ ∈ R are of M type p. By multiplier theorems (see e.g. Runst and Sickel [43,Theorem 1,p. 190]) one sees that If d p < 2(1 − 1 α ) then Theorem 2.1 and Remark 2.2 give existence and uniqueness of the mild solution of (4) in L 0 (ID([0, T ]; L p (O)). Example 2.3. Fix p ∈ (1, 2] and d ∈ IN. Let (Ω; F, P) be a complete measurable probability space with usual filtration (F t ) t≥0 . Let θ : B(R) → R + be a symmetric Lévy measure such that be a Poisson random measure over (Ω; F, P) with characteristic measure ν defined by where λ d denotes the Lebegue measure in R d . We consider the following SPDE    du(t, ξ) = ∆u(t, ξ) and lim ξ→±∞ u(t, ξ) = 0, t ≥ 0. We will show, that if α < 2 d + 1 is satisfied, then for any T > 0 Theorem 2.1 will give existence and uniqueness for Problem (5).
In particular, let us define the function . The latter holds, since for some β > 0, the dual of W −γ p (R d ) can be continuously embedded in the Hölder spaceC β (R d ) (see e.g. Triebel [45,Chapter 4.6.1]). Let j be the embedding of Z in W −γ p (R d ). Then A = j −1 • f (A) and therefore A ∈ B(Z).
Let the measure σ : ∂U → R + be defined by σ(A) := λ(f −1 (A)), A ∈ B(∂U ), and λ denotes the Lebesgue measure. Now, the characteristic measure ν : B(Z) → R is defined by Let η : B(B)×B(R + ) → R + be the corresponding Poisson random measure. By Linde [32, Theorem 6.2.8] the process is well defined and coincides with the space-time white noise defined above (for definition see e.g. Bié [8]). We are interested in the problem which is equivalent to Problem (5). Theorem 2.1 gives existence and uniqueness of the solution u to Problem (6) such that In fact, one can show that there exists constant C 1 and C 2 , such that for all Therefore, the conditions of Theorem 2.1 are satisfied if A short calculation shows, that above is satisfied, if p < 2 d + 1. This is a condition which coincides with the condition of Bié [8]. In particular, for all β > n p , the unique solution u to Problem (6) exists and satisfies u, φ ∈ ID((0, T ]; R) for all φ ∈ W β p (R d ).

Poisson random measures and stochastic integration
3.1. Poisson random measures. Let (Ω, F, (F t ) t∈[0,T ] , P) be a filtered probability space. Let Z be an arbitrary separable Banach space with σ-algebra Z. A point process with state space Z is a sequence of Z × R + -valued random variables (Z i , T i ), i = 1, 2, . . . such that for each i, Z i is F T i -measurable. Given a point process, one usually works with the associated random measure η defined by A point process is called a Poisson point process with characteristic measure ν on (Z, Z), iff for each Borel set A ∈ Z with ν(A) < ∞ and for each t the counting process of the set A, i.e. the random variable N t (A) = η(A × [0, t]), has a Poisson distribution with parameter ν(A)t, i.e.
It follows, that the random measure η associated to a Poisson point process has independent increments and that η(A × [0, t]) and η(B × [0, t]) are independent for all A, B ∈ Z, A ∩ B = ∅, and t ≥ 0.
Let Z be a separable Banach space with Borel σ algebra Z. Starting with a measure ν on Z, one may ask under what conditions one can construct a Poisson random measure η on Z×B(R + ), such that ν is the characteristic measure of η. If the measure ν is finite and satisfies ν({0}) = 0, then the Poisson random measure η : Z×B(R + ) → R + exists. Moreover, one can show, that if ν ∈ L sym (Z) (see Definition 2.2), then the Poisson random measure η : Z×B(R + ) → R + exists.
Some useful properties of symmetric Lévy measures are stated in the following remark. Let Z be a separable Banach space with norm |·|, let Z be the Borel-σ algebra on Z and let Z be the topological dual of Z. Let ν ∈ L(Z) be a Lévy measure. Then the following holds true.
A typical example of Poisson random measure is provided by an α-stable Poisson random measure, 0 < α < 2.
Definition 3.1. A probability measure µ ∈ P(E) is said to be stable iff for each a, b > 0 there exists some c > 0 and an element z ∈ E, such that for all independent random variables X and Y with law µ we have The measure µ is called strictly stable if for all a > 0 and b > 0 one can choose z = 0 in (8).
Example 3.1. Let E be a separable Banach space and U = {x ∈ E | |x| ≤ 1}. Let σ : B(∂U ) → R + be an arbitrary finite measure and ν : B(E) → R + be defined by Then ν : B(E) → R + is a Lévy measure. Let η be the corresponding Poisson random measure. If E is of type p, p > α (for definition see the next Section), then the process L = (L t ) t≥0 defined by is an E-valued, unique, α-stable symmetric Lévy-process. On the other hand it is known from Linde [32, Theorem 6.2.8] that the Lévy measure of an α-stable random variable is given by its distribution on ∂U , i.e. by the measure σ : B(∂U ) → R + .
3.2. M type p Banach spaces: a short account. A stochastic integral is defined with respect to its integrator. If the integrator is of finite variation, then the integral is defined as a Stieltjes integral in a pathwise sense. If the integrator is a square integrable martingale, Itô's extension procedure yields to an integral defined on all previsible square integrable processes. Let Z and E be two separable Banach spaces with Borel σ algebras Z and E. Fix 1 ≤ p < 2. Let η : Z×B(R + ) → R + be a Poisson random measure with characteristic measure ν ∈ L sym (Z). We will define an integral with respect to ν, i.e.
with h : Ω×R + → L(Z, E) is a càglàd predictable step function, such that T 0 Z |h(s, z)| p ν(dz)ds < ∞. The question which needs to be answered is under which conditions on the underlying Banach space E and on the integrator, the stochastic integral in (9) can be extended to the set of all functions with finite L p -mean. First, we consider the case where h is a deterministic function, secondly we consider the case, where h is a random function.
Definition 3.2. (Linde [32]) Let E be a Banach space and 0 < p ≤ 2 be fixed. Let   [19]) Let 1 ≤ p ≤ 2 and let E be a separable Banach space. An E-valued process X = (X(t)) t≥0 is said to be of integrable p-variation iff for any T > 0 one can find a constant C(T ) such that for any partition 0 = t 0 < t 1 < · · · < t n = T one has Example 3.5. Let E be a Hilbert space and (W (t)) t≥0 be an E-valued Wiener process with covariance operator Q. If Q is of trace class, then the Wiener process (W (t)) t≥0 is of integrable 2-variation.
Example 3.6. Let E be a Hilbert space and M = (M t ) t≥0 be an E-valued stationary Lévy process with bounded second moment. Then the Lévy-Khintchine formula implies that M is of integrable 2-variation.
Example 3.7. Let E be a Hilbert space. Let ν ∈ L sym (E) such that E |z| 2 ν(dz) < ∞. Then one can construct a unique Poisson random measure η : B(E)×B([0, T ]) → R + such that ν is the characteristic measure of η. Moreover, the process L t = t 0+ E z η(dz, ds) is of integrable 2-variation.
A Hilbert space is a Banach space of type 2. Let p ∈ (0, 2]. In case the underlying Banach space E is of type p and the Lévy measure is p integrable, the Example 3.7 can be transferred to E.  [17], in the proof of the implication of (ii) ⇒ (i) on p. 129 of Proposition 2.3, [19] or Hamedani and Mandrekar [25]) Let E and Z be separable Banach spaces, E of type p, p ∈ (1, 2]. Let ν ∈ L sym (Z, B(Z)) and h ∈ L(Z, E), such that Let η be a Poisson random measure on B(Z)×B(R + ) with characteristic measure ν. Then a Lévy process L = (L(t)) t≥0 exists such that Moreover, L is a martingale and is of integrable p-variation.

Remark 3.2.
To be more precise, Dettweiler shows in his paper [17] By means of this inequality one can extend the stochastic integral in p mean (see e.g. Chapter 3.3, or Woyczyński [48, Theorem 2.2], Dettweiler [19]).
To deal with moments of higher order, stronger inequalities are needed. In fact using the techniques described in Burkholder [14] one can prove a generalized version of an inequality of Burkholder type. In particular, let E be a Banach space of M type p, 1 ≤ p ≤ 2. Then it can be shown, that there exists a constant C < ∞, such that we have for all discrete E-valued martingales M = (M 1 , M 2 , . . .) and 1 ≤ r < ∞ ( see e.g. Assuad [3], Brzeźniak [ Remark 3.5. An interested reader can consult the following articles: Burkholder [13,15], Pisier [38,39] and Woyczyński [48,49] for the connection of Banach spaces of M type p and their geometric properties. Brzeźniak [10], Dettweiler [19,20,17] Neidhardt [36] and Rüdiger [42] for the connection between Banach spaces of M type p and stochastic integration. Linde [32], and the articles Dettweiler [17,18], Gine and Arujo [24], Hamedani and Mandrekar [25] for the connection between Lévy processes and Banach spaces of type p.

The Stochastic
Integral in M type p Banach spaces. In the following let p ∈ (1, 2] be fixed. Let Z be a separable Banach space and E be a M type p Banach space. The Borel σ algebras of Z and E are denoted by Z and E respectively.
As mentioned in the introduction the stochastic integral will first be defined on the set of predictable and simple function. To be precise a process h : where 0 = t 0 < · · · < t n = T is a partition of [0, T ] consisting of stopping times, and Let η : Z×B([0, T ]) → R + be a Poisson random measure with characteristic measure ν ∈ L sym (Z, Z). The stochastic integral with respect to η is a linear operator I : S → ID defined by where h ∈ S has the representation (12).
The next step is to extend the stochastic integral to the set IL p (ν), where Since IL p (ν) is separable, by standard arguments, one can show, that S ∩ IL p (ν) is dense in IL p (ν). Let ID p the Skorohod space equipped with the following norm Note, the Skorohod space ID([0, T ]; E) topologized with uniform convergence is a complete metric space, but not separable. Analysing the proof of completeness of L p spaces over an arbitrary measurable set, one can see that the space ID p is a complete normed space. By means of the following proposition, it can be shown that the stochastic integral defined in (13) is a continuous operator from S ∩ IL p (ν) into ID p (see e.g. Woyczyński [48, Theorem 2.2] or Dettweiler [19]).  (12) we have Proof. Let h = {h(s, z), 0 ≤ s ≤ T, z ∈ Z} be a simple, predictable process written in the form of (12). Now, the stochastic integral is defined by the Riemann integral (see (13) Thus, we can apply the generalized Burkholder inequality (11) to get Moreover, the process defined by is a martingale. In fact the integrability conditions are given by (15). Further, let 0 ≤ s < t ≤ T and let j ∈ {1, . . . , n} be this index for which t j ≤ s < t j+1 holds. Then But ν is symmetric. Thus it follows and we can apply the Doob's maximal inequality for martingales to get The Jensen inequality yields The tower property of the conditional expectation leads to Inserting (18) in equation (17) yields which can be written as In case of r > p, we can only apply the martingale inequality (11) and have to stop before equation (16).
By means of Proposition 3.2 the stochastic integral is a continuous operator from S ∩ IL p (ν) into ID p . The next point is to reassure, that S ∩ IL p (ν) is dense in IL p (ν). But in fact IL p (ν) equipped with the predictable σ algebra is a measurable separable metric space and by a modification of the Proof of Proposition I.4.7, Da Prato and Zabczyk [16] one can show that S∩IL p (ν) is dense in IL p (ν). Since ID p is complete, there exists a bounded linear operator I : IL p (ν) → ID p , which is an extension of the operator introduced in (13). In the next Proposition we will show, that the inequalities in Proposition 3.2 are preserved.
Proof. To show that S ∩ IL p (ν) is dense in IL p (ν), one can modify the Proof of Proposition I.4.7 of Da Prato and Zabczyk [16]. Thus there exists a sequence where the step functions h j can be written as a sum of the following type -measurable random variables. The stochastic integral for h j defined in (13) is given by .
The inequalities of Proposition (3.3) can be extended to higher order moments.
Let E and Z be two separable Banach spaces, E be of M type p. Let Z and E be the Borel σ-algebras on Z and E. Let η be Poisson random measure on Z with characteristic measure ν ∈ L sym (Z, Z). Let h ∈ IL p (ν). Let q = p n for some n ∈ IN and suppose, that holds. Then there exists some constant C < ∞ such that Proof. The proof is a generalization of the proofs of Bass (14) we have Let us define The process X t is a real valued semimartingale and by condition (21) of finite variation. Moreover, it only has positive jumps, i.e. it is a subordinator. The associated random measure Let γ X be the compensator of η X , i.e. the unique predictable random measure on Since (21) holds, L(t) (0) is a real-valued martingale, only has positive jumps and is of finite variation. Note, L (0) t has a characteristic function of the following form E exp iξL Since η and η X are Poisson random measures and by the uniqueness of the compensator we infer that

Simple calculations leads to
In case n = 2, we have By the definition of the compensator, in particular since By the discussion above, i.e. relation (23) we have Thus, the proposition is proven, provided n = 2. In case n > 2, we have to continue. In particular, let z p r (η X − γ X )(dz; ds) for r = 1, . . . , n.
Since (21) and (22) holds, L(t) (r) is a real-valued martingale, has only positive jumps and is of finite variation for r = 1, . . . , n. Moreover, since R is of M type 2, R is also of M type p for all p ∈ [1, 2]. Using inequality (14) we have Using simple calculations we get That means we have Note, since R is of M type p, we have by Proposition (3.3) Note, that R is also of type 1, that means of M type 1. Thus we have Substitution of (26) to (25) gives Iteration of the calculation (24) and substitution of (27) leads to Assumption (22) and interpolation yields Remark 3.6. By Remark 3.3 it follows, that Proposition 3.3 and Proposition 3.1 remain valid also if η is a compensated Poisson random measure with characteristic measure ν such that ν is a Lévy measure on (Z, Z).

Proof of Existence and Uniqueness of Solutions
The proof of existence and uniqueness is based on the Banach fixed point theorem for contractions (see e.g. Zeidler [50, Theorem 1.A, Chapter 1.1]). We will first prove part (a) and then secondly sketch the proof of part (b) and (c).
For p ≤ r < ∞ and δ ∈ R let V r,δ := ϕ ∈ L 0 (Ω; V δ ), such that equipped with norm Let δ f and δ g be given. We say the constants δ f , δ g , δ and γ satisfy the assumption (A) iff δ g q < 1 and δ f < 1, is a metrizable topological space and the resulting metric is complete (see e.g. [23, Theorem 5.6, Chapter 3]). Moreover, by Theorem 1.7, Chapter 3 in [23], and since V −δ is separable, L 0 (Ω; ID([0, T ]; V −δ )) is also a complete separable metric space with respect to the Prohorov metric (for the definition see e.g. [23,Chapter 3.1] or the Appendix, Chapter A.1). Further, the completion of V ID q,q (T ) with respect to the norm given in (1) Firstly we will show that for all ϕ ∈ V q,−γ the operator ). (2) Secondly, we will show that there exists a constantT > 0 such that for all ϕ ∈ V q,−γ the operator K ϕ has a unique fixed point x in V q,q (T ) and x ∈ L 0 (Ω; ID([0,T ]; V −δ ). This will be shown in the following two substeps: (i) there exists constantsT > 0 and 0 < k < 1, such that for all ϕ ∈ V q,−γ the operator K ϕ : V ID q,q (T ) → V q,q (T ) is Lipschitz continuous with respect to the norm in V q,q (T ) with Lipschitz constant k.
(ii) For all ϕ ∈ V q,−γ the sequence x (n) n∈IN defined by x (n) = K ϕ x (n−1) , n ≥ 1 and  and for all ϕ ∈ V q,−γ the operator K ϕ : V ID q,q → V p,∞,−γ is bounded and Lipschitz continuous. In particular, there exists some constant and there exists a constant C < ∞, such that Proof. Inequality (31) follows from a sequence of calculations: The generalized Burkholder inequality implies The Lipschitz continuity, the Hölder inequality and the Jensen inequality give where r satisfies (δ g − γ)p < 1 − 1 r and δ f − γ < 1 − 1 r . Since q satisfies the assumptions above, inequality (31) follows. Inequality (32) follows from similar calculation: Again, the Lipschitz continuity, the Hölder inequality and the Jensen inequality give r and δ f − γ < 1 − 1 r . Since q satisfies the assumptions above, (32) follows.
Proof. A short calculation shows, Thus, we have Since γ < δ, we have for the first term

The Minkowski inequality yields for
The Lipschitz property of f and the Hölder inequality lead to The Jensen inequality and the fact that δ > δ f − 1 + 1 q give The Burkholder inequality gives for the third term The Lipschitz property of g gives The Minkowski inequality gives for the fourth term The Hölder inequality gives for q = q The Jensen and the Burkholder inequality give The Lipschitz property yields Collecting all together gives the assertion.

Proof of step (1)-(i):
In this section we will show, that under assumptions (A), (B) and (C) and for all ϕ ∈ V q,−γ the operator K ϕ maps V ID q,q (T ) into V q,q (T ). Note that for 0 < γ < 1 Using Remark B.1 we have Furthermore we have Therefore, by the Young inequality we infer that Next, the Lipschitz condition of f , i.e. (2), implies that By Proposition 3.1 we get By the Lipschitz condition of g, i.e. (4), we infer that Again, the Young inequality yields Summing up, we have proved that the operator K ϕ maps V q,q (T ) into V q,q (T ). Proof of step (1)-(ii). In order to show K ϕ u belongs to ID([0, ∞); V −δ ), we will show, that the set {K ϕ u} satisfies the Aldou's condition (see Definition A.2) and the compact containment condition (see Definition A.3).
Proof that the Aldou's condition is satisfied: We will show, that {K ϕ u} satisfies the assumptions of Theorem A.1 with β = p. In particular, we will show that there exists some ρ > 0 such that for all t ∈ [0, T ] and all θ > 0 we have By (31) there exists some C 1 , C 2 < ∞ such that Therefore there exists some constant C < ∞ such that C θ ρ + θ δ is an upper bound for the RHS of (38). Moreover C θ ρ + θ δ → 0 as θ → 0 and the assumptions of Lemma A.1 are satisfied.
Proof. To prove inequality (38), we first note that Using the same calculations as in (33), (34) and (35) we obtain where η • θ t (ω) := η(θ t • ω), and θ t is the usual shift operator defined by θ t ω(s) := ω(t + s). Following (36) and (37) we get Proof that the compact containment condition is satisfied: We will show A.3 by Lemma A.1. By Remark B.2 the space V −δ is compactly embedded in V −δ for allδ < δ. Let δ > 0 be chosen accordingly Moreover, by Claim 4.2 there exists some constant By Lemma A.1 the compact containment condition follows. Proof of step (2)-(i) Next, we will show, that there exists some function C : [0, 1] → R + such that and C(T ) → 0 as T → 0. Thus, we can find aT , such that C(T ) < 1.
In particular, there exists a constant 0 < k < 1 such that for all ϕ ∈ V q,−γ the operator Proof. Similarly to calculations (33), (34) and (35) the Lipschitz continuity of f , i.e. (2), leads to where the constant C(q, T ) tends to zero as T → 0. Next we have Again, the Young inequality yields Summing up we have proved where C(T ) tends to zero as T → 0.
Proof of step (2)-(ii) Let u (n) n≥0 be the sequence defined by u (n) := K ϕ u (n−1) , n ≥ 1 and u (0) (t) = S(t)ϕ and let u the fixed point of K ϕ . We have to show, that u (n) n≥1 is tight in L 0 (Ω; ID([0,T ]; V −δ )). But first we will show the following.
• Under the assumptions (A) and (C) the set u (n) n≥0 is bounded in V q,q (T ). By (2)-(i) we know, that K ϕ : V q,q (T ) → V q,q (T ) is a strict contraction. Therefore there exists a constant 0 < k < 1, such that • If the assumptions (A), (B) and (C) are satisfied then it follows from (31) that • Since |||u (n) −u ||| q,q → 0 as n → ∞ it follows from (32) that the sequence as n → ∞. Therefore, the Chebyscheff inequality shows, that for any finite set as n → ∞.
Proof that the Aldou's condition is satisfied: Tracing the calculations in (1)-(ii), we see that there exists some ρ > 0 such that for all n ∈ IN But from the consideration before we know sup n≥1 |||u (n) ||| q,q < ∞ and sup n≥1 |||K ϕ u (n) ||| p,∞,−γ < ∞ Thus, there exists some constants C < ∞ and ρ > 0 such that This proves that the assumptions of Lemma A.1 are satisfied.
Proof that the compact containment condition is satisfied: The embedding Vδ → V −δ is compact for allδ < δ (see Remark B.2). Moreover, by Claim 4.2 it follows Since the set {u (n) | n ∈ IN} is bounded in V q,q (T ), the compact containment condition follows. Proof of step (3) LetT be so small that C(T ) < 1, where C : [0, 1] → R + is the constant from (41). Then for all ϕ ∈ V q,−γ there exists a unique fixed point u ∈ V q,q (T ) such that K ϕ u = u . Let u 0 be the fixed point of K u 0 in V q,q (T ). By (2)-(ii), the fixed point belongs to V ID q,q (T ). If then Step (2) can be repeated. But, by the following calculation we can show (42): The first summand is obviously finite, because ϕ ∈ V q,−γ . The second summand is finite, because of (B). In particular In the case of the third summand, we study only the worst case if l = n. By the Lipschitz condition of g, i.e. (4), we have Because of (A), we have γ ≥ max(δ f − 1 + 1 q , δ g ) and we can write E|u 0 (s−)| q ds .

Thus, we have
E|u 0 (T )| q −γ ≤ C(T ) 1 + |||u 0 ||| q q,q . Therefore, there exists a unique fixed point u of K u 0 (T ) in V ID q,q (T ). Let u 1 := u . Using the same calculations as above we can show that E|u 1 (T )| q −γ < ∞. Hence, we can repeat the same argument. SinceT only depends on δ g , δ f , p and q, we only need to repeat the procedure a finite number of times. In particular, there exists a constant m, such that T ≤T m. Let u i be the unique fixed point of (0), it follows that K u 0 u = u and therefore that u is a solution of Problem (1).

Proof of
For p ≤ r < ∞ let V r,−γ be given by definition (29). In the proof we will consider the case q < ∞, the case q = ∞ is similar. To be precise Part (c) can be proven by tracing the proof of part (b), setting q = ∞ and γ = 0. Let δ > 0 be arbitrary. Let For ϕ ∈ V p,−γ let K ϕ : V ID p,q (T ) → V ID p,q (T ) be the same operator as in (30), in particular As before existence and uniqueness of the solution to (1) will be shown by fixed point arguments. But first, we will introduce the conditions on δ f , δ σ , δ and γ.
(A) iff δ g p < 1 and δ f < 1, Analogously to part (a), the proof of part (b) will rely on the following steps: (1) Firstly, we will show, that the operator K ϕ maps V ID p,q (T ) into V ID p,q (T ), in particular we will show that . ( ). (2) Secondly, we will show that there exists some constantT > 0, such that for all ϕ ∈ V p,−γ the operator K ϕ has a unique fixed point x inV ID p,q (T ) and x belongs to L 0 (Ω; ID([0,T ]; V −δ ). This will be shown in the following two steps: (i) there exists a constantT > 0 and a constant 0 < k < 1, such that for all ϕ ∈ V p,−γ the operator K ϕ : V p,q (T ) → V p,q (T ) is Lipschitz continuous with constant k. (ii) For all ϕ ∈ V p,−γ the sequence x (n) n∈IN , defined by x (n) = K ϕ x (n−1) , n ≥ 1 and x (0) (t) = S t ϕ, is tight in L 0 (Ω; ID([0,T ]; V −δ ). Applying the Banach fixed point theorem, there exists a unique x * ∈ V p,q (T ), such that K ϕ x * = x * and K ϕ (n) y → x * for all y ∈ V q,q (T ). From (2)  and for all ϕ ∈ V p,−γ the operator K ϕ : V ID p,q → V p,∞,−γ is bounded and Lipschitz continuous. In particular, there exists some constant C 1 , C 2 < ∞, such that and there exists a constant C < ∞, such that Proof. Inequality (43) follows from the same sequence of calculations as (31). In particular, The Hölder inequality gives where q satisfies (δ g − γ)p < 1 − 1 q and δ f − γ < 1 − 1 q . Inequality (44) follows from similar calculations. Then, there exists a constant C 1 , C 2 , C 3 < ∞ such that for all ϕ ∈ V q,−γ and all u ∈ V q,q (T ) we have Proof. The proof is similar to the proof of Claim 4.2. We have Since γ < δ, we have for the first term The Minkowski inequality yields for = δ − δ f Let 1 , 2 > 0 and 1 + 2 = . Then, the Lipschitz property of f and the Hölder inequality lead to The Jensen inequality, the Hölder inequality and the fact that δ > δ f − 1 + 1 q give The Burkholder inequality gives for the third term The Lipschitz property of g gives The Minkowski inequality gives for the fourth term ds Let 1 , 2 > 0 and 1 + 2 = . The Hölder inequality gives for p = p The Jensen, the Burkholder inequality and the Lipschitz property of g give Again the Hölder inequality yields to Collecting all together gives the assertion.

Proof of part (1)-(i):
In this section we will show, that under conditions (A) and (C) and for all ϕ ∈ V p,−γ the operator K ϕ : V ID p,q (T ) → V p,q (T ) is a bounded operator. Note we have for 0 < γ < 1 Using Remark B.1 we have Let c 1 , c 2 ≥ 0 such that c 1 + c 2 = 1 and p conjugate to p. Then we have by the Hölder inequality A second application of the Hölder inequality yields Therefore, by the Young inequality we infer Next, the Lipschitz condition of f , i.e. (2), implies that By Proposition 3.1 we get Again, the Young inequality yields The Lipschitz condition of g, i.e. (4), leads to Summing up, we have proved that the operator maps V p,q (T ) into V p,q (T ), i.e.
Remark 4.1. Note, that using the Lipschitz conditions (2) and (4) the equation (48) can be written as Proof of part (1)-(ii). Fix u ∈ V ID p,q (T ). We have to show that K ϕ u belongs to the Skorohod space ID([0, T ]; V −δ ), in particular that {K ϕ u} satisfies the Aldou's condition A.2 and the compact containment condition A.3 in Appendix A. Since V −δ is compactly embedded in V −δ for allδ < δ, Claim (4.4) implies the compact containment condition. The Aldou's condition will be shown by proving the assumptions of Theorem A.1 with β = p. In particular, we will prove that there exists some ρ > 0 such that for all 0 ≤ t ≤ T and all θ we have Since (43) holds, there exists some C < ∞ such that 2C θ ρ + θ δ is an upper bound for the RHS of (49). Moreover, 2C θ ρ + θ δ → 0 as θ → 0, the assumptions of Lemma A.1 are satisfied. To prove inequality in (49), we first note that for r < q such that 1 − δ f ≤ 1 r and 1 p − δ g ≤ 1 r we have .
Using the same calculation as for (40) we have • If the assumptions (A), (B) and (C) are satisfied, then it follows from (43) that for all • Since for all ϕ ∈ V p,−γ we have |||u (n) − u ||| p,q → 0 as n → ∞ it follows from (44) that as n → ∞. Therefore, the Chebyscheff inequality shows, that for any finite set as n → ∞.
By Theorem 3.7.8 of Ethier and Kurtz [23] it remains to show, that the set {u (n) , n ∈ IN} is tight in L 0 (Ω; ID([0,T ]; V −γ−δ )). This implies that u (n) converges in distribution to u and u ∈ L 0 (Ω; ID([0,T ]; V −γ−δ )). Proof that the Aldou's condition is satisfied: Tracing the calculation in (b)-(1)-(ii), we can see that there exists some ρ > 0 such that for all n ∈ IN t+θ] u (n−1) ||| p p,q + C 2 θ pδ |||1 (0,t] K ϕ u (n−1) ||| p p,∞,−γ , But (43) implies, that there exists some constant C < ∞ such that the RHS is bounded by C θ ρ + θ pδ . Therefore, the assumptions of Lemma A.1 are satisfied. Proof that the compact containment condition is satisfied: In order to prove the compact containment condition we use again the fact that the embedding Vδ → V −δ is compact for allδ < δ (see Remark B.2). Claim 4.4 implies Since the set {u (n) | n ∈ IN} is bounded in V q,q (T ), i.e. there exists some C < ∞ such that |||u (n) ||| q,q ≤ C, the compact containment condition follows. Proof of part (3) This part can be shown by the same consideration as in (a)-(3).

A.s. Regularity Results -Proof of Remark 2.2
Let Z be a separable Banach space, Z the Borel-σ algebra and η : Z×B(R + ) → R + a Poisson random measure on Z × R + with characteristic measure ν : Z → R + ∈ L sym (Z). Let N t (A), A ∈ Z, be the counting process defined on page 1503. Let (Ω, F, P) be a complete probability space and (F t ) t≥0 be the right continuous filtration induced by η. That means, the smallest filtration, such that the counting measure N t (A) is F t -measurable for all s ≤ t ≤ T and A ∈ Z. Let Z 0 = {z ∈ Z | |z| ≤ 1}, , and . Since for A, B ∈ Z, A ∩ B = ∅, the random variables η(A × (s 1 , s 2 ]) and η(A × (s 1 , s 2 ]) are independent for all 0 < s 1 ≤ s 2 ≤ T , the two filtration F 0 t and F C t are independent. Let us define the two probability spaces (Ω 0 , F 0 , {F t } t , P 0 ) and (Ω C , F C , {F C t } t , P C ), where Ω 0 := Ω, Ω C := Ω, F 0 := ∧ t F 0 t , F C := ∧ t F C t , P 0 ( · ) := P( · |F C ), and P C ( · ) := P( · |F 0 ). From the independence of F 0 and F C it follows, that P is the product of P 0 and P C .
We have to show, that lim n→∞ū

Appendix A. The Skorohod Space
For an introduction to the Skorohod space we refer to Ethier and Kurtz [23] and Jacod and Shiryaev [27]. The results stated in this chapter are taken from Ethier and Kurtz [23].
Let  If X is finite dimensional, then tightness of the sequence {x (n) | n ∈ IN} can be shown by the Aldou's condition.
Note, in the definition above, only τ n , n ∈ IN are stopping times, θ is a constant. In case of E being infinite dimensional we have to add to the Aldou's condition a compact containment condition (see e.g. Ethier and Kurtz [23,Chapter 3]) .
Definition A.3. Let {x (n) | n ∈ IN} be a sequence of stochastic processes with sample paths belonging to ID([0, T ]; R) and x (n) ∈ L 0 (Ω (n) , F (n) , P (n) ), n ∈ IN. Let (F (n) t ) be the natural filtration induced by x (n) , n ∈ IN. We say this sequence {x (n) | n ∈ IN} satisfies the compact containment condition, iff for each > 0 and every rational t > 0, there exists a compact subset K ,t ⊂ X, such that Let X be a separable Banach space with norm | · |. The Aldou's condition is often difficult to verify directly. Thus, to show that {x (n) | n ∈ IN} satisfies the Aldou's condition A.2, one can show certain integrability conditions given the Lemma below (see Ethier and Kurtz [23, Chapter 3.8]).
Theorem A.1. (see Ethier and Kurtz [23, Theorem 8.6, Remark 8.7 in Chapter 3]) Let X be a separable Banach space and let {x (n) | n ∈ IN} be a sequence of stochastic processes with sample paths in ID([0, T ]; X), such that x (n) ∈ L 0 (Ω (n) , F (n) , P (n) ), n ∈ IN. Let (F (n) t ) be the natural filtration induced by x (n) , n ∈ IN. Then {x (n) | n ∈ IN} satisfies the Aldou's condition, if there exists an p > 0 and a family f (n) (δ) : 0 < δ < 1, n ∈ IN of nonnegative random variables, such that for all t ∈ [0, T ] and h ∈ (0, δ] Similarly, the following lemma gives an integrability condition which implies A.3. Lemma A.1. Let {x (n) | n ∈ IN} be a sequence of stochastic processes with sample path in ID([0, T ]; R), such that x (n) ∈ L 0 (Ω (n) , F (n) , P (n) ), n ∈ IN. Let Γ be a subspace of E with norm | · | Γ , such that the embedding Γ → E is compact (see Remark B.2). Then the compact containment condition A.3 is satisfied, if some p > 0 exists, such that where the path γ is a piecewise smooth part Σ \ R + going from ∞e −iδ to ∞e iδ for some δ > 0.
Definition B.5. Let A be a sectorial operator. For every α > 0 we define For α = 0, A α = I.
Remark B.1. In the following we will use certain interpolation inequalities. In particular let E be a separable Banach space, A be a operator generating an analytic semigroup (exp(−ωt)S t ) t≥0 of contraction on E. Then there exists a constant C < ∞, such that we have (see e.g. (see e.g. I would like to thank the anonymous referee for his carefull proof reading and helpful remarks.