ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8],...

52
ZiF: Research Group STOCHASTIC MODELING IN THE SCIENCES: STOCHASTIC PARTIAL DIFFERENTIAL EQUATIONS AND RANDOM MEDIA 1 April 2005 – 15 September 2005 Preprint No. 13/2005 EXISTENCE, UNIQUENESS AND REGULARITY OF PARABOLIC SPDES DRIVEN BY POISSON RANDOM MEASURE Erika Hausenblas University of Salzburg Department of Mathematics [email protected]

Transcript of ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8],...

Page 1: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

ZiF: Research Group

STOCHASTIC MODELING IN THE SCIENCES: STOCHASTIC PARTIAL DIFFERENTIAL EQUATIONS

AND RANDOM MEDIA

1 April 2005 – 15 September 2005

Preprint No. 13/2005

EXISTENCE, UNIQUENESS AND REGULARITY OF PARABOLIC SPDES DRIVEN BY POISSON RANDOM MEASURE

Erika Hausenblas

University of Salzburg Department of Mathematics [email protected]

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EXISTENCE, UNIQUENESS AND REGULARITY OFPARABOLIC SPDES DRIVEN BY POISSON RANDOM

MEASURE

ERIKA HAUSENBLAS

Abstract. In this paper we investigate SPDEs in certain Banach spacesdriven by a Poisson random measure. We show existence and uniquenessof the solution, investigate certain integrability properties and verify thecadlag property.

1. Introduction

The classical Ito stochastic integral with respect to Brownian motion hasbeen generalized in several directions. One of these directions is to definethe stochastic integral in the Lp–mean, 1 < p ≤ 2 (see Bichteler [5, 6]).This can be easily done if the integrator is of integrable p–variation. Theworks of Dettweiler [18], Gine and Marcus [23], Dang Hung Thang [43] haveextended this procedure to certain types of Banach spaces in which mostof the probabilistic theorems necessary for defining a stochastic integral arevalid.

If the integrator is a Poisson random measure it is quite natural to definethe stochastic integral in Lp-mean. In the case of SPDEs, examples canbe covered which are not easily treated using Hilbert space or M–type 2Banach 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 Borelσ-algebra E . Let A be an infinitesimal generator of an analytic semigroupS 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 Levymeasure ν : Z → R

+ satisfying certain moment conditions. Assume thatf : E → E, and g : E × Z → E are Lipschitz continuous and measurable

Date: 21st May 2005.1991 Mathematics Subject Classification. Primary 60H15; Secondary 60G57.This work was supported by the Austrian Academy of Science, APART 700.

1

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2 ERIKA HAUSENBLAS

functions. We consider the following SPDE written in the Ito-form{u(t) dt = (Au(t−) + f(u(t−))) dt +

∫Z g(u(t−); z)η(dz; dt),

u(0) = u0 ∈ E.(1)

By a mild solution of equation (1) we understand a predictable E-valuedcadlag process u satisfying the integral equation

u(t) = Stu0 +∫ t

0+St−sf(u(s−)) ds +∫ t

0+

∫Z

St−sg(u(s−); z) η(ds, dz), a.s. , t ≥ 0.

In Theorem 2.1 we find conditions which guarantee existence and uniquenessof the solution u of problem (1).

In contrast to the Wiener process, the Levy process

L(t) =∫ t

0+

∫Z

z η(dz; ds), t ≥,

itself is a.s. discontinuous, however L = {L(t), t ≥ 0} is cadlag. ParabolicSPDEs driven by the Gaussian white noise were introduced and discussedinitially by Walsh [45, 46], where he also mentioned as an example the ca-ble equation driven by a Poisson random measure. Kallianpur and Xiong[27, 28] showed existence and uniqueness for equation (1) in the space of dis-tributions, while Albeverio, Wu and Zhang [1] investigated equation (1) andshowed existence and uniqueness in Hilbert spaces under the L2–integrabilitycondition of the Poisson random measure under some suitable hypothesis.Existence and uniqueness of SPDEs driven by time and space Poisson ran-dom measure are considered by, among others, Applebaum and Wu [2], Bie[7], Knoche [29], Mueller [33], and Mytnik [34]. Stochastic integration inBanach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18],Kussmaul [30], Neidhardt [35], Rudiger [41], in which article Banach spacesof M–type 1 and M–type 2 are considered. SPDEs in M -type 2 Banachspaces driven by Wiener noise are handled by Brzezniak [9, 10], Elworthyand Brzezniak [11].

We extend the Ito stochastic integral in the Lp–mean, 1 ≤ p ≤ 2 to awide class of Banach spaces. This allows us to investigate the existence anduniqueness 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 maintheorem and some examples. In section three we recall some results about

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SPDEs driven by Poisson Random Measure 21st May 2005 3

Poisson random measures, Levy processes, and stochastic integration. Wethen prove our results in sections four and five.

2. The main result

There exist many connections between the validity of certain theorems forBanach space valued processes and the geometric structure of the underlyingBanach space. We omit a detailed introduction to this topic as this wouldexceed the scope of the paper. A short summary of stochastic integrationin Banach spaces is given in Chapter 3.2.

Definition 2.1. (see Pisier [38]) Let 1 ≤ p ≤ 2. A Banach space E is ofM type p, iff there exists a constant C = C(E, p) > 0 such that for anyE-valued discrete martingale (M0,M1,M2, . . .) with M0 = 0 the followinginequality holds

supn≥1

E|Mn|p ≤ C∑n≥1

E|Mn − Mn−1|p.

Remark 2.1. This definition is equivalent to uniform p-smoothability. Forliterature on this subject see e.g. Brzezniak [9], Burkholder [12], Dettweiler[17, 18], Pisier [38], Rudiger [41] and Woyczynski [47, 48].

Example 2.1. Let O be a bounded domain in Rd and p > 1. Then Lp(O)

is of M -type p ∧ 2 (see e.g. [47, Chapter 2, Example 2.2]).

Definition 2.2. (see Linde [31, Chapter 5.4]) Let E be a separable Banachspace and E′ be the topological dual of E. Let E be the Borel-σ-algebra ofE. A Borel-measure ν : E → R

+ is called a Levy measure if it is σ–finite,ν({0}) = 0, and the function

E′ � a �→ exp(∫

E(ei〈x,a〉 − 1) ν(dx)

)∈ C

is a characteristic function of a Radon measure on E. If in addition ν(A) =ν(−A) for all A ∈ E, then ν is called symmetric Levy measure. Lsym(E)denotes the set of all symmetric Levy measures on (E, E).

Moreover, for δ ∈ R we denote by Vδ the domain of the fractional powerof −A, i.e. Vδ := D((−A)δ), equipped with the norm | · |δ := |(−A)δ · | (seeAppendix 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 the Borel σ–algebras. Let A

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4 ERIKA HAUSENBLAS

be an infinitesimal generator of a compact, analytic semigroup on E. Let(Ω,F , P ) be a probability space with given right-continuous filtration (Ft)t≥0

and η : Z×B(R+) → R be a Poisson random measure with characteristicmeasure ν ∈ Lsym(Z) over (Ω,F , P). Let p, q ∈ [1,∞) be two constantssuch that 1 < p ≤ 2 and p < q. Let δf and δg be two constants andf : E → V−δf

and g : E → L(Z, V−δg ) two mappings satisfying the followinggrowth condition, i.e.

|f(x)|−δf≤ C1(1 + |x|), x ∈ E,(2) ∫

Z|g(x; z)|pl

−δgν(dz) ≤ C2(1 + |x|pl

), x ∈ E, l = 1, . . . , n,(3)

and are Lipschitz, i.e.

|f(x) − f(y)|−δf≤ C3|x − y|, x, y ∈ E,(4) ∫

Z|g(x; z) − g(y; z)|pl

−δgν(dz) ≤

C4 |x − y|pl, x, y ∈ E, l = 1, . . . , n,(5)

where C1, C2, C3 and C4 are some constants. Let γ ≥ 0 be fixed. Then thefollowing holds:

a.) Let q < ∞. Assume that the initial condition u0 satisfies(i) E|u0|q−γ < ∞

and the constants δf , δg and γ satisfy the following conditions(ii) δgq < 1, and δf < 1,(iii) (δg − γ)p < 1 − 1

q , and δf − γ < 1 − 1q ,

(iv) 0 ≤ γq < 1 .Then there exists a unique mild solution to Problem (1) such thatfor any T > 0 ∫ T

0E|u(s)|q ds < ∞,

and for all δ > 0

u ∈ L0(Ω; ID([0, T ];V−(γ+δ))) ∩ L0(Ω; ID((0, T ];V−γ)),

where

L0(Ω; ID((0, T ];V−γ )) = ∪s>0L0(Ω; ID([s, T ];V−γ)).

b.) Let q < ∞. Assume that the initial condition u0 satisfies(v) E|u0|p−γ < ∞,

and the constants δf , δg and γ satisfy the following conditions(vi) δgp < 1, and δf < 1,

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SPDEs driven by Poisson Random Measure 21st May 2005 5

(vii) (δg − γ)p < 1 − 1q , and δf − γ < 1 − 1

q ,(viii) 0 ≤ γq < 1.Then there exists a unique mild solution to Problem (1), such thatfor any T > 0 ∫ T

0(E|u(s)|p) q

p ds < ∞,

and for all δ > 0

u ∈ L0(Ω; ID([0, T ];V−(γ+δ))) ∩ L0(Ω; ID((0, T ];V−γ)).

c.) Let q = ∞ and γ = 0. Assume in addition to (vi) that the initialcondition u0 satisfies(ix) E|u0|p < ∞.Then there exists a unique mild solution to Problem (1), such thatfor any T > 0

sup0≤s≤T

E|u(s)|p < ∞,

and for all δ > 0

u ∈ L0(Ω; ID([0, T ];V−δ))) ∩ L0(Ω; ID((0, T ];E)).

Remark 2.2. (see Section 5) Using the same idea as Ikeda and Watanabe[25, Chapter 4] and the amalgamation procedure of Elworthy [20, ChapterIII.6], one can show the cadlag property for solutions to equation (1) underthe hypothesis of Theorem 2.1, with no integrability condition on the jumpssize. In particular, conditions (3) and (5) can be replaced by∫

{z∈Z||z|≤1}|g(x; z) − g(y; z)|p−δg

ν(dz) ≤ C1(|x − y|p), x, y ∈ E,

|g(x; �) − g(y; �)|L(Z,E) ≤ C2 |x − y|, x, y ∈ E.

where C1 and C2 are some generic constants.

Remark 2.3. Let 1 ≤ p ≤ 2. Let E and Z be two separable Banach spaces,E is of M type p. Let ν : B(Z) → R

+ be a not necessarily symmetric Levymeasure and η : B(E)×B(R+) → R

+ be a Poisson random measure withcharacteristic measure ν. Let γ(t) : B(Z)×B(R+) → R

+ be the compensatorof η, i.e. the unique predictable random measure such that

Eη(A × I) − γ(A, I) = 0

for each A ∈ B(Z) and I ∈ B(R+). Let η := η−γ be the compensated Poissonrandom measure. Then by small modification (see Remark 3.4 and Remark

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6 ERIKA HAUSENBLAS

3.7) it can be shown, that the Theorem 2.1 holds also for the following SPDE{u(t) dt = (Au(t−) + f(u(t−))) dt +

∫Z g(u(t−); z)η(dz; dt),

u(0) = u0.

Example 2.2. Let 1 < α < p ≤ 2. Take any p ∈ (α,∞) and put E =Lp(O), where O a smooth domain in R

d and U = {x ∈ E | |x| ≤ 1}. Letσ : ∂U → R

+ be a finite measure. Let η : B(E)×B(R+) → B(E) a Poissonrandom measure with characteristic measure ν : B(E) → R given by (seeExample 3.1)

ν(B) =∫

R+

∫∂U

χB(sz)σ(dz) s−1−α ds, B ∈ B(E).

Then the process

Lt =∫ t

0+

∫E

z η(dz; ds), t ≥ 0,

is an E–valued, α-stable symmetric process and the process

LUt =

∫ t

0+

∫U

z η(dz; ds), t ≥ 0,

is of integrable p variation. Our interest is the existence and uniqueness ofthe solution (u(t))t≥0 to⎧⎨

⎩du(t, ξ) = Δu(t−, ξ) dt +

∫E z(ξ) u(t−, ξ) η(dz; dt),

ξ ∈ O, t ≥ 0,u(0, ξ) = u0(ξ) ξ ∈ O.

(6)

Let E = W 2(γ+δ),p(O) and B = V−(δ+γ) = Lp(O). First, note that byExamples 3.9 and 3.10 the spaces W ϑ,p(O), ϑ ∈ R are of M–type p. Bymultiplier theorems (see e.g. Runst and Sickel [42, Theorem 1, p. 190]) onesees that

|z u|B ≤ C |z|B |u|W

dp ,p

(O).

By embedding theorems follows that if γ ≥ d2p then E ↪→ W

dp,p(O). Choose

q ≥ p such that γ < 1q . Theorem 2.1 and Remark 2.2 give existence and

uniqueness of the mild solution of (6) in L0(ID([0, T ];Lp(O)) for d = 1.

Example 2.3. Let O = [0, 1] be a bounded interval. Let α ∈ (1, 2) andp ∈ (α, 2]. Choose γ > 0 such that γ > 1− 1

p . Let δ > 0 and δg be such that

12

< δg <1p.

Let E = W−α+2(δg+δ),p(O) and Z = B1p−1

p,∞ (O) be the Besov space (see e.g.Triebel [44, Chapter 2.3.2]). Let B := V−δg = W−α,p(O) and let B∗ be the

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SPDEs driven by Poisson Random Measure 21st May 2005 7

topological dual of B, i.e. B∗ = W α,q(O) ∼ W β,p(O), where q is given by1p + 1

q = 1 and β = 2p −α−d. The fact 1

2 < δg implies E ↪→ B∗ continuously.Let U be the unit ball in Z, i.e. U = {x ∈ B | |x| ≤ 1}. By Runst and

Sickel [42, p.34, Remark 3] we have δξ ∈ Z, where δξ denotes the Deltadistribution at ξ ∈ O. Moreover, there exists a constant c > 0 such that|δξ|Z = c for all ξ ∈ O and therefore 1

c δξ ∈ U .First, note that if A ∈ B(O), then A = {δξ | ξ ∈ A} ∈ B(Z). In partic-

ular, the function f : [0, 1] → Z : ξ �→ δξ is bounded and for γ > 1 − 1p the

function f : [0, 1] → W−γp ([0, 1]) : ξ �→ δξ is continuous. The latter holds,

since for some β > 0, the dual of W−γp ([0, 1]) can be continuously embedded

in the Holder space Cβ([0, 1]) (see e.g. Triebel [44, Chapter 4.6.1]). Let j

be the embedding of Z in W−γp ([0, 1]). Then A = j−1 ◦ f(A) and therefore

A ∈ B(Z).Let the measure σ : ∂U → R

+ be defined in the following way: If A ∈ B(∂U)is of the form A = 1

c ∪ξ∈A δξ for some A ∈ B([0, 1]), we put σ(A) = 12λ(A),

where λ denotes the Lebesgue measure. Similarly, if A = 1c ∪ξ∈A (−δξ)

for some A ∈ B([0, 1]), we put σ(A) = 12λ(A). In particular, σ(A) :=

λ({f−1 (A ∩ f([0, 1]))

}, where f : [0, 1] → Z : ξ �→ 1

c δξ. Now, the charac-teristic measure ν : B(Z) → R is defined by

ν(B) =∫

R+

∫∂U

χB(tx)σ(dx)t−1−α dt, B ∈ B(Z).

Let η : B(B)×B(R+) → R+ be the corresponding Poisson random measure.

By Linde [31, Theorem 6.2.8] the process

Lt =∫ t

0

∫Z

zη(dx, ds), t ≥ 0,

is the space-time α-stable white noise (for definition see e.g. Mytnik [34]).We are interested in the problem

{du(t) = Δu(t−) dt +

∫Z u(t−) z η(dz, dt), t ≥ 0,

u(0) = u0 ∈ B.(7)

Theorem 2.1 and Remark 2.2 give existence and uniqueness of the solutionu to Problem (7) in L0(Ω; ID((0, T ];B)). Such a unique solution satisfies〈u, φ〉 ∈ ID((0, T ]; R) for all φ ∈ B∗.

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8 ERIKA HAUSENBLAS

3. Poisson random measures and stochastic integration

3.1. Poisson random measures. Let (Ω,F , (Ft)t∈[0,T ], P) be a filteredprobability 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 (Zi, Ti), i = 1, 2, . . . such that for each i Zi is FTi-measurable. Given a point process, one usually works with the associatedrandom measure η defined by

η(A × [0, t])(ω) =∑Ti≤t

1A(Zi(ω)), A ∈ Z, t ≥ 0, ω ∈ Ω.

A point process is called a Poisson point process with characteristic mea-sure 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, the random variable Nt(A) = η(A× [0, t])has a Poisson distribution with parameter ν(A)t, i.e.

P(Nt(A) = k) = exp(−tν(A))(ν(A)t)k

k!.(8)

It follows, that the random measure η associated to a Poisson point processhas independent increments, and that η(A × [0, t]) and η(B × [0, t]) areindependent for all A,B ∈ Z, A ∩ B = ∅, and t ≥ 0.

Let Z be a separable Banach space with Borel σ algebra Z. Starting witha measure ν on Z, one may ask under what conditions one can construct aPoisson random measure η on Z×B(R+), such that ν is the characteristicmeasure of η. If the measure ν is finite and satisfies ν({0}) = 0, then thePoisson random measure η : Z×B(R+) → R

+ exists. Moreover, one canshow, that if ν ∈ Lsym(Z) (see Definition 2.2), then the Poisson randommeasure η : Z×B(R+) → R

+ exists.Some useful properties of symmetric Levy measures are stated in the

following remark.

Remark 3.1. (See also Linde [31, Proposition 5.4.5]) Let Z be a separableBanach 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 Levy measure. Then thefollowing holds true.

• For each δ > 0, ν{|x| > δ} < ∞.• sup|a|≤1

∫|x|≤1 |〈x, a〉|2 dν(x) < ∞, where 〈a, x〉 := a(x), a ∈ Z ′.

• If σ ≤ ν, then σ is also a Levy measure.

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SPDEs driven by Poisson Random Measure 21st May 2005 9

A typical example of Poisson random measure is provided by an α–stablePoisson random measure, 0 < α < 2.

Definition 3.1. A probability measure μ ∈ P(E) is said to be stable iff foreach a, b > 0 there exists some c > 0 and an element z ∈ E, such that forall independent random variables X and Y with law μ we have

L(aX + bY ) = L(cX + z).(9)

The measure μ is called strictly stable if for all a > 0 and b > 0 one canchoose z = 0 in (9). Moreover, μ is called α–stable iff (9) holds with

c = (aα + bα)1α .

Example 3.1. Let E be a separable Banach space, E := B(E) and U ={x ∈ E | |x| ≤ 1}. Let σ : B(∂U) → R

+ be an arbitrary finite measure andν : E → R

+ be defined by

ν(B) =∫

R+

∫∂U

χB(sz)σ(dz) s−1−α ds, B ∈ B(E).

Then ν : E → R+ is a Levy measure. Let η be the corresponding Poisson

random measure. If E is of type p, p > α (for definition see the nextSection), then the process L = (Lt)t≥0 defined by

Lt =∫ t

0+

∫E

z η(dz; ds), t ≥ 0,

is an E–valued, unique, α-stable symmetric Levy-process. On the other handit is known from Linde [31, Theorem 6.2.8] that the Levy measure of an α-stable random variable is given by its distribution on ∂U , i.e. by the measureσ : B(∂U) → R

+.

3.2. M-type p Banach spaces: a short account. A stochastic integral isdefined 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 theintegrator is a square integrable martingale, Ito’s extension procedure yieldsto 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 . Fix1 ≤ p < 2. Let η : E×B(R+) → R

+ be a Poisson random measure withsymmetric characteristic measure ν. We will define an integral with respectto ν, i.e. ∫ t

0+

∫E

h(s, z) η(dz; ds),(10)

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10 ERIKA HAUSENBLAS

whith h : Ω×R+ → L(Z,E) is a caglad predictable step function, such that∫ T

0

∫Z |h(s, z)|p ν(dz) ds < ∞. The question which has to be answered is

under which conditions on the underlying Banach space E and the integra-tor, the stochastic integral in (10) can be extended to the set of all functionswhith finite Lp-mean. First, we consider the case where h is a deterministicfunction, secondly we consider the case, where h is a random function.

Here and hereafter {εn}n∈IN denotes a sequence of {1,−1}-valued randomvariables such that

P (εi = ±1) =12.

Definition 3.2. (Linde [31]) Let E be a Banach space and 0 < p ≤ 2 befixed. Let x = {xi}n∈IN be a sequence in E. A Banach space E is of type p

(R-type p) iff x ∈ lp(E) implies that the series∞∑i=1

εixi

is a.s. convergent in E.

Proposition 3.1. (see Linde [Proposition 3.5.1][31]) The Banach space E

is of type p iff for some (each) r ∈ (0,∞) there exists a constant C =C(E, p) > 0 such that for all sequences x = {xi}n∈IN, for all sequences{εn}n∈IN and all N ∈ IN{

E

∣∣∣∣∣N∑

i=1

εixi

∣∣∣∣∣r} 1

r

≤ C

{N∑

i=1

|xi|p} 1

p

.

Example 3.2. Let (M,M, P) be a probability space. Then Lp(M,M, P) isof type p, 1 ≤ p ≤ 2 (see Pisier [38, p.186]).

Example 3.3. Let E be of type p, 0 < p ≤ 2. Then E is of type q, 0 < q ≤ p

(see Linde [31, Chapter 3, Theorem 3.5.2]).

Example 3.4. Assume E is of type p. Then any closed subspace of E isalso of type p (this follows by Pisier [38, Theorem 4.5]).

Definition 3.3. (see Dettweiler [18]) Let 1 ≤ p ≤ 2 and E be a sepa-rable Banach space. An E-valued process X = (X(t))t≥0 is said to be ofp-integrable variation iff for any T > 0 one can find a constant C(T ) suchthat for any partition 0 = t0 < t1 < · · · < tn = T one has

n∑i=1

E |X(ti+1) − X(ti)|p ≤ C(T ).

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SPDEs driven by Poisson Random Measure 21st May 2005 11

Example 3.5. Let E be a Hilbert space and (W (t))t≥0 be an E-valuedWiener process with covariance operator Q. If Q is of trace class, thenthe Wiener process (W (t))t≥0 is of 2-integrable variation.

Example 3.6. Let E be a Hilbert space and M = (Mt)t≥0 be an E-valuedstationary Levy process with bounded second moment. Then the Levy–Khintchineformula implies that M is of integrable 2-variation.

Example 3.7. Let E be a Hilbert space. Let ν ∈ Lsym(E) such that∫E |z|2ν(dz) < ∞. Then one can construct a unique Poisson random mea-

sure η : B(E)×B([0, T ]) → R+ such that ν is the characteristic measure of

η. Moreover, the process Lt =∫ t0+

∫E z η(dz, ds) is of integrable 2-variation.

A Hilbert space is of type 2. Let p ∈ (0, 2]. In case the underlying Banachspace E is of type p and the Levy measure is p integrable, the Example 3.7can be transferred to E.

Example 3.8. (see Theorem 2.1 in Dettweiler [16] resp. page 129, (ii) ⇒(i), [18, Proposition 2.3], Hamedani and Mandrekar [24]) Let E and Z beseparable Banach spaces, E of type p, p ∈ (1, 2] with Borel σ–algebras E andZ respectively. Let ν ∈ Lsym(Z,Z) and h ∈ L(Z,E), such that∫

Z|h(z)|pE ν(dz) < ∞.

Let η be a Poisson random measure on Z×B(R+) with characteristic mea-sure ν. Then a Levy process L = (L(t))t≥0 exists such that

L(t) d=∫ t

0+

∫Z

h(z) η(dz, ds).

In addtion, L is a martingale and is of p-integrable variation.

Remark 3.2. The contrary is in general not true. Here the cotype propertyof E is essential (see e.d. [18, Proposition 2.2])

Remark 3.3. In fact, Dettweiler shows in his paper [16] the following. LetE be a separable Banach space of type p, p ∈ (1, 2] with Borel σ–algebra E.Then there exists a constant C = C(E, p) such that for each Poisson randommeasure η : E×B([0, T ]) → R

+ with characteristic measure ν ∈ Lsym(E) andeach I ∈ B([0, T ]) we have

E|∫

Ezν(z, I)|p ≤ C

∫E|h(z)|pE ν(dz)λ(I),

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12 ERIKA HAUSENBLAS

where λ denotes the Lebesgue measure. If E is not of type p, then theinequality above does not hold.

Remark 3.4. Let E be a Banach spaces of type p, 1 ≤ p ≤ 2. In Example3.3 we assumed that the characteristic measure ν is symmetric. Consideringa compensated Poisson random measure, this condition can be weakend. Inparticular, from the proof of Proposition 2.5 from Dettweiler [18] followsthat there exists a constant C = C(E, p) such that for each Poisson randommeasure η : B(E)×B(R+) → R

+ with characteristic measure ν : B(E) →R

+, where ν is a Levy measure we have

E

∣∣∣∣∫ t

0

∫E

h(z) (μ − γ)(ds, dz)∣∣∣∣p ≤ C

∫E|h(z)|p ν(dz).

Above, γ denotes the compensator of η. It follows, that the integral∫ t

0

∫E

h(z) (μ − γ)(ds, dz)

is also of integrable p–variation.

If h is deterministic, E a Banach space of type p and the integrator ofintegrable p variation, one can extend the stochastic integral to the class ofall h : [0, T ] → L(Z,E) such that

∫ T0

∫Z |h(z)|p ν(dz) ds < ∞, see e.g. Pisier

[38, Proposition 4.3], Rosinski [40]. If h is random the underlying Banachspace has additional to be UMD.

Definition 3.4. A Banach space is said to be UMD (Unconditional Martin-gale Differences) if for each 1 < r < ∞ there exists a constant C(E, r) < ∞,such that for each E-valued martingale M = (M0,M1, . . .), each {−1, 1}–valued sequence (ε0, ε1, . . .) and each positive integer n ∈ IN

E

∣∣∣∣∣n∑

k=1

εk (Mk − Mk−1)

∣∣∣∣∣r

E

≤ C(E, r) E |Mn − M0|rE .

Remark 3.5. Let 1 < p ≤ 2. A Banach space E which is UMD and of typep is called of M -type p and is uniformly p-smoothable (see Brzezniak [10]).

Example 3.9. (Woyczynski [47, p. 247, Example 2.2]) Let O be a boundeddomain. Then the space Lp(O) is of M -type p∧ 2 (see e.g. [47, Chapter 2]).

Example 3.10. (Brzezniak [9, p. 10]) Let 0 < p ≤ 2. Let E be of M -typep and A : E → E an operator with domain D(A). If A−1 is bounded, thenD(A) is isomorphic to E and therefore of M–type p.

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SPDEs driven by Poisson Random Measure 21st May 2005 13

Example 3.11. (Brzezniak [9, Appendix A, Theorem A.4]) Assume E1 andE2 are a Banach space of M -type p, where E2 is continuously and denselyembedded in E1. Then for any ϑ ∈ (0, 1) the complex interpolation space[E1, E2]ϑ and the real interpolation space (E1, E2)ϑ,p are of M–type p.

Example 3.12. Let (S,S) be a measurable space. Then L∞(S), L1(S) arenot UMD.

Suppose E is of M -type p. Then there exists a constant C = C(E; p),such that for each discrete E-valued martingale M = (M1,M2, . . .) (see e.g.Pisier [38, Chapter 6])

supn≥1

E|Mn|pE ≤ C∑n≥1

E|Mn − Mn−1|pE .

By means of this inequality one can extend the stochastic integral in p mean(see e.g. next Chapter, or Woyczynski [47, Theorem 2.2], Dettweiler [18]).

To deal with moments of higher order, stronger inequalities are needed.In fact one can prove using the techniques described in Burkholder [13] ageneralized version of an inequality of Burkholder type. In particular, letE 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 = (M1,M2, . . .) and 1 ≤ r < ∞ ( see e.g. Assuad [3],Brzezniak [9, Proposition 2.1], Pisier [37, Remark 3.3 on page 346] and [38,Chapter 6])

E supn≥1

|Mn|rE ≤ CE

⎡⎣∑

n≥1

|Mn−1 − Mn|pE

⎤⎦

rp

.(11)

Remark 3.6. An interested reader can consult the following articles: Burk-holder [12, 14], Pisier [37, 38] and Woyczynski [47, 48] for the connectionof uniformly p-smoothable Banach spaces and their geometric properties.Brzezniak [9], Dettweiler [18, 19, 16] Neidhardt [35] and Rudiger [41] forthe connection between uniformly p-smoothable Banach spaces and stochasticintegration. The book of Linde [31], and the articles Dettweiler [16, 17], Gineand Arujo [23], Hamedani and Mandrekar [24] for the connection betweenLevy processes and p Banach spaces.

3.3. The Stochastic Integral in M–type p Banach spaces. In thefollowing let p ∈ (1, 2] be fixed. Let Z be a separable Banach space and E

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14 ERIKA HAUSENBLAS

be a M–type p Banach space. The Borel σ algebras of Z and E are denotedby Z and E respectively.

As mentioned in the introduction the stochastic integral will be first de-fined on the set of predictable and simple function. To be precise a processh : [0, T ] → L(Z,E) is said to be simple predictable if h has a representation

h(s, z) =n∑

i=1

1(ti−1,ti](s)Hi−1(z), s ≥ 0, z ∈ Z,(12)

where 0 = t0 < · · · < tn = T is a partition of [0, T ] by stopping times,and Hi ∈ L(Z,E), i = 0, . . . , n, are Fti-measurable random variables. Thecollection of simple predictable processes h : [0, T ] → L(Z,E) is denotedby S. Let ID := ID([0, T ];E) be the Skorohod space of all adapted cadlagfunctions h : [0, T ] → E, endowed by the Skorohod topology (see AppendixA).

Let η : Z×B([0, T ]) → R+ be a Poisson random measure with character-

istic measure ν ∈ Lsym(Z,Z). The stochastic integral with respect to η isthe following linear operator I : S → ID by letting

I(h)(t) =∫ t

0+

∫Z

h(s, z)η(dz, ds) :=

n∑i=1

∫Z

Hi−1(z)η(dz × (ti−1 ∧ t, ti ∧ t]), t > 0, ,(13)

where h ∈ S has the representation (12).The next step is to extend the stochastic integral to the set ILp(ν), where

ILp(ν) =

{h : Ω × R

+ → L(Z,E), h is a predictable,

caglad process, such that∫ t

0

∫Z

E |h(ω, s, z)|p ν(dz) ds < ∞}

,

equipped with norm

|u|pILp(ν) :=∫ t

0

∫Z

E |h(ω, s, z)|p ν(dz).

Since ILp(ν) is separable, by standard arguments, one can show, that S ∩ILp(ν) is dense in ILp(ν). Let IDp the Skorohod space equipped with thefollowing norm

|u|pIDp:= E sup

0≤t≤T|u(t)|p, u ∈ IDp.

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SPDEs driven by Poisson Random Measure 21st May 2005 15

Note, the Skorohod space ID([0, T ];E) topologized with uniform convergenceis a complete metric space, but not separable. Analysing the proof of com-pleteness of Lp spaces over an arbitrary measurable set, on can see that thespace IDp is a complete normed space. By means of the following proposition,it can be shown that the stochastic integral defined in (13) is a continuousoperator from Sp ∩ ILp(ν) into IDp (see e.g. Woyczynski [47, Theorem 2.2] orDettweiler [18]).

Proposition 3.2. Let 1 < p ≤ 2. Assume Z and E are a separable Banachspaces, E is of M type p. Let Z and E be the Borel σ–algebras and η

be a Poisson random measure on Z × B([0, T ]) with characteristic measureν ∈ Lsym(Z). Let h : Ω× [0, T ] → L(Z,E) be a function belonging to S withrepresentation (12). Then we have for T > 0 and 0 < r < ∞

E sup0<t≤T

∣∣∣∣∫ t

0+

∫Z

h(σ, z) η(dz; dσ)∣∣∣∣r ≤

Cp E

(∫ T

0

∫Z|h(s, z)|p η(dz; ds)

) rp

and for 0 < r ≤ p

E sup0<t≤T

∣∣∣∣∫ t

0+

∫Z

h(σ, z) η(dz; dσ)∣∣∣∣r ≤(14)

Cp

(∫ T

0

∫Z

E |h(s, z)|p ν(dz) ds

) rp

.

Proof. Let h = {h(s, z), 0 ≤ s ≤ T, z ∈ Z} be simple, predictable process,in particular h can be written as (see (12))

h(s, z) =n∑

i=1

1(ti−1,ti](s)Hi−1(z), s ≥ 0, z ∈ Z

where 0 = t0 < · · · < tn = T is a partition of stopping times with maximaldiameter |π| := max{ti − ti−1}. Moreover, Hi : Z → E, i = 0, . . . , n, areFti-measurable random variables. Now, the stochastic integral is defined bythe Riemann integral (see 13)∫ t

0+

∫Z

h(s, z)η(dz, ds) =n∑

i=1

∫Z

Hi−1(z)η(dz × (ti−1 ∧ t, ti ∧ t]).

First, let us fix r ≤ p. The sequence {∫Z Hi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t]), i =1, . . . , n} is a sequence of martingale differences. In particular

E

∫Z

Hi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t]) = 0

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16 ERIKA HAUSENBLAS

and because of Remark 3.3

E

∣∣∣∣∫

ZHi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p = C(ti∧t−ti−1∧t)∫

ZE|Hi−1(z)|pν(dz).

Therefore we can apply inequality (11) and obtain

E sup0≤t≤T

∣∣∣∣∣n∑

i=1

∫Z

Hi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣∣r

C E

∣∣∣∣∣n∑

i=1

∣∣∣∣∫

ZHi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p∣∣∣∣∣

rp

.

The Jensen inequality yields

E sup0≤t≤T

∣∣∣∣∣n∑

i=1

∫Z

Hi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣∣r

≤ C

(n∑

i=1

E

∣∣∣∣∫

ZHi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p) r

p

.(15)

The tower property of the conditional expectation leads to

E sup0≤t≤T

∣∣∣∣∣n∑

i=1

∫Z

Hi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣∣r

≤ C

(n∑

i=1

E

[E

[∣∣∣∣∫

ZHi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p | Fti−1

]]) rp

.(16)

The random variable Hi−1 is Fti−1 -measurable. Therefore by Remark 3.3we have

E

[∣∣∣∣∫

ZHi−1(z)η(dz; (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p | Fti−1

]=∫

Z|Hi−1(z)|p ν(dz) (ti ∧ t − ti−1 ∧ t).(17)

Inserting (17) in equation (16)(n∑

i=1

E

∣∣∣∣∫

ZHi(z)η(dz × (ti−1 ∧ t, ti ∧ t])

∣∣∣∣p) r

p

C

(n∑

i=1

∫Z

(ti − ti−1) E

∫Z|Hi−1(z)|pν(dz)

) rp

,(18)

which can be written as(∫ t

0

∫Z

E |h(s, z)|p ν(dz) ds

) pr

.

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SPDEs driven by Poisson Random Measure 21st May 2005 17

In case of r > p, we can only apply the martingale inequality (11) and have tostop before equation (15). Since for all 1 ≤ i ≤ n,

∫Z Hi−1(z)η(dz× (ti−1, t])

is martingale for ti−1 < t ≤ ti, we know∣∣∣∣∫ t

0+h(s, z)η(dz; ds)

∣∣∣∣is a submartingale and Doob’s maximal inequality yields the assertion. �

By means of Proposition 3.2 the stochastic integral is a continuous oper-ator from S∩ ILp(ν) into IDp. The next point is to reassure, that S∩ ILp(ν) isdense in ILp(ν). But in fact ILp(ν) equipped with the predictable σ algebra

P := σ (h : [0, T ] × Z → E,h is (Ft)–adapted, and cadlag)

is a measurable separable metric space and by a modification of the Proof ofProposition I.4.7, Da Prato and Zabczyk [15] one can show that S∩ ILp(ν) isdense in ILp(ν). Since IDp is complete, a bounded linear operator I : ILp(ν) →IDp

u exists, which is an extension of the operator introduced in (13). In thenext Proposition we will show, that the inequalities in Proposition 3.2 arepreserved.

Proposition 3.3. (see e.g. Dettweiler [18, Theorem 3.1]) Let 1 < p ≤ 2.Assume Z and E are a separable Banach spaces, E is of M type p. Let Z andE be the Borel σ–algebras and η be a Poisson random measure on Z×B(R+)with characteristic measure ν ∈ Lsym(Z). Let h : Ω × R

+ × Z → E be afunction belonging to ILp(ν). Then we have for 0 < r < ∞

E sup0<s≤t

∣∣∣∣∫ s

0+

∫Z

h(σ, z) η(dz; dσ)∣∣∣∣r ≤

CpE

(∫ t

0

∫Z|h(s, z)|p η(dz; ds)

) rp

and for 0 < r ≤ p

E sup0<s≤t

∣∣∣∣∫ t

0+

∫Z

h(σ, z) η(dz; dσ)∣∣∣∣r ≤(19)

Cp

(∫ t

0

∫Z

E |h(s, z)|p ν(dz) ds

) rp

.

Proof. To show that S∩ILp(ν) is dense in ILp(ν), one can modify the Proof ofProposition I.4.7 of Da Prato and Zabczyk [15]. Thus there exists a sequence{hj(s, z) | j ∈ IN} in S ∩ ILp(ν), such that∫ T

0+

∫Z

E∣∣hj(s, z) − h(s, z)

∣∣p ν(dz) ds −→ 0 as j → ∞,

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18 ERIKA HAUSENBLAS

where the step functions hj can be written as a sum of the following type

hj(s, z) =nj∑i=1

1(tji−1,tji ]

(s)Hji−1(z), s ≥ 0, z ∈ Z

where 0 = tj1 < · · · < tjnj = T is a sequence of partitions of [0, T ], such

that the maximal diameter |πj | := max{tji − tji−1} tends to zero as j → ∞and Hj

i : Z → E, i = 0, . . . , nj are Ftji

-measurable random variables. The

stochastic integral for hj defined in (13) is given by∫ t

0+

∫Z

hj(s, z)η(dz, ds) =nj∑i=1

∫Z

Hji−1(z)η(dz × (tji−1 ∧ t, tji ∧ t]).

Proposition 3.2 yields

E sup0≤t≤T

∣∣∣∣∫ t

0+hj(s, z)η(dz; ds)

∣∣∣∣r ≤ C

(∫ t

0

∫Z

E |hj(s, z)|p ν(dz) ds

) pr

.

Since

E sup0≤t≤T

∣∣∣∣∫ t

0+hj(s, z)η(dz; ds)

∣∣∣∣r ≤ C

(∫ t

0

∫Z

E |hj(s, z)|p ν(dz) ds

) pr

≤ C

(∫ t

0

∫Z

E |h(s, z)|p ν(dz) ds

) pr

it follows by the Lebesgue’s dominated convergence theorem, that for allt > 0 ∫ t

0+hj(s, z)η(dz; ds) −→

∫ t

0+h(s, z)η(dz; ds) as j → ∞

and

E sup0≤t≤T

∣∣∣∣∫ t

0+h(s, z)η(dz; ds)

∣∣∣∣p ≤ C

(∫ t

0

∫Z

E |h(s, z)|p ν(dz) ds

) pr

.

The inequalities of Proposition (3.3) can be extended to higher ordermoments.

Corollary 3.1. Let 1 ≤ p ≤ 2. 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 bePoisson random measure on Z with characteristic measure ν ∈ Lsym(Z,Z).Let h ∈ ILp(ν). Let q = pn for some n ∈ IN and suppose, that

E

(∫ T

0

∫Z|h(s, z)|p ν(dz) ds

) qp

< ∞,(20)

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SPDEs driven by Poisson Random Measure 21st May 2005 19

and ∫ T

0

∫Z

E|h(s, z)|q ν(dz) ds < ∞(21)

holds. Then

E sup0<s≤t

∣∣∣∣∫ t

0+

∫Z

h(s, z)η(dz; ds)∣∣∣∣q ≤

C

n∑l=1

(∫ t

0

∫Z

E|h(s, z)|plν(dz) ds

)pn−l

,

where C is some generic constants.

Proof. The proof is a generalization of the proofs of Bass and Cranston [4,Lemma 5.2] or Protter and Talay [39, Lemma 4.1]. By Proposition 3.3 andinequality (14) we have

E sup0≤s≤t

∣∣∣∣∫ s

0+

∫Z

h(r, z)η(dz; dr)∣∣∣∣q ≤ C E

(∫ t

0+

∫Z|h(s, z)|p η(dz; ds)

) qp

.

Let us define

Xt :=∫ t

0+

∫Z|h(s, z)|p η(dz; ds), t ≥ 0.

The process Xt is a real valued semimartingale and by condition (20) of finitevariation. Moreover, it has only positive jumps, i.e. it is a subordinator. Theassociated random measure ηX : B(R)×B(R+) → R

+ is given by (see Jacodand Shiryaev [26, Proposition 2.1.16])

ηX(ω; dt; dx) =∑

s

1{ΔXs(ω)=0}δ(s,ΔXs(ω))(dx; dt).

Let γX be the compensator of ηX , i.e. the unique predictable random mea-sure on B(R) × B(R+), such that for all A ∈ B(R) the process

∫ t0+(ηX −

γX)(A, ds) is a local martingal. Let us define

L(t)(0) :=∫ t

0+

∫R

z (ηX − γX)(dz; ds), t ≥ 0.

Since (20) holds, L(t)(0) is a real–valued martingale, has only positive jumpsand is of finite variation. Note, L

(0)t has a characteristic function of the

following form E exp(iξL

(0)t

)= exp(ψt(ξ)), where

ψt(ξ) =∫

R

(eiξζ − 1 − iξ

)νt(dξ),

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20 ERIKA HAUSENBLAS

where νt : R → R+. The uniqueness of the characteristic function gives for

A ∈ B(R)

νt(A) = ν {z ∈ E | h(t, z) ∈ A} .

Since η and ηX are Poisson random measures and by the uniqueness of thecompensator we infer that

γX(dt, ds) = νXt (ds) × dt.(22)

Since E is of M–type p, it follows from Proposition (3.3)

E sup0<s≤t

∣∣∣∣∫ t

0+

∫Z

h(s; z)η(dz; ds)∣∣∣∣p

n

≤ C E

(∫ t

0+

∫Z|h(s; z)|p η(dz; ds)

)pn−1

.

Simple calculations leads to

E sup0<s≤t

∣∣∣∣∫ t

0+

∫Z

h(s; z)η(dz; ds)∣∣∣∣p

n

≤ C E

(∫ t

0+

∫R

z ηX(dz; ds))pn−1

≤ C

(E

(∫ t

0+

∫R

z (ηX − γX)(dz; ds))pn−1

+ E

(∫ t

0+

∫R

z γX(dz; ds))pn−1)

≤ C

(E|L(t)(0)|pn−1

+ E

(∫ t

0+

∫R

z νXt (dz) ds

)pn−1)

≤ C

(E|L(t)(0)|pn−1

+ E

(∫ t

0+

∫Z|h(s; z)|p ν(dz)ds

)pn−1).

In case n = 2, we have

E

∣∣∣L(t)(0)∣∣∣p = E

∣∣∣∣∫ t

0+

∫R

z (ηX − γX)(dz; ds)∣∣∣∣p

= E

∫ t

0+

∫R

|z|p ηX(dz; ds).

By the definition of the compensator, in particular since∫ t0+

(ηX − γX

)(A, ds)

is a martingale, we can continue

E

∣∣∣L(t)(0)∣∣∣p ≤ E

∫ t

0+

∫R

|z|p γX(dz; ds)

= E

∫ t

0+

∫R

|z|p νXt (dz) ds.

By the discussion above, i.e. relation (22) we have

E

∣∣∣L(t)(0)∣∣∣p ≤ E

∫ t

0+

∫|z|p2

ν(dz) ds.

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SPDEs driven by Poisson Random Measure 21st May 2005 21

Thus, the proposition is proven, provided n = 2. In case n > 2, we have tocontinue. In particular, let

L(t)(r) :=∫ t

0+

∫R

zpr(ηX − γX)(dz; ds) for r = 1, . . . , n.

Since (20) and (21) holds, L(t)(r) is a real–valued martingale, has only pos-itive jumps and is of finite variation for r = 1, . . . , n. Moreover, since R isof M -type 2, R is also of M -type p for all p ∈ [1, 2]. Using inequality (14)we have

E

∣∣∣L(t)(r)∣∣∣pm

≤ C E

(∫ t

0

∫R

zpr(ηX − γX) (dz; ds)

)pm

≤ C E

(∫ t

0

∫R

zpr+1ηX(dz; ds)

)pm−1

.

Using simple calculations we get

E

∣∣∣L(t)(r)∣∣∣pm

C E

(∫ t

0

∫R

zpr+1(ηX − γX)(dz; ds) +

∫ t

0

∫R

zpr+1γX(dz; ds)

)pm−1

≤ C1 E

(∫ t

0

∫R

zpr+1(ηX − γX)(dz; ds)

)pm−1

+ C2 E

(∫ t

0

∫R

zpr+1γX(dz; ds)

)pm−1

≤ C1 E

∣∣∣L(t)(r+1)∣∣∣pm−1

+ C2 E

(∫ t

0

∫R

zpr+1νX

t (dz) ds

)pm−1

≤ C1 E

∣∣∣L(t)(r+1)∣∣∣pm−1

+ C2E

(∫ t

0+

∫E|h(s; z)|pr+2

ν(dz)ds

)pm−1

.

That means we have

E

∣∣∣L(t)(r)∣∣∣pm

C1 E

∣∣∣L(t)(r+1)∣∣∣pm−1

+ C2E

(∫ t

0+

∫E|h(s; z)|pr+2

ν(dz)ds

)pm−1

.(23)

Note, since R is of M type p, we have by inequality (14)

E

∣∣∣L(t)(r)∣∣∣p = E

∣∣∣∣∫ t

0+

∫R

zpr(ηX − γX)(dz; ds)

∣∣∣∣p≤ C E

∫ t

0+

∫R

zpr+1ηX(dz; ds).(24)

Page 23: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

22 ERIKA HAUSENBLAS

Note, that R is also of type 1, that means of M -type 1. Thus we have

E

∫ t

0+

∫R

zpr+1ηX(dz; ds) ≤ E

∫ t

0+

∫R

zpr+1νX

t (dz) ds

≤∫ t

0+

∫R

E |h(s; z)|pr+2

E ν(dz) ds.(25)

Substitution of (25) to (24) gives

E

∣∣∣L(t)(r)∣∣∣p ≤ C E

∫ t

0+

∫Z|h(s, z)|pr+2

E ν(dz)ds

≤ C

∫ t

0

∫Z

E |h(s, z)|pr+2

E ν(dz)ds.(26)

Iteration of the calculation (23) and substitution of (26) leads to

E

∣∣∣L(t)(r)∣∣∣pm

C0 E

∣∣∣L(t)(r+1)∣∣∣pm−2

+2∑

i=1

CiE

(∫ t

0+

∫E|h(s; z)|pr+i

ν(dz)ds

)pm−i+1

≤ C

m∑k=1

E

(∫ t

0+

∫Z|h(s, z)|pr+1+k

ν(z)ds

)m−k

Assumption (21) and interpolation yields

E

∣∣∣L(t)(r)∣∣∣pm

< ∞.

Remark 3.7. By Remark 3.4 it follows, that Proposition 3.3 and Proposi-tion 3.1 remains valid also if η is a compensated Poisson random measurewith characteristic measure ν such that ν is a Levy measure on (Z, Z).

4. Proof of Existence and Uniqueness of Solutions

The proof of existence and uniqueness is based on the Banach fixed pointtheorem for contractions (see e.g. Zeidler [49, Theorem 1.A, Chapter 1.1]).We first will proof part (a) and secondly we sketch the proof of part (b) and(c).

4.1. Proof of Part (a) of Theorem 2.1. We denote by Vq,q(T ) the com-plete Banach space of all V−(γ−δ)-valued measurable, predictable processesu defined on the time interval [0, T ] equipped with the norm

|||u|||q,q =[∫ T

0E|u(s)|q ds

]1q

< ∞, u ∈ Vq,q(T ).

Page 24: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

SPDEs driven by Poisson Random Measure 21st May 2005 23

Let ϕ be a F0-measurable random variable belonging to V−γ . Let δf and δg

be given. We say a the random variable ϕ satisfies the assumption

(A) iff E|ϕ|q−γ < ∞,We say the constants δf , δg and γ satisfies the assumption

(B) iff δgq < 1, and δf < 1,(C) iff (δg − γ)p < 1 − 1

q , and δf − γ < 1 − 1q ,

(D) iff 0 ≤ γq < 1.

Furthermore, let

VIDq,q(T ) := Vq,q(T ) ∩ L0

(Ω; ID([0, T ];V−(γ+δ))

).

Since V−(γ+δ) is a separable Banach space, ID([0, T ];V−(γ+δ)) is metrizableand the resulting metric is complete (Ethier and Kurtz [22, Theorem 5.6,Chapter 3]). Moreover, see Ethier and Kurtz [22, Theorem 1.7, Chapter3]), L0

(Ω; ID([0, T ];V−(γ+δ))

)is also a complete separable metric space with

respect to the Prohorov metric (for definition see e.g. Ethier and Kurtz [22,Chapter 3.1] or the Appendix, Chapter A.1). Further, let VID

q,q(T ) be theclosure of Vq,q(T )-norm. Note, since ID([0, T ];V−(γ+δ)) is not a topologicalvector space VID

q,q(T ) ⊃ VIDq,q(T ) but not necessarily VID

q,q(T ) = VIDq,q(T ). Let

Kϕ : VIDq,q(T ) → VID

q,q(T ) be the following transformation

(Kϕu)(t) = Stϕ +∫ t

0+St−sf(u(s−)) ds +∫ t

0+

∫Z

St−sg(u(s−); z) η(dz; dz)

:= Stϕ + K1u + K2u, t ∈ [0, T ], u ∈ VIDq,q(T ),(27)

The proof of existence and uniqueness of the solution under the assumptions(A), (B), (C) and (D) is divided into the following steps:

(1) First we show that Kϕ maps VIDq,q(T ) into VID

q,q(T ). Thus, we showthat(i) Kϕ maps VID

q,q(T ) into Vq,q(T ).(ii) Kϕ maps VID

q,q(T ) into L0(Ω; ID([0, T ];V−(γ+δ))

).

(2) Secondly, we will show that there exists a constant T > 0 dependingonly on f and g, such that Kϕ has a unique fix point in Vq,q(T ) andthat this fix point belongs to L0(Ω; ID([0, T ];V−(γ+δ)). This will beshown in the following two substeps:

Page 25: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

24 ERIKA HAUSENBLAS

(i) there exists a constant T > 0, such that for all ϕ satisfyingassumption (A) the operator Kϕ : VID

q,q(T ) → Vq,q(T ) is a strictcontraction with respect to the norm in Vq,q(T )

(ii) For all ϕ satisfying assumption (A) the sequence{x(n)

}n∈IN

defined by x(n) = Kϕx(n−1), n ≥ 1 and x(0)(t) = Stϕ, is tight inL0(Ω; ID([0, T ];V−(γ+δ)).

Applying the Banach fix point theorem (see e.g. Zeidler [49, Theorem1.A]), there exists a unique x∗ ∈ VID

q,q(T ), such that Kϕx∗ = x∗

and Kϕ(n)y → x∗ for all y ∈ VID

q,q(T ). From (2)-(ii), it follows thatthere exists a unique fixed point x∗ belongs to L0(Ω; ID([0, T ];E),i.e. VID

q,q(T ).(3) By step (2) u0 ∈ VID

q,q(T ) that is a unique fixed point of Kϕ. Usingthe same argument as in step (2) on VID

q,q(u(T ); T ), we have a uniquefix point u1 ∈ VID

q,q(u(T ); T ), provided u(T ) satisfies the assumption(A). This argument can be shifted to [T , 2T ]. Since T does notdepend on ϕ, we can find solutions u0, u1, . . ., um, respective, on[0, T ], [T , 2T ], . . . , [mT , T ∧ (m + 1)T ], respective, provided ui−1(iT )satisfies the assumption (A), i = 1, . . . ,m. Finally, we have to gluetogether all the solutions.

Proof of step (1)-(i): In this section we will show, that under assumptions(A), (B), and (D) the operator Kϕ maps VID

q,q(T ) into Vq,q(T ). Note that for0 < γ < 1

q

|||Stϕ|||qq,q =∫ T

0E|Stϕ|q ≤

∫ T

0|St|qL(V−γ ,E) dt |ϕ|q−γ .

Using Remark B.1 we have

|||Stϕ|||qq,q ≤∫ T

0C t−qγ dt |ϕ|q−γ

≤ C(T ) E |ϕ|q−γ .

Page 26: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

SPDEs driven by Poisson Random Measure 21st May 2005 25

Furthermore we have

|||K1u|||qq,q ≤ E

[∫ T

0

∣∣∣∣∫ t

0+St−sf(u(s−)) ds

∣∣∣∣q dt

]

≤ E

∫ T

0

(∫ t

0+|St−sf(u(s−))| ds

)q

dt

≤ E

∫ T

0

(∫ t

0+|St−s|L(V−δf

,E) |f(u(s−))|−δfds

)q

dt

≤ E

∫ T

0

(∫ t

0+(t − s)−δf |f(u(s−))|−δf

ds

)q

dt.(28)

Therefore, by the Young inequality we infer that

|||K1u|||qq,q ≤(∫ T

0+(T − t)−δf dt

)q

E

∫ T

0+|f(u(t−))|q−δf

dt.(29)

Next, the growth condition of f , i.e. (2), implies that

|||K1u|||qq,q ≤ C T (1−δf )q

(1 − δf )1q′

E

∫ T

0+(1 + |u(t−))|q) dt

≤ C T1−δf

q′

(1 − δf )q(T + |||u|||qq,q

).(30)

By Proposition 3.1 we get for K2

|||K2u|||qq,q = E

∫ T

0

∣∣∣∣∫ t

0+

∫Z

St−sg(u(s−); z) η(ds, dz)∣∣∣∣q dt

=∫ T

0E

∣∣∣∣∫ t

0+

∫Z

St−sg(u(s−); z) η(ds, dz)∣∣∣∣q dt

≤ C

∫ T

0

n∑l=1

(E

∫ t

0+

∫Z|St−sg(u(s−); z)|pl

ν(dz) ds

)pn−l

dt

≤ C

∫ T

0

n∑l=1

(E

∫ t

0+

∫Z(t − s)−δgpl |g(u(s−); z)|pl

−δgν(dz) ds

)pn−l

dt.(31)

The growth condition of g, i.e. (3), leads to

|||K2u|||qq,q ≤ C

∫ T

0

n∑l=1

(E

∫ t

0+(t − s)−δgpl

(1 + |u(s−))|pl)ds

)pn−l

.(32)

Again, the Young inequality yields

|||K2u|||qq,q ≤ Cn∑

l=1

(∫ T

0t−δgpl

dt

)pn−l

E

∫ t

0(1 + |u(s−))|q)ds.

Page 27: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

26 ERIKA HAUSENBLAS

Summing up, we have proved that the operator Kϕ maps Vq(T ) into Vq(T )and

|||Kϕu|||q,q ≤ C1(T )|u0|−γ + C2(T )|||u|||q,q + C3(T ), u ∈ Vq,q(T ).

Proof of step (1)-(ii). Fix u ∈ VIDq,q(T ). By a modification of the Proof of

Proposition I.4.7, Da Prato and Zabczyk [15] one can show that there existsa sequence {u(n))}n≥1 of simple random variables in ID([0, T ];V−(δ+γ)), suchthat we have ∫ T

0|u(s, ω) − u(n)(s, ω)|qE ds → 0 a.s.(33)

This sequence has the following properties:

(a) There exists N ∈ IN, such that {u(n)}n≥N is bounded in Vr, r(T ),r < q.

(b) The sequence {Kϕu(n)}n≥1 converges to Kϕu in Vp,∞,−γ(T ), i.e.sup0≤t≤T E

∣∣Kϕu(n)(t) −Kϕu(t)∣∣p−γ

→ 0 as n → ∞.

Proof. Ad (a): This follows by∫ T0 E|u(s)|q ds < ∞ and (33).

Ad (b): This property follows from a sequence of simple calculations:

E

∣∣∣Kϕu(n)(t) −Kϕu(t)∣∣∣p−γ

≤ E

∫ t

0

∣∣∣St−s

(f(u(s−)) − f(u(n)(s−))

)ds

∣∣∣p−γ

+ E

∣∣∣∣∫ t

0+

∫Z

St−s

(g(u(s−); z) − g(u(n)(s−); z)

)η(ds, dz)

∣∣∣∣p−γ

≤ E

(∫ t

0+

∣∣∣St−s

(f(u(s−)) − f(u(n)(s−))

)∣∣∣−γ

ds

)p

+ E

∫ t

0+

∫Z

∣∣∣St−s

(g(u(s−); z) − g(u(n)(s−); z)

)∣∣∣p−γ

ν(dz) ds

≤ E

(∫ t

0+|St−s|L(V−δf

,V−γ)

∣∣∣f(u(s−)) − f(u(n)(s−))∣∣∣−δf

ds

)p

+ E

∫ t

0+

∫Z|St−s|pL(V−δg ,V−γ)

∣∣∣g(u(s−); z) − g(u(n)(s−); z)∣∣∣p−δg

ν(dz) ds

≤ E

(∫ t

0+(t − s)−δf +γ

∣∣∣f(u(s−)) − f(u(n)(s−))∣∣∣−δf

ds

)p

dt

+ E

∫ t

0+

∫Z(t − s)(−δg+γ)p

∣∣∣g(u(s−); z) − g(u(n)(s−); z)∣∣∣p−δg

ν(dz) ds.

Page 28: ZiF: Research Group - uni-bielefeld.de · Banach spaces is done e.g. by Brooks and Dinculeanu [8], Dettweiler [18], Kussmaul [30], Neidhardt [35], R¨udiger [41], in which article

SPDEs driven by Poisson Random Measure 21st May 2005 27

The Lipschitz continuity, the Holder inequality and the Jensen inequalitygive

E

∣∣∣Kϕu(n)(t) −Kϕu(t)∣∣∣p ≤ E

(∫ t

0+(t − s)−δf +γ

∣∣∣u(s−) − u(n)(s−)∣∣∣ ds

)p

+E

∫ t

0+(t − s)(γ−δg)p

∣∣∣u(s−) − u(n)(s−)∣∣∣p ds

≤ C tp(1− 1r−(δf−γ))

(E

∫ t

0+

∣∣∣u(s−) − u(n)(s−)∣∣∣r ds

) pr

+ tp(1− 1r−(δg−γ)p)

(E

∫ t

0+

∣∣∣u(s−) − u(n)(s−)∣∣∣r ds

)≤ C(t)|||u − u(n)|||pr,r,

where r satisfies (δg−γ)p < 1− 1r , and δf −γ < 1− 1

r . By (a) and assumption(C) the assertions are proved. �

By (b) it can be shown, that for any finite set {t1, . . . , tk} ⊂ [0, T ], we have(Kϕu(n)(t1), . . . ,Kϕu(n)(tk)

)→

(Kϕu(t1), . . . ,Kϕu(tk)

),

as n → ∞.Applying Theorem 3.7.8 - (b) of Ethier and Kurtz [22] it remains to show,

that the sequence {Kϕu(n)}n is tight in ID([0, T ];V−(γ+δ)). Then it followsKϕu(n) → Kϕu in distribution as n tends to infinity.

In order to show {Kϕu(n)}n∈IN is a tight sequence in ID([0,∞);V−(γ+δ)),we will show, that {Kϕu(n)}n∈IN satisfies the Aldous condition (see Defini-tion A.2) and the compact containment condition (see Definition A.3).Proof that the Aldous condition is satisfied: We will show that thesequence Kϕu(n) satisfies the Aldous condition, see Definition A.2. In par-ticular, we will show, that the sequence satisfy the assumptions of TheoremA.1 with β = p. That means we will prove that for all t ∈ [0, T ] and allθ > 0 we have

E

[∣∣∣Kϕu(n)(t + θ) −Kϕu(n)(t)∣∣∣p−(γ+δ)

| F (n)t

]≤

θρ|||1(t,t+θ]Kϕu(n)|||pq,q,−γ + C2θpδ|||1(0,t]Kϕu(n)|||pp,∞,−γ.(34)

where ρ = min(1 − pq − (δg − γ − δ)p, p − 1− (δf − γ − δ)p). By (a) and (b)

we have for n ≥ N

|||1(0,t]Kϕu(n)|||p,∞,−γ ≤ |||Kϕu(n)|||p,∞,−γ ≤ C |||u(n)|||r,r ≤ R.

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28 ERIKA HAUSENBLAS

where R is a constant independent of n. Therefore 2C(θρ + θδ

)is an upper

bound for the RHS of (34). Moreover 2C(θρ + θδ

)tends to zero as θ → 0.

Therefore the assumptions of Lemma A.1 are satisfied.To prove inequality (34), we first note that

Kϕu(n)(t + θ) −Kϕu(n)(t) = (St+θ − St)ϕ+∫ t+θ

0+St+θ−σf(u(n)(σ−)) dσ +

∫ t+θ

0+St+θ−σg(u(n)(σ−); z) η(dz; dσ)

−∫ t

0+St−σf(u(n)(σ−)) dσ −

∫ t

0+St−σg(u(n)(σ−); z) η(dz; dσ)

= (Sθ − I) Stϕ + (Sθ − I)∫ t+θ

0+St−σf(u(n)(σ−)) dσ

+ (Sθ − I)∫ t+θ

0+St−σg(u(n)(σ−); z) η(dz; dσ)

+∫ t+θ

t+St+θ−σf(u(n)(σ−)) dσ +

∫ t+θ

t+St+θ−σg(u(n)(σ−); z) η(dz; dσ)

= (Sθ − I) Stϕ + (Sθ − I)(Kϕu(n)

)(t + θ)

+∫ t+θ

t+St+θ−σf(u(n)(σ−)) dσ︸ ︷︷ ︸

=:f1(t,θ)

+∫ t+θ

t+St+θ−σg(u(n)(σ−); z) η(dz; dσ)︸ ︷︷ ︸

=:f2(t,θ)

.

Hence Sθ − I =∫ θ0 ASr dr (see Pazy [36, Theorem 1.2.4-(b)]) and Remark

B.1 we have

|(Sθ − I)x|−(γ+δ) ≤∫ θ

0|ASrx|−(γ+δ) dr

≤ C

∫ θ

0r−1+δ |x|−γ dr ≤ C θδ |x|−γ .

Therefore

|(Sθ − I)Stϕ|−(γ+δ) ≤ C θδ |Stϕ|−γ ≤ C θδ |ϕ|−γ(35)

and

E

∣∣∣(Sθ − I)(Kϕu(n)

)(t + θ)

∣∣∣p−(γ+δ)

≤ θpδE

∣∣∣(Kϕu(n))

(t + θ)∣∣∣p−γ

.(36)

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SPDEs driven by Poisson Random Measure 21st May 2005 29

By the same calculations as in (28), (29) and (30) we obtain

E |f1(t, θ)|p−(γ+δ) = C E

∣∣∣∣∫ θ

0+Sθ−σf(u(n)(t + σ−)) dσ

∣∣∣∣p−(γ+δ)

C E

(∫ θ

0+(θ − σ)−(δf−(γ+δ))

∣∣∣f(u(n)(t + σ−))∣∣∣−δf

)p

≤ C θp

“1− 1

q−(δf−γ−δ)

”×

{∫ θ

0E

∣∣∣u(n)(t + σ)∣∣∣q dσ

} pq

.

Note, that

E |f2(t, θ)|p−(γ+δ) ≤ C E

∣∣∣∣∫ θ

0+

∫Z

Sθ−σg(u(n)((σ−) + t); z) η ◦ θt(dz; dσ)∣∣∣∣p−(γ+δ)

,

where η ◦ θt(ω) := η(θt ◦ ω), and θt is the usual shift operator defined byθtω(s) := ω(t + s). Following (31) and (32) we get

E |f2(t, θ)|p−(γ+δ) ≤ C θ1− p

q−p(δg−γ−δ)

{∫ θ

0E

∣∣∣u(n)(t + σ)∣∣∣q dσ

} pq

.

Proof that the compact containment condition is satisfied: Next,we have to prove the compact containment condition, see A.3, is satisfied.This is done by showing that the assumptions of Lemma A.1 are satisfied.Note, by Remark B.2 that if δ > 0, the space V−γ is compactly embeddedin V−(γ+δ). Hence, if we proof that

sup0<s≤t

E|Kϕu(n)(s)|p−γ ≤ C(1 + |||u(n)|||pr,r + E|ϕ|p−γ

),

for some r < q, then the compact containment condition follows. Indeed,by (b) and the assumption (A) the RHS is uniformly bounded in n. Firstnote, sup0<s≤t |Stϕ|−γ ≤ C |ϕ|−γ . Following the calculation on page 25 weobtain

E|u(t) − Stϕ|p−γ ≤ C1 E

(∫ t

0+

∣∣∣St−sf(u(n)(s−))∣∣∣−γ

ds

)p

+

C2 E

∫ t

0+

∫Z|St−sg(u(n)(s−); z)|p−γ η(dz; ds).(37)

The first summand on the right hand side of (37) is finite since

E

(∫ t

0+

∣∣∣St−sf(u(n)(s−))∣∣∣−γ

ds

)p

≤ C E

(∫ t

0+(t − s)−(δf−γ)

∣∣∣f(u(n)(s−))∣∣∣−δf

ds

)p

≤ C

(∫ t

0(t − s)−(δf−γ)r′ ds

) pr′

(1 + E

∫ t

0+

∣∣∣u(n)(s−)∣∣∣r ds

) pr

,(38)

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30 ERIKA HAUSENBLAS

where 1r′ + 1

r = 1 and r large enough, such that δf − γ < 1r . Because of (C)

the quantity C(t)(1 + |||u(n)|||pr,r) is an upper bound of the first summand.Analogous we have for the second summand of on the right hand side of(37)

E

∫ t

0+

∫Z

∣∣∣St−sg(u(n)(s−); z)∣∣∣p−γ

η(dz; ds) ≤ E

∫ t

0+

∫Z

∣∣∣St−sg(u(n)(s−); z)∣∣∣p−γ

ν(dz) ds

Next the growth condition of g, i.e. (3), leads

E

∫ t

0+

∫Z

∣∣∣St−sg(u(n)(s−); z)∣∣∣p−γ

η(dz; dz) ≤

C E

∫ t

0+

|St−s|pL(V−δg ,V−γ)

∫Z|g(u(n)(s−); z)|p−δg

ν(dz) ds

≤ C

∫ t

0|St−s|pL(V−δg ,V−γ)

(1 + E|u(n)(s−)|p

)ds.(39)

The Holder inequality gives for pr′ + p

r = 1 and r large enough such thatp(δg − γ) < 1

r holds

E

∫ t

0+

∫Z

∣∣∣St−sg(u(n)(s−); z)∣∣∣p−γ

η(dz; dz) ≤

C

(∫ t

0|St−s|r′L(V−δg ,V−γ) ds

) pr′

(1 +

∫ t

0E|u(n)(s)|r ds

) pr

,

Because of assumption (C) we have

E

∫ t

0+

∫Z|St−sg(u(n)(s−); z)|p−γ η(dz; dz) ≤ C(t)

(1 + |||u(n)|||pr,r

).

Proof of step (2)-(i) Next, we will show, that there exists a constantT > 0 independent from ϕ, such that Kϕ : VID

q,q(T ) → VIDq,q(T ) is a strict

contraction. Similarly to calculations (28), (29) and (30) the Lipschitz con-tinuity of f , i.e. (4), leads to

|||K1u −K1v|||qq,q ≤ C(q, T ) |||u − v|||qq,q, u, v ∈ VIDq,q(T ),

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SPDEs driven by Poisson Random Measure 21st May 2005 31

where the constant C(q, T ) tends to zero as T → 0 and only depends inaddition on δf and q. Next we have

|||K2u −K2v|||qq,q = E

∫ T

0|K2u(t) −K2v(t)|q dt

≤ C

∫ T

0

∣∣∣∣∫ t

0

∫Z

St−s (g(u(s−); z) − g(v(s−); z)) η(dz; ds)∣∣∣∣q dt

≤ C

∫ T

0E

n∑l=1(∫ t

0+

∫Z|St−s (g(u(s−); z) − g(v(s−); z)) |pl

ν(dz) ds

)pn−l

dt

≤ C

∫ T

0E

n∑l=1(∫ t

0+

∫Z(t − s)−δgpl|g(u(s−); z) − g(v(s−); z)|pl

−δgν(dz) ds

)pn−l

dt

≤ C E

n∑l=1

∫ T

0(∫ t

0+

∫Z(t − s)−δgpl|u(s−) − v(s−)|pl

h(z) ν(dz) ds

)pn−l

dt

≤ C E

n∑l=1

∫ T

0(∫ t

0+

∫Z(t − s)−δgpl|u(s−) − v(s−)|pl

ν(dz) ds

)pn−l

dt.

Again, the Young inequality yields

|||K2u −K2v|||qq,q ≤ C T (1−δgq)q

(1 − δgq)qE

∫ T

0+|u(s−) − v(s−)|q ds.

Summing up we have proved

|||Kϕu −Kϕv|||q,q ≤ C(T ) |||u − v|||q,q.

Moreover, the constant C(T ) tends to zero as T → 0 and only depends inaddition on δf , δg, q qnd p.Thus, we can find a T , such that

C(T ) < 1.

Proof of step (2)-(ii) Let{u(n)

}n≥0

be the sequence defined by u(n) :=Kϕu(n−1), n ≥ 1 and u(0)(t) = S(t)ϕ. We have to show, that

{u(n)

}n≥1

istight in L0(Ω; ID([0, T ];V−(γ+δ))). But first we will show the following.

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32 ERIKA HAUSENBLAS

• Under the assumptions (A), (B) and (D), the set{u(n)

}n≥0

is boundedin Vq,q(T ). By (2)-(i) we know, that Kϕ : V ID

q,q(T ) → V IDq,q(T ) is a strict

contraction, i.e. there exists a constant 0 < k < 1, such that

|||Kϕ(u − v)|||q,q ≤ k |||u − v|||q,q

for all u, v ∈ V IDq,q(T ). Hence

|||u(n)|||q,q = |||n∑

j=1

u(j) − u(j−1)|||q,q ≤

≤ 11 − k

(|||u(1)|||q,q + |||u(0)|||q,q

).

Since Kϕ : V IDq,q(T ) → V ID

q,q(T ) is bounded, |||u(1)|||q,q = |||Kϕu(0)|||q,q ≤C |||u(0)|||q,q. Moreover, since u(0)(t) = S(t)ϕ we have by RemarkB.1

|||u(0)|||qq,q = C

∫ T

0t−qγ dt E|ϕ|q−γ =

C T 1−qγ

1 − qγE|ϕ|q−γ .

Thus, the sequence{u(n)

}n≥0

is bounded in VIDq,q(T ) by

C T1q−γ

1 − k

[E|ϕ|q−γ

]1q .

• Under the assumptions (A), (B), (C) and (D) the sequence{u(n)

}n≥0

is uniformly bounded in VIDp,∞,−γ(T ). First, one can show under the

condition (B)-(δ), that

|||u(n)|||p,∞,−γ = |||Kϕu(n−1)|||p,∞,−γ ≤ [E|ϕ|q−γ

] 1q + |||u(n−1)|||q,q.

Secondly, by above it follows, that the RHS is uniformly bounded.

Proof that the Aldous condition is satisfied: Tracing the calculationsin (1)-(ii), we see that

{u(n)

}n≥0

satisfies the Lemma A.1 with assumptions

E

[∣∣∣u(n)(t + θ) − u(n)(t)∣∣∣p−(γ+δ)

| F (n)t

]≤

θρ|||1(t,t+θ]u(n−1)|||pq,q,−γ + C2θ

pδ|||1(0,t]Kϕu(n−1)|||pp,∞,−γ.

where ρ = min(1− pq − (δg −γ−δ)p, p−1− (δf −γ−δ)p). Since

{u(n)

}n≥0

isuniformly bounded in Vq,q(T ) and in Vp,∞,−γ(T ), there exists a constant 0 <

R < ∞, such that |||u(n)|||q,q, |||u(n)|||p,∞,−γ < R for all n ∈ IN. Furthermore,since Kϕu(n) = u(n+1), we infer that

E

[∣∣∣u(n)(t + θ)− u(n)(t)∣∣∣p−(γ+δ)

| F (n)t

]≤ R

(θρ + θpδ

)

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SPDEs driven by Poisson Random Measure 21st May 2005 33

and the assumptions of Lemma A.1 are satisfied.Proof that the compact containment condition is satisfied: By thesame calculation as in (1)-(ii), we can show that there exists a r < q, suchthat

sup0≤s≤t

E|u(n)(s)|p−γ ≤ C(1 + |||u(n−1)|||pr,r + E|ϕ|p−γ

).

Moreover, the sequence{u(n)

}n≥0

is uniformly bounded in Vr,r(T ). Thus,there exists a constant R, such that

supn≥1

sup0≤s≤t

E|u(n)(s)|pδ ≤ C (1 + R) .

Hence the compact containment condition follows.Proof of step (3) Since Kϕ is a strict contraction, for all φ there existsa unique fix point u� in VID

q,q(T , ϕ). By (2)-(ii), the fix point belongs toVID

q,q(φ, T ). Let us denote the fix point by u0, i.e. u0 := u�. Since T doesnot depend on ϕ, we can repeat the step (2). Set ϕ := u0(T ). If ϕ satisfiesassumption (A) we have again a unique fix point u� in VID

q,q(T ) of Kϕ. But

E|ϕ|q−γ = E|u(T )|q−γ

≤ C

(|ST ϕ|q−γ + E

(∫ T

0

∣∣ST−sf(u(s−))∣∣−γ

ds

)q

+

n∑l=1

E

(∫ T

0+

∫Z|ST−sg(u(s−); z)|pl

−γ ν(dz; ds)

)pn−l⎞⎠ .

The first summand is obviously finite, because of (A). The second summandis finite, because of (C). In particular

E

(∫ T

0

∣∣ST−sf(u(s−))∣∣−γ

ds

)q

≤ C E

(∫ T

0(T − s)−(δf−γ) |f(u(s−))|−δf

ds

)q

≤ C

(∫ T

0(T − s)−(δf−γ)q′′ ds

) qq′′

(1 + E

∫ T

0|u(s−)|q ds

)

≤ C(T )(1 + |||u|||qq,q

).

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34 ERIKA HAUSENBLAS

In the case of the third summand, we study only the worst when l = n. Bythe growth condition of g, i.e. (3) we have

E

∫ T1

0+

∫Z

∣∣ST−sg(u(s−); z)∣∣q−γ

ν(dz)ds =∫ T

0

∫Z

E∣∣ST−sg(u(s−); z)

∣∣q−γ

ν(dz)ds

C

∫ T

0|ST−s|qL(V−δg ,V−γ) E

∫Z|g(u(s−); z)|q−δg

ν(dz) ds

≤ C

∫ T

0|ST−s|qL(V−δg ,V−γ) (1 + E|u(s−)|q) ds.

Because of (B)-(δ), γ ≥ max(δf − 1 + 1q , δg), so that we can write∫ T

0

∫Z

E|ST−sg(u(s−); z)|q−γ ν(dz)ds ≤ C(T )

(1 +

∫ T

0E|u(s−)|q ds

).

Thus, we have

E|ϕ|q−γ ≤ C(T )(1 + |||u|||qq,q

).

Therefore we have shown, that there exists a unique fixed point u in V IDq,q(u1(T ), T ).

Let us call this fixed point u1. Hence by the same calculation as above,u1(T ) satisfies the assumption (A), we can repeat the same argument. SinceT only depends on δg, δf , p and q, we need only to repeat the procedure afinite number of times. In particular, there exists a constant m, such thatT ≤ Tm. Let ui be the unique fix point in V ID

q,q(ui−1(T ), T ). Let be

u(t) := 1t=0ϕ +(m−1)∑

i=0

1(iT ,(i+1)T ](t)ui(t − iT ).

Since ui ∈ ID([(i − 1)T , iT ];V−(γ+δ)) for all i = 1, . . . m − 1, it follows thatu(t) ∈ ID([0, T ];V−(γ+δ)). Since Kϕui = ui on [(i − 1)T , iT ], and ui(iT ) =ui+1(0), it follows Kϕu(t) = u(t) and therefore, that u(t) is a solution ofProblem (1).

Remark 4.1. Tracing the calculation in (2)-(ii) one can show if

δf < 1 − 1q, and δgp < 1 − 1

q,

then the unique solution belongs to L0 (Ω; ID((0, T ];E)). where

L0 (Ω; ID((0, T ];E)) := ∪r>0L0 (Ω; ID([r, T ];E)) .

This can be done, by showing, that the sequence u(n) defined by u(n) =Kϕu(n−1) and u(0) := Kϕφ is tight in L0 (Ω; ID((0, T ];E)) for all u ∈ VID

q,q(T ).

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SPDEs driven by Poisson Random Measure 21st May 2005 35

4.2. Proof of Part (b) in Theorem 2.1. Tracing the proof of part a,part b can be shown. Let us define the space for p < q < ∞

Vp,q(T ) :=

{u : Ω × [0, T ] → E, u be an adapted process,

such that∫ T

0(E|u(s)|p) q

p ds < ∞}

,

and for q = ∞

Vp,∞(T ) :=

{u : Ω × [0, T ] → E, u be an adapted process, u(0) = ϕ,

such that sup0≤s≤T

E|u(s)|q < ∞}

,

equipped with norm

|||u|||p,q :=

⎧⎨⎩

[∫ T0 (E|u(s)|p) q

p ds] 1

q if p < q < ∞,

sup0≤s≤T [E|u(s)|q] 1q if q = ∞.

In the proof we will consider the case q < ∞, the case q = ∞ is similarly.To be precise Part (c) can be proven by tracing the proof of part (b), settingq = ∞ and γ = 0. Let γ ≥ 0 and δ > 0 arbitrary small. Let

VIDp,q(T ) := Vp,q(T ) ∩ L0

(Ω; ID([0, T ];V−(δ+γ))

).

Let ϕ be a F0–measurable random variable, and Kϕ : VIDp,q(T ) → VID

p,q(T ) bethe same operator as in (27), in particular

(Kϕu)(t) = Stϕ +∫ t

0+St−sf(u(s−)) ds +∫ t

0+

∫Z

St−sg(u(s−); z) η(dz; dz)

:= Stϕ + K1u + K2u, t ∈ [0, T ], u ∈ Vq,q(T ).

As before existence and uniqueness of the solution to (1) will be shown byfixed point arguments. But before, we will introduce the conditions on ϕ,δf , δσ and γ. We say a the random variable ϕ satisfies assumption

(A) iff E|ϕ|p < ∞,We say the constants δf , δg and γ satisfies assumption

(B) iff δgp < 1, and δf < 1,(C) iff δgp < 1 − 1

q , and δf < 1 − 1q .

Analogously to part (a), the proof of part (b) will rely on the following steps:

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36 ERIKA HAUSENBLAS

(1) The operator Kϕ maps VIDp,q(T ) into VID

p,q(T ) . Thus, we have to showthat(i) that the operator Kϕ maps VID

p,q(T ) into Vp,q(T ).(ii) and that the operator Kϕ maps VID

q,q(T ) into L0(Ω; ID([0, T ];V−(γ+δ))

).

(2) There exists some constant T > 0 only depending on f and g, suchthat the operator Kϕ has a unique fix point in VID

p,q(T ) and that thisfix point belongs to L0(Ω; ID([0, T ];V−(γ+δ)). This will be shown inthe following two steps:(i) there exists a constant T > 0, such that for all ϕ satisfying

(A) the operator Kϕ : Vp,q(T ) → Vp,q(T ) is a contraction withrespect to the norm in Vp,q(T )

(ii) For all ϕ satisfying assumption (A) the sequence{x(n)

}n∈IN

,where x(n) = Kϕx(n−1), n ≥ 1 and x(0)(t) = Stϕ, is tight inL0(Ω; ID([0, T ];V−(γ+δ)).

Applying the Banach fix point theorem (see e.g. Zeidler [49, Theorem1.A]), there exists a unique x∗ ∈ VID

p,q(T ), such that Kϕx∗ = x∗ andKϕ

(n)y → x∗ for all y ∈ VIDq,q(T ). From (2)-(ii), it follows that the fix

point x∗ belongs to L0(Ω; ID([0, T ];V−(δ+γ)), i.e. to VIDp,q(T ).

(3) By (2) we have a unique fix point u in VIDp,q(T ). Using the same

arguments as in (2), we have a unique fix point, i.e. fix point, inVID

p,q(u(T ); T ), provided u(T ) satisfies the assumption (A). This so-lution can be shifted to [T , 2T ]. Since T does not depend on ϕ, wecan find solutions on [0, T ], [T , 2T ], . . . , [mT , T ∧ (m+1)T ], providedu(iT ) satisfies the assumption (A), i = 1, . . . ,m−1. Finally, we haveto glue together all the solutions.

Proof of part (1)-(i): In this section we will show, that under condi-tion (A), (B), and (D) the operator Kϕ : VID

p,q(T ) → Vp,q(T ) is a boundedoperator. Note we have for 0 < γ < 1

q

|||Stϕ|||qp,q =∫ T

0[E|Stϕ|p]

qp ≤

∫ T

0|St|qL(V−γ ,E) dt

[E|ϕ|p−γ

] qp

Using B.1 we have

|||Stϕ|||qp,q ≤∫ T

0C t−qγ dt

[E|ϕ|p−γ

] qp

≤ C T 1−qγ[E|ϕ|p−γ

] qp .(40)

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SPDEs driven by Poisson Random Measure 21st May 2005 37

Let c1, c2 ≥ 0 such that c1 + c2 = 1 and p′ conjugate to p. Then we have bythe Holder inequality

|||K1u|||qp,q ≤∫ T

0

[E

∣∣∣∣∫ t

0+St−sf(u(s−)) ds

∣∣∣∣p] q

p

dt

≤∫ T

0

[E

(∫ t

0+|St−sf(u(s−))| ds

)p] qp

dt

≤∫ T

0

[E

(∫ t

0+(t − s)−(c1+c2)δf |f(u(s−))|−δf

ds

)p] pq

dt

≤∫ T

0

[(∫ t

0(t − s)−c1p′δf ds

) pp′

E

(∫ t

0(t − s)−c2δf p |f(u(s−))|p−δf

ds

)] qp

dt.

A second application of the Holder inequality yields

|||K1u|||qp,q ≤

≤ C T (1−c1δf p′) qp′

∫ T

0

[(∫ t

0(t − s)−c2δf p

E |f(u(s−))|p−δfds

)] qp

dt.

The Young inequality for convolution yields

|||K1u|||qp,q ≤

≤ C T (1−c1δf p′) qp′

(∫ T

0(T − s)−c2δf p ds

) qp(∫ T

0

[E |f(u(s−))|p−δf

] qp

ds

)

≤ C T

“1p′ +

1p−(c1+c2)δf

”q(∫ T

0[(1 + E |u(s−))|p)] q

p ds

).

Next, the growth condition of f , i.e. (2), implies

|||K1u|||qp,q ≤ ≤ C T (1−δf)q(|||u|||qp,q

).(41)

By Proposition 3.1 we get for K2

|||K2u|||qp,q =∫ T

0

[E

∣∣∣∣∫ t

0+

∫Z

St−sg(u(s−); z) η(ds, dz)∣∣∣∣p] q

p

dt

≤ C

∫ T

0

[∫ t

0

∫Z

E|St−sg(u(s−); z)|pν(dz) ds

] qp

dt

Remark B.1 yields to

|||K2u|||qp,q ≤ C

∫ T

0

[∫ t

0(t − s)−δgp

∫Z

E|g(u(s−); z)|p−δgν(dz) ds

] qp

dt

The Young inequality for convolution leads to

|||K2u|||qp,q ≤ C

(∫ T

0s−δgp ds

) qp

(∫ T

0

(∫Z

E|g(u(s−); z)|p−δgν(dz)

) qp

ds

).

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38 ERIKA HAUSENBLAS

The growth condition of g, i.e. (3), leads to

|||K2u|||qp,q ≤ C T (1−pδg) qp

(∫ T

0(1 + E|u(s)|p) q

p ds

)≤ C T

(1−pδg) qp(1 + |||u|||qp,q

).(42)

Collecting then all together, i.e. (40), (41) and (42), gives boundedness ofthe operator Kϕ : Vp,q(T ) → Vp,q(T ), i.e.

|||Kϕu|||p,q ≤ C1(T )|u0|−γ + C2(T )|||u|||q,q + C3(T ).(43)

Remark 4.2. Note, that using the Lipschitz conditions (4) and (5) insteadof the grow conditions (2) and (3) the equation (43) can be written as

|||Kϕu|||p,q ≤ C1 T1q−γ |u0|−γ + C2 T ρ|||u|||q,q.

where ρ = min(1 − δf , 1p − δg).

Proof of part (1)-(ii). Fix u ∈ VIDp,q(T ). We have to show, that Kϕu

belongs to the Skorohod space ID([0, T ];V−(γ+δ)). By the same arguments asin (a)-(1)-(ii), there exists a sequence

{u(n)

}n∈IN

of simple random variablesbelonging to ID([0, T ];V−(γ+δ)), such that∫ T

0|u(s, ω) − u(n)(s, ω)|q ds → 0 a.s.

Similarly to (a)-(1)-(ii) the sequence satisfies the following points:

(a) For all r < q there exists an N > 0 and R > 0 such that for alln ∈ IN |||u(n)|||p,r ≤ R < ∞.

(b) There exists some r < q such that we have for all r ≤ r′ < q

|||Kϕu(n)|||p,∞,−γ ≤ [E|ϕ|p−γ

] 1p + |||u(n)|||p,r′ .

(c) The quantity sup0≤t≤T E∣∣Kϕu(n)(t) −Kϕu(t)

∣∣p converges to zero asn tends to infinity.

Tightness is shown again by verifying the Aldous condition A.2 and thecompact containment condition A.3 in Appendix A. Proof of the Aldouscondition: We will show, that the sequence satisfy the assumptions ofTheorem A.1 with β = p. In particular, we will prove that for all times0 ≤ t ≤ T and all θ we have

E

[∣∣∣Kϕu(n)(t + θ)−Kϕu(n)(t)∣∣∣p−(γ+δ)

| F (n)t

]≤

θρ|||1(t,t+θ]u(n)|||pp,r + C2θ

pδ|||1(0,t]Kϕu(n)|||pp,∞.(44)

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SPDEs driven by Poisson Random Measure 21st May 2005 39

where ρ = min(1 − pq − (δg − γ − δ)p, p − 1 − (δf − γ − δ)p). Note, that we

have by (a) for the quantity

|||1(t,t+θ]u(n)|||p,r ≤ |||u(n)|||p,r ≤ C < ∞.

Moreover,

|||1(0,t]Kϕu(n)|||p,∞,−γ ≤ |||Kϕu(n)|||p,∞,−γ ≤ |||u(n)|||p,r ≤ C < ∞.

Therefore we can give an upper bound for the RHS of (44) 2C(θρ + θδ

),

which tends to zero for θ tends to zero, and the assumptions of Lemmas A.1are satisfied.In order to show the inequality in (44), note that we have for r < q satisfying1 − δf ≤ 1

r and 1p − δg ≤ 1

r

Kϕu(n)(t + θ) −Kϕu(n)(t) = (Sθ − I)(Kϕu(n)

)(t + θ)

+∫ t+θ

tSt+θ−σf(u(n)(σ−)) dσ︸ ︷︷ ︸

=:f1(t,θ)

+∫ t+θ

tSt+θ−σg(u(n)(σ−); z) η(dz; dσ)︸ ︷︷ ︸

=:f2(t,θ)

.

By the same calculation as done for (36) we have and therefore

E

∣∣∣(Sθ − I)(Kϕu(n)

)(t + θ)

∣∣∣p−(γ+δ)

≤ θpδE

∣∣∣(Kϕu(n))

(t + θ)∣∣∣p−γ

.

Let c1, c2 ≥ 0, c1p′(δf − γ) < 1 and p′ conjugate to p. Then, we obtain by

the Holder inequality

E |f1(t, θ)|p−(γ+δ) ≤

C E

(∫ θ

0(θ − σ)−(δf−(γ+δ))

∣∣∣f(u(n)(t + σ−))∣∣∣−δf

)p

≤ C

(∫ θ

0(θ − σ)−c1p′(δf−(γ+δ)) dσ

) pp′

×(∫ θ

0(θ − σ)−c2p(δf−(γ+δ))

E

∣∣∣f(u(n)(t + σ))∣∣∣p−δf

).

The Holder inequality leads for r′ conjugate to qr to

E |f1(t, θ)|p−(γ+δ) ≤ C θp( 1

p′ −c1(δf−(γ+δ)))θ

1r′−c2p(δf−(γ+δ))

×[∫ θ

0

(E

∣∣∣f(u(n)(t + σ))∣∣∣p−δf

) rp

] pr

.

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40 ERIKA HAUSENBLAS

The growth condition on f , i.e. (2) yields

E |f1(t, θ)|p−(γ+δ) ≤ C θp(1− 1r(δf−(γ+δ)))

[∫ θ

0

(1 + E

∣∣∣u(n)(t + σ)∣∣∣p) r

pdσ

] pr

≤ C θp(1− 1r(δf−(γ+δ)))

[1 + |||u(n)1[t,t+θ]|||pp,r

].

Analogously to the proof of (a)-(1)-(ii) we have

E |f2(t, θ)|p−(γ+δ) ≤ C E

∣∣∣∣∫ θ

0

∫Z

Sθ−σg(u(n)((σ−) + t); z) η ◦ θt(dz; dσ)∣∣∣∣p−(γ+δ)

,

where η ◦ θt(ω) := η(θt ◦ ω), and θt is the usual shift operator defined byθtω(s) := ω(t + s). Following the calculations of (b)-(1)-(i) we get

E |f2(t, θ)|p−(γ+δ) ≤ C θp

“1p− 1

r−(δg−(γ+δ))

” [1 + |||u(n)1[t,t+θ]|||pp,r

].

Proof of the compact containment condition: Analogously to (a)-(1)-(ii) we have to show

sup0<s≤t

E|u(n)(s)|p−γ ≤ C(1 + |||u(n)|||qp,r + E|ϕ|p−γ

),(45)

the compact containment condition follows. In particular, by (b) and theassumption (A) the RHS is uniformly bounded in n. But tracing the calcu-lation in (b)-(1)-(i) for q = ∞, inequality (45) follows.

Proof of part (2)-(i) Next, we have to show, that a constant T > 0independent from ϕ exists, that Kϕ : VID

p,q(T ) → VIDp,q(T ) is a contraction,

where Kϕ is defined by (27). But thanks to remark (4.2) we have

|||Kϕu|||p,q ≤ C1 T1q−γ |u0|−γ + C2 T ρ|||u|||q,q.

where ρ = min(1 − δf , 1p − δg). Thus, we can find a constant T , such that

Kϕ is a contraction on VIDp,q(T ).

Proof of part (2)-(ii) Let{u(n)

}n≥0

be the sequence defined by u(n) :=Kϕu(n−1) and u(0)(t) = S(t)ϕ. We have to show, that

{u(n)

}n≥1

is tight inL0(Ω; ID([0, T ];V−(γ+δ))), but first we show the following:

• Under the assumptions (A), (B) and (D), the set{u(n)

}n≥0

is boundedin Vq,q(T ) (see (a)-(2)-(ii)).

• Under the assumptions (A), (B), (C) and (D) the sequence{u(n)

}n≥0

is uniformly bounded in VIDp,∞,−γ(T ) (see (a)-(2)-(ii)).

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SPDEs driven by Poisson Random Measure 21st May 2005 41

Proof of the Aldous condition: Tracing the calculation in (b)-(1)-(ii),we can see that

{u(n)

}n≥0

satisfies the condition of Lemma A.1 with

E

[∣∣∣u(n)(t + θ)− u(n)(t)∣∣∣p−(γ+δ)

| F (n)t

]≤

θρ|||1(t,t+θ]u(n−1)|||pp,q + C2θ

pδ|||1(0,t]Kϕu(n−1)|||pp,∞,−γ,

where ρ = min(1− pq − (δg − γ − δ)p, p− 1− (δf − γ − δ)p), and by the same

consideration as in (a)-(2)-(ii) the assumptions of Lemma A.1 are satisfied.Proof of the compact containment condition: By the same calculationas in (1)-(ii), we can show that

sup0≤s≤t

E|u(n)(s)|p−γ ≤ C(1 + |||u(n−1)|||pq,q + E|ϕ|p−γ

).

The sequence{u(n)

}n≥0

is uniformly bounded in Vq,q(T ), there exists aconstant R, such that

sup0≤s≤t

E|u(n)(s)|pδ ≤ C (1 + R) .

Thus, the compact containment condition follows.Proof of part (3) This part can be shown by the same consideration as in(a)-(3).

5. 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

+ be a Poisson random measure on Z × R+ with characteristic

measure ν : Z → R+ ∈ Lsym(Z). Let Nt(A), A ∈ Z, be the counting process

defined on page 8. Let (Ω,F , P) be a complete probability space and (Ft)t≥0

be the right continuous filtration induced by η. That means, the smallestfiltration, such that the counting measure Nt(A) is Ft-measurable for alls ≤ t ≤ T and A ∈ Z. Let Z0 = {z ∈ Z | |z| ≤ 1},

F0t := σ (Ns(A); A ⊂ Z0, 0 ≤ s ≤ t) ,

and

FCt := σ (Ns(A); A ⊂ Z \ Z0, 0 ≤ s ≤ t) .

Since for A,B ∈ Z, A∩B = ∅, the processes η(A×(s1, s2]) and η(A×(s1, s2])are independent for all 0 < s1 ≤ s2 ≤ T , the two filtration F0

t and FCt are

independent. Let us define the two probability spaces (Ω0,F0, {Ft}t, P0) and(ΩC ,FC , {FC

t }t, PC), where Ω0 := Ω, ΩC := Ω, F0 := ∧tF0t , FC := ∧tFC

t ,P0( · ) := P( · |FC), and PC( · ) := P( · |F0). From the independence of F0

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42 ERIKA HAUSENBLAS

and FC follows, that P is the product of P0 and PC .

Let E be a separable Banach space of M type p, 1 < p ≤ 2. Further weassume that the mapping g and f satisfies the hypothesis of Theorem 2.1-(c)with δf and δg. We consider the following SPDE⎧⎨

⎩u(t) dt = (Au(t) + f(u(t−))) dt

+∫Z0 g(u(t−); z)η(dz; dt), t ≥ 0

u(0) = u0 ∈ E,(46)

Given δ > 0, by Theorem 2.1–(c) it follows that Problem (46) has a uniquesolution u(1) belonging to

L0(Ω0; ID((0, T ];V−δ)

)such that sup0≤s≤T E|u(s)|p < ∞. Let τ1 be the following FC

t –stopping time

τ1 := inft>0

{Nt(Z \ Z0) > 0}and τ1 = τ1∧T . Note, τ1 is an exponential distributed random variable withparameter Cν := ν(Z\Z0) over (ΩC ,FC , PC), independent from (Ω0,F0, P0).Let u(1)(t) = u(1)(t) for 0 ≤ t < τ1 and u(1)(t) = 0 for t ≥ τ1. Next,E|u(1)(τ1)|p can be written as

E|u(1)(τ1)| = E

(E

[|u(1)(s)|p | τ1 = s

]).

Since τ1 is independent from (Ω0;F0; P0) we have

E|u(1)(τ1)| =∫ τ1

0P

C (τ1 = s) E0 |u(1)(s)|p ds

=∫ T

0Cνe

−sCν ds sup0≤s≤T

E0|u(1)(s)|p + e−TCν E0 |u(1)(T )|p,(47)

where E0 denotes the expectation value with respect to P0. The underlyingPoisson random measure is time homogeneous, therefore we can introducethe shift operator. We assume that a quadruple

T = (Ω,F , P, ϑ) ,

is given where (Ω,F , P, ) is a probability space and ϑ : R × Ω � (t, ω) �→ϑtω ∈ Ω is a measurable map such that for all t, s ∈ R, ϑt+s = ϑt ◦ ϑs. Letu(2) be a solution to⎧⎪⎪⎪⎨

⎪⎪⎪⎩du(t) = (Au(t) + f(u(t−))) dt +∫

Z0 g(z;u(t−)) (η ◦ ϑτ1) (dz; dt), t ≥ 0,

u(0) = u(1)(τ1−) +∫Z g(z;u(1)(τ1−)) η(dz; {τ1}).

(48)

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SPDEs driven by Poisson Random Measure 21st May 2005 43

By the calculation (47) the assumption of Theorem 2.1-(c) are satisfied.Therefore the solution u(2) exists, is unique and belongs to

L0(Ω0; ID([0, T − τ1];V−δ)).

Let τ2 be the FCt –stopping time

τ2 := inft>τ1

{Nt(Z \ Z0) > 1}

and τ2 = τ2 ∧ T . Let us define u(2) by

u(2)(t) :={

u(1)(t), for 0 ≤ t < τ1,

u(2)(t − τ1), for τ1 ≤ t < τ2.

Obviously, we have u(2) ∈ L0(Ω0; ID([0, τ2);V−δ)). Repeating the step andtaking into account that ν(Z \Z0) < ∞, we get a countable set of stoppingtimes {τn | n ∈ IN}, where

τn := inft>τn−1

{Nt(Z \ Z0) > n − 1}

and τn = τn ∧ T . Moreover, for each n we can define the process u(n) ∈ID([0, τn, Vδ) by

u(n)(t) :={

u(n−1)(t), for 0 ≤ t < τn−1,

u(n)(t − τn−1), for τn−1 ≤ t < τn,

where u(n) is the solution to Problem⎧⎪⎪⎪⎨⎪⎪⎪⎩

du(t) = (Au(t) + f(u(t−))) dt +∫Z0 g(z;u(t−))

(η ◦ ϑτn−1

)(dz; dt), t ≥ 0,

u(0) = u(n−1)(τn−1−) +∫Z g(z;u(n−1)(τn−1−)) η(dz; {τn−1}).

We have to show, that limn→∞ u(n) exists and belongs to L0 (Ω; ID([0, T );Vδ)).This can be done by the amalgamation procedure of Elworthy [20, ChapterIII.6]. The family of stopping times is a ordered by its natural order. Tobe precise τn ≤ τm holds a.s. as n ≤ m. The stopping time T is an upperbound of the family of stopping times, i.e. τn ≤ T and P(T − τn ≥ δ) tendsto zero as n tends to infinity. Moreover, for each n ∈ IN, we have an adaptedprocess u(n) belonging to L0(ID([0, τn);V−δ)). By Elworthy [20, Chapter III,Lemma 6B, p. 43], the process u : [0, T ) → V−δ exists and u|[0,τn) = u(n) forall n ∈ IN. Moreover, tracing the proof of Elworthy [20, Chapter III, Lemma6B, p. 43] on can show, that a version of the process u exists, such that theversion belongs to L0(ID([0, T );V−δ)).

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44 ERIKA HAUSENBLAS

Appendix A. The Skorohod Space

For a introduction to the Skorohod space we refer to Ethier and Kurtz[22] and Jacod and Shiryaev [26]. The results stated in this chapter is takenfrom Ethier and Kurtz [22].

Let E be a separable Banach space. The space ID([0, T ];E) denotes thespace of all right continuous functions x : [0, T ] → E with left limits. SinceID([0, T ];E) is complete but not separable in the uniform topology, we endowID([0, T ];E) with the Skorohod topology, which is characterized as follows:a sequence (x(n))n∈IN ⊂ ID([0, T ];E) converges to x ∈ ID([0, T ];E) iff thereis a sequence {λn}n∈IN ⊂ Λ :=

{λ : R → R, λ(0) = 0, λ(T ) = T, and λ is

strictly increasing}, such that

• sup0≤s≤T |λn(s) − s| → 0 as n → ∞.• sup0≤s≤T |x(n) ◦ λn(s) − x(s)| → 0 as n → ∞.

The Skorohod topology is metrizable, and the resulting metric space is sep-arable and complete. Nevertheless, ID([0, T ];E) is not a topological vectorspace (see Jacod and Shiryaev [26, Remark VI.1.22]).

Let {x(n)}n∈IN be a sequence of processes defined on the filtered prob-ability space (Ω(n),F (n),

(F (n)

t

)0≤t≤T

, P(n)), such that x(n) ∈ ID([0, T ];E)

for all n ∈ IN. Let L0(Ω; ID([0, T ];E)) be the space of all random variablesx : Ω → ID([0, T ];E) topologized by convergence in distribution. Note,if the underlying metric space is separable, convergence in distribution isequivalent to convergence in the Prohorov metric.

Definition A.1. Let (S, d) be a metric space, S denoting the Borel σ al-gebra of S and P(S) is the family of Borel probability measures on S. TheProhorov ρ metric on P(S) is given by

ρ(P,Q) := inf {ε > 0 : P (F ) ≤ Q(F ε) + ε for all F ∈ C} ,

where C is the collection of closed subsets of S and

F ε ={

x ∈ S : infy∈F

d(x, y) < ε

}.

Now, under which conditions the sequence{x(n)

}n∈IN

, x(n) ∈ L0(Ω(n); ID([0, T ];E))converges in distribution to a limit x ∈ L0(Ω; ID([0, T ];E))? By Theorem3.7.8-(b) of Ethier and Kurtz [22] it remains to show, that

• {x(n)

}n∈IN

is tight,

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SPDEs driven by Poisson Random Measure 21st May 2005 45

• there exists a dense set D ⊂ [0, T ] such that for any finite set{t1, . . . , tk} ⊂ D we have(

x(n)(t1), . . . , x(n)(tk)) → (

x(t1), . . . , x(tk))

in distribution as n → ∞.

In case E is finite dimensional, tightness of the sequence {x(n)}n can beshown by the Aldou’s condition.

Definition A.2. Let {x(n)}n∈IN be a sequence of stochastic processes withsample path in ID([0, T ]; R), such that x(n) ∈ L0(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 Aldou’s condition, iff for all ε > 0 and δ > 0 thereexists θ > 0 and n0 ∈ IN such that for any family {τn}n∈IN, where τn is a(F (n)

t )–stopping time on Ω(n) with τn ≤ T , we have

supn≥n0

suph≤θ

P(n)

(|x(n)

τn− x

(n)(τn+h)∧T | ≥ δ

)≤ ε.

Note, in the definition above, only τn n ∈ IN are stopping times, θ is aconstant. In case of E being infinite dimensional we have to add to theAldou’s condition a compact containment condition (see e.g. Ethier andKurtz [22, Chapter 3]) .

Definition A.3. Let {x(n)}n∈IN be a sequence of stochastic processes withsample path in ID([0, T ]; R), such that x(n) ∈ L0(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 ε > 0and every rational t > 0, there exists a compact subset Kε,t ⊂ E, such that

infn

P(n)(x(n)(t) ∈ Kε,t) ≥ 1 − ε.

Remark A.1. In fact Condition A.3 is weaker than the following condition:For each ε > 0 and every rational t > 0, there exists a compact subsetKε,t ⊂ E, such that

infn

P(n)

(x(n)(s) ∈ Kε,t ∀ s ∈ [0, t]

)≥ 1 − ε.(49)

The stronger condition (49) follows from A.3 (see Ethier and Kurtz [22,Remark 3.7.3]) and is sufficient to show relative compactness.

Let E be a separable Banach space with norm | · |. The Aldou’s conditionis often difficult to verify directly. Thus, to show, that {x(n)}n∈IN satisfies

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46 ERIKA HAUSENBLAS

the Aldous condition A.2, one can show certain integrability conditions giventhe Lemma below (see Ethier and Kurtz [22, Chapter 3.8]).

Theorem A.1. (see Ethier and Kurtz [22, Theorem 8.6, Remark 8.7 inChapter 3]) Let E be a separable Banach space with norm | · |E, and let{x(n)}n∈IN be a sequence of stochastic processes with sample path in ID([0, T ]; R),such that x(n) ∈ L0(Ω(n),F (n), P(n)), n ∈ IN. Let (F (n)

t ) be the natural filtra-tion 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

}of nonnegative

random variables, such that for 0 ≤ t ≤ T and 0 ≤ h ≤ δ arbitrary

E

[∣∣∣x(n)(t + h) − x(n)(t)∣∣∣p | F (n)

t

]≤ E

[f(n)(δ) | F (n)

t

],

and supn∈IN E[f(n)(δ) | Ft

] → 0 as δ → 0.

Similarly, the following lemma gives an intergrability condition whichimplies A.3.

Lemma A.1. let {x(n)}n∈IN be a sequence of stochastic processes with sam-ple path in ID([0, T ]; R), such that x(n) ∈ L0(Ω(n),F (n), P(n)), n ∈ IN. LetΓ be a subspace of E with norm | · |Γ, such that the embedding Γ ↪→ E iscompact. Then the compact containment condition A.3 is satisfied, if somep > 1 exists, such that

supn∈IN

E(n)

∣∣∣x(n)(t)∣∣∣pΓ≤ C < ∞.

Appendix B. Analytic semigroups and fractional powers of

operators

For an introduction to analytic semigroup we refer to Engel and Nagel[21] or Pazy [36], for interpolation theory e.g. the appunti of Lunardi [32].

B.1. Analytic Semigroups. An important subclass of semigroups are so–called analytic semigroups. For any ω ∈ R and θ ∈ (0, π) let

Σθ,ω = {λ ∈ C \ {ω} | | arg(λ − ω)| ≤ θ} .

Definition B.1. (Engel and Nagel [21, Definition II.4.2]) A closed linearoperator (A,D(A)) with dense domain D(A) in a Banach space E is calledsectorial it there exist a 0 < δ ≤ π

2 and a ω ∈ R such that the sector Σπ2+δ,ω

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SPDEs driven by Poisson Random Measure 21st May 2005 47

is contained in the resolvent set ρ(A), and if for each ε ∈ (0, δ) there existsMε ≥ 1 such that

|R(λ : A)| ≤ Mε

|λ − ω| ∀0 �= λ ∈ Σπ2+δ−ε,ω,

where R(λ : A) := (λ − A)−1 denotes the resolvent of the operator A. If(A,D(A)) is sectorial with angle δ and ω ∈ R, we say (A,D(A)) ∈ H(ω, δ).

For sectorial operators and appropriate paths γ in the complex plane, theassociated semigroup can defined via the Cauchy integral formula.

Definition B.2. (Engel and Nagel [21, Definition II.4.2]) Let (A,D(A)) bea sectorial operator of angle δ. Define T (0) := I and operators T (z), forz ∈ σδ, by

T (z) :=1

2πi

∫γR(λ;A) dλ.

where γ is any piecewise smooth curve in Σπ2+δ going from ∞e−i(π

2+δ′) to

∞ei(π2+δ′) for some δ′ ∈ (| arg(z)|, δ)).

It can be shown (see e.g. [21, Proposition II.4.3] or [36, Theorem 2.5.2]),thatthe in Definition B.1 defined semigroup of an sectorial operator is analytic.

Definition B.3. (Engel and Nagel [21, Definition II.4.5]) A family of oper-ators (T (z))z∈Σδ∪{0} ⊂ L(E) is called an analytic semigroup if

(i) T (0) = I and T (z1 + z2) = T (z1)T (z2) for all z1, z2 ∈ Σδ.(ii) The map z → T (z) is analytic in Σδ.(iii) limz→0,z∈Σδ′ T (z)x = x for all x ∈ E and 0 < δ′ < δ.

B.2. Fractional Power of Operators. For a sectorial operator A, where(A,D(A)) ∈ H(0, δ), 0 < δ < π

2 arbitrary and α > 0 one can define fractionalpowers of the operator.

Definition B.4. Let σ ⊂ C be an open sector, such that R+ ⊂ Σ ⊂ ρ(A).

For α > 0 the bounded linear operator (−A)α is defined by

(−A)α :=1

2πi

∫γλ−αR(λ,A)dλ,

where the path γ is a piecewise smooth part Σ \ R+ going from ∞e−iδ to

∞eiδ for some δ > 0.

Definition B.5. Let A be a sectorial operator. For every α > 0 we define

Aα = ((−A)−α)−1.

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48 ERIKA HAUSENBLAS

For α = 0, Aα = I.

In order to study regularity property of solution of some Cauchy problemit is convenient to introduce several scales of subspaces of E, e.g. domainof fractional powers of operators. In particular, Vα := D((−A)α) equippedwith norm | · |α := |(I − A)α · |.

Remark B.1. In the following we will use certain interpolation inequality.In particular let E be a separable Banach space, A be a operator generatingan analytic semigroup (exp(−ωt)St)t≥0 of contraction on E. Then thereexists a constant M ≥ 1, such that we have (see e.g. (see e.g. [21, TheoremII.4.6] or Pazy [36, Theorem 2.6.3])

|(−A)αS(t)x|α ≤ C exp(ωt) t−α.

Remark B.2. Let E be a separable Banach space and A : E → E a sectorialoperator. If the embedding V1 = D(A) ↪→ E is compact, then the embeddingVδ ↪→ E is compact as well. This can be shown by interpolation, i.e. (seePazy [36, Theorem 2.6.10])

|x|δ ≤ C|x|1−δ|x|δ1 δ ∈ (0, 1).

If xn is a bounded sequence in E, we know xn is compact in V1. Thatmeans there exists some subsequence nk, such that xnk

converges to some x

in V1 as k tends to infinity. By the inequality above, we have for all δ > 0|x − xnk

|δ → 0 as k tends to infinity. Thus, xnkconverges to x also in Vδ.

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