6. Outlook 66
8.2. Tightness and construction of the solution
of the previous lemma. Define
Z(t, x) = Z t
0
Z
Rq
pt−s(x−y)σ(s, y, u(s, y))W(ds dy) +
Z t 0
Z
Rq
pt−s(x−y)b(s, y, u(s, y))dyds
for t ≥ 0, x ∈ Rq. Then, for T, R > 0 and 0 ≤ t ≤ t0 ≤ T, x, x0 ∈ Rq such that
|x−x0| ≤R, as well as p∈[2,∞) andξ ∈(0,1−α/2), E
h|Z(t, x)−Z(t0, x0)|pe−λ|x|i
≤C(T, λ, p) 1 + sup
0≤t≤T
sup
z∈Rq
Gλ/(p+1),pu(s, z)
!
×
|t−t0|ξp/2+|x−x0|ξp .
Proof. The difficult part of the proof can be found in [MPS06]. We content ourselves here with the statement that Lemma 8.1.2 and Lemma 5.2 (a) of [MPS06], combined with the growth bound on bgive sufficient estimates for
Y(t, x) = Z t
0
Z
Rq
pt−s(x−y)b(s, y, u(s, y))dyds.
One could remark that as there is a Riemann integral it is not surprising to have very high H¨older-regularity, even better than the one proposed in the lemma for this integral.
8.2. Tightness and construction of the solution
We give the proof of Theorem 5.2.1 and decompose it into several steps. First we define a family of approximate mild solutions to (5.1), show that it has a limit point (at least one) and then show that any limit point is a mild solution. Finally, the specific filtration chosen is considered.
Step 1: Family of approximate solutions
Let (σn)n be a sequence of Lipschitz-continuous functions with
• |σn(u)| ≤c5.2(1 +|u|) and
• σn→σ uniformly on compacta.
86 Existence and regularity An example of such a sequence is constructed in [MP11] (proof of Theorem 1.1).
Furthermore define for any x∈Rq: Step 2: Tightness of um,n
Tightness in C(R+, Ctem) can be shown using Lemma 6.3 (ii) of [Shi94]. The esti-mates obtained in Lemma 8.1.4 do not depend on the specific value of the Lipschitz-constant forσn only on the constant of the growth bound.
Step 3: The limit points are mild solutions
By Skorohod’s representation theorem (cf. Theorem 3.1.8 of [EK86]) extract an a.s. converging subsequence on a joint probability space. Let u be an almost sure limit point ofum,non (Ω,F,P),whereF is assumed to beP-complete. That means there is a sequence (n(l), m(l))l s.t. both m(l) andn(l) tend to infinity asl → ∞ and
um(l),n(l) →u inC(R+, Ctem) almost sure.
In order to simplify notation we will write ul:=um(l),n(l).
Lemma 8.2.1. Then u is a mild solution of the limiting equation:
u(t, x) =
8.2 Tightness and construction of the solution 87 The superscript stands for completion with respect to P.
The last point is shown in Step 4. Before giving a proof we state some lemmas.
Lemma 8.2.2. For anyλ >0, T >0and solutionsul, uto the SPDE with diffusion coefficient σl, σ, respectively, the following holds
(a) supl≥1supt≤Tsupx∈Rde−λ|x|E[|ul(t, x)|k]< c(λ, T)<∞, k∈N (b) supt≤T supx∈Rde−λ|x|E[|u(t, x)|k]<∞, k∈N
(c) The family(e−λ|x||ul(t, x)|k)t≤T ,x∈Rd,l∈Nis uniformly integrable for anyk∈N. Proof. The first line (a) follows by Lemma 8.1.3, since we never used the Lipschitz-coefficient in the proof (consider line (131) in [MPS06] for the explicit statement).
The proof of (b) is taken from Lemma 6.19 of [Z¨ah04]:
sup
The last assertion (c) follows by the uniform bound on higher moments of order 2k and using (a), when replacing λbyλ/2 there.
Define the cut-off functionsFN :R→Rfor any natural number N ∈N: FN(x) =−N, x≤ −N; FN(x) =x, −N < x < N; FN(x) =N, x≥N.
Then
88 Existence and regularity for N → ∞. The same holds true for σ and u without the superscripts.
Proof. WriteAN for the expression inside the modulus in the lemma.
E[|AN|2]≤E[ By Lemma 5.1 in [MPS06] it is true that
Z t 0
Z
R2d
|pt−s(x−y)pt−s(x−z)|eλ(|z|+|y|)k(y, z) dy dz ds
constitutes a finite measure on R+×R2q.Using Lemma 8.2.2 we also know that sup
l 1{ul(s, y)> N} →0 a.s. as N → ∞.
Hence, we can use a standard argument for uniformly integrable random variables (e.g. [Kle08, Thm 6.24(iii)]) and see that
E[|AN|2]→0 asN → ∞.
The fact that the same holds true in the case without the superscriptsl can be shown just the same way.
Now we try to put things together to show that u is a mild solution of the SPDE. We need to show (8.6). We do that by showing that the second moment
8.2 Tightness and construction of the solution 89 of the differences from left and right hand side is zero. Using that um(l) are mild solutions, consider where FN was the cut-off function defined above. If for l → ∞ all expressions vanish, we know that u is a mild solution. This looks complicated, but reduces to two problems since
• I1 goes to zero since ((um(t, x))2)m is a uniformly integrable family (Lemma 8.2.2) and combined with its almost sure convergence we haveL2-convergence (as l→ ∞).
Therefore given an >0 we can bound each of the previousIi by 7 when choosing l sufficiently large, say l≥L1 and N sufficiently large, say N ≥N1 (keep that N
90 Existence and regularity fixed). We have to deal withI4 and I5.
Before going ahead we define two integrals M1(t, x) :=
The trick somehow needs to be that we have some information about the closeness of σ(um(l)) and σ(u). Asσ is a continuous function it is uniformly continuous on the compactum [−N, N] and for any 4>0 we can find aδ4(4)>0 such that
u, u0 ∈[−N, N] and |u−u0|< δ4 ⇒ |σ(u)−σ(u0)|< 4. (8.16) Let us take 24∈(0,36c
5.2(1+N)M1).
Hence, the only thing which is required is that we need to know about the close-ness of u and um(l). As we have almost sure convergence in the path space with values in Ctem we also have almost sure convergence on any compact set, e.g.
{(t, x) : t ≤ T,|x| ≤ K1}, where K1 is the constant from above. Hence, let an
8.2 Tightness and construction of the solution 91 Then we we can start putting the things together
E[| The first two summands are similar and we will only consider the first one:
E[| now use (8.18) and (8.15) to bound that by
≤24c5.2(1 +N)M1(t, x) +2M1(t, x)c25.2(1 +N)2
<
9 ∀l≥L3.
So, we are left to deal with the last integral in (8.19):
E[| where again we used (8.15). Hence, the sum of the three integrals in (8.19) is at most/6.
92 Existence and regularity Therefore we have shown that for the givenwe could choose (in a cascade of many i) an l≥L1∨L2∨L3 such that the expression in (8.10) is bounded by and we see that u is a mild solution to the SPDE.
Step 4: u is adapted
It remains to show that u is adapted to the filtration chosen. We provide the following proposition:
Proposition 8.2.4. Let Xn, W, X be stochastic processes with values in a Polish space (E,E), s.t. Ftn=T
s>tσ(Wr, Xrn:r ≤s) is the right-continuous filtration of the pair (Xn, W), n ∈ N, t ≥ 0. Assume (Xn, W) → (X, W) P-almost surely on a complete probability space (Ω,F, P) and W is an Ftn-Brownian Motion for each n∈N. Define Ft=W
nFtn∨F¯0, t≥0.Then, (a) Xt∈ Ft, t≥0.
(b) (Wt) is an Ft-Brownian Motion, i.e. Wt ∈ Ft, Wt+u−Wt is independent of Ft (and the distribution is Gaussian).
Proof. (a) Let A∈ E. Then,
{Xt∈A}={Xt∈A,limXn=X} ∪ {Xt∈A,limXn6=X}
={limXtn∈A,limXn=X} ∪C
= ({limXtn∈A} ∩Cc)∪C
={limXtn∈A} ∪C ∈_
n
Ftn∨F¯0 =Ft, with obvious notation forC.
(b) Wt∈ Ft by definition. And for measurable functions f, g:E→R. E[f(Wt+u−Wt)g(Xtn)] =E[E[f(Wt+u−Wt)|Ftn]g(Xtn)] = 0, for all n∈N,t, u≥0,since we require W to be a Ftn -Brownian Motion.
Remark 8.2.5. The proof of Proposition 8.2.4 suggests, that the limit point X is adapted to Ft, a filtration which can be chosen to be the one of the noise ˙W, if Xn are strong solutions. But keep in mind, that after having used the Skorohod representation the probability space has changed. So we do not have a strong solution in general.