71.3. STOCHASTIC DIFFERENTIAL EQUATIONS 2433

+CE(∫ t

0∥σ (s,v1,ω)−σ (s,v2,ω)∥2 ds

)Now using the Lipschitz condition on σ , this simplifies further to give an inequality of theform

E

(sup

s∈[0,t]|u1 (s)−u2 (s)|2

)≤C

∫ t

0E

(sup

r∈[0,s]|v1 (r)− v2 (r)|2

)ds

Letting T v = u where u is defined from v in the integral equation 71.3.17, the aboveinequality implies that

E

(sup

s∈[0,t]|T nv1 (s)−T nv2 (s)|2H

)≤C

∫ t

0E

(sup

r∈[0,s]

∣∣T n−1v1 (r)−T n−1v2 (r)∣∣2)ds

≤C2∫ t

0

∫ s

0E

(sup

r1∈[0,r]

∣∣T n−2v1 (r1)−T n−2v2 (r1)∣∣2 )drds

One can iterate this, eventually finding that

E

(sup

s∈[0,t]|T nv1 (s)−T nv2 (s)|2H

)

≤ Cn∫ t

0

∫ t1

0· · ·∫ tn−1

0dtn−1 · · ·dtE

(sup

s∈[0,t]|v1 (s)− v2 (s)|2H

)

=CnT n

(n!)E

(sup

s∈[0,t]|v1 (s)− v2 (s)|2H

)In particular, one could take t = T . This shows that for all n large enough, T n is a contrac-tion map on L2 (Ω,C ([0,T ] ;H)). Therefore, picking v∈ L2 (Ω,C ([0,T ] ;H)) , such that v isalso progressively measurable,

{T kv

}∞

k=1 converges in L2 (Ω,C ([0,T ] ;H)) to the uniquefixed point of T denoted as u. Thus T u = u in L2 (Ω;C ([0,T ] ;H)) . That is,∫

Ω

supt|T u−u|2 dP = 0

It follows that there is a set of measure zero such that for ω not in this set,

u(t,ω)−u0(ω)+∫ t

0N (s,u(s,ω),u(s−h,ω) ,w(s,ω) ,ω)ds

=∫ t

0f (s,ω)ds+

∫ t

0σ (s,u,ω)dW. (71.3.18)

The function u is progressively measurable because each T nv is progressively measurableand there exists a subsequence still indexed with n such that for ω off a set of measure zero,T nv(·,ω)→ u(·,ω) in C ([0,T ] ;H).

Note that the fixed point of T is unique in the space L2 (Ω;C ([0,T ] ;H)) and so anysolution to the integral equation in this space must equal this one. Hence, there exists a setof measure zero such that for ω off this set, the two solutions are equal for all t.