2498 CHAPTER 74. A MORE ATTRACTIVE VERSION

Also, each Φ

(tk

j

),Φ(

tkj−1

)is in Lp (Ω;E). One can also assume that Φ(0) = 0. The mesh

points{

tkj

}mk

j=0can be chosen to miss a given set of measure zero. In addition to this, we

can assume that ∣∣∣tkj − tk

j−1

∣∣∣= 2−nk

except for the case where j = 1 or j = mnk when this might not be so. In the case of the lastsubinterval defined by the partition, we can assume∣∣∣tk

m− tkm−1

∣∣∣= ∣∣∣T − tkm−1

∣∣∣≥ 2−(nk+1)

74.1 The SituationNow consider the following situation. There are real separable Banach spaces V,W suchthat W is a Hilbert space and

V ⊆W, W ′ ⊆V ′

where V is dense in W . Also let B ∈L (W,W ′) satisfy

⟨Bw,w⟩ ≥ 0, ⟨Bu,v⟩= ⟨Bv,u⟩

Note that B does not need to be one to one. Also allowed is the case where B is theRiesz map. It could also happen that V = W . Assume that B = B(ω) where B is F0measurable into L (W,W ′). This dependence on ω will be suppressed in the interest ofsimpler notation. For convenience, assume ∥B(ω)∥ is bounded. This is assumed mainlyso that an estimate can be made on ⟨BX0,X0⟩ for X0 given in L2 (Ω) . It probably suffices tosimply give an estimate on ∥⟨BX0,X0⟩∥L1(Ω) along with something else on the Ito integral.However, it seems at this time like this is more trouble than it is worth.

Situation 74.1.1 Let X have values in V and satisfy the following

BX (t) = BX0 +∫ t

0Y (s)ds+BM (t) , (74.1.1)

X0 ∈ L2 (Ω;W ) and is F0 measurable. Here M (t) is a continuous L2 martingale havingvalues in W. By this is meant that limt→0+ ∥M (t)∥L2(Ω) = 0 and for each ω, limt→0+ M (t)=

0, ∥M∥2W ∈ L2 ([0,T ]×Ω) . Assume that d [M] = kdm for k ∈ L1 ([0,T ]×Ω), that is, the

measure determined by the quadratic variation for the martingale is absolutely continuouswith respect to Lebesgue measure as just described.

Assume Y satisfiesY ∈ K′ ≡ Lp′ ([0,T ]×Ω;V ′

),

the σ algebra of measurable sets defining K′ will be the progressively measurable sets.Here 1/p′+1/p = 1, p > 1.

Also the sense in which the equation holds is as follows. For a.e. ω, the equation holdsin V ′ for all t ∈ [0,T ]. Thus we are considering a particular representative X for which thishappens. Also it is only assumed that BX (t) = B(X (t)) for a.e. t. Thus BX is the name