65.13. TAKING OUT A LINEAR TRANSFORMATION 2261

≤C∥Φ∥α

∞

∫Ω

(∫ t

sdτ

)α/2

dP =C∥Φ∥α

∞|t− s|α/2

By the Kolmogorov Čentsov theorem, Theorem 62.2.2, this shows that t →∫ t

0 ΦdW isHolder continuous with exponent

γ <(α/2)−1

α=

12− 1

α

Since α > 2 is arbitrary, this shows that for any γ < 1/2, the stochastic integral is Holdercontinuous with exponent γ . This is exactly the same kind of continuity possessed by theWiener process.

Theorem 65.12.1 Suppose Φ ∈ L∞([0,T ]×Ω,L2

(Q1/2U,H

))and is progressively mea-

surable. Then if γ < 1/2, there exists a set of measure zero such that off this set,

t→∫ t

0ΦdW

is Holder continuous with exponent γ .

65.13 Taking Out A Linear TransformationWhen is

L∫ T

aΦdW =

∫ T

aLΦdW?

It is assumed L ∈L (H,H1) where H1 is another separable real Hilbert space. First of all,here is a lemma which shows

∫ ta LΦdW at least makes sense.

Proposition 65.13.1 Suppose Φ is L2(Q1/2U,H

)progressively measurable and

P([∫ T

a||Φ||2

L2(Q1/2U,H) ds < ∞

])= 1.

Then the same is true of LΦ. Furthermore, for each t ∈ [a,T ]∫ t

aLΦdW = L

∫ t

aΦdW

Proof: First note that if Φ ∈L2(Q1/2U,H

), then LΦ ∈L2

(Q1/2U,H1

)and that the

map Φ→ LΦ is continuous. It follows LΦ is L2(Q1/2U,H1

)progressively measurable.

All that remains is to check the appropriate integral.∫ T

a||LΦ||2

L2(Q1/2U,H1)dt ≤

∫ T

a||L||2 ||Φ||2

L2(Q1/2U,H) dt

and so this proves LΦ satisfies the same conditions as Φ, being stochastically square inte-grable.