Topics In Analysis

64.8 Localization For Elementary Functions

It is desirable to extend everything to stochastically square integrable functions. This will involve localization using a suitable stopping time. First it is necessary to understand localization for elementary functions. As above, we are in the situation described by the following diagram.

U ↓ Q1 ∕2 U1 ⊇ JQ1 ∕2U ←J Q1 ∕2U 1−1 Φ ↘ ↓ Φ n H

The elementary functions

have values in ℒ₀ meaning they are restrictions of functions in ℒ to JQ^1∕2U and converge to Φ ∘ J⁻¹ in

( ( ) ) L2 [a,T ]× Ω;ℒ2 JQ1 ∕2U,H

where Φ ∈ L²

is given. Let

n∑−1 Φ(t) ≡ Φ (tk)X (tk,tk+1](t) k=0

be an elementary function. In particular, let Φ

be ℱ_tk measurable as a map into ℒ, and has finitely many values. As just mentioned, the topic of interest is the elementary functions Φ_n in the above diagram. Thus Φ will be one of these elementary functions.

Let τ be a stopping time having values from the set of mesh points

for the elementary function. Then from the definition of the integral for elementary functions,

∫ t∧τ n∑−1 ΦdW ≡ Φ (tk)(W (t∧ τ ∧ tk+1) − W (t∧τ ∧ tk)) a k=0

If ω is such that τ

= t_j, then to get something nonzero, you must have t_j > t_k so k ≤ j − 1. Thus the above on the right reduces to

j∑−1 Φ (tk)(W (t∧ tk+1)− W (t∧ tk)) k=0

It clearly is 0 if j = 0. Define ∑ _k=0⁻¹ ≡ 0. Thus the integral equals

∑n j−∑ 1 X[τ=tj] Φ(tk)(W (t∧ tk+1)− W (t∧ tk)) j=0 k=0

Interchanging the order of summation, k ≤ j − 1 so j ≥ k + 1 and this equals

Therefore

∫ t∧τ ∫ tn∑−1 ΦdW = X [τ>tk]Φ (tk)X (tk,tk+1]dW (64.8.15) a a k=0

(64.8.15)

Now observe

The last step occurs because of the following reasoning. The k^th term of the sum in the middle expression above equals Φ if and only if t > t_k and τ ≥ t. If the two conditions do not hold, then the k^th term equals 0. As to the third line, if τ > t_k and t ∈ (t_k,t_k+1], then τ ≥ t_k+1 ≥ t which is the same as the situation in the second line. The term equals Φ. Note that X

[τ>tk]

is ℱ_tk measurable, because is the complement of . Therefore, this is an elementary function. Thus, from 64.8.15 -64.8.16, X

[a,τ]

Φ is an elementary function and

∫ t∧τ ∫ tn∑−1 ∫ t ΦdW = X[τ>tk]Φ (tk)X(tk,tk+1](t)dW = X[a,τ](t)Φ (t)dW a a k=0 a

From Proposition 64.1.5, if you have Φ,Ψ two of these elementary functions

( || || ) ||||∫ t ∫ t ||||2 E || a X[a,τ](t)Φ (t)dW − a X [a,τ](t)Ψ (t)dW || = H