43.2. AN IMPORTANT FORMULA 1423

about. If it is not so, what is the space of initial values? How can you give this space anorm? What are its properties? It turns out that if V is a closed subspace of the Sobolevspace, W 1,p (Ω) which contains W 1,p

0 (Ω) for p≥ 2 and H = L2 (Ω) the answer to the abovequestion is yes. Not surprisingly, there are many generalizations of the above ideas.

43.2 An Important FormulaIt is not necessary to have p > 2 in order to do the sort of thing just described. Here is amajor result which will have a much more difficult stochastic version presented later. Firstis a simple version of an approximation theorem of Doob.

Lemma 43.2.1 Let Y : [0,T ]→ E, be B ([0,T ]) measurable and suppose

Y ∈ Lp (0,T ;E)≡ K, p≥ 1

Then there exists a sequence of nested partitions, Pk ⊆Pk+1,

Pk ≡{

tk0 , · · · , tk

mk

}such that the step functions given by

Y rk (t) ≡

mk

∑j=1

Y(

tkj

)X[tk

j−1,tkj )(t)

Y lk (t) ≡

mk

∑j=1

Y(

tkj−1

)X(tk

j−1,tkj ](t)

both converge to Y in K as k→ ∞ and

limk→∞

max{∣∣∣tk

j − tkj+1

∣∣∣ : j ∈ {0, · · · ,mk}}= 0.

Also, each Y(

tkj

),Y(

tkj−1

)is in E. One can also assume that Y (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)

Proof: For t ∈ R let γn (t) ≡ k/2n,δ n (t) ≡ (k+1)/2n, where t ∈ (k/2n,(k+1)/2n],

and 2−n < T/4. Also suppose Y is defined to equal 0 on [0,T ]C. Then t → ∥Y (t)∥ is inLp (R). Therefore by continuity of translation, as n→ ∞ it follows that for t ∈ [0,T ] ,∫

R||Y (γn (t)+ s)−Y (t + s)||pE ds→ 0