1174 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

Then letting µ denote Lebesgue measure,

µ ([mn (s)> λ ])≤ 1λ

∫ T

0mn (s)ds.

It follows there exists a subsequence nk such that

µ

([mnk (s)>

1k

])< 2−k

Hence by the Borel Cantelli lemma, there exists a set of measure zero N such that for s /∈N,

mnk (s)≤ 1/k

for all k sufficiently large. Pick such an s. Then consider t→Φ

((γnk

(t− s)+ s)+)

. For

nk, t→(

γnk(t− s)+ s

)+has jumps at points of the form 0, s+ l2−nk where l is an integer.

Thus Pnk consists of points of [0,T ] which are of this form and these partitions are nested.

Define Φlk (0) ≡ 0, Φl

k (t) ≡ Φ

((γnk

(t− s)+ s)+)

. Now suppose N1 is a set of measure

zero. Can s be chosen such that all jumps for all partitions occur off N1? Let (a,b) bean interval contained in [0,T ]. Let S j be the points of (a,b) which are translations of themeasure zero set N1 by t l

j for some j. Thus S j has measure 0. Now pick s∈ (a,b)\∪ jS j. Toget the other sequence of step functions, the right step functions, just use a similar argumentwith δ n in place of γn. Just apply the argument to a subsequence of nk so that the same scan hold for both.

Theorem 34.3.2 Let V ⊆H =H ′⊆V ′ be a Gelfand triple and suppose Y ∈ Lp′ (0,T ;V ′)≡K′ and

X (t) = X0 +∫ t

0Y (s)ds in V ′ (34.3.16)

where X0 ∈ H, and it is known that X ∈ Lp (0,T,V ) ≡ K for p > 1. Then t → X (t) is inC ([0,T ] ,H) and also

12|X (t)|2H =

12|X0|2H +

∫ t

0⟨Y (s) ,X (s)⟩ds

Proof: By Lemma 34.3.1, there exists a sequence of uniform partitions{

tnk

}mnk=0 =

Pn,Pn ⊆Pn+1, of [0,T ] such that the step functions

mn−1

∑k=0

X (tnk )X(tn

k ,tnk+1]

(t) ≡ X l (t)

mn−1

∑k=0

X(tnk+1)X(tn

k ,tnk+1]

(t) ≡ X r (t)

converge to X in K and in L2 ([0,T ] ,H).