34.4. THE IMPLICIT CASE 1179

34.4 The Implicit CaseThe above theorem can be generalized to the case where the formula is of the form

BX (t) = BX0 +∫ t

0Y (s)ds

This involves an operator B ∈L (W,W ′) and B satisfies

⟨Bx,x⟩ ≥ 0, ⟨Bx,y⟩= ⟨By,x⟩

forV ⊆W,W ′ ⊆V ′

Where V is dense in the Banach space W . Before giving the theorem, here is a technicallemma. First is one which is not so technical.

Lemma 34.4.1 Let V be a separable Banach space. Then there exists {gk}∞

k=1 which arelinearly independent and whose span is dense in V .

Proof: Let { fk} be a countable dense subset. Thus their span is dense. Delete fk1 suchthat k1 is the first index such that fk is in the span of the other vectors. That is, it is the firstwhich is a finite linear combination of the others. If no such vector exists, then you havewhat is wanted. Next delete fk2 where k2 is the next for which fk is a linear combinationof the others. Continue. The remaining vectors must be linearly independent. If not, therewould be a first which is a linear combination of the others. Say fm. But the process wouldhave eliminated it at the mth step.

Lemma 34.4.2 Suppose V,W are separable Banach spaces such that V is dense in W andB ∈L (W,W ′) satisfies

⟨Bx,x⟩ ≥ 0, ⟨Bx,y⟩= ⟨By,x⟩ ,B ̸= 0.

Then there exists a countable set {ei} of vectors in V such that⟨Bei,e j

⟩= δ i j

and for each x ∈W,

⟨Bx,x⟩=∞

∑i=1|⟨Bx,ei⟩|2 ,

and also

Bx =∞

∑i=1⟨Bx,ei⟩Bei,

the series converging in W ′. If B = B(ω) and B is F measurable into L (W,W ′) and ifthe ei = ei (ω) are as described above, then these ei are measurable into V . If t→ B(t,ω)is C1 ([0,T ] ,L (W,W ′)) and if for each w ∈W,⟨

B′ (t,ω)w,w⟩≤ kw,ω (t)⟨B(t,ω)w,w⟩

Where kw,ω ∈ L1 ([0,T ]) , then the vectors ei (t) can be chosen to also be right continuousfunctions of t.