1180 CHAPTER 34. GELFAND TRIPLES AND RELATED STUFF

In the case of dependence on t, the extra condition is trivial if ⟨B(t,ω)x,x⟩ ≥ δ ∥w∥2W

for example. This includes the usual case of evolution equations where W = H = H ′ =W ′.It also includes the case where B does not depend on t.

Proof: Let {gk}∞

k=1 be linearly independent vectors of V whose span is dense in V .This is possible because V is separable. Thus, their span is also dense in W . Let n1 be thefirst index such that ⟨Bgn1 ,gn1⟩ ̸= 0.

Claim: If there is no such index, then B = 0.Proof of claim: First note that if there is no such first index, then if x = ∑

ki=1 aigi

|⟨Bx,x⟩| =

∣∣∣∣∣∑i̸= jaia j

⟨Bgi,g j

⟩∣∣∣∣∣≤∑i ̸= j|ai|∣∣a j∣∣ ∣∣⟨Bgi,g j

⟩∣∣≤ ∑

i̸= j|ai|∣∣a j∣∣⟨Bgi,gi⟩1/2 ⟨Bg j,g j

⟩1/2= 0

Therefore, if x is given, you could take xk in the span of {g1, · · · ,gk} such that ∥xk− x∥W →0. Then

|⟨Bx,y⟩|= limk→∞

|⟨Bxk,y⟩| ≤ limk→∞

⟨Bxk,xk⟩1/2 ⟨By,y⟩1/2 = 0

because ⟨Bxk,xk⟩ is zero by what was just shown. Hence the conclusion of the lemma istrivially true. Just pick e1 = g1 and let {e1} be your set of vectors.

Thus assume there is such a first index. Let

e1 ≡gn1

⟨Bgn1 ,gn1⟩1/2

Then ⟨Be1,e1⟩= 1. Now if you have constructed e j for j ≤ k,

e j ∈ span(gn1 , · · · ,gnk

),⟨Bei,e j

⟩= δ i j,

gn j+1 being the first in the list{

g j}

for which⟨Bgn j+1 −

j

∑i=1

⟨Bgn j+1 ,ei

⟩Bei,gn j+1 −

j

∑i=1

⟨Bgn j,ei

⟩ei

⟩̸= 0,

andspan

(gn1 , · · · ,gnk

)= span(e1, · · · ,ek) ,

let gnk+1 be such that gnk+1 is the first in the list {gn} nk+1 > nk such that⟨Bgnk+1 −

k

∑i=1

⟨Bgnk+1 ,ei

⟩Bei,gnk+1 −

k

∑i=1

⟨Bgnk+1 ,ei

⟩ei

⟩̸= 0

Note the difference between this and the Gram Schmidt process. Here you don’t necessarilyuse all of the gk due to the possible degeneracy of B.

Claim: If there is no such first gnk+1 , then B(span(ei, · · · ,ek)) = BW so in this case,{Bei}k

i=1 is actually a basis for BW .