2044 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Proof: By Lemma 61.11.1, let {z1, · · · ,zn} be a basis for Fn where ∪∞n=1Fn is dense in E.

Then let α1 be such that e1≡α1z1 ∈B(0,1) . Thus β 1e1 ∈B(0,1) whenever |β 1| ≤ 1. Sup-pose α i has been chosen for i = 1,2, · · · ,n such that for all β ∈ Dn ≡ {α ∈ Fn : |α| ≤ 1} ,it follows

n

∑k=1

β kαkzk ∈ B(0,1) .

Then

Cn ≡

{n

∑k=1

β kαkzk : β ∈ Dn

}is a compact subset of B(0,1) and so it is at a positive distance from the complement ofB(0,1) ,δ . Now let 0 < αn+1 < δ/ ||zn+1|| . Then for β ∈ Dn+1,

n

∑k=1

β kαkzk ∈Cn

and so ∣∣∣∣∣∣∣∣∣∣n+1

∑k=1

β kαkzk−n

∑k=1

β kαkzk

∣∣∣∣∣∣∣∣∣∣ =

∣∣∣∣β n+1αn+1zn+1∣∣∣∣

< ||αn+1zn+1||< δ

which showsn+1

∑k=1

β kαkzk ∈ B(0,1) .

This proves the lemma. Let ek ≡ αkzk.Now the main result is the following. It says that any separable Banach space is the

upper third of some abstract Wiener space.

Theorem 61.11.3 Let E be a real separable Banach space with norm ||·||. Then thereexists a separable Hilbert space, H such that H is dense in E and the inclusion map iscontinuous. Furthermore, if ν is the Gaussian measure defined earlier on the cylinder setsof H, ||·|| is Gross measurable.

Proof: Let {ek} be the points of E described in Lemma 61.11.2. Then let H0 denotethe subspace of all finite linear combinations of the {ek}. It follows H0 is dense in E. Nextdecree that {ek} is an orthonormal basis for H0. Thus for

n

∑k=1

ckek,n

∑j=1

d jek ∈ H0,

(n

∑k=1

ckek,n

∑j=1

d je j

)H0

≡n

∑k=1

ckdk