60.11 Existence Of Abstract Wiener Spaces
It turns out that if E is a separable Banach space, then it is the top third of an abstract
Wiener space. This is what will be shown in this section. Therefore, it follows from
the above that there exists a Gaussian measure on E which is the law of an
a.e. convergent series as discussed above. First recall Lemma 15.4.2 on Page
Lemma 60.11.1 Let E be a separable Banach space. Then there exists an
increasing sequence of subspaces,
equals 1 for all n if the dimension of E is infinite. Also ∪n=1∞Fn is dense in E.
Lemma 60.11.2 Let E be a separable Banach space. Then there exists a sequence
of points of E such that whenever
1 for β ∈ Fn,
the unit ball in E.
Proof: By Lemma 60.11.1, let
be a basis for
dense in E.
Then let α1
be such that e1 ≡ α1z1 ∈ B
Thus β1e1 ∈ B
has been chosen for i
such that for all
β ∈ Dn ≡
is a compact subset of B
and so it is at a positive distance from the complement of
Now let 0 < αn+1 < δ∕
Then for β ∈ Dn+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 60.11.3 Let E be a real separable Banach space with norm
there exists a separable Hilbert space, H such that H is dense in E and the inclusion
map is continuous. Furthermore, if ν is the Gaussian measure defined earlier on the
cylinder sets of H,
is Gross measurable.
be the points of
described in Lemma 60.11.2
. Then let H0
denote the subspace of all finite linear combinations of the
. It follows
is dense in E.
Next decree that
is an orthonormal basis for
this being well defined because the
are independent. Let the norm on
be denoted by
. Let H1
be the completion of H0
with respect to this
I want to show that
is stronger than
. Suppose then that
It follows then from the definition of
and so from the construction of the ek, it follows that
Stated more simply, this has just shown that if h ∈ H0 then since
It follows that the completion of H0 must lie in E because this shows that every Cauchy
sequence in H0 is a Cauchy sequence in E. Thus H1 embedds continuously into E and is
dense in E. Denote its norm by
Now consider the nuclear operator,
where each λk > 0 and ∑
kλk < ∞. This operator is clearly one to one. Also it is clear
the operator is Hilbert Schmidt because ∑
kλk2 < ∞. Let
and for x ∈ H, define
Since each ek is in H it follows that H is dense in E. Note also that H ⊆ H1 because A
maps H1 to H1.
and the series converges in H1 because
Also H is a Hilbert space with inner product given by
H is complete because if
is a Cauchy sequence in
this is the same as
being a Cauchy sequence in
which implies A−1xn → y
for some y ∈ H1
. Then it
For x ∈ H ⊆ H1,
and so the embedding of H into E is continuous. Why is
a measurable norm on
Note first that for x ∈ H ⊆ H1,
Therefore, if it can be shown A is a Hilbert Schmidt operator on H, the desired
measurability will follow from Lemma 60.9.11 on Page 6869.
Claim: A is a Hilbert Schmidt operator on H.
Proof of the claim: From the definition of the inner product in H, it follows an
orthonormal basis for H is
This is because
To show that A is Hilbert Schmidt, it suffices to show that
because this is the definition of an operator being Hilbert Schmidt. However, the above
This proves the claim.
Now consider 60.11.49. By Lemma 60.9.11, it follows the norm
Gross measurable on H
is also Gross measurable because it is smaller.
This proves the theorem.
Using Theorem 60.11.3 and Theorem 60.10.1 this proves most of the following
Corollary 60.11.4 Let E be any real separable Banach space. Then there exists a
⊆ E such that for any
a sequence of independent random variables
such that ℒ
, it follows
converges a.e. and its law is a Gaussian measure defined on ℬ
E ≤ λk where ∑
kλk < ∞.
Proof: From the proof of Theorem 60.11.3 a basis for H is
is a sequence of independent
random variables, then
to a random variable whose law is Gaussian. Also from
the proof of Theorem 60.10.1
, each ek
in that proof has the property that
then you could consider β ≡
and from the
construction of the
you would need 1ek ∈ B
which is a contradiction. Thus
and changing the notation, replacing λkek
this proves the