2028 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Proof: By Lemma 61.8.2 which comes from Kolmogorov’s extension theorem, thereexists a probability space (Ω,F ,P) and a sequence {ξ i} of independent random variableswhich are normally distributed with mean 0 and variance 1. Then let

X (ω)≡ m+∞

∑j=1

√λ jξ j (ω)e j

where the{

e j}

are the eigenvectors of Q such that Qe j = λ je j. The series in the aboveconverges in L2 (Ω;U) because∣∣∣∣∣

∣∣∣∣∣ n

∑j=m

√λ jξ je j

∣∣∣∣∣∣∣∣∣∣2

L2(Ω;U)

=∫

Ω

n

∑j=m

λ jξ2j (ω)dP =

n

∑j=m

λ j

and so the partial sums form a Cauchy sequence in L2 (Ω;U).Now if h ∈U, I need to show that ω → (X (ω) ,h) is normally distributed. From this it

will follow that L (X) is Gaussian. A subsequence{m+

nk

∑j=1

√λ jξ j (ω)e j

}≡{

Snk (ω)}

of the above sequence converges pointwise a.e. to X .

E (exp(it (X ,h)))

= limk→∞

E(exp(it(Snk ,h

)))= exp(it (m,h)) lim

k→∞E

(exp

(it

nk

∑j=1

√λ jξ j (ω)(e j,h)

))Since the ξ j are independent,

= exp(it (m,h)) limk→∞

nk

∏j=1

E(

exp(

it√

λ j (e j,h))

ξ j

)

= exp(it (m,h)) limk→∞

nk

∏j=1

e−12 t2λ j(e j ,h)

2

= exp(it (m,h)) limk→∞

exp

(−1

2t2

nk

∑j=1

λ j (e j,h)2

). (61.8.40)

Now

(Qh,h) =

(Q

∞

∑k=1

(ek,h)ek,∞

∑j=1

(e j,h)e j

)

=

(∞

∑k=1

(ek,h)λ kek,∞

∑j=1

(e j,h)e j

)

=∞

∑j=1

λ j (e j,h)2