61.8. GAUSSIAN MEASURES FOR A SEPARABLE HILBERT SPACE 2025

First, the above equals∫U(x,h)(x−m,g)dµ−

∫U(m,h)(x−m,g)dµ (x)

=∫

U(x,h)(x−m,g)dµ (61.8.37)

because from the first part,∫U(x−m,g)dµ (x) =

∫U(x,g)dµ (x)− (m,g)U = 0.

Now by the first part, the term in 61.8.37 is∫U(x,h)(x,g)dµ (x)− (m,g)

∫U(x,h)dµ (x)

=∫

U(x,h)(x,g)dµ (x)− (m,g)(m,h) .

Thus|(Qh,g)| ≤

∫U|(x,h)(x,g)|dµ (x)+ ||m||2 ||h|| ||g||

and since the random variables h′ and g′ given by x→ (x,h) and x→ (x,g) respectivelyare given to be normally distributed with variance σ2 (h) and σ2 (g) respectively, the aboveintegral is finite. Also for all h, ∫

U|(x,h)|2 dµ (x)< ∞

because the random variable h′ is given to be normally distributed. Therefore from Lemma61.8.6, there exists some constant C such that

|(Qh,g)| ≤C ||h|| ||g||

which shows Q is continuous as desired.Why is σ2 (h) = (Qh,h)? This follows because from the above

(Qh,h) ≡∫

U(h,x−m)2 dµ (x)

=∫

U((x,h)− (h,m))2 dµ (x) =

∫R(t− (h,m))2 dλ h′ (t)

=1√

2πσ2 (h)

∫R(t− (h,m))2 e

− 12σ2(h)

(t−(h,m))2

dt = σ2 (h)

from a standard result for the normal distribution function which follows from an easychange of variables argument.

Why must Q have finite trace? For h ∈U, it follows from the above that h′ is normallydistributed with mean (h,m) and variance (Qh,h). Therefore, the characteristic function ofh′ is known. In fact ∫

Ueit(x,h)dµ (x) = eit(h,m)e−

12 t2(Qh,h)