2024 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

This proves the lemma.Now continue with the proof of the theorem. I need to identify m and Q. It is assumed

µ is Gaussian. Recall this means h′ is normally distributed for each h ∈U . Then using

|x| ≤ |x−m(h)|+ |m(h)|∫U|(x,h)U |dµ (x) =

∫R|x|dλ h′ (x)

=1√

2πσ2 (h)

∫R|x|e−

12σ2 (x−m(h))2

dx

≤ 1√2πσ2 (h)

∫R|x−m(h)|e−

12σ2 (x−m(h))2

dx

+ |m(h)|

Then using the Cauchy Schwarz inequality, with respect to the probability measure

1√2πσ2 (h)

e−1

2σ2 (x−m(h))2dx,

≤ 1√2πσ2 (h)

(∫R|x−m(h)|2 e−

12σ2 (x−m(h))2

dx)1/2

+ |m(h)|< ∞

Thus by Lemma 61.8.6

h→∫

U(x,h)dµ (x)

is a continuous linear transformation and so by the Riesz representation theorem, thereexists a unique m ∈U such that

(h,m)U =∫

U(h,x)dµ (x)

Also the above says (h,m) is the mean of the random variable x→ (x,h) so in the above,

m(h) = (h,m)U .

Next it is necessary to find Q. To do this let Q be given by 61.8.34. Thus

(Qh,g) ≡∫

U((x,h)− (m,h))((x,g)− (m,g))dµ (x)

=∫

U(x−m,h)(x−m,g)dµ (x)

It is clear Q is linear and the above is a bilinear form (The integral makes sense because ofthe assumption that h′,g′ are normally distributed.) but is it continuous? Does (Qh,h) =σ2 (h)?