61.8. GAUSSIAN MEASURES FOR A SEPARABLE HILBERT SPACE 2027

Then from 61.8.38 it follows that for h ∈ A,

38=

78− 1

2≤ 7

8− 1

2(Qch,h)≤ e−

12 (Qh,h)

Therefore, for such h,83≥ e

12 (Qh,h) ≥ 1+

12(Qh,h)

and so for h ∈ A,

(Qh,h)≤(

83−1)

2 =103

Now let h be arbitrary. Then for each ε > 0

h

ε +√(Qch,h)

∈ A

and so (Q

(h

ε +√(Qch,h)

),

h

ε +√(Qch,h)

)≤ 10

3

which implies

(Qh,h)≤ 103

(ε +√(Qch,h)

)2

Since ε is arbitrary,

(Qh,h)≤ 103(Qch,h) . (61.8.39)

However, Qc has finite trace. To see this, let {ek} be an orthonormal basis in U . Then

∑k(Qcek,ek) = ∑

k

∫[||x−m||≤c]

|(x−m,ek)|2 dµ (x)

=∫[||x−m||≤c]

∑k|(x−m,ek)|2 dµ (x) =

∫[||x−m||≤c]

||x−m||2 dµ (x)≤ c2

It follows from 61.8.39 Q that Q must also have finite trace.That µ is uniquely determined by m and Q follows from Theorem 59.12.9. This proves

the theorem.Suppose you have a given Q having finite trace and m ∈U. Does there exist a Gaussian

measure on B (U) having these as the covariance and mean respectively?

Proposition 61.8.7 Let U be a real separable Hilbert space and let m∈U and Q be a pos-itive, symmetric operator defined on U which has finite trace. Then there exists a Gaussianmeasure with mean m and covariance Q.