61.8. GAUSSIAN MEASURES FOR A SEPARABLE HILBERT SPACE 2023

Proof: I need to show that for each h∈Uk, the integral in 61.8.36 exists. From this it isobvious it is k− linear, meaning linear in each argument. Then it is shown it is continuous.

First note|(h1,x) · · ·(hk,x)| ≤ |(h1,x)|k + · · ·+ |(hk,x)|k

This follows from observing that one of∣∣(h j,x)

∣∣ is largest. Then the left side is smaller

than∣∣(h j,x)

∣∣k. Therefore, the above inequality is valid. This inequality shows the integralin 61.8.36 makes sense.

I need to establish an estimate of the form∫U|(x,h)|k dµ (x)<C < ∞

for every h ∈U such that ||h|| is small enough.Let

Un ≡{

z ∈U :∫

U|(x,z)|k dµ (x)≤ n

}Then by assumption U = ∪∞

n=1Un and it is also clear from Fatou’s lemma that each Un isclosed. Therefore, by the Bair category theorem, at least one of these Un0 contains an openball, B(z0,r) . Then letting |y|< r,∫

U|(x,z0 + y)|k dµ (x) ,

∫U|(x,z0)|k dµ (x)≤ n0,

and so for such y,∫U|(x,y)|k dµ =

∫U|(x,z0 + y)− (x,z0)|k dµ

≤∫

U2k |(x,z0 + y)|k +2k |(x,z0)|k dµ (x)

≤ 2k (n0 +n0) = 2k+1n0.

It follows that for arbitrary nonzero y ∈U

∫U

∣∣∣∣(x,(r/2)y||y||

)∣∣∣∣k dµ ≤ 2k+1n0

and so ∫U|(x,y)|k dµ ≤

(2k+2/r

)n0 ||y||k ≡C ||y||k .

Thus by Holder’s inequality,

∫U|(h1,x) · · ·(hk,x)|dµ (x) ≤

k

∏j=1

(∫U

∣∣(h j,x)∣∣k dµ (x)

)1/k

≤ Ck

∏j=1

∣∣∣∣h j∣∣∣∣