61.7. GAUSSIAN MEASURES 2017

Now since X and Y are independent,

E(

eith◦X ′)

= E(

eith(

1√2

)(X+Y )

)= E

(eith

(1√2

)X)

E(

eith(

1√2

)Y)

the product of two characteristic functions of two random variables, 1√2X and 1√

2Y. The

variance of these two random variables which are normally distributed with zero mean is12 σ2 and so

E(

eith◦X ′)= e−

12 (

12 σ2)e−

12 (

12 σ2) = e−

12 σ2

= E(

eith◦X).

Similar reasoning shows E(

eith◦Y ′)= E

(eith◦Y ) = E

(eith◦X) . Letting t = 1, this yields

61.7.26. This proves the lemma.With this preparation, here is an incredible theorem due to Fernique.

Theorem 61.7.5 Let µ be a symmetric Gaussian measure on B (E) where E is a realseparable Banach space. Then for λ sufficiently small and positive,∫

Eeλ ||x||2dµ < ∞.

More specifically, if λ and r are chosen such that

ln

µ ([x : ||x||> r])

µ

(B(0,r)

)+25λ r2 <−1,

then ∫E

eλ ||x||2dµ ≤ exp(λ r2)+ e2

e2−1.

Proof: Let X ,Y be independent random variables having values in E such that L (X) =L (Y ) = µ . Then by Lemma 61.7.4

1√2(X−Y ) ,

1√2(X +Y )

are also independent and have the same law. Now let 0≤ s≤ t and use independence of theabove random variables along with the fact they have the same law as X and Y to obtain

P(||X || ≤ s, ||Y ||> t) = P(||X || ≤ s)P(||Y ||> t)

= P(∣∣∣∣∣∣∣∣ 1√

2(X−Y )

∣∣∣∣∣∣∣∣≤ s)

P(∣∣∣∣∣∣∣∣ 1√

2(X +Y )

∣∣∣∣∣∣∣∣> t)

= P(∣∣∣∣∣∣∣∣ 1√

2(X−Y )

∣∣∣∣∣∣∣∣≤ s,∣∣∣∣∣∣∣∣ 1√

2(X +Y )

∣∣∣∣∣∣∣∣> t)

≤ P(

1√2|||X ||− ||Y ||| ≤ s,

1√2(||X ||+ ||Y ||)> t

).