2038 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

Let P denote an orthogonal projection in F ({ek}) such that PPN = 0. Thus P is the pro-jection on to span(ei1 , · · · ,eim) where each ik > N. Then

ν ({x ∈ H : ||Px||> ε})= ν ({x ∈ H : |APx|> ε})

Now Px = ∑mj=1(x,ei j

)ei j and the above reduces to

ν

({x ∈ H :

∣∣∣∣∣ m

∑j=1

(x,ei j

)Aei j

∣∣∣∣∣> ε

})≤

ν

x ∈ H :

(m

∑j=1

∣∣(x,ei j

)∣∣2)1/2( m

∑j=1

∣∣Aei j

∣∣2)1/2

> ε



≤ ν

x ∈ H :

(m

∑j=1

∣∣(x,ei j

)∣∣2)1/2

α1/2 > ε



= ν

x ∈ H :

(m

∑j=1

∣∣(x,ei j

)∣∣2)1/2

>ε

α1/2



= ν

({x ∈ H : ((x,ei1) , · · · ,(x,eim)) ∈ B

(0,

ε

α1/2

)C})

≤ ν

({x ∈ H : max

{∣∣(x,ei j

)∣∣}> ε√

mα1/2

})This is no larger than

1(√2π)m

∫|t1|> ε√

m√

α

∫|t2|> ε√

m√

α

· · ·∫|tm|> ε√

m√

α

e−|t|2/2dtm · · ·dt1

=

2∫

∞

ε/(√

mα1/2) e−t2/2dt√

2π

m

which by Jensen’s inequality is no larger than

2∫

∞

ε/(√

mα1/2) e−mt2/2dt√

2π=

2 1√m

∫∞

ε/(α1/2) e−t2/2dt√

2π

≤2∫

∞

ε/(ε/r) e−t2/2dt√

2π

=2∫

∞

r e−t2/2dt√2π

< ε

By 61.9.42. This proves the lemma.