2502 CHAPTER 74. A MORE ATTRACTIVE VERSION

Proof: This follows from Lemma 66.0.20. This can be seen because, thanks to the fact

that BXlσ k

qk is bounded, the function BX l

k is in the set G described there. This is a placewhere we use that d [M] = kdt.

74.3 The Main EstimateThe argument will be based on a formula which follows in the next lemma.

Lemma 74.3.1 In Situation 74.1.1 the following formula holds for a.e. ω for 0 < s < t. Inthe following, ⟨·, ·⟩ denotes the duality pairing between V,V ′.

⟨BX (t) ,X (t)⟩= ⟨BX (s) ,X (s)⟩+

+2∫ t

s⟨Y (u) ,X (t)⟩du+ ⟨B(M (t)−M (s)) ,M (t)−M (s)⟩

−⟨BX (t)−BX (s)− (M (t)−M (s)) ,X (t)−X (s)− (M (t)−M (s))⟩

+2⟨BX (s) ,M (t)−M (s)⟩ (74.3.6)

Also for t > 0

⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+2∫ t

0⟨Y (u) ,X (t)⟩du+2⟨BX0,M (t)⟩+

⟨BM (t) ,M (t)⟩−⟨BX (t)−BX0−BM (t) ,X (t)−X0−M (t)⟩ (74.3.7)

Proof: From the formula which is assumed to hold,

BX (t) = BX0 +∫ t

0Y (u)du+BM (t)

BX (s) = BX0 +∫ s

0Y (u)du+BM (s)

Then

BM (t)−BM (s)+∫ t

sY (u)du = BX (t)−BX (s)

It follows that⟨B(M (t)−M (s)) ,M (t)−M (s)⟩−

⟨BX (t)−BX (s)− (M (t)−M (s)) ,X (t)−X (s)− (M (t)−M (s))⟩

+2⟨BX (s) ,M (t)−M (s)⟩

= ⟨B(M (t)−M (s)) ,M (t)−M (s)⟩−⟨BX (t)−BX (s) ,X (t)−X (s)⟩+2⟨BX (t)−BX (s) ,M (t)−M (s)⟩

−⟨B(M (t)−M (s)) ,M (t)−M (s)⟩+2⟨BX (s) ,M (t)−M (s)⟩