2506 CHAPTER 74. A MORE ATTRACTIVE VERSION

Therefore,

suptm∈Pk

⟨BX (tm) ,X (tm)⟩ ≤ ⟨BX0,X0⟩+ e(k)+2∫ T

0|⟨Y (u) ,X r

k (u)⟩|du+

+2 suptm∈Pk

∣∣∣∣∫ tm

0

⟨BX l

k ,dM⟩∣∣∣∣

+m−1

∑j=0

⟨B(M(t j+1

)−M (t j)

),M(t j+1

)−M (t j)

⟩(74.3.12)

where there are mk +1 points in Pk. Consider that last term. It is no larger than

∥B∥m−1

∑j=0

∥∥M(t j+1

)−M (t j)

∥∥2

Say the last point in the partition is tp = T and consider the sum

p−1

∑j=0

∥∥M(t j+1∧ t

)−M (t j ∧ t)

∥∥2=

p−1

∑j=0

∥∥Mt j+1 −Mt j∥∥2

(t)

=p−1

∑j=0

[Mt j+1 −Mt j

](t)+N j (t) =

p−1

∑j=0

[M]t j+1 (t)− [M]t j (t)+N j (t)

for N j a martingale which equals 0 for t ≤ t j. Now when you put in t = tm, this becomes

m−1

∑j=0

[M]t j+1 (tm)− [M]t j (tm)+N j (tm)

Thus the expectation of that last term in 74.3.12 is no larger than

∥B∥E

(m−1

∑j=0

[M]t j+1 (tm)− [M]t j (tm)

)= ∥B∥E ([M] (tm))

The next task is to take the expectation of both sides of 74.3.12. Of course there isa small problem with things not being in L1. Hence it is appropriate to localize with thestopping time σ k

q defined in 74.2.3. That is, we obtain all of the above with X replaced with

Xσkq , stopping the original integral equation by introducing X[0,σ k

q]in the integrals. Then

carry out the following argument and pass to a limit as q→ ∞. In fact σ kq = ∞ if q is large

enough. Then carry out everything with Xσ kq . We don’t write it, but this is what is being

done in the following argument.

E

(sup

tm∈Pk

⟨BX (tm) ,X (tm)⟩)≤ E (⟨BX0,X0⟩)+E (|e(k)|)+2∥Y∥K′ ∥X

rk∥K