2260 CHAPTER 65. STOCHASTIC INTEGRATION
Now consider
P
(∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣2 > λ
)
≤E(∣∣∣∫ T
0 X[0,τnp]ΦndW −∫ T
0 X[0,τ p]ΦdW∣∣∣2)
λ
and so, there exists a subsequence, still denoted as n such that
P
(∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣2 > 1n
)< 2−k
It follows that N can be enlarged so that for ω /∈ Np∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣2 ≤ 1n
for all n large enough. Now obtain a succession of subsequences for p = 1,2, · · · , each asubsequence of the preceeding one such that the above convergence takes place and let Ninclude ∪pNp. Then for ω /∈ N, and letting n denote the diagonal sequence, it follows thatfor all p,
limn→∞
∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣= 0
For ω /∈ N, there is a p such that τ p = ∞. Then this means∫ T
0 ∥Φ∥2 ds < p. It follows that
the same is true for Φn for all n large enough. Hence τnp = ∞ also. Thus, for large enoughn, ∣∣∣∣∫ T
0ΦndW −
∫ T
0ΦdW
∣∣∣∣= ∣∣∣∣∫ T
0X[0,τnp]ΦndW −
∫ T
0X[0,τ p]ΦdW
∣∣∣∣and the latter was just shown to converge to 0.
65.12 The Holder Continuity Of The IntegralLet Φ ∈ L2
([0,T ]×Ω,L2
(Q1/2U,H
)). Then you can consider the stochastic integral as
described above and it yields a continuous function off a set of measure zero. What ifΦ ∈ L∞
([0,T ]×Ω,L2
(Q1/2U,H
))? Can you say more? The short answer is yes. You
obtain a Holder condition in addition to continuity. This is a consequence of the BurkolderDavis Gundy inequality and Corollary 65.11.1 above. Let α > 2. Let ∥Φ∥
∞denote the
norm in L∞([0,T ]×Ω,L2
(Q1/2U,H
)). By the Burkholder Davis Gundy inequality,∫
Ω
(∣∣∣∣∫ t
sΦdW
∣∣∣∣)α
dP≤
∫Ω
(sup
r∈[s,t]
∣∣∣∣∫ r
sΦdW
∣∣∣∣)α
dP≤C∫
Ω
(∫ t
s∥Φ∥2 dτ
)α/2
dP