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Lemma 66.0.9 The above definition is well defined. Also,∫ t
0 (Y,dMτ p) is a continuousmartingale. The inequality
E
(∣∣∣∣∫ t
0(Y,dMτ p)
∣∣∣∣2)≤ E
(∫ t
0∥Y∥2
H d [M]τ p
)is also valid. For any sequence of elementary functions {Y n} ,∥Y n (t)∥M∗ ∈ L2 (Ω) ,
∥Y n−Y∥L2(Ω;L2([0,T ];H,d[Mτ p ]))→ 0
there exists a subsequence, still denoted as {Y n} of elementary functions for which∫ t
0(Y n,dMτ p)
converges uniformly to∫ t
0 (Y,dMτ p) on [0,T ] for ω off some set of measure zero.
Proof: First of all, why does the limit even exist? From Lemma 66.0.6,
E
(∣∣∣∣∫ t
0(Y n,dMτ p)−
∫ t
0(Y m,dMτ p)
∣∣∣∣2)≤ E
(∫ T
0∥Y n−Y m∥2
H d [M]τ p
)which converges to 0 as n,m→∞ by definition of Y ∈G . This also shows that the definitionis well defined and that the same thing is obtained from any other sequence converging toY .
{∫ t0 (Y
n,dMτ p)}
is a Cauchy sequence in L2 (Ω). Hence it converges to somethingN (t) ∈ L2 (Ω) . This is a martingale because if A ∈Fs,s < t∫
AN (t)dP = lim
n→∞
∫A
∫ t
0(Y n,dMτ p)dP
= limn→∞
∫A
∫ s
0(Y n,dMτ p)dP =
∫A
N (s)dP
Since A is arbitrary, this shows that E (N (t) |Fs) = N (s) . Then
N (t)≡∫ t
0(Y n,dMτ p)
In fact, this has a continuous version off a set of measure zero.These are martingales and so actually, by maximal theorems,
P
(sup
t∈[0,T ]
∣∣∣∣∫ t
0(Y n,dMτ p)−
∫ t
0(Y m,dMτ p)
∣∣∣∣2 > λ
)
≤ 1λ
E
(∣∣∣∣∫ T
0(Y n,dMτ p)−
∫ T
0(Y m,dMτ p)
∣∣∣∣2)
≤ 1λ
E(∫ T
0∥Y n−Y m∥2
H d [M]τ p
)