2450 CHAPTER 72. THE HARD ITO FORMULA

and for a.e. ω, this is just the sup of continuous functions of t. Therefore, for given ω offa set of measure zero,

t→ |X (t)|2

is lower semicontinuous. Hence letting t ∈ [0,T ] and t j→ t where t j ∈ D,

|X (t)|2 ≤ lim infj→∞

∣∣X (t j)∣∣2

so it follows for a.e. ω

supt∈[0,T ]

|X (t)|2 ≤ supt∈D|X (t)|2 ≤ sup

t∈[0,T ]|X (t)|2

Hence

E

(sup

t∈[0,T ]|X (t)|2

)≤C

(||Y ||K′ , ||X ||K , ||Z||J ,∥X0∥L2(Ω,H)

). (72.4.9)

Note the above shows that for a.e. ω, supt∈[0,T ] |X (t)|H < ∞ so that for such ω,X (t)has values in H. Note that we began by assuming it had a representative with values in Halthough the equation only held in V ′. Say

|X (t,ω)| ≤C (ω) .

Hence if v ∈V, then for a.e. ω

limt→s

(X (t) ,v) = limt→s⟨X (t) ,v⟩= ⟨X (s) ,v⟩= (X (s) ,v)

Therefore, since for such ω, |X (t,ω)| is bounded, the above holds for all h ∈H in place ofv as well. Therefore, for a.e. ω, t→ X (t,ω) is weakly continuous with values in H.

Eventually, it is shown that in fact, the function t → X (t,ω) is continuous with valuesin H.

This lemma also provides a way to simplify one of the formulas derived earlier in thecase that X0 ∈ Lp (Ω,V ). Refer to 72.4.8. One term there is |X (t1)−X0−M (t1)|2 .

|X (t1)−X0−M (t1)|2 ≤ 2 |X (t1)−X0|2 +2 |M (t1)|2

It was shown above that 2 |M (t1)|2→ 0 a.e. and also in L1 (Ω) as k→ ∞. Apply the abovelemma to |X (t)−X0|2 using [0, t1] instead of [0,T ] . The new X0 equals 0. Then from theestimate 72.4.9, it follows that

E(|X (t1)−X0|2

)→ 0

as k→ ∞. Taking a subsequence, we could also assume that |X (t1)−X0|2 → 0 a.e. ω ask→ ∞. Then, using this subsequence, it would follow from 72.4.8,

|X (tm)|2−|X0|2 = 2∫ tm

0⟨Y (u) ,X r

k (u)⟩du+ e(k)