73.5. A SIMPLIFICATION OF THE FORMULA 2479

73.5 A Simplification Of The FormulaThis lemma also provides a way to simplify one of the formulas derived earlier in the casethat X0 ∈ Lp (Ω,V ) so that X−X0 ∈ Lp ([0,T ]×Ω,V ). Refer to 73.4.11. One term there is

⟨B(X (t1)−X0−M (t1)) ,X (t1)−X0−M (t1)⟩

Also,⟨B(X (t1)−X0−M (t1)) ,X (t1)−X0−M (t1)⟩

≤ 2⟨B(X (t1)−X0) ,X (t1)−X0⟩+2⟨BM (t1) ,M (t1)⟩

It was observed above that 2⟨BM (t1) ,M (t1)⟩ → 0 a.e. and also in L1 (Ω) as k→∞. Applythe above lemma to ⟨B(X (t1)−X0) ,X (t1)−X0⟩ using [0, t1] instead of [0,T ] . The new X0equals 0. Then from the estimate 73.4.8, it follows that

E (⟨B(X (t1)−X0) ,X (t1)−X0⟩)→ 0

as k→ ∞. Taking a subsequence, we could also assume that

⟨B(X (t1)−X0) ,X (t1)−X0⟩ → 0

a.e. ω as k→ ∞. Then, using this subsequence, it would follow from 73.4.11,

⟨BX (tm) ,X (tm)⟩−⟨BX0,X0⟩= e(k)+2∫ tm

0⟨Y (u) ,X r

k (u)⟩du+

+2∫ tm

0

(Z ◦ J−1)∗BX l

k ◦ JdW

+m−1

∑j=0

⟨B(M(t j+1

)−M (t j)

),M(t j+1

)−M (t j)

⟩−

m−1

∑j=1

⟨B(∆X (t j)−∆M (t j)) ,∆X (t j)−∆M (t j)

⟩(73.5.18)

where e(k)→ 0 in L1 (Ω) and a.e. ω and

∆X (t j)≡ X(t j+1

)−X (t j)

∆M (t j) being defined similarly. Note how this eliminated the need to consider the term

⟨B(X (t1)−X0−M (t1)) ,X (t1)−X0−M (t1)⟩

in passing to a limit. This is a very desirable thing to be able to conclude.Can you obtain something similar even in case X0 is not assumed to be in Lp (Ω,V )?

Let Z0k ∈ Lp (Ω,V )∩ L2 (Ω,W ) ,Z0k → X0 in L2 (Ω,W ) . Then from the usual argumentsinvolving the Cauchy Schwarz inequality,

⟨B(X (t1)−X0) ,X (t1)−X0⟩1/2 ≤ ⟨B(X (t1)−Z0k) ,X (t1)−Z0k⟩1/2

+⟨B(Z0k−X0) ,Z0k−X0⟩1/2

73.5. A SIMPLIFICATION OF THE FORMULA 247973.5 A Simplification Of The FormulaThis lemma also provides a way to simplify one of the formulas derived earlier in the casethat Xp € L? (Q,V) so that X — Xo € L? ([0,T] x Q,V). Refer to 73.4.11. One term there is(B(X (t1) — Xo —M (t1)) ,X (t1) —Xo — M (t1))Also,(B(X (t1) —Xo —M (t1)) ,X (t1) —Xo0 — M (t1)))-<2(B(X (t1) — Xo) ,X (t1) —Xo) +2 (BM (11) ,M (t1))It was observed above that 2 (BM (t;),M(t,)) + 0 ae. and also in L! (Q) as k —+ 0. Applythe above lemma to (B(X (t1) — Xo) ,X (t1) — Xo) using [0,1] instead of [0,7]. The new Xoequals 0. Then from the estimate 73.4.8, it follows thatE ((B(X (t1) — Xo) ,X (t1) —Xo)) >0as k — co, Taking a subsequence, we could also assume that(B(X (t1) — Xo) ,X (1) —X0) +0a.e. @ as k — oo, Then, using this subsequence, it would follow from 73.4.11,(BX (tm) ,X (tn)) — (BXo,Xo) =e (k) #2 [ "OY (u) .XP (u)) du+2" (ZoJ~!)* BX} oJaW0m—1+ » (B (M (ti+1) —M (t;)) »M (t)41) —M(t;))j=m—1— )) (B(AX (t;) — AM (1;)) ,AX (tj) — AM (1;)) (73.5.18)j=lwhere e (k) + 0 in L! (Q) and a.e. @ andAX (tj) =X (41) —X (tj)AM (t;) being defined similarly. Note how this eliminated the need to consider the term(B(X (t1) —Xo —M (t1)) ,X (1) —Xo -M (t1))in passing to a limit. This is a very desirable thing to be able to conclude.Can you obtain something similar even in case Xp is not assumed to be in L? (Q,V)?Let Zo, € L? (Q,V) NL? (Q,W) ,Zox — Xo in L?(Q,W). Then from the usual argumentsinvolving the Cauchy Schwarz inequality,(B(X (t1) —Xo) .X (t1) Xo)? < (B(X (t1) — Zon) .X (t1) — Zon)!”+ (B(Zox —Xo) -Zox —Xo)"/