Kenneth Kuttler

2458 CHAPTER 72. THE HARD ITO FORMULA

continuous into H. Therefore, from the uniform convexity of the norm in H it followst→ X (t) is continuous.This is very easy to see in Hilbert space. Say an ⇀ a and |an|→ |a| .From the parallelogram identity.

|an−a|2 + |an +a|2 = 2 |an|2 +2 |a|2

so|an−a|2 = 2 |an|2 +2 |a|2−

(|an|2 +2(an,a)+ |a|2

)Then taking limsup both sides,

0≤ lim supn→∞

|an−a|2 ≤ 2 |a|2 +2 |a|2−(|a|2 +2(a,a)+ |a|2

)= 0.

Of course this fact also holds in any uniformly convex Banach space.Now consider the last claim. If the last term in 72.6.17 were a martingale, then there

would be nothing to prove. This is because if M (t) is a martingale which equals 0 whent = 0, then

E (M (t)) = E (E (M (t) |F0)) = E (M (0)) = 0.

However, that last term is unfortunately only a local martingale. One can obtain a localizingsequence as follows.

τn (ω)≡ inf{t : |X (t,ω)|> n}

where as usual inf( /0) ≡ ∞. This is all right because it was shown above that t → X (t,ω)is continuous into H for a.e. ω . Then stopping both processes on the two sides of 72.6.17with τn,

|X (t ∧ τn)|2 = |X0|2 +∫ t∧τn

(2⟨Y (s) ,X (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds

+2∫ t∧τn

0R((

Z (s)◦ J−1)∗X (s))◦ JdW (s)

Now from Lemma 65.10.5,

|X (t ∧ τn)|2 = |X0|2 +∫ t

0X[0,τn] (s)

(2⟨Y (s) ,X (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds

+2∫ t

0X[0,τn] (s)R

((Z (s)◦ J−1)∗X (s)

)◦ JdW (s)

That last term is now a martingale and so you can take the expectation of both sides. Thisgives

E(|X (t ∧ τn)|2

)= E

(|X0|2

)+E(∫ t

0X[0,τn] (s)

(2⟨Y (s) , X̄ (s)⟩+ ||Z (s)||2

L2(Q1/2U,H)

)ds)

Letting n→∞ and using the dominated convergence theorem and τn→∞ yields the desiredresult.

2458 CHAPTER 72. THE HARD ITO FORMULAcontinuous into H. Therefore, from the uniform convexity of the norm in H it followst —> X (t) is continuous.This is very easy to see in Hilbert space. Say a, — a and |a,| — |al.From the parallelogram identity.lan — al? + |an tal? =2 an)? +2 lal?solan — a)? =2|ag|? +2 [al — (lanl? +2 (an,a) + lal)Then taking lim sup both sides,0 < lim sup ay —a|” < 2|a)’ +2lal? — (lal? +2(a,a) + al?) =0.n-eoOf course this fact also holds in any uniformly convex Banach space.Now consider the last claim. If the last term in 72.6.17 were a martingale, then therewould be nothing to prove. This is because if M(t) is a martingale which equals 0 whent = 0, thenE(M(t)) =E(E(M (t)|-¥o)) = E(M (0)) =0.However, that last term is unfortunately only a local martingale. One can obtain a localizingsequence as follows.T (@) = inf {t : |X (t,@)| > n}where as usual inf (0) = 0. This is all right because it was shown above that t > X (t, @)is continuous into H for a.e. @. Then stopping both processes on the two sides of 72.6.17with Ty,K(At)P=[Xol+ [” (27).X) HIZOEoreun)) @42 [a ((Z(s) oJ”')*X(s)) oJdW (s)Now from Lemma 65.10.5,X (CA tH)? = [XO + f° Fo.e (8) (207 (9) X(9)) + IZOIIZa(ov2v.m))42 [ Poni )# (Zio) od)’ X(s)) ofa (s)That last term is now a martingale and so you can take the expectation of both sides. ThisgivesE (ix (1 t)|”) =E ([x0l”)+E (jf; 20,1] (8) (2 (Y (s) ,X (s)) + IZ(s)Il-4(o120.m)) i)Letting n — ~ and using the dominated convergence theorem and T,, — 9 yields the desiredresult. Jf