79.1. THE CASE OF UNIQUENESS 2691

Thus {Mk} is a Cauchy sequence in M1T and so there is a continuous martingale M such

that

limk→∞

E(

supt|Mk (t)−M (t)|

)= 0

Taking a further subsequence if needed, one can also have

P(

supt|Mk (t)−M (t)|> 1

k

)≤ 1

2k

and so by the Borel Cantelli lemma, there is a set of measure zero such that off this set,supt |Mk (t)−M (t)| converges to 0. Hence for such ω,M∗k (T ) is bounded independent ofk. Thus for ω off a set of measure zero, 79.1.14 implies that for such ω,

sups∈[0,T ]

⟨Bur,ur⟩(s)+∫ T

0∥ur (s)∥p

V ds≤C (ω)

where C (ω) does not depend on the index r, this for the subsequence just described whichwill be the sequence of interest in what follows. Using the boundedness assumption for A,one also obtains an estimate of the form

sups∈[0,T ]

⟨Bur,ur⟩(s)+∫ T

0∥ur (s)∥p

V ds+∫ T

0∥zr∥p′

V ′ ≤C (ω) (79.1.18)

The idea here is to take weak limits converging to a function u and then identify z(·,ω)as being in A(u,ω) but this will involve a difficulty. It will require a use of the aboveIto formula and this will need u to be progressively measurable. By uniqueness, it wouldseem that this could be concluded by arguing that one does not need to take a subsequencedue to uniqueness but the problem is that we won’t know the limit of the sequence is asolution unless we use the Ito formula. This is why we make the extra assumption that forzi (·,ω) ∈ A(ui,ω) and for all λ large enough,

⟨λBu1 (t)+ z1 (t)− (λBu2 (t)+ z2 (t)) ,u1 (t)−u2 (t)⟩ ≥ δ ∥u1 (t)−u2 (t)∥α

V̂ , α ≥ 1(79.1.19)

where here V̂ will be a Banach space such that V is dense in V̂ and the embedding iscontinuous. As mentioned, this is not surprising in the case of most interest where there isa Gelfand triple and B = I and A is defined pointwise with no memory terms involving timeintegrals. Then using the integral equation for r = p,q, p < q along with the conclusion ofthe Ito formula above,

E (⟨B(un−um) ,un−um⟩(t))+E(∫ t

0∥un−um∥α

V̂ ds)

≤ E(∫ t

0∥B∥∥Φn−Φm∥2

L2ds)≡ e(m,n)

Hence, the right side converges to 0 as m,n→∞ from the dominated convergence theorem.In particular,

E(∫ T

0∥un−um∥α

V̂ ds)≤ E

(∫ T

0∥B∥∥Φn−Φm∥2

L2ds)≡ e(m,n) (79.1.20)

79.1. THE CASE OF UNIQUENESS 2691Thus {M,} is a Cauchy sequence in M}. and so there is a continuous martingale M suchthatlim E (sup |M;. (t) MC) =0tkoTaking a further subsequence if needed, one can also haveP (sup|ma(n)—M() > i) S xand so by the Borel Cantelli lemma, there is a set of measure zero such that off this set,sup, |M; (t) — M (t)| converges to 0. Hence for such @,M; (T) is bounded independent ofk. Thus for @ off a set of measure zero, 79.1.14 implies that for such @,Tsup (Burt) (8) + [ |jur(s)|Ke ds <C(@)s€[0,7] 0where C(@) does not depend on the index r, this for the subsequence just described whichwill be the sequence of interest in what follows. Using the boundedness assumption for A,one also obtains an estimate of the formT T ,sup (Bu,,u,)(s)+ | lu, (s)ds+ | lene, <C(o) (79.1.18)se [0,7] 0 0The idea here is to take weak limits converging to a function u and then identify z(-, @)as being in A(u,q@) but this will involve a difficulty. It will require a use of the aboveIto formula and this will need u to be progressively measurable. By uniqueness, it wouldseem that this could be concluded by arguing that one does not need to take a subsequencedue to uniqueness but the problem is that we won’t know the limit of the sequence is asolution unless we use the Ito formula. This is why we make the extra assumption that forz(-,@) € A (u;,@) and for all A large enough,(ABuy (t) +21 (¢) — (A Bua (t) +22 (t) ,u1 (¢) — ua (t)) = 6 llr (4) — 2 (1)6, a>1(79.1.19)where here V will be a Banach space such that V is dense in V and the embedding iscontinuous. As mentioned, this is not surprising in the case of most interest where there isa Gelfand triple and B = J and A is defined pointwise with no memory terms involving timeintegrals. Then using the integral equation for r = p,q, p < g along with the conclusion ofthe Ito formula above,? d s)tE(B (ln ~ tn) ta ~ tn) ()) +E (ftHence, the right side converges to 0 as m,n — oo from the dominated convergence theorem.In particular,T TE (/ Intnl <E (/ ||B|| |, Gnas) = e(m,n) (79.1.20)t< £([ [alle —@nl,as) =e(mn)