79.1. THE CASE OF UNIQUENESS 2687

Theorem 79.1.6 In Situation 79.1.5, for ω off a set of measure zero, for every t ∈ NCω , the

measure of Nω equalling 0,

⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+∫ t

02⟨Y (s) ,X (s)⟩ds+

∫ t

0⟨BZ,Z⟩L2

ds+2M (t) (79.1.10)

where M (t) is a stochastic integral and a local martingale equal to 0 when t = 0. Also,there exists a unique continuous, progressively measurable function denoted as ⟨BX ,X⟩such that it equals ⟨BX (t) ,X (t)⟩ for a.e. t and ⟨BX ,X⟩(t) equals the right side of theabove for all t. In addition to this,

E (⟨BX ,X⟩(t)) =

E (⟨BX0,X0⟩)+E(∫ t

0

(2⟨Y (s) ,X (s)⟩+ ⟨BZ,Z⟩L2

)ds)

(79.1.11)

Also the quadratic variation of M (t) in 79.1.10 is dominated by

C∫ t

0∥Z∥2

L2∥BX∥2

W ′ ds (79.1.12)

for a suitable constant C. Also t → BX (t) is continuous with values in W ′ for t ∈ NCω . In

fact, this martingale can be written as∫ t

0

(Z ◦ J−1)∗BX ◦ JdW

That ugly integral displayed above can be written in the form∫ t

0⟨BX ,dN⟩

where N (t) =∫ t

0 Z (s)dW .Now we consider the meaning of the symbol ⟨BZ,Z⟩L2

. You begin with a completeorthonormal set {gk} in Q1/2U. Then to say that Z has values in L2

(Q1/2U ;W

)is to say

that ∑ j ∑i (Z (gi) ,e j)2 = ∑i ∥Z (gi)∥2

W < ∞ where{

e j}

is an orthonormal basis in W. Youcan let it be the one used earlier where each is actually in V or even in E. Then the symbolmeans (

R−1BZ,Z)L2

where R is the Riesz map from the Hilbert space W to its dual space. Thus it equals

∑i

(R−1BZ (gi) ,Z (gi)

)W = ∑

i⟨BZ (gi) ,Z (gi)⟩

so it is seen to be nonnegative.

79.1. THE CASE OF UNIQUENESS 2687Theorem 79.1.6 In Situation 79.1.5, for @ off a set of measure zero, for every t € NS, themeasure of Nq equalling 0,(BX (t) ,X (t)) = (BXo, Xo) + [2 (s),X (s)) ds+[ (BZ,Z) y,ds+2M (t) (79.1.10)where M (t) is a stochastic integral and a local martingale equal to 0 when t = 0. Also,there exists a unique continuous, progressively measurable function denoted as (BX ,X)such that it equals (BX (t),X (t)) for ae. t and (BX,X)(t) equals the right side of theabove for all t. In addition to this,E ((BX,X)(t)) =E((BXo,X0)) +E (/' (2 (Y (s),X(s)) + (BZ.Z) ~,) as) (79.1.11)Also the quadratic variation of M (t) in 79.1.10 is dominated by2 2Ch, |Z ||"v, ||BX || ds (79.1.12)for a suitable constant C. Also t + BX (t) is continuous with values in W' for t € NS. Infact, this martingale can be written ast[ (ZoJ~!)* BX oJdW0That ugly integral displayed above can be written in the form| ' (BX dN)0where N (t) = fj Z(s) dW.Now we consider the meaning of the symbol (BZ, Z) y,, You begin with a completeorthonormal set {g,} in Q'/?U. Then to say that Z has values in % (Q'/?U;W) is to saythat Y/Y; (Z(g:),e;)” = Lil|Z (gs) lw < ee where {e;} is an orthonormal basis in W. Youcan let it be the one used earlier where each is actually in V or even in E. Then the symbolmeans(R-'BZ,Z) v,where R is the Riesz map from the Hilbert space W to its dual space. Thus it equalsY" (R'BZ (gi) .Z(gi)) y = Y, (BZ (gi) .Z(g:))i iso it is seen to be nonnegative.