2382 CHAPTER 69. GELFAND TRIPLES

Therefore, this term is dominated by an expression of the form

mk−1

∑j=0

(∫ t j+1

t j

Y (u)du,X(t j+1

)−X (t j)

)

=mk−1

∑j=0

⟨∫ t j+1

t j

Y (u)du,X(t j+1

)−X (t j)

⟩

=mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X

(t j+1

)−X (t j)

⟩du

=mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X

(t j+1

)⟩−

mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X (t j)

⟩=

∫ T

0⟨Y (u) ,X r (u)⟩du−

∫ T

0

⟨Y (u) ,X l (u)

⟩du

However, both X r and X l converge to X in K = Lp (0,T,V ). Therefore, this term mustconverge to 0. Passing to a limit, it follows that for all t ∈ D, the desired formula holds.Thus, for such t ∈ D,

⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+2∫ t

0⟨Y (u) ,X (u)⟩du

It remains to verify that this holds for all t. Let t /∈ D and let t (k) ∈Pk be the largestpoint of Pk which is less than t. Suppose t (m)≤ t (k) so that m≤ k. Then

BX (t (m)) = BX0 +∫ t(m)

0Y (s)ds,

a similar formula for X (t (k)) . Thus for t > t (m) ,

BX (t)−BX (t (m)) =∫ t

t(m)Y (s)ds

which is the same sort of thing already looked at except that it starts at t (m) rather than at0 and X0 = 0. Therefore,

⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩

= 2∫ t(k)

t(m)⟨Y (s) ,X (s)−X (t (m))⟩ds

Thus, for m≤ k

limm,k→∞

⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩= 0 (69.4.26)