43.3. THE IMPLICIT CASE 1439

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 (43.3.19)

Hence {BX (t (k))}∞

k=1 is a convergent sequence in W ′ because

|⟨B(X (t (k))−X (t (m))) ,y⟩|≤ ⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩1/2 ⟨By,y⟩1/2

≤ ⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩1/2 ∥B∥1/2 ∥y∥W

Does it converge to BX (t)? Let ξ (t) ∈W ′ be what it does converge to. Let v ∈V. Then

⟨ξ (t) ,v⟩= limk→∞

⟨BX (t (k)) ,v⟩= limk→∞

⟨BX (t (k)) ,v⟩= ⟨BX (t) ,v⟩

because it is known that t → BX (t) is continuous into V ′. It is also known that BX (t) ∈W ′ ⊆V ′ and that the BX (t) for t ∈ [0,T ] are uniformly bounded in W ′. Therefore, since Vis dense in W, it follows that ξ (t) = BX (t).

Now for every t ∈ D, it was shown above that

⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+2∫ t

0⟨Y (u) ,X (u)⟩du

Also it was just shown that BX (t (k))→ BX (t) . Then

|⟨BX (t (k)) ,X (t (k))⟩−⟨BX (t) ,X (t)⟩|

≤ |⟨BX (t (k)) ,X (t (k))−X (t)⟩|+ |⟨BX (t (k))−BX (t) ,X (t)⟩|

Then the second term converges to 0. The first equals

|⟨BX (t (k))−BX (t) ,X (t (k))⟩|≤ ⟨B(X (t (k))−X (t)) ,X (t (k))−X (t)⟩1/2 ⟨BX (t (k)) ,X (t (k))⟩1/2

From the above, this is dominated by an expression of the form

⟨B(X (t (k))−X (t)) ,X (t (k))−X (t)⟩1/2 C

Then using the lower semicontinuity of t → ⟨B(X (t (k))−X (t)) ,X (t (k))−X (t)⟩ whichfollows from the above, this is no larger than

lim infm→∞⟨B(X (t (k))−X (t (m))) ,X (t (k))−X (t (m))⟩1/2 C < ε