43.3. THE IMPLICIT CASE 1437

≤ ⟨BX0,X0⟩+2(∫ T

0∥Y (u)∥p′

V ′ du)1/p′(∫ T

0∥X r

k (u)∥pV du

)1/p

≤ C (∥Y∥K′ ,∥X∥K)

because these partitions are chosen such that

limk→∞

(∫ T

0∥X r

k (u)∥pV

)1/p

=

(∫ T

0∥X (u)∥p

V

)1/p

and so these are bounded. This has shown that for the dense subset of [0,T ] , D≡ ∪kPk,

supt∈D⟨BX (t) ,X (t)⟩<C (∥Y∥K′ ,∥X∥K)

From Lemma 43.3.1 above, there exists {ei} ⊆V such that⟨Bei,e j

⟩= δ i j and

⟨BX (t) ,X (t)⟩=∞

∑k=1|⟨BX (t) ,ei⟩|2 = sup

m

m

∑k=1|⟨BX (t) ,ei⟩|2

Since each ei ∈V, and since t→ BX (t) is continuous into V ′ thanks to the formula 43.3.16,it follows that t→ ∑

mk=1 |⟨BX (t) ,ei⟩| is continuous and so t→ ⟨BX (t) ,X (t)⟩ is the sup of

continuous functions. Therefore, this function of t is lower semicontinuous. Since D isdense in [0,T ] , it follows that for all t,

⟨BX (t) ,X (t)⟩ ≤C (∥Y∥K′ ,∥X∥K)

It only remains to verify the claim about weak continuity.Consider now the claim that t→ BX (t) is weakly continuous. Letting v ∈V,

limt→s⟨BX (t) ,v⟩= ⟨BX (s) ,v⟩= ⟨BX (s) ,v⟩ (43.3.18)

The limit follows from the formula 43.3.16 which implies t→ BX (t) is continuous into V ′.Now

∥BX (t)∥= sup∥v∥≤1

|⟨BX (t) ,v⟩| ≤ ⟨Bv,v⟩1/2 ⟨BX (t) ,X (t)⟩1/2

which was shown to be bounded for t ∈ [0,T ]. Now let w ∈W . Then

|⟨BX (t) ,w⟩−⟨BX (s) ,w⟩| ≤ |⟨BX (t)−BX (s) ,w− v⟩|+ |⟨BX (t)−BX (s) ,v⟩|

Then the first term is less than ε if v is close enough to w and the second converges to 0 so43.3.18 holds for all v ∈W and so this shows the weak continuity.

Now pick t ∈ D, the union of all the mesh points. Then for all k large enough, t ∈Pk.Say t = tm. From Lemma 43.3.3,

−m−1

∑j=0

⟨B(X(t j+1

)−X (t j)

),(X(t j+1

)−X (t j)

)⟩=