69.3. AN IMPORTANT FORMULA 2371

≤ |X0|2 +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|X (t)|<C (∥Y∥K′ ,∥X∥K)

Now let {gk}∞

k=1 be linearly independent vectors of V whose span is dense in V . This ispossible because V is separable. Then let

{e j}∞

j=1 be an orthonormal basis for H such thatek ∈ span(g1, . . . ,gk) and each gk ∈ span(e1, . . . ,ek) . This is done with the Gram Schmidtprocess. Then it follows that span({ek}∞

k=1) is dense in V . I claim

|y|2H =∞

∑j=1

∣∣⟨y,e j⟩∣∣2 .

This is certainly true if y ∈ H because⟨y,e j

⟩= (y,e j)H

If y /∈ H, then the series must diverge since otherwise, you could consider the infinite sum

∞

∑j=1

⟨y,e j

⟩e j ∈ H

because ∣∣∣∣∣ q

∑j=p

⟨y,e j

⟩e j

∣∣∣∣∣2

=q

∑j=p

∣∣⟨y,e j⟩∣∣2→ 0 as p,q→ ∞.

Letting z = ∑∞j=1⟨y,e j

⟩e j, it follows that

⟨y,e j

⟩is the jth Fourier coefficient of z and that

⟨z− y,v⟩= 0

for all v ∈ span({ek}∞

k=1) which is dense in V. Therefore, z = y in V ′ and so y ∈ H.It follows

|X (t)|2 = supn

n

∑j=1

∣∣⟨X (t) ,e j⟩∣∣2

which is just the sup of continuous functions of t. Therefore, t→ |X (t)|2 is lower semicon-tinuous. It follows that for any t, letting t j→ t for t j ∈ D,

|X (t)|2 ≤ lim infj→∞

∣∣X (t j)∣∣2 ≤C (∥Y∥K′ ,∥X∥K)