2370 CHAPTER 69. GELFAND TRIPLES

Lemma 69.3.3 Let s < t. Then for X ,Y satisfying 69.3.15

|X (t)|2 = |X (s)|2 +2∫ t

s⟨Y (u) ,X (t)⟩du−|X (t)−X (s)|2 (69.3.16)

Proof: It follows from the following computations

X (t)−X (s) =∫ t

sY (u)du

−|X (t)−X (s)|2 =−|X (t)|2 +2(X (t) ,X (s))−|X (s)|2

= −|X (t)|2 +2(

X (t) ,X (t)−∫ t

sY (u)du

)−|X (s)|2

= −|X (t)|2 +2 |X (t)|2−2⟨∫ t

sY (u)du,X (t)

⟩−|X (s)|2

Hence|X (t)|2 = |X (s)|2 +2

∫ t

s⟨Y (u) ,X (t)⟩du−|X (t)−X (s)|2

Lemma 69.3.4 In the above situation,

supt∈[0,T ]

|X (t)|H ≤C (∥Y∥K′ ,∥X∥K)

Also, t→ X (t) is weakly continuous with values in H.

Proof: From the above formula applied to the kth partition of [0,T ] described above,

|X (tm)|2−|X0|2 =m−1

∑j=0

∣∣X (t j+1)∣∣2− ∣∣X (t j)

∣∣2

=m−1

∑j=0

2∫ t j+1

t j

⟨Y (u) ,X

(t j+1

)⟩du−

∣∣X (t j+1)−X (t j)

∣∣2H

=m−1

∑j=0

2∫ t j+1

t j

⟨Y (u) ,X rk (u)⟩du−

∣∣X (t j+1)−X (t j)

∣∣2H

Thus, discarding the negative terms and denoting by Pk the kth of these partitions,

supt j∈Pk

∣∣X (t j)∣∣2H ≤ |X0|2 +2

∫ T

0|⟨Y (u) ,X r

k (u)⟩|du

≤ |X0|2 +2∫ T

0∥Y (u)∥V ′ ∥X

rk (u)∥V du