76.3. THE EXISTENCE OF APPROXIMATE SOLUTIONS 2545

Consider the other inequality. Let ∥z∥V ≤ 1.Then

|⟨A(t,u+q(t,ω) ,ω) ,z⟩| ≤ k∥u+q(t,ω)∥p−1 + c1/p′ (t,ω)

Since p≥ 2, a convexity argument shows that

⟨A(t,u+q(t,ω) ,ω) ,z⟩ ≤ k(

2p−2 ∥u∥p−1 +2p−2 ∥q(t,ω)∥p−1)+ c1/p′ (t,ω)

= 2p−2k∥u∥p−1 +(c̄(t,ω))1/p′

where c̄ ∈ L1 ([0,T ]×Ω). Thus the same two inequalities continue to hold.In what follows, c≥ 0 and is in L1 ([0,T ]×Ω) , the σ algebra being B ([0,T ])×FT .

⟨A(t,u,ω) ,u⟩V ≥ δ ∥u∥pV − c(t,ω) (76.3.14)

∥A(t,u,ω)∥V ′ ≤ k∥u∥p−1V + c1/p′ (t,ω) (76.3.15)

Letting Ā be defined above in 76.3.13,

Ā(t,u,ω)≡ A(t,u+q,ω)≡ Ā(ω)(t,u)

Assume the following pathwise uniqueness condition which is the hypothesis of the fol-lowing lemma.

Lemma 76.3.3 Suppose it is true that whenever u,v ∈ Vω and

Bu(t)−Bv(t)+∫ t

0A(u)−A(v) = 0 (76.3.16)

it follows that u = v. Then if

(Bu)′+ Ā(ω)u = f in V ′ω , Bu(0) = Bu0

(Bv)′+ Ā(ω)v = f in V ′ω , Bv(0) = Bu0 (76.3.17)

it follows that u = v in Vω . Here u0 ∈W.

Proof: If (Bu)′+ Ā(ω)u = f and (Bv)′+ Ā(ω)v = f , then

Bu(t)−Bv(t)+∫ t

0A(u+q)−A(v+q)ds = 0

Hence

B(u(t)+q(t))−B(v(t)+q(t))+∫ t

0A(u+q)−A(v+q)ds = 0

and so u+q = v+q showing that u = v.We give the following measurability lemma.