76.7. OTHER EXAMPLES, INCLUSIONS 2587

It follows from 76.7.64 that for all ω /∈ N and adjusting the constant,

⟨Bun (t) ,un (t)⟩+∫ T

0∥un∥p

V ds≤C (ω) (76.7.65)

for all n, where C (ω) depends only on ω .For ω /∈ N, the above estimate implies there exists a further subsequence, still called n

such thatBun→ Bu weak ∗ in L∞

(0,T,W ′

)un→ u weak ∗ in L∞ (0,T,H)

un→ u weakly in Vω

From the integral equation solved and the assumption that A is bounded, it can also beassumed that(

B(

un−∫ (·)

0ΨdW

))′→(

B(

u−∫ (·)

0ΨdW

))′weakly in V ′ω (76.7.66)

Bu(0) = Bu0

Aun→ ξ weakly in V ′ω

It is known that un is bounded in Vω . Also it is known that(

B(

un−∫ (·)

0 ΨdW))′

is

bounded in V ′ω . Therefore, B(

un−∫ (·)

0 ΨdW)

satisfies a Holder condition into V ′. Since Ψ

is in L∞,∫ (·)

0 ΨdW satisfies a Holder condition, and so Bun satisfies a Holder condition intoV ′ while Bun is bounded in W ′

ω . By compactness of the embedding of V into W, it followsthat W ′ embeds compactly into V ′. This is sufficient to conclude that {Bun} is precompactin W ′

ω . The proof is similar to one given by Lions. [91] page 57. See Theorem 69.5.6.Since B is the Riesz map, this implies that {un} is precompact in Wω and hence in Hω .

Therefore, one can take a further subsequence and conclude that

un→ u strongly in Hω ≡ L2 ([0,T ] ,L2 (U))

Therefore, a further subsequence, still denoted by n satisfies

un (t)→ u(t) in L2 (U) for a.e. t

We can also assume that

Jn (un)→ ζ weak ∗ in L∞ (0,T,L∞ (U))

From the integral equation solved,⟨(B(

un−∫ (·)

0ΨdW

))′,un−u

⟩Vω

+⟨A(t,un)+h(t,ω)Jn (un) ,un−u⟩= 0 (76.7.67)