79.4. STOCHASTIC INCLUSIONS WITHOUT UNIQUENESS ?? 2717

it follows from 79.4.57 that∣∣∣∣unk (t,ω)

∣∣∣∣V and consequently

∣∣∣∣znk (t,ω)∣∣∣∣

V ′ are bounded.Thus, denoting this subsequence with n, there is a further subsequence for which un (t,ω)→ψ (t,ω) weakly in V which was what was claimed.

But also, it follows from Lemma 79.4.6 that for ω /∈ N,∥∥∥∥un (t,ω)−u0 (ω)+∫ t

0(zn (s,ω)− f (s,ω))ds−

∫ t

0ΦdW

∥∥∥∥U ′→ 0

where n doesn’t depend on (t,ω).By convexity, Lemma 79.4.6, and weak semicontinuity considerations, it must be the

case that ∥∥∥∥ψ (t,ω)−u0 (ω)+∫ t

0(z(s,ω)− f (s,ω))ds−

∫ t

0ΦdW

∥∥∥∥U ′

≤ lim infn→∞

∥∥∥∥ un (t,ω)−u0 (ω)+∫ t

0 (zn (s,ω)− f (s,ω))ds−∫ t

0 ΦdW

∥∥∥∥U ′

= 0

Here n = n(t,ω) is a subsequence. But of course, this requires ψ (t,ω) = u(t,ω) in U ′

thanks to 79.4.60 and so in fact, un (t,ω)→ u(t,ω) weakly.Now, 79.4.62 and the limit conditions for pseudomonotone operators imply that the

liminf condition holds. There exists z∞ ∈ A(u(t,ω) , t,ω) such that

lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩

≥ ⟨z∞,u(t,ω)−u(t,ω)⟩= 0> lim

k→∞⟨znk (t,ω) ,unk (t,ω)−u(t,ω)⟩

= lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩

which is a contradiction. This completes the proof of the claim.It follows from this claim that for given ω off a set of measure zero and t /∈Mω ,

lim infn→∞⟨zn (t,ω) ,un (t,ω)−u(t,ω)⟩ ≥ 0. (79.4.65)

Also, it is assumed thatlim sup

n→∞

⟨zn,un−u⟩V ′,V ≤ 0.

This continues holding for subsequences. From the estimates,∫Ω

∫ T

0

(b3 ||un (t,ω)||pV −b4 (t,ω)−λ |un (t,ω)|2H

)dtdP

≤∫

Ω

∫ T

0∥u(t,ω)∥V

(∥un (t,ω)∥p−1

V b1 +b2

)dtdP

so it is routine to get ∥un∥V is bounded. This follows from the assumptions, in particular79.4.57.