2644 CHAPTER 77. STOCHASTIC INCLUSIONS

this, you consider Ĝ(y)≡ G(x+ y) . Then Ĝ is also maximal monotone and so there existsa solution to

0 ∈ F (x̂)+µp−1Ĝ(x̂) = F (x̂)+µ

p−1G(x+ x̂)

Now let xµ = x+ x̂ so x̂ = xµ − x. Hence

0 ∈ F(xµ − x

)+µ

p−1Gxµ

The symbol limsupn,n→∞ amn means limN→∞

(supm≥N,n≥N amn

).

Lemma 77.8.2 Suppose limsupn,n→∞ amn ≤ 0. Then limsupm→∞ (limsupn→∞ amn)≤ 0.

Proof: Suppose the first inequality. Then for ε > 0, there exists N such that if n,m areboth as large as N, then amn ≤ ε. Thus supn≥N amn ≤ ε provided m ≥ N also. Hence forsuch m,

limn→∞

(supn≥N

amn

)≤ ε

for each m ≥ N. It follows limsupm→∞ (limsupn→∞ amn) ≤ ε. Since ε is arbitrary, thisproves the lemma.

Then here is a simple observation.

Lemma 77.8.3 Let G : D(G)⊆X→P (X ′) where X is a Banach space be maximal mono-tone and let vn ∈ Gun and

un→ u, vn→ v weakly.

Also suppose thatlim sup

m,n→∞

⟨vn− vm,un−um⟩ ≤ 0

orlim sup

n→∞

⟨vn− v,un−u⟩ ≤ 0

Then [u,v] ∈ G (G) and ⟨vn,un⟩ → ⟨v,u⟩.

Proof: By monotonicity,

0 ≥ lim supm,n→∞

⟨vn− vm,un−um⟩

≥ lim infm,n→∞

⟨vn− vm,un−um⟩ ≥ 0

and solim

m,n→∞⟨vn− vm,un−um⟩= 0

Suppose then that ⟨vn,un⟩ fails to converge to ⟨v,u⟩. Then there is a subsequence, stilldenoted with subscript n such that ⟨vn,un⟩ → µ ̸= ⟨v,u⟩. Let ε > 0. Then there exists Msuch that if n,m > M, then

|⟨vn,un⟩−µ|< ε, |⟨vn− vm,un−um⟩|< ε