894 CHAPTER 25. NONLINEAR OPERATORS

To see this, you consider Â(y)≡ A(x+ y) . Then  is also maximal monotone and so thereexists a solution to

0 ∈ F (x̂)+λ Â(x̂) = F (x̂)+λA(x+ x̂)

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

0 ∈ F (xλ − x)+λAxλ

Here you could have F the duality map for any given p > 1.The symbol limsupn,n→∞ amn means limN→∞

(supm≥N,n≥N amn

). Then here is a simple

observation.

Lemma 25.7.33 Suppose limsupn,n→∞ amn ≤ 0. Then

lim supm→∞

(lim sup

n→∞

amn

)≤ 0.

Proof: There exists N such that if both m,n≥ N,amn ≤ ε . Then

lim supn→∞

amn = lim supn→∞,n>N

amn ≤ ε

Thus also

lim supm→∞

(lim sup

n→∞

amn

)= lim sup

m→∞,m≥N

(lim sup

n→∞

amn

)≤ ε.

The argument will be based on the following lemma.

Lemma 25.7.34 Let A : D(A)→P (X ′) be maximal monotone and let vn ∈ Aun 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 (A) and ⟨vn,un⟩ → ⟨v,u⟩.

Proof: By monotonicity,

limm,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