25.7. MAXIMAL MONOTONE OPERATORS 901
Proof: Let α = limsupn→∞
⟨Aλ n xn,xn
⟩. It is finite because the expression is bounded
independent of n. Then
lim supm→∞
(lim sup
n→∞
( ⟨Aλ n xn,xn
⟩+⟨Aλ mxm,xm
⟩−[⟨
Aλ nxn,xm⟩+⟨Aλ mxm,xn
⟩] ))≤ 0
Thuslim sup
m→∞
(α +
⟨Aλ m xm,xm
⟩−[⟨x∗,xm⟩+
⟨Aλ mxm,x
⟩])≤ 0
and so2α−2⟨x∗,x⟩ ≤ 0
The next simple observation is that∥∥Aλ nxn∥∥= ∥∥∥λ
−(p−1)n F
(Jλ nxn− xn
)∥∥∥≤C
due to the weak convergence. Hence λ−(p−1)n
∥∥Jλ nxn− xn∥∥p−1 ≤C and so∥∥Jλ nxn− xn
∥∥≤ λ nC1/(p−1). (25.7.59)
Thus if [u,u∗] ∈ G (A) ,
lim infn→∞
⟨Aλ nxn−u∗,xn−u
⟩= lim inf
n→∞
⟨Aλ nxn−u∗,Jλ nxn−u
⟩≥ 0
because Aλ x ∈ AJλ x. However, the left side satisfies
0 ≤ lim infn→∞
⟨Aλ nxn−u∗,xn−u
⟩≤ lim sup
n→∞
⟨Aλ nxn−u∗,xn−u
⟩= lim sup
n→∞
[⟨Aλ nxn,xn
⟩−⟨Aλ nxn,u
⟩−⟨u∗,xn⟩+ ⟨u∗,u⟩
]= α−⟨x∗,u⟩−⟨u∗,x⟩+ ⟨u∗,u⟩ ≤ ⟨x∗,x⟩−⟨x∗,u⟩−⟨u∗,x⟩+ ⟨u∗,u⟩= ⟨x∗−u∗,x−u⟩
and this shows that [x,x∗] ∈ G (A) since [u,u∗] was arbitrary.Next let [u,u∗] ∈ G (A). Then thanks to 25.7.59,
0 ≤ lim infn→∞
⟨Aλ nxn−u∗,Jλ nxn−u
⟩= lim inf
n→∞
⟨Aλ nxn−u∗,xn−u
⟩≤ lim sup
n→∞
⟨Aλ nxn−u∗,xn−u
⟩= lim sup
n→∞
(⟨Aλ nxn,xn
⟩−⟨Aλ nxn,u
⟩−⟨u∗,xn⟩+ ⟨u∗,u⟩
)= lim sup
n→∞
⟨Aλ nxn,xn
⟩−⟨x∗,u⟩−⟨u∗,x⟩+ ⟨u∗,u⟩
≤ ⟨x∗,x⟩−⟨x∗,u⟩−⟨u∗,x⟩+ ⟨u∗,u⟩= ⟨x∗−u∗,x−u⟩
In particular, you could let [u,u∗] = [x,x∗] and conclude that
limn→∞
⟨Aλ nxn− x∗,xn− x
⟩= lim
n→∞
(⟨Aλ nxn,xn
⟩−⟨Aλ nxn,x
⟩+ ⟨x∗,x⟩−⟨x∗,xn⟩
)= lim
n→∞
⟨Aλ nxn,xn
⟩−⟨x∗,x⟩+ ⟨x∗,x⟩−⟨x∗,x⟩= 0