25.7. MAXIMAL MONOTONE OPERATORS 879

Then from 25.7.48⟨v,u− xn⟩+ ⟨Bxn,u⟩ ≥ ⟨Bxn,xn⟩

Then it follows that

⟨v,u− xn⟩+ ⟨Bxn,u⟩−⟨Bxn,x⟩ ≥ ⟨Bxn,xn− x⟩

It follows that

lim supn→∞

⟨Bxn,xn− x⟩ ≤ ⟨v,u− x⟩+ ⟨y,u⟩−⟨y,x⟩

= ⟨v+ y,u− x⟩

Claim: limsupn→∞ ⟨Bxn,xn− x⟩ ≤ 0.Proof of claim: This is so if ⟨v+ y,u− x⟩ ≤ 0 for some [u,v] ∈ G

(Â).If ⟨v+ y,u− x⟩

is greater than 0 for all [u,v] , then since  is maximal, it would follow that −y ∈ Âx. Nowconsider 25.7.49.

⟨v,u− x⟩ ≥ lim supn→∞

⟨Bxn,xn⟩−⟨y,u⟩

Since x ∈ D(Â), you could put in u = x in the above and obtain

0≥ lim supn→∞

⟨Bxn,xn⟩−⟨y,x⟩= lim supn→∞

⟨Bxn,xn− x⟩

which shows the claim is true.Since B is monotone and hemicontinuous, it satisfies the pseudomonotone condition,

Theorem 25.1.4. Hence for any z,

⟨y,x− z⟩ ≥ lim supn→∞

⟨Bxn,xn− x⟩+ lim supn→∞

⟨Bxn,x− z⟩

≥ lim supn→∞

(⟨Bxn,xn− x⟩+ ⟨Bxn,x− z⟩)

≥ lim infn→∞

(⟨Bxn,xn− z⟩)≥ ⟨Bx,x− z⟩

Since z is arbitrary, this shows that y = Bx. It follows from 25.7.48 that for any [u,v] ∈G(Â),

⟨Bxn + v,u− xn⟩= ⟨Bxn + v,u− x⟩+ ⟨Bxn + v,x− xn⟩ ≥ 0

⟨Bxn + v,u− x⟩ ≥ ⟨Bxn,xn− x⟩ ≥ ⟨Bx,xn− x⟩

Now take a limit of both sides and use the fact that y = Bx to obtain

⟨Bx+ v,u− x⟩ ≥ 0

for all [u,v] ∈ G(Â). Here  extends A on ∪nXn. Why does it follow from this that there

exists an x such that the inequality holds for all [u,v] ∈ G (A)?Let V be a finite dimensional subspace.

KV ≡{

x ∈ K : ⟨Bx+ v,u− x⟩X ′,X ≥ 0 for all [u,v] ∈ G (A) ,u ∈V}.