25.7. MAXIMAL MONOTONE OPERATORS 887

Theorem 25.7.25 Suppose A : X →P (X ′) is maximal monotone. That is, D(A) = X.Then A is pseudomonotone.

Proof: Consider the first condition. Say x∗i ∈ Ax. Let u∗ ∈ Au. For λ ∈ [0,1] ,

⟨λx∗1 +(1−λ )x∗2−u∗,x−u⟩= λ ⟨x∗1−u∗,x−u⟩+(1−λ )⟨x∗2−u∗,x−u⟩ ≥ 0

and so, since [u,u∗] is arbitrary, it follows that λx∗1 +(1−λ )x∗2 ∈ Ax. Thus Ax is convex.Is it closed? Say x∗n ∈ Ax and x∗n→ x∗. Is it the case that x∗ ∈ D(A)? Let [u,u∗] ∈ G (A) bearbitrary. Then

⟨x∗−u∗,x−u⟩= limn→∞⟨x∗n−u∗,xn−u⟩ ≥ 0

and so Ax is also closed.Consider the second condition. It is to show that if xn → x in V a finite dimensional

subspace and if U is a weakly open set containing 0, then eventually Axn⊆Ax+U. Supposethen that this is not the case. Then there exists x∗n outside of Ax+U but in Axn. Since A islocally bounded at x, it follows that the ∥x∗n∥ are bounded. Thus there is a subsequence, stilldenoted as xn and x∗n such that x∗n→ x∗ weakly and x∗ /∈ Ax+U. Now let [u,u∗] ∈ G (A) .

⟨x∗−u∗,x−u⟩= limn→∞⟨x∗n−u∗,xn−u⟩ ≥ 0

and since [u,u∗] is arbitrary, it follows that x∗ ∈ Ax and so is inside Ax+U . Thus the secondcondition holds also.

Consider the third. Say xk→ x weakly and letting x∗k ∈ Axk,suppose

lim supk→∞

⟨x∗k ,xk− x⟩ ≤ 0,

Is it the case that there exists x∗ (y) ∈ Ax such that

lim infk→∞

⟨x∗k ,xk− y⟩ ≥ ⟨x∗ (y) ,x− y⟩?

The proof goes just like it did earlier in the case of single valued pseudomonotone operators.It is just a little more complicated. First, let x∗ ∈ Ax.

⟨x∗k− x∗,xk− x⟩ ≥ 0

and solim inf

k→∞

⟨x∗k ,xk− x⟩ ≥ lim infk→∞

⟨x∗,xk− x⟩= 0≥ limsup⟨x∗k ,xk− x⟩

Thuslimk→∞

⟨x∗k ,xk− x⟩= 0.

Now let x∗t ∈ A(x+ t (y− x)) , t ∈ (0,1) , where here y is arbitrary. Then

⟨x∗n− x∗t ,xn− x+ t (x− y)⟩ ≥ 0