25.7. MAXIMAL MONOTONE OPERATORS 877

from which it follows from coercivity of B that the xn are bounded independent of n. Say∥xn∥ <C. Then there is a subsequence still denoted as xn such that xn→ x ∈ K, thanks tothe assumption that K is closed and convex. Let [u,v] ∈ G (A) . Then for all n large enough∥u∥< n and so

⟨Bxn + v,u− xn⟩ ≥ 0

Then letting n→ ∞ and using the continuity of B,

⟨Bx+ v,u− x⟩ ≥ 0

Since [u,v] was arbitrary, this proves the lemma.

Observation 25.7.8 If you have a monotone set valued function, then its graph can alwaysbe considered a subset of the graph of a maximal monotone graph. If A is monotone, thenlet F be G (B) such that G (B)⊇G (A) and B is monotone. Partially order by set inclusion.Then let C be a maximal chain. Let G

(Â)= ∪C . If [xi,yi] ∈ G

(Â), then both are in some

B ∈ C . Hence (y1− y2,x1− x2)≥ 0 so monotone and must be maximal monotone becauseif ⟨z− v,x−u⟩ ≥ 0 for all [u,v] ∈ G

(Â)

and [x,z] /∈ Â, then you could include this orderedpair and contradict maximality of the chain C .

Next is an interesting theorem which comes from this lemma. It is an infinite dimen-sional version of the above lemma.

Theorem 25.7.9 Let X be a reflexive Banach space and let K be a closed convex subset ofX. Let A,B be monotone such that

1. D(A)⊆ K,0 ∈ D(A) .

2. B is single valued, hemicontinuous, bounded and coercive mapping X to X ′.

Then there exists x ∈ K such that

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

Before giving the proof, here is an easy lemma.

Lemma 25.7.10 Let E be finite dimensional and let B : E→ E ′ be monotone and hemicon-tinuous. Then B is continuous.

Proof: The space can be considered a finite dimensional Hilbert space (Rn) and so weakand strong convergence are exactly the same. First it is desired to show that B is bounded.Suppose it is not. Then there exists ∥xk∥E = 1 but ∥Bxn∥E ′ → ∞. Since finite dimensional,there is a subsequence still denoted as xk such that xk→ x,∥x∥E = 1.

⟨Bxk−Bx,xk− x⟩ ≥ 0

Hence ⟨Bxk−Bx∥Bxk∥E ′

,xk− x⟩≥ 0