828 CHAPTER 25. NONLINEAR OPERATORS

since L is bounded, there is a further subsequence, still called n such that

Mun = L(un−u0)⇀ L(u−u0) = Mu.

Since M is monotone,⟨Mun−Mu,un−u⟩ ≥ 0

Thus⟨Mun,un⟩−⟨Mun,u⟩−⟨Mu,un⟩+ ⟨Mu,u⟩ ≥ 0

and so⟨Mun,un⟩ ≥ ⟨Mun,u⟩+ ⟨Mu,un−u⟩

Hence with this further subsequence, the limsup is no larger and so

⟨ξ ,u⟩ ≥ lim supn→∞

⟨Aun +Mun,un⟩

≥ lim supn→∞

(⟨Aun,un⟩+ ⟨Mun,u⟩+ ⟨Mu,un−u⟩)

= lim supn→∞

⟨Aun,un⟩+ limn→∞

(⟨Mun,u⟩+ ⟨M (u) ,un−u⟩)≤ ⟨ξ ,u⟩

and solim sup

n→∞

⟨Aun,un⟩ ≤ ⟨ξ −Mu,u⟩

It follows since A is type M that Au = ξ −Mu, which contradicts the assumption thatξ ̸= Au+Mu.

The following is Browder’s lemma. It is a very interesting application of the Brouwerfixed point theorem.

Lemma 25.1.11 (Browder) Let K be a convex closed and bounded set in Rn and let A :K→ Rn be continuous and f ∈ Rn. Then there exists x ∈ K such that for all y ∈ K,

(f−Ax,y−x)Rn ≤ 0

If K is convex, closed, bounded subset of V a finite dimensional vector space, then the sameconclusion holds. If f ∈V ′, there exists x ∈ K such that for all y ∈ K,

⟨ f −Ax,y− x⟩V ′,V ≤ 0

Proof: Let PK denote the projection onto K. Thus PK is Lipschitz continuous.

x→ PK (f−Ax+x)

is a continuous map from K to K. By the Brouwer fixed point theorem, it has a fixed pointx ∈ K. Therefore, for all y ∈ K,

(f−Ax+x−x,y−x) = (f−Ax,y−x)≤ 0