75.1. AN ASSORTMENT OF NONLINEAR OPERATORS 2529

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 75.1.10 (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)≤ 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

From this lemma, there is an interesting theorem on surjectivity.

Proposition 75.1.11 Let A : Rn→ Rn be continuous and coercive,

lim|x|→∞

(A(x+x0) ,x)|x|

= ∞

for some x0. Then for all f ∈ Rn, there exists x ∈ Rn such that Ax = f.

Proof: Define the closed convex sets Bn ≡ B(x0,n). By Browder’s lemma, there existsxn such that

(f−Axn,y−xn)≤ 0

for all y ∈ Bn. Then taking y = x0, it follows from the coercivity condition that the xn−x0are bounded. It follows that for large n, xn is an interior point of Bn. Therefore,

(f−Axn,z)≤ 0

for all z in some open ball centered at x0. Hence f = Axn.

Lemma 75.1.12 Let A : V →V ′ be type M and bounded and suppose V is reflexive or V isseparable. Then A is demicontinuous.

Proof: Suppose un → u and Aun fails to converge weakly to Au. Then there is a fur-ther subsequence, still denoted as un such that Aun ⇀ ζ ̸= Au. Then thanks to the strongconvergence, you have

lim supn→∞

⟨Aun,un⟩= ⟨ζ ,un⟩

which implies ζ = Au after all.With these lemmas and the above proposition, there is a very interesting surjectivity

result.