25.1. SOME NONLINEAR SINGLE VALUED OPERATORS 829

As to the second claim. Consider the following diagram.

Rn θ∗← V ′

Rn θ→ V

where

θ (x) =n

∑i=1

xivi

Thus θ and θ∗ are both continuous linear and one to one and onto. Hence there is x∈ θ

−1Ka closed convex and bounded subset of Rn such that x = θ

−1u,u ∈ K, and(θ∗ f −θ

∗Aθ(θ−1u),θ−1y−θ

−1u)Rn ≡ ⟨ f −Au,y−u⟩V ′,V ≤ 0

for all y ∈ K.From this lemma, there is an interesting theorem on surjectivity.

Proposition 25.1.12 Let A : V →V ′ be continuous and coercive,

lim∥v∥→∞

⟨A(v+ v0) ,v⟩∥v∥V

= ∞

for some v0. Then for all f ∈V ′, there exists v ∈V such that Av = f .

Proof: Define the closed convex sets Bn ≡ B(v0,n). By Browder’s lemma, there existsxn such that

( f −Avn,y− vn)≤ 0

for all y ∈ Bn. Then taking y = v0,

⟨Avn,vn− v0⟩ ≤ ⟨ f ,vn− v0⟩

letting wn = vn− v0,

⟨A(wn + v0) ,wn⟩ ≤ ⟨ f ,wn⟩

and so⟨A(wn + v0) ,wn⟩

∥wn∥≤ ∥ f∥

which implies that the ∥wn∥ and hence the ∥vn∥ are bounded. It follows that for large n, vnis an interior point of Bn. Therefore,

⟨ f −Avn,z⟩V ′,V ≤ 0

for all z in some open ball centered at v0. Hence f −Avn = 0.

Lemma 25.1.13 Let A : V →V ′ be type M and bounded and suppose V is reflexive or V isseparable. Then A is demicontinuous.