830 CHAPTER 25. NONLINEAR OPERATORS

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 ,u⟩

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

result.

Theorem 25.1.14 Let A : V →V ′ be type M, bounded, and coercive

lim∥u∥→∞

⟨A(u+u0) ,u⟩∥u∥

= ∞, (25.1.4)

for some u0, where V is a separable reflexive Banach space. Then A is surjective.

Proof: Since V is separable, there exists an increasing sequence of finite dimensionalsubspaces {Vn} such that ∪nVn = V and each Vn contains u0. Say span(v1, · · · ,vn) = Vn.Then consider the following diagram.

V ′ni∗← V ′

Vni→ V

The map i is the inclusion map. Consider the map i∗Ai. By Lemma 25.1.13 this map iscontinuous.

⟨i∗Ai(v+u0) ,v⟩V ′nVn

∥v∥=⟨A(v+u0) ,v⟩V ′,V

∥v∥

Hence i∗Ai is coercive. Let f ∈V ′. Then from Proposition 25.1.12, there exists xn such that

i∗Aivn = i∗ f

In other words,⟨Avn,y⟩V ′V = ⟨ f ,y⟩V ′V (25.1.5)

for all y ∈Vn. Letting y≡ vn−u0 ≡ wn,

⟨A(wn +u0) ,wn⟩= ⟨ f ,wn⟩

Then from the coercivity condition 25.1.4, the wn are bounded independent of n. Hence thisis also true of the vn. Since V is reflexive, there is a subsequence, still called {vn} whichconverges weakly to v ∈V. Since A is bounded, it can also be assumed that Avn ⇀ ζ ∈V ′.Then

lim supn→∞

⟨Avn,vn⟩= lim supn→∞

⟨ f ,vn⟩= ⟨ f ,v⟩

Also, passing to the limit in 25.1.5,

⟨ζ ,y⟩= ⟨ f ,y⟩