2530 CHAPTER 75. SOME NONLINEAR OPERATORS

Theorem 75.1.13 Let A : V →V ′ be type M, bounded, and coercive

lim∥u∥→∞

⟨A(u+u0) ,u⟩∥u∥

= ∞, (75.1.4)

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 . Say span(v1, · · · ,vn) =Vn. Then consider the follow-ing diagram.

Rn θ∗← V ′n

i∗← V ′

Rn θ→ Vni→ V

Here the map θ is the one which does the following.

θ (x) =n

∑i=1

xivi.

The map i is the inclusion map. Consider the map θ∗i∗Aiθ . By Lemma 75.1.12 this map

is continuous. The map θ is continuous, one to one, and onto. Thus its inverse is alsocontinuous. Let x0 correspond to u0. Then for some constant C,

(θ ∗i∗Aiθ (x+x0) ,x)|x|

≥ ⟨Aiθ (x+x0) , iθx⟩C∥iθx∥V

and to say |x| →∞ is the same as saying that ∥iθx∥V →∞. Hence θ∗i∗Aiθ is coercive. Let

f ∈V ′. Then from 75.1.11, there exists xn such that

θ∗i∗Aiθxn = θ

∗i∗f

Thus, i∗Aiθxn = i∗ f and this implies that for vn ≡ θxn,

i∗Aivn = i∗ f

In other words,⟨Avn,y⟩= ⟨ f ,y⟩ (75.1.5)

for all y ∈ Vn. Then from the coercivity condition 75.1.4, the vn are bounded independentof n. Since V is reflexive, there is a subsequence, still called {vn} which converges weaklyto 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 75.1.5,

⟨ζ ,y⟩= ⟨ f ,y⟩

for any y ∈Vn, this for any n. Since the union of these Vn is dense, it follows that the aboveequation holds for all y ∈V. Therefore, f = ζ and so

lim supn→∞

⟨Avn,vn⟩= lim supn→∞

⟨ f ,vn⟩= ⟨ f ,v⟩= ⟨ζ ,v⟩

Since A is type M,Av = ζ = f