2532 CHAPTER 75. SOME NONLINEAR OPERATORS

and so for arbitrary y ̸= 0,

|⟨Fx,y⟩| = ||y||∣∣∣∣⟨Fx,

y||y||

⟩∣∣∣∣≤ ||y|| ||x||p/p′

= |⟨Fy,y⟩|1/p |⟨Fx,x⟩|1/p′

Next we can show that F is monotone.

⟨Fx−Fy,x− y⟩ = ⟨Fx,x⟩−⟨Fx,y⟩−⟨Fy,x⟩+ ⟨Fy,y⟩≥ ||x||p + ||y||p−||y|| ||x||p/p′ −||y||p/p′ ||x||

≥ ||x||p + ||y||p−(||y||p

p+||x||p

p′

)−(||y||p

p′+||x||p

p

)= 0

Next it can be shown that F is hemicontinuous. By the construction, F (x+ ty) isbounded as t→ 0. Let t→ 0 be a subsequence such that

F (x+ ty)→ ξ weak ∗

Then we ask: Does ξ do what it needs to do in order to be F (x)? The answer is yes. Firstof all ||F (x+ ty)||= ||x+ ty||p−1→ ||x||p−1 . The set{

x∗ : ||x∗|| ≤ ||x||p−1 + ε

}is closed and convex and so it is weak ∗ closed as well. For all small enough t, it followsF (x+ ty) is in this set. Therefore, the weak limit is also in this set and it follows ||ξ || ≤||x||p−1 + ε. Since ε is arbitrary, it follows ||ξ || ≤ ||x||p−1 . Is ξ (x) = ||x||p? We have

||x||p = limt→0||x+ ty||p = lim

t→0⟨F (x+ ty) ,x+ ty⟩

= limt→0⟨F (x+ ty) ,x⟩= ⟨ξ ,x⟩

and so, ξ does what it needs to do to be F (x). This would be clear if ||ξ || = ||x||p−1 .

However, |⟨ξ ,x⟩|= ||x||p and so ||ξ || ≥∣∣∣⟨ξ , x

||x||

⟩∣∣∣= ||x||p−1 . Thus ||ξ ||= ||x||p−1 which

shows ξ does everyting it needs to do to equal F (x) and so it is F (x) . Since this conclusionfollows for any convergent sequence, it follows that F (x+ ty) converges to F (x) weaklyas t → 0. This is what it means to be hemicontinuous. This proves the following theorem.One can show also that F is demicontinuous which means strongly convergent sequencesgo to weakly convergent sequences. Here is a proof for the case where p = 2. You canclearly do the same thing for arbitrary p.

Lemma 75.2.2 Let F be a duality map for p = 2 where X ,X ′ are reflexive and have strictlyconvex norms. (If X is reflexive, there is always an equivalent strictly convex norm [8].)Then F is demicontinuous.