25.7. MAXIMAL MONOTONE OPERATORS 897

Proof: 1.) It is clear that these are single valued operators. What about the asser-tion that they are bounded? Let y∗ ∈ Axλ such that the inclusion defining xλ becomes anequality. Thus

F (xλ − x)+λp−1y∗ = 0

Then let x0 ∈ D(A) be given.

⟨F (xλ − x) ,xλ − x⟩+λp−1 ⟨y∗,xλ − x0⟩+λ

p−1 ⟨y∗,x0− x⟩= 0

Then by monotonicity of A,

∥xλ − x∥p +λp−1 ⟨y∗0,xλ − x0⟩+λ

p−1 ⟨y∗,x0− x⟩ ≤ 0

It follows that

∥xλ − x∥p ≤ λp−1 ∥y∗0∥∥xλ − x0∥+λ

p−1 ∥y∗∥∥x0− x∥

Hence if x is in a bounded set, it follows the resulting xλ = Jλ x remain in a bounded set.Now from the definition of Aλ , it follows that this is also a bounded operator.

Why is Aλ monotone?

0 ≤ ⟨Aλ x−Aλ y,x− y⟩= ⟨Aλ x−Aλ y,x− Jλ x− (y− Jλ y)⟩+⟨Aλ x−Aλ y,Jλ x− Jλ y⟩

=⟨

λ−(p−1)F (Jλ x− x)−λ

−(p−1)F (Jλ y− y) ,Jλ x− x− (Jλ y− y)⟩

+⟨Aλ x−Aλ y,Jλ x− Jλ y⟩

and both terms are nonnegative, the first because F is monotone so indeed Aλ is monotone.2.) What of the demicontinuity of Aλ ? This one is really tricky. Suppose xn→ x. Does

it follow that Aλ xn→ Aλ x weakly? The proof will be based on a pair of equations. Theseare

limm,n→∞

⟨F (Jλ xn− xn)−F (Jλ xm− xm) ,Jλ xn− xn− (Jλ xm− xm)⟩= 0

andlim

m,n→∞⟨Aλ (xn)−Aλ (xm) ,Jλ xn− Jλ xm⟩= 0

When these have been established, Lemma 25.7.34 is used to get the desired result fora subsequence. It will be shown that every sequence has a subsequence which gives theright sort of weak convergence and from this the desired weak convergence of Aλ xn to Aλ xfollows.

0 ∈ F (Jλ xn− xn)+λp−1A(Jλ xn)

0 ∈ F (Jλ x− x)+λp−1A(Jλ x)

−λ−(p−1)F (Jλ x− x) ≡ Aλ (x) ∈ A(Jλ x)

−λ−(p−1)F (Jλ xn− xn) ≡ Aλ (xn) ∈ A(Jλ xn)