25.7. MAXIMAL MONOTONE OPERATORS 899

As noted at the beginning, there is only one solution to this inclusion for a given x and it isa = Jλ x. This has shown that in terms of weak convergence,

Aλ (xn)→−λ−(p−1)b =−λ

−(p−1)F (a− x) =−λ−(p−1)F (Jλ x− x)≡ Aλ (x)

This has shown that Aλ is demicontinuous. Also it has shown that Jλ is also demicontinu-ous. (This result is a lot nicer in Hilbert space. )

3.) Why is ∥Aλ (x)∥ ≤ |Ax| whenever x ∈ D(A)?

Aλ (x) =−λ−(p−1)F (Jλ x− x)

where 0 ∈ F (Jλ x− x) + λp−1A(Jλ x) . Therefore, Aλ (x) ∈ A(Jλ x) . Then letting [u,v] ∈

G (A) ,0≤ ⟨v−Aλ (x) ,u− Jλ x⟩

In particular, if y ∈ Ax

0≤ ⟨y−Aλ (x) ,x− Jλ x⟩=⟨

y+λ−(p−1)F (Jλ x− x) ,x− Jλ x

⟩Hence

λ−(p−1) ∥Jλ x− x∥p ≤ ∥y∥∥Jλ x− x∥

and soλ−(p−1) ∥Jλ x− x∥p−1 = λ

−(p−1) ∥F (Jλ x− x)∥= ∥Aλ (x)∥ ≤ ∥y∥

and since y ∈ Ax is arbitrary, ∥Aλ (x)∥ ≤ |Ax| ≡ inf{∥y∥ : y ∈ Ax}.Next consider the claim that for all x ∈ conv(D(A)), it follows that

limλ→0

Jλ (x) = x.

Let [u,v] ∈ G (A) and x is arbitrary.

0≤ ⟨v−Aλ (x) ,u− Jλ x⟩=⟨

v+λ−(p−1)F (Jλ x− x) ,u− Jλ x

⟩=⟨

v+λ−(p−1)F (Jλ x− x) ,u− x

⟩+⟨

v+λ−(p−1)F (Jλ x− x) ,x− Jλ x

⟩Thus

∥Jλ x− x∥p ≤ λp−1 ⟨v,u− x⟩+ ⟨F (Jλ x− x) ,u− x⟩+λ

p−1 ⟨v,x− Jλ x⟩ (25.7.58)

for x arbitrary and u anything in D(A) . This implies that the above inequality 25.7.58 holdsfor any u ∈ conv(D(A)). Say u = xn ∈ conv(D(A)) where xn→ x. Then

∥Jλ x− x∥p ≤ λp−1 ⟨v,xn− x⟩+ ⟨F (Jλ x− x) ,xn− x⟩+λ

p−1 ⟨v,x− Jλ x⟩

≤ λp−1 ∥v∥∥xn− x∥+∥Jλ x− x∥p−1 ∥xn− x∥+λ

p−1 ∥v∥∥Jλ x− x∥

You have something like this: yλ = ∥Jλ x− x∥ ,an = ∥xn− x∥ ,

ypλ≤ λ

p−1 ∥v∥an + yp−1λ

an +λp−1 ∥v∥yλ , yλ ≥ 0