1398 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

=λ

2

(|Aλ y|2−|Aλ x|2

)+φ (Jλ y)−φ (Jλ x)

≥ λ

2

(|Aλ y|2−|Aλ x|2

)+(Aλ x,Jλ y− Jλ x)

=λ

2

(|Aλ y|2−|Aλ x|2

)+(Aλ x,y−λAλ y− (x−λAλ x))

=λ

2

(|Aλ y|2−|Aλ x|2

)+(Aλ x,y− x)+λ (Aλ x,Aλ x−Aλ y)

≥ λ

2

(|Aλ y|2−|Aλ x|2

)+λ |Aλ x|2− λ

2|Aλ x|2− λ

2|Aλ y|2 +(Aλ x,y− x)

= (Aλ x,y− x) = (Aλ x−Aλ y,y− x)+(Aλ y,y− x) (42.5.36)

Then it follows that

−(Aλ x−Aλ y,y− x)≥ φ λ (x)−φ λ (y)− (Aλ y,x− y)

However, Aλ is Lipschitz continuous with constant 1/λ and so

1λ|x− y|2 ≥ φ λ (x)−φ λ (y)− (Aλ y,x− y) (42.5.37)

Then switching x,y in the equation 42.5.37,

1λ|x− y|2 ≥ φ λ (y)−φ λ (x)− (Aλ x,y− x) (42.5.38)

But also that term on the end in 42.5.37 equals (Aλ y,y− x)≥ (Aλ x,y− x) and so it is alsothe case that

1λ|x− y|2 ≥ φ λ (x)−φ λ (y)+(Aλ x,y− x)

= −(φ λ (y)−φ λ (x)− (Aλ x,y− x)) (42.5.39)

From 42.5.38 and 42.5.39 it follows that

1λ|x− y|2 ≥ |φ λ (y)−φ λ (x)− (Aλ x,y− x)|

which shows that Dφ λ (x) = Aλ x. This proves the differentiability part.Next recall that for any maximal monotone operatior A, if you have x ∈ D(A), then

limλ→0

Jλ x = x

Recall why this was so. If x ∈ D(A) , then

x− Jλ x ∈ λAx