918 CHAPTER 25. NONLINEAR OPERATORS

Using the fact that if ψ (x) = ∥x∥2 , then ∂ψ (x) = Fx, and that Aλ x ∈ ∂φ (Jλ x) ,

≥ λ−1 ⟨F (x− Jλ (x)) ,(y− Jλ y)− (x− Jλ x)⟩+ ⟨Aλ x,Jλ (y)− Jλ (x)⟩

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

Hence(φ λ (y)−φ λ (x))−⟨Aλ x,y− x⟩ ≥ 0

Also from 25.7.81

12λ∥y− Jλ y∥2− 1

2λ∥x− Jλ x∥2 =−

(1

2λ∥x− Jλ x∥2− 1

2λ∥y− Jλ y∥2

)

≤− 1λ⟨F (y− Jλ y) ,(x− Jλ x)− (y− Jλ y)⟩= ⟨Aλ y,(y− Jλ y)− (x− Jλ x)⟩

Similarly, from 25.7.81,

φ (Jλ (y))−φ (Jλ (x)) =−(φ (Jλ (x))−φ (Jλ (y)))

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

It follows that

⟨Aλ (y) ,Jλ (y)− Jλ (x)⟩+ ⟨Aλ y,(y− Jλ y)− (x− Jλ x)⟩≥ (φ λ (y)−φ λ (x))≥ ⟨Aλ x,y− x⟩

and so⟨Aλ (y) ,y− x⟩ ≥ (φ λ (y)−φ λ (x))≥ ⟨Aλ x,y− x⟩

Therefore,

⟨Aλ (y)−Aλ (x) ,y− x⟩ ≥ (φ λ (y)−φ λ (x))−⟨Aλ x,y− x⟩ ≥ 0

Next let y = x+ tz for t > 0. Then

t ⟨Aλ (x+ tz)−Aλ (x) ,z⟩ ≥ (φ λ (x+ tz)−φ λ (x))− t ⟨Aλ x,z⟩ ≥ 0

Using the demicontinuity of Aλ , you can divide by t and pass to a limit to obtain

limt↓0

φ λ (x+ tz)−φ λ (x)t

= ⟨Aλ x,z⟩ .

A much better theorem is available in case X = X ′ = H a Hilbert space. In this case φ λ

is also Frechet differentiable. See Theorem 42.5.1 which is presented later. Everything ismuch nicer in the Hilbert space setting because F is just replaced with the identity and theapproximations are defined more easily.

0 ∈ Jλ x− x+λAJλ x,

x ∈ Jλ x+λAJλ x = (I +λA)Jλ x

Jλ x = (I +λA)−1 x