25.7. MAXIMAL MONOTONE OPERATORS 911

Is it the case that F+∂φ is onto? First of all, is ∂φ monotone? Let x∗ ∈ ∂φ (x) ,y∗ ∈ ∂φ (y) .Then

φ (y)−φ (x) ≥ ⟨x∗,y− x⟩φ (x)−φ (y) ≥ ⟨y∗,x− y⟩

Hence adding these yields

⟨y∗− x∗,x− y⟩ ≤ 0, ⟨y∗− x∗,y− x⟩ ≥ 0.

Yes, ∂φ is certainly monotone. Is it maximal monotone?

Theorem 25.7.50 Let φ be convex, proper, and lower semicontinuous on X where X is areflexive Banach space having strictly convex norm. Then ∂φ is maximal monotone.

Proof: It is necessary to show that F +∂φ is onto. To do this, let

ψ (x)≡ 12∥x∥2 +φ (x)−⟨y∗,x⟩

where y∗ is a given element of X ′ and the idea is to show that y∗ ∈ F (x)+∂φ (x) for somex. Then by separation theorems, φ (x) ≥ b+ ⟨z∗,x⟩ for some b,z∗. Hence it is clear that ψ

is convex, lower semicontinuous and coercive in the sense that

lim∥x∥→∞

ψ (x) = ∞

It follows that any minimizing sequence for ψ is bounded. Hence by the weak lowersemicontinuity, this function has a minimum at x0 say. Thus

12∥x0∥2 +φ (x0)−⟨y∗,x0⟩ ≤

12∥x∥2 +φ (x)−⟨y∗,x⟩

for all x. Then12∥x0∥2− 1

2∥x∥2 + ⟨y∗,x− x0⟩ ≤ φ (x)−φ (x0)

Now from Theorem 25.7.48,

⟨F (x) ,x0− x⟩ ≤ 12∥x0∥2− 1

2∥x∥2

and so, the above reduces to

⟨F (x) ,x0− x⟩+ ⟨y∗,x− x0⟩ ≤ φ (x)−φ (x0)

Next let x = x0 + t (z− x0) , t ∈ (0,1) , where z is arbitary. Then

−t ⟨F (x0 + t (z− x0)) ,z− x0⟩+ t ⟨y∗,z− x0⟩ ≤ φ (x0 + t (z− x0))−φ (x0)

and so, by convexity,

−t ⟨F (x0 + t (z− x0)) ,z− x0⟩+ t ⟨y∗,z− x0⟩ ≤ (1− t)φ (x0)+ tφ (z)−φ (x0)