842 CHAPTER 25. NONLINEAR OPERATORS

Proof: First suppose the condition. Then for y ∈ K, it follows from the above proposi-tion about the subgradient,

12∥x− y∥2− 1

2∥x−w∥2 ≥ ⟨F (x−w) ,w− y⟩ ≥ 0

and so since this holds for all y it follows that

∥x− y∥ ≥ ∥x−w∥

for all y which says that w = Px.Next, using the subgradient idea again, for θ ∈ [0,1] , suppose w = Px then for y ∈ K

arbitrary,

0≥ 12∥x−w∥2− 1

2∥x− (w+θ (y−w))∥2 ≥ ⟨F (x− (w+θ (y−w))) ,θ (y− x)⟩

Now divide by θ and let θ ↓ 0 and use the hemicontinuity of F given above. Then

0≥ ⟨F (x−w) ,y− x⟩

Definition 25.3.4 An operator of penalization is an operator f : X → X ′ such that f = 0on K, f is monotone and nonzero off K as well as demicontinuous. (Strong convergencegoes to weak convergence.) Actually, in applications, it is usually easy to give an ad hocdescription of an appropriate penalization operator.

Proposition 25.3.5 Let K be a closed convex nonempty subset of X a reflexive Banachspace such that X ,X ′ have strictly convex norms. Then

f (x)≡ F (x−Px)

is an operator of penalization. Here P is the projection onto K. This operator of penaliza-tion is demicontinuous.

Proof: First, observe that f (x) is 0 on K and nonzero off K. Why is it monotone?

⟨F (x−Px)−F (x1−Px1) ,x− x1⟩

= ⟨F (x−Px)−F (x1−Px1) ,x−Px− (x1−Px1)⟩+⟨F (x−Px)−F (x1−Px1) ,Px−Px1⟩

The first term is ≥ 0 because F is monotone. As to the second, it equals

⟨F (x−Px) ,Px−Px1⟩+ ⟨F (x1−Px1) ,Px1−Px⟩

and both of these are ≥ 0 because of Proposition 25.3.3 which characterizes the projectionmap.