25.3. PENALIZATON AND PROJECTION OPERATORS 841
25.3 Penalizaton And Projection OperatorsIn this section, X will be a reflexive Banach space such that X , X ′ has a strictly convexnorm. Let K be a closed convex set in X . Then the following lemma is obtained.
Lemma 25.3.1 Let K be closed and convex nonempty subset of X a reflexive Banach spacewhich has strictly convex norm. Then there exists a projection map P such that Px ∈ K andfor all y ∈ K,
∥y− x∥ ≥ ∥x−Px∥
Proof: Let {yn} be a minimizing sequence for y→∥y− x∥ for y ∈ K. Thus
d ≡ inf{∥y− x∥ : y ∈ K}= limn→∞∥yn− x∥
Then obviously {yn} is bounded. Hence there is a subsequence, still denoted by n such thatyn→ w ∈ K. Then
∥w− x∥ ≤ lim infn→∞∥yn− x∥= d
How many closest points to x are there? Suppose w1 is another one. Then∥∥∥∥w1 +w2− x∥∥∥∥= ∥∥∥∥w1− x+w− x
2
∥∥∥∥< ∥∥∥∥w1− x2
∥∥∥∥+∥∥∥∥w− x2
∥∥∥∥= d
contradicting the assumption that both w,w1 are closest points to x. Therefore, Px consistsof a single point.
Denote by F the duality map such that ⟨Fx,x⟩ = ∥x∥2. This is described earlier butthere is also a very nice treatment which is somewhat different in [13]. Everything can begeneralized and is in [91] but here I will only consider this case. First here is a useful result.
Proposition 25.3.2 Let F be the duality map just described. Let φ (x)≡ ∥x∥2
2 . Then F (x) =∂φ (x) .
Proof: This follows from
⟨Fx,y− x⟩ ≤ ⟨Fx,y⟩−⟨Fx,x⟩ ≤ ⟨Fx,x⟩1/2 ⟨Fy,y⟩1/2−⟨Fx,x⟩
≤ ⟨Fy,y⟩2− ⟨Fx,x⟩
2=∥y∥2
2− ∥x∥
2
2.
Next is a really nice result about the characterization of Px in terms of F .
Proposition 25.3.3 Let K be a nonempty closed convex set in X a reflexive Banach spacein which both X ,X ′ have strictly convex norms. Then w ∈ K is equal to Px if and only if
⟨F (x−w) ,y−w⟩ ≤ 0
for every y ∈ K.