914 CHAPTER 25. NONLINEAR OPERATORS

and soφ 1 (x)+φ 2 (x)− (φ 1 (x)+φ 2 (x))+1≤ x∗ (x− x)

≤ φ 1 (x)+φ 2 (x)− (φ 1 (x)+φ 2 (x)),

a contradiction. Therefore, β ≥ 0.Now suppose β = 0. Letting

a = φ 1 (x)− x∗ (x− x)−φ 1 (x)+1,

(x,a) ∈ int(C1) ,

and so there exists an open set U containing 0 and η > 0 such that

x+U× (a−η ,a+η)⊆C1.

Therefore, 25.7.74 applied to (x+ z,a) ∈C1 and (x,φ 2 (x)−φ 2 (x)) ∈C2 for z ∈U yields

w∗ (x+ z)≥ w∗ (x)

for all z ∈U . Hence w∗ (z) = 0 on U which implies w∗ = 0, contradicting 25.7.73. Thisproves the claim.

Now with the claim, it follows β > 0 and so, letting z∗ = w∗/β , 25.7.74 and Lemma18.2.15 implies

z∗ (y)+a≥ z∗ (y1)+a1 (25.7.75)

whenever (y,a) ∈C1 and (y1,a1) ∈C2. In particular,

(y,φ 1 (y)−φ 1 (x)− x∗ (y− x)) ∈C1 (25.7.76)

becauseφ 1 (y)−φ 1 (x)− x∗ (y− x)≤ φ 1 (y)− x∗ (y− x)−φ 1 (x)

and(y1,φ 2 (x)−φ 2 (y1)) ∈C2. (25.7.77)

by similar reasoning so letting y = x,

z∗ (x)+

 =0︷ ︸︸ ︷φ 1 (x)− x∗ (x− x)−φ 1 (x)

≥ z∗ (y1)+φ 2 (x)−φ 2 (y1).

Therefore,z∗ (y1− x)≤ φ 2 (y1)−φ 2 (x)

for all y1 and so z∗ ∈ ∂φ 2 (x). Now let y1 = x in 25.7.77 and using 25.7.75 and 25.7.76, itfollows

z∗ (y)+φ 1 (y)− x∗ (y− x)−φ 1 (x)≥ z∗ (x)

φ 1 (y)−φ 1 (x)≥ x∗ (y− x)− z∗ (y− x)

and so x∗− z∗ ∈ ∂φ 1 (x) so x∗ = z∗+(x∗− z∗) ∈ ∂φ 2 (x)+∂φ 1 (x) .