25.7. MAXIMAL MONOTONE OPERATORS 913

≤ tφ (y1)+(1− t)φ (y2)− (tφ 1 (x)+(1− t)φ (x))

−(tx∗ (y1− x)+(1− t)x∗ (y2− x))

≤ ta+(1− t)a = a

so C1 is indeed convex. The case of C2 is similar.C1 is nonempty because it contains (x,φ 1 (x)−φ 1 (x)− x∗ (x− x)) since

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

C2 is also nonempty because it contains (x,φ 2 (x)−φ 2 (x)) since

φ 2 (x)−φ 2 (x)≤ φ 2 (x)−φ 2 (x)

In addition to this,

(x,φ 1 (x)− x∗ (x− x)−φ 1 (x)+1) ∈ int(C1)

due to the assumed continuity of φ 1 at x and so int(C1) ̸= /0. If (y,a) ∈ int(C1) then

φ 1 (y)− x∗ (y− x)−φ 1 (x)≤ a− ε

whenever ε is small enough. Therefore, if (y,a) is also in C2, the assumption that x∗ ∈∂ (φ 1 +φ 2)(x) implies

a− ε ≥ φ 1 (y)− x∗ (y− x)−φ 1 (x)≥ φ 2 (x)−φ 2 (y)≥ a,

a contradiction. Therefore int(C1) ∩C2 = /0 and so by Theorem 18.2.14, there exists(w∗,β ) ∈ X ′×R with

(w∗,β ) ̸= (0,0) , (25.7.73)

andw∗ (y)+βa≥ w∗ (y1)+βa1, (25.7.74)

whenever (y,a) ∈C1 and (y1,a1) ∈C2.Claim: β > 0.Proof of claim: If β < 0 let

a = φ 1 (x)− x∗ (x− x)−φ 1 (x)+1,

a1 = φ 2 (x)−φ 2 (x) , and y = y1 = x.

Then from 25.7.74

β (φ 1 (x)− x∗ (x− x)−φ 1 (x)+1)≥ β (φ 2 (x)−φ 2 (x)) .

Dividing by β yields

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