1388 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Proof: 42.3.22 is obvious so we only need to show 42.3.23. Suppose x is as described.It is clear 42.3.23 holds whenever x /∈ dom(φ 1)∩ dom(φ 2) since then both sides equal /0.Therefore, assume

x ∈ dom(φ 1)∩dom(φ 2)

in what follows. Let x∗ ∈ δ (φ 1 +φ 2)(x). Is x∗ is the sum of an element of δφ 1 (x) andδφ 2 (x)? Does there exist x∗1 and x∗2 such that for every y,

x∗ (y− x) = x∗1 (y− x)+ x∗2 (y− x)

≤ φ 1 (y)−φ 1 (x)+φ 2 (y)−φ 2 (x)?

If so, thenφ 1 (y)−φ 1 (x)− x∗ (y− x)≥ φ 2 (x)−φ 2 (y) .

DefineC1 ≡ {(y,a) ∈ X×R : φ 1 (y)−φ 1 (x)− x∗ (y− x)≤ a},

C2 ≡ {(y,a) ∈ X×R : a≤ φ 2 (x)−φ 2 (y)}.

I will show int(C1)∩C2 = /0 and then by Theorem 18.2.14 there exists an element of X ′

which does something interesting.Both C1 and C2 are convex and nonempty. C1 is nonempty because it contains

(x,φ 1 (x)−φ 1 (x)− x∗ (x− x))

sinceφ 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∗ ∈δ (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) , (42.3.24)

andw∗ (y)+βa≥ w∗ (y1)+βa1, (42.3.25)