912 CHAPTER 25. NONLINEAR OPERATORS

t ⟨y∗,z− x0⟩ ≤ t (φ (z)−φ (x0))+ t ⟨F (x0 + t (z− x0)) ,z− x0⟩

Now cancel the t on both sides to obtain

⟨y∗,z− x0⟩ ≤ (φ (z)−φ (x0))+ ⟨F (x0 + t (z− x0)) ,z− x0⟩

By the fact that F is hemicontinuous, actually demicontinuous, one can let t ↓ 0 and obtain

⟨y∗,z− x0⟩ ≤ (φ (z)−φ (x0))+ ⟨F (x0) ,z− x0⟩

This says that y∗−F (x0) ∈ ∂φ (x0) from the definition of what ∂φ (x0) means.There is a much harder approach to this theorem which is based on a theorem about

when the subgradient of a sum equals the sum of the subgradients. This major theorem isgiven next. Much of the above is in [13] but I don’t remember where I found the followingproof.

Theorem 25.7.51 Let φ 1 and φ 2 be convex, l.s.c. and proper having values in (−∞,∞].Then

∂ (λφ i)(x) = λ∂φ i (x) , ∂ (φ 1 +φ 2)(x)⊇ ∂φ 1 (x)+∂φ 2 (x) (25.7.71)

if λ > 0. If there exists x ∈ dom(φ 1)∩ dom(φ 2) and φ 1 is continuous at x then for allx ∈ X,

∂ (φ 1 +φ 2)(x) = ∂φ 1 (x)+∂φ 2 (x). (25.7.72)

Proof: 25.7.71 is obvious so we only need to show 25.7.72. Suppose x is as described.It is clear 25.7.72 holds whenever x /∈ dom(φ 1)∩ dom(φ 2) since then ∂ (φ 1 +φ 2) = /0.Therefore, assume

x ∈ dom(φ 1)∩dom(φ 2)

in what follows. Let x∗ ∈ ∂ (φ 1 +φ 2)(x). Is x∗ is the sum of an element of ∂φ 1 (x) and∂φ 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. Say y1,y2 ∈C1 and t ∈ [0,1] . Then

φ 1 ((ty1)+(1− t)y2)−φ 1 (x)− x∗ (((ty1)+(1− t)y2)− x)