25.7. MAXIMAL MONOTONE OPERATORS 909

Definition 25.7.47 Let φ : X → (−∞,∞] be convex lower semicontinuous, and proper.Then

∂φ (x)≡ {x∗ : φ (y)−φ (x)≥ ⟨x∗,y− x⟩ for all y}

The domain of ∂φ , denoted as D(∂φ) is just the set of all x for which ∂φ (x) ̸= /0. Note thatD(∂φ)⊆ D(φ) since if x /∈ D(φ) , the defining inequality could not hold for all y becausethe left side would be −∞ for some y.

Theorem 25.7.48 For X a real Banach space, let φ (x) ≡ 12 ||x||

2. Then F (x) = ∂φ (x).Here F was the set valued map satisfying x∗ ∈ Fx means

∥x∗∥= ∥Fx∥ ,⟨Fx,x⟩= ∥x∥2 .

Proof: Let x∗ ∈ F (x). Then

⟨x∗,y− x⟩ = ⟨x∗,y⟩−⟨x∗,x⟩

≤ ||x|| ||y||− ||x||2 ≤ 12||y||2− 1

2||x||2.

This shows F (x)⊆ ∂φ (x).Now let x∗ ∈ ∂φ (x). Then for all t ∈ R,

⟨x∗, ty⟩= ⟨x∗,(ty+ x)− x⟩ ≤ 12

(||x+ ty||2−||x||2

). (25.7.66)

Now if t > 0, divide both sides by t. This yields

⟨x∗,y⟩ ≤ 12t

((∥x∥+ t ∥y∥)2−∥x∥2

)=

12t

(2t ||x|| ||y||+ t2 ||y||2

)Letting t→ 0,

⟨x∗,y⟩ ≤ ||x|| ||y|| . (25.7.67)

Next suppose t =−s, where s > 0 in 25.7.66. Then, since when you divide by a negative,you reverse the inequality, for s > 0

⟨x∗,y⟩ ≥ 12s

[||x||2−||x− sy||2

]≥

12s

[||x− sy||2−2 ||x− sy|| ||sy|| + ||sy||2−||x− sy||

]2. (25.7.68)

=12s

[−2 ||x− sy|| ||sy||+ ||sy||2

](25.7.69)

Taking a limit as s→ 0 yields⟨x∗,y⟩ ≥ −||x|| ||y||. (25.7.70)

It follows from 25.7.70 and 25.7.67 that

|⟨x∗,y⟩| ≤ ||x|| ||y||