1382 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Lemma 42.3.7 Let φ : X → (−∞,∞] be convex and lower semicontinuous and φ (x) < ∞

for some x. (proper). Then if β < φ (x0) so that (x0,β ) is not in epi(φ) , it follows thatthere exists δ > 0 and z∗ ∈ X ′ such that for all y,

z∗ (y− x0)+β +δ < φ (y) , all y ∈ X

Proof: Let C = epi(φ)∩ (X×R). Then C is a closed convex nonempty set and it doesnot contain the point (x0,β ). Let β̂ > β be slightly larger so that also

(x0, β̂

)/∈C. Thus

there exists y∗ ∈ X ′ and α ∈ R such that for some ĉ, and all y ∈ X ,

y∗ (x0)+αβ̂ > ĉ > y∗ (y)+αφ (y)

for all y ∈ X . Now you can’t have α ≥ 0 because

α

(β̂ −φ (y)

)> y∗ (y− x0)

and you can let y = x0 to have

α

<0︷ ︸︸ ︷

β̂ −φ (x0)

> 0

Hence α < 0 and so, dividing by it yields that for all y ∈ X ,

x∗ (x0)+ β̂ < c < x∗ (y)+φ (y)

where x∗ = y∗/α, ĉ/α ≡ c. Then

(−x∗)(y− x0)+β +(

β̂ −β

)< c− x∗ (y)< φ (y)

(−x∗)(y− x0)+β +δ < φ (y) , δ ≡ β̂ −β

Let z∗ =−x∗.

Theorem 42.3.8 φ∗∗ (x) ≤ φ (x) for all x and if φ is convex and l.s.c., φ

∗∗ (x) = φ (x) forall x ∈ X.

Proof:

φ∗∗ (x)≡ sup

x∗ (x)−

φ∗(x∗)︷ ︸︸ ︷sup{x∗ (y)−φ (y) : y ∈ X} : x∗ ∈ X ′

≤ sup{x∗ (x)− (x∗ (x)−φ (x))}= φ (x).

Next suppose φ is convex and l.s.c. If φ∗∗ (x0) < φ (x0), then using Lemma 42.3.7,

there exists x∗0,δ > 0 such that for all y ∈ X ,

(x∗0)(y− x0)+φ∗∗ (x0)+δ < φ (y)