1384 CHAPTER 42. MAXIMAL MONOTONE OPERATORS, HILBERT SPACE

Theorem 42.3.10 Suppose φ is convex, l.s.c. (lower semicontinuous or in other wordshaving a closed epigraph), and proper. Then

y∗ ∈ δφ (x) if and only if x ∈ δφ∗ (y∗)

where this last expression means

(z∗− y∗)(x)≤ φ∗ (z∗)−φ

∗ (y∗)

for all z∗and in this case,y∗ (x) = φ

∗ (y∗)+φ (x).

Proof: If y∗ ∈ δφ (x) then y∗ (z− x)≤ φ (z)−φ (x) and so

y∗ (z)−φ (z)≤ y∗ (x)−φ (x)

for all z ∈ X . Therefore,

φ∗ (y∗)≤ y∗ (x)−φ (x)≤ φ

∗ (y∗).

Hencey∗ (x) = φ

∗ (y∗)+φ (x). (42.3.20)

Now if z∗ ∈ X ′ is arbitrary, 42.3.20 shows

(z∗− y∗)(x) = z∗ (x)− y∗ (x) = z∗ (x)−φ (x)−φ∗ (y∗)≤ φ

∗ (z∗)−φ∗ (y∗)

and this shows x ∈ δφ∗ (y∗).

Now suppose x ∈ δφ∗ (y∗). Then for z∗ ∈ X ′,

(z∗− y∗)(x)≤ φ∗ (z∗)−φ

∗ (y∗)

soz∗ (x)−φ

∗ (z∗)≤ y∗ (x)−φ∗ (y∗)

and so, taking sup over all z∗, and using Theorem 42.3.8,

φ∗∗ (x) = φ (x)≤ y∗ (x)−φ

∗ (y∗)≤ φ∗∗ (x) .

Thus

y∗ (x) = φ∗ (y∗)+φ

∗∗ (x) = φ∗ (y∗)+φ (x)≥

≤φ∗(y∗)︷ ︸︸ ︷y∗ (z)−φ (z)+φ (x)

for all z ∈ X and this implies for all z ∈ X ,

φ (z)−φ (x)≥ y∗ (z− x)

so y∗ ∈ δφ (x) and this proves the theorem.