42.3. SUBGRADIENTS 1381

This function, φ∗, defined above, is called the conjugate function of φ or the polar of

φ . Note φ∗ (x∗) ̸=−∞ because φ (y)< ∞ for some y.

Theorem 42.3.4 Let X be a real Banach space. Then φ∗ is convex and l.s.c.

Proof: Let λ ∈ [0,1]. Then

φ∗ (λx∗+(1−λ )y∗) = sup{(λx∗+(1−λ )y∗)(y)−φ (y) : y ∈ X}

sup{λ (x∗ (y)−φ (y))+(1−λ )(y∗ (y)−φ (y)) : y ∈ X}

≤ λφ∗ (x∗)+(1−λ )φ

∗ (y∗).

It remains to show the function is l.s.c. Consider fy (x∗)≡ x∗ (y)−φ (y). Then fy is obvi-ously convex. Also to say that (x,α) ∈ epi(φ ∗) is to say that α ≥ x∗ (y)− φ (y) for all y.Thus

epi(φ ∗) = ∩y∈X epi( fy).

Therefore, if epi( fy) is closed, this will prove the theorem. If (x∗,a) /∈ epi( fy), then a <x∗ (y)−φ (y) and, by continuity, for b close enough to a and y∗ close enough to x∗ then

b < y∗ (y)−φ (y) , (y∗,b) /∈ epi( fy)

Thus epi( fy) is closed.Note this theorem holds with no change in the proof if X is only a locally convex

topological vector space and X ′ is given the weak ∗ topology.

Definition 42.3.5 We define φ∗∗ on X by

φ∗∗ (x)≡ sup

{x∗ (x)−φ

∗ (x∗) ,x∗ ∈ X ′}.

The following lemma comes from separation theorems. First is a simple observation.

Observation 42.3.6 f ∈ (X×R)′ if and only if there exists x∗ ∈ X ′ and α ∈ R such thatf (x,λ ) = x∗ (x) + λα . To get x∗, you can simply define x∗ (x) ≡ f (x,0) and to get α

you just let αλ ≡ f (0,λ ) . Why does such an α exist? You know that f (0,aλ +bδ ) =a f (0,α)+ b f (0,δ ) and so in fact λ → f (0,λ ) satisfies the Cauchy functional equationg(x+ y) = g(x)+ g(y) and is continuous so there is only one thing it can be and that isf (0,λ ) = αλ for some α .

This picture illustrates the conclusion of the following lemma.

epi(φ)

(x0,β )

β + ⟨z∗,y− x0⟩+δ < φ(y)