Nonlinear Equations - UFRJ

74 [CH. 6: EXPONENTIAL SUMS AND SPARSE POLYNOMIAL SYSTEMS
Proposition 6.6. Let f : E → R be given. Then,
1. f ∗ is convex. In part. U ∗ is convex.
2. For all x ∈ U, α ∈ U ∗ ,
f ∗ (α) + f(x) ≥ α(x).
3. If furthermore f is convex then f ∗∗
|U ≡ f.
Proof. Let (α 0 , β 0 ), (α 1 , β 1 ) ∈ Epi f ∗ This means that β i ≥ f ∗ (α i ),
i = 1, 2 so
β i ≥ α i (x) − f(x) ∀x ∈ U.
Hence, if t ∈ [0, 1],
(1 − t)β 0 + tβ 1 ≥ ((1 − t)α 0 + tα 1 )(x) − f(x) ∀x ∈ U
and ((1 − t)α 0 + tα 1 , (1 − t)β 0 + tβ 1 ) ∈ Epi f ∗.
Item 2 follows directly from the definition.
Let x ∈ U. By Lemma 6.3, there is a separating hyperplane
between (x, f(x)) and the interior of Epi f . Namely, there are α, β so
that for all y ∈ U, for all z with z > f(y),
α(y) + βz < α(x) + βf(x).
Since x ∈ U, β < 0 and we may scale coefficients so that β = −1.
Under this convention,
with equality when x = y. Thus,
α(x − y) − f(x) + f(y) ≥ 0
f ∗∗ (x) = sup α(x) − f ∗ (α)
α
= sup
α
= sup f(x)
α
= f(x)
inf α(x − y) + f(y)
y