= Solution
The <Legendre-Fenchel transform> and <biconjugate> are
$$
f^*(p)=\sup_x\{\langle p,x\rangle-f(x)\},\qquad
f^{**}(x)=\sup_p\{\langle p,x\rangle-f^*(p)\}.
$$
The <Fenchel-Moreau theorem> says that every <proper convex function> which is <lower semicontinuous> equals its <biconjugate>. More generally, if a proper extended-real $f$ has an <affine minorant>, then
$$
\boxed{f^{**}=\operatorname{cl\,conv}f,}
$$
where the right side is the greatest <lower semicontinuous> <convex> minorant. This is <biconjugation as closed convexification>. The affine-minorant hypothesis ensures that the closed convexification is proper; an unqualified statement including arbitrary improper functions would need separate conventions.
Here are the essential proof steps. The <Fenchel–Young inequality> gives $f^{**}\leq f$. Every term in the supremum defining $f^{**}$ is an <affine minorant> of $f$, and conversely any affine minorant $\langle p,x\rangle+a\leq f(x)$ has $a\leq-f^*(p)$. Thus $f^{**}$ is exactly the supremum of all affine minorants, hence is convex and lower semicontinuous.
Put $g=\operatorname{cl\,conv}f$. Its <epigraph> is the closed convex hull of the epigraph of $f$. Applying the <half-space representation of a closed convex set> in $\mathbb R^{n+1}$ recovers that epigraph from its containing half-spaces. A containing half-space written
$$
\langle v,x\rangle+q t\leq a
$$
has $q\leq0$, since epigraphs extend upwards. If $q<0$, it is precisely the epigraph inequality of an affine minorant, $t\geq(\langle v,x\rangle-a)/(-q)$.
Vertical half-spaces with $q=0$ must also be accounted for. Choose one affine minorant $\ell_0$ of $g$, which exists because $g$ is proper and closed: strictly separate $(x_0,g(x_0)-1)$ from its epigraph for a finite domain point $x_0$; the separating coefficient of $t$ cannot be zero, since that would not distinguish points with the same $x_0$. If a vertical containing inequality is $\langle v,x\rangle\leq a$, then
$$
\ell_0(x)+L(\langle v,x\rangle-a),\qquad L\geq0,
$$
is still an affine minorant on the domain of $g$. At a point violating the vertical inequality, these minorants tend to infinity as $L\to\infty$. Thus vertical domain restrictions are also recovered by the supremum of affine minorants.
Consequently $g$ equals that supremum. Every affine minorant of $g$ is below $f$, while the epigraph of any affine minorant of $f$ contains its closed convex epigraph hull. Therefore $f$ and $g$ have the same affine minorants, completing $g=f^{**}$. \b[The theorem in the previous solution supplies the geometric separation step behind biconjugation.]
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