Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 62 / 1

Let $\mathcal{H}$ be the family of closed half-spaces containing $C$. Certainly $C\subseteq {\bigcap }_{H\in \mathcal{H}}H$. To prove the reverse inclusion, exclude an arbitrary $z\notin C$.

Suppose first that $C$ is nonempty. Its Euclidean projection onto a convex set $p$ exists: minimizing the continuous squared distance may be restricted to a sufficiently large closed ball, whose intersection with the closed set is compact. Put $d=z-p\ne 0$. For every $x\in C$, convexity puts $p+t(x-p)$ in $C$ for $0\le t\le 1$. Minimality of $p$ givesThus the closed half-spacecontains $C$, while $⟨d,z⟩=⟨d,p⟩+\Vert d{\Vert }^2$ excludes $z$. Since every point outside $C$ is excluded by some member of $\mathcal{H}$,This is the half-space representation of a closed convex set. If $C=\varnothing$, every point can again be excluded by a containing half-space, so the intersection is empty. If $C={\mathbb{R}}^n$, no proper half-space contains it and the empty intersection is, by convention, ${\mathbb{R}}^n$.

The Legendre-Fenchel transform and biconjugate areThe 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, thenwhere 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 $f^{\ast \ast }\le f$. Every term in the supremum defining $f^{\ast \ast }$ is an affine minorant of $f$, and conversely any affine minorant $⟨p,x⟩+a\le f(x)$ has $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is exactly the supremum of all affine minorants, hence is convex and lower semicontinuous.

Put $g=\mathrm{clconv}f$. 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 writtenhas $q\le 0$, since epigraphs extend upwards. If $q<0$, it is precisely the epigraph inequality of an affine minorant, $t\ge (⟨v,x⟩-a)/(-q)$.

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 $⟨v,x⟩\le a$, thenis 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 $g=f^{\ast \ast }$. The theorem in the previous solution supplies the geometric separation step behind biconjugation.

Let $C=\mathrm{conv}\{a,b\}$. Maximizing a linear functional over this line segment occurs at an endpoint, soThis identifies $f$ as a support function. Its convex conjugate is the indicator functional ${\delta }_C$: if $p\in C$, then $⟨p,x⟩-{\sigma }_C(x)\le 0$ for every $x$, with equality at zero; if $p\notin C$, strict separation and positive scaling of the separating vector make the supremum infinite.

For a convex function, equality in the Fenchel–Young inequality characterizes its subdifferential. HenceThis is support-function subgradients as exposed faces. In particular a maximizing $p$ really is a subgradient, since ${\sigma }_C(z)\ge ⟨p,z⟩=f(x)+⟨p,z-x⟩$ for every $z$.

Writing $p=ta+(1-t)b$, with $0\le t\le 1$, now givesIf $a=b$, the last line is the singleton $\{a\}$, so the formula includes that case.