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For a proper extended-real function with an affine minorant, biconjugate gives its greatest lower-semicontinuous convex minorant. Geometric separation of its epigraph turns containing closed half-spaces into affine lower bounds. Vertical half-spaces are recovered by adding multiples of a domain-separating inequality to one fixed affine minorant.

The study of convex sets, their convex hulls, supporting closed half-spaces, and inequalities for their sizes.

A facet of a full-dimensional convex polytope in ${\mathbb{R}}^n$ is a face of dimension $n-1$. Each irredundant supporting closed half-space defines one facet.

A closed convex set in a finite-dimensional Euclidean space is the intersection of all containing closed half-spaces. For a point outside a nonempty set, its Euclidean projection onto a convex set gives a separating normal. Applied to an epigraph, this turns geometric separation into affine lower bounds and is a key step in the Fenchel-Moreau theorem.

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, ${\mathbb{R}}^n$.

If $C={\mathbb{R}}^n$, the family of proper containing half-spaces is empty and its intersection is, by convention, ${\mathbb{R}}^n$. If $C=\varnothing$, every closed half-space contains it and their intersection is empty. Now suppose $C$ is a nonempty proper closed convex set. Take $x\notin C$ and let $z$ be its Euclidean projection onto a convex set. This projection exists: a minimizing sequence can be restricted to a bounded ball, and closedness gives attainment. It is unique by convexity and strict convexity of squared distance.

For $y\in C$, the segment $z+t(y-z)$ remains in $C$ for $0\le t\le 1$. Minimality at $t=0$ impliesThus the closed half-space $H_x=\{y:⟨x-z,y-z⟩\le 0\}$ contains $C$ but excludes $x$, since $⟨x-z,x-z⟩>0$. Every point outside $C$ is excluded by at least one containing half-space. The reverse inclusion is immediate, givingThis is the half-space representation of a closed convex set. The argument gives an explicit separating hyperplane rather than just citing a Hahn-Banach separation theorem.

Let $m$ be the number of facets of the convex polytope $P$. Its facet description can be writtenBecause $r^{-1}{B}^n\subset P$, each supporting closed half-space has $h_j\ge 1/r$. For any $\theta \in S^{n-1}$, follow the ray $t\theta$ to its last point in $P$. Its radius $t_{\theta }$ is at most one because $P\subset B^n$. At least one facet is active there, givingThus the unit sphere is covered by the $m$ spherical caps centred at $u_j$ with angular radius $\rho =\arccos (1/r)$. Since $r\ge \sqrt{2}$, this radius lies in $[\pi /4,\pi /2)$ and the spherical cap area upper bound applies. Subadditivity of surface area givesUsing $\log (1-x)\le -x$ for $0<x<1$,The facet lower bound for ball approximations is exponential in the dimension.

Choose a maximal subset $U\subset S^{n-1}$ whose distinct points have Euclidean distance greater than one. The volumetric bound for Euclidean metric nets makes the construction finite: the open Euclidean balls of radius $1/2$ about the points of $U$ are disjoint and all lie in $(3/2)B^n$, soMaximality means that $U$ is a metric net of radius one on the unit sphere. Thus every $\theta \in S^{n-1}$ has some $u\in U$ with $\Vert u-\theta \Vert \le 1$. Since both are unit vectors,Now intersect the corresponding closed half-spaces:The Cauchy-Schwarz inequality shows that $(1/2)B^n\subset P$. Conversely, write any nonzero $x\in P$ as $x=t\theta$ and choose $u$ as above. Then $t/2\le t(u\cdot \theta )=u\cdot x\le 1/2$, so $t\le 1$. Hence $P\subset B^n$, which also proves boundedness. This convex polytope has at most $|U|\le 3^n$ facets, since redundant inequalities can only reduce their number. It meets the required inclusions with $r=2$.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

The origin belongs to $S$, so $S$ is nonempty. Each displayed inequality defines a closed half-space, and every half-space is a convex set. An intersection of convex sets is convex: if $x,y\in S$ and $0\le t\le 1$, each linear inequality remains valid for $(1-t)x+ty$. Thus $S$ is a nonempty convex polytope.