Gruber P. Convex and Discrete Geometry

248 Convex Polytopes
Convex cones which are polyhedra are called polyhedral convex cones. These
may be represented as convex V-cones and convex H-cones. Clearly, polyhedral convex
cones are closed.
Support hyperplanes and the notion of (bounded or unbounded) faces, in particular
of vertices, edges and facets, of convex polyhedra are introduced as in the case
of convex polytopes.
We now show that the notions of convex V- and H-polyhedra coincide, as in the
case of convex polytopes.
Theorem 14.3. Let P ⊆ E d . Then the following statements are equivalent:
(i) P is a convex V-polyhedron.
(ii) P is a convex H-polyhedron.
Proof. The theorem will be proved in several steps:
(1) Let C ⊆ E d . Then the following are equivalent:
(i) C is a pointed convex V-cone with apex o.
(ii) C is a pointed convex H-cone with apex o.
(i)⇒(ii) Let C ={µ1y1 +···+µn yn : µ j ≥ 0} = � {µT : µ ≥ 0}, where
T ={µ1y1 +···+µn yn : µ j ≥ 0, µ1 +···+µn = 1} is a convex polytope. If
o ∈ T ,itmustbeavertexofT . To see this, note that C is pointed, hence o is an
extreme point of C and thus of T and therefore a vertex of T . Hence o is among the
points y1,...,yn. Removingo, the convex hull of the remaining points is a convex
polytope U such that C = � {µU : µ ≥ 0} and o �∈ U.LetH be a hyperplane which
strictly separates o and U and let V be the radial projection of U into H with centre
o. V is a convex V-polytope and thus a convex H-polytope in H by Theorem 14.2.
Since C = � {µV : µ ≥ 0}, we easily see that C is a convex H-polyhedron.
(ii)⇒(i) Let K ={x :|xi| ≤1}. Since C is pointed, o is an extreme point of the
convex polytope C ∩ K and thus a vertex. Hence there is a support hyperplane H
of C ∩ K with (C ∩ K ) ∩ H ={o}. Since C is a cone with apex o, it follows that
C ∩ H ={o}. Letp ∈ C \{o}. Then the H-polyhedron C ∩ (H + p) is bounded.
Otherwise it contains a ray p + S with o ∈ S ⊆ H, say. Since C is a cone with
apex o, wehaveµ(p + S) = µp + S ⊆ C for all µ > 0. Since C is closed,
also S = 0p + S ⊆ C, in contradiction to C ∩ H ={o}. Since C ∩ (H + p) is
bounded, it is an H- and thus a V-polytope by Theorem 14.2, say C ∩ (H + p) =
{µ1y1 +···+µn yn : µ j ≥ 0, µ1 +···+µn = 1} with suitable y1,...,yn ∈ C.
Since C is a convex cone with apex o and C ∩ H ={o}, each ray in C with endpoint
o meets H + p. Together this shows that
C = � � µ � C ∩ (H + p) � : µ ≥ 0 � ={µ1y1 +···+µn yn : µ j ≥ 0}.
The proof of (1) is complete.
(2) Let C ⊆ E d . Then the following are equivalent:
(i) C is a convex V-cone with apex o.
(ii) C is a convex H-cone with apex o.