Gruber P. Convex and Discrete Geometry

Convex Polytopes and Faces
14 Preliminaries and the Face Lattice 245
A convex polytope P in E d is the convex hull of a finite, possibly empty, set in
E d .IfP = conv{x1,...,xn}, then the extreme points of P are among the points
x1,...,xn by Minkowski’s theorem 5.5 on extreme points. Thus, a convex polytope
has only finitely many extreme points. If, conversely, a convex body has only finitely
many extreme points, then it is the convex hull of these, again by Minkowski’s theorem.
Hence, the convex polytopes are precisely the convex bodies with finitely many
extreme points. The intersection of P with a support hyperplane H is a face of P. It
is not difficult to prove that
(1) H ∩ P = conv(H ∩{x1,...,xn}).
This shows that a face of P is again a convex polytope. The faces of dimension 0 are
the vertices of P. These are precisely the extreme, actually the exposed points of P.
The faces of dimension 1 are the edges of P and the faces of dimension dim P −1the
facets. The empty set ∅ and P itself are called the improper faces, all other faces are
proper. Since a face of P is the convex hull of those points among x1,...,xn which
are contained in it, P has only finitely many faces. Since each boundary point of P
is contained in a support hyperplane by Theorem 4.1, bd P is the union of all proper
faces of P. ByF(P) we mean the family of all faces of P, including ∅ and P. The
space of all convex polytopes in E d is denoted by P = P(E d ) and Pp = Pp(E d ) is
its sub-space consisting of all proper convex polytopes, that is those with non-empty
interior.
Gauss’s Theorem on the Zeros of the Derivative of a Polynomial
In an appendix to his third proof of the fundamental theorem of algebra, Gauss [363]
proved the following result.
Theorem 14.1. Let p be a polynomial in one complex variable. Then the zeros of its
derivative p ′ are contained in the convex polygon determined by the zeros of p.
Proof. Let z1,...,zn ∈ C be the zeros of p, each written according to its multiplicity.
Then
p(z) = a(z − z1) ···(z − zn) for z ∈ C
with suitable a ∈ C. Letz �= z1,...,zn. Dividing the derivative p ′ (z) by p(z)
implies that
(2)
p ′ (z)
p(z)
1
= +···+
z − z1
1
z − zn
= ¯z −¯z1
¯z −¯zn
+···+
|z − z1| 2 |z − zn|
Assume now that z is a zero of p ′ . We have to show that z ∈ conv{z1,...,zn}. Ifz
is equal to one of z1,...,zn, this holds trivially. Otherwise (2) shows that
z =
1
|z−z1| 2 z1 +···+ 1
|z−zn| 2 zn
1
|z−z1| 2 +···+ 1
|z−zn| 2
Thus z is a convex combination of z1,...,zn. ⊓⊔
.
2 .