Gruber P. Convex and Discrete Geometry

6 Mixed Volumes and Quermassintegrals 97
pn(λ1,...,λd) → p(λ1,...,λd) as n →∞
for each d-tuple λ1,...,λd ≥ 0.
Assume that all coefficients ai1...idn and ai1...id of pn and p, respectively,
are symmetric in their indices i1,...,id. Then
ai1...id n → ai1...id for all i1,...,id ∈{1,...,d}.
To show the theorem, consider sequences (C1n), . . . , (Cdn) in C converging to
C1,...,Cd ∈ C say, respectively. Since for λ1,...,λd ≥ 0, we have λ1C1n +···+
λdCdn → λ1C1 +···+λdCd, the continuity of volume, which will be proved in
Theorem 7.5, yields
V (λ1C1n +···+λdCdn) → V (λ1C1 +···+λdCd)
for λ1,...,λd ≥ 0.
Now apply Minkowski’s mixed volume Theorem 6.5 and the above remark for
a12...dn = V (C1n,...,Cdn) and a12...d = V (C1,...,Cd). ⊓⊔
Monotony
Mixed volumes are non-decreasing in their entries.
Theorem 6.9. Let C1,...,Cd, D1,..., Dd ∈ C such that C1 ⊆ D1,...,Cd ⊆ Dd.
Then V (C1,...,Cd) ≤ V (D1,...,Dd).
As examples show, equality does not mean that necessarily Ci = Di for all i. The
proof of the theorem is based on several lemmas which are of interest themselves.
Let v(·) denote (d −1)-dimensional volume. (For (d −1)-dimensional polytopes and
thus for facets of d-dimensional polytopes it coincides with elementary area measure
for polytopes, see Sect. 16.1.)
Lemma 6.4. Let C ∈ C and P ∈ P. For each facet F of P let u F be the exterior unit
normal vector of F. Then
(3) V (C, P,...,P) = 1 �
hC(u F)v(F).
d
F facet of P
Proof. First, the following will be shown:
(4) Let G be a face of P with dim G ≤ d − 2. Then
V (G + εB d ) = O(ε 2 ) as ε →+0.
We may assume that G ⊆ E d−2 , where E d is represented in the form E d = E d−2 ×
E 2 .LetB d−2 , B 2 be the corresponding unit balls. Choose ϱ>0 so large that G +
εB d−2 ⊆ ϱB d−2 for 0 < ε ≤ 1. Then G + εB d ⊆ (G + εB d−2 ) × εB 2 ⊆
ϱB d−2 × εB 2 and thus V (G + εB d ) = O(ε 2 ) as ε →+0.