Gruber P. Convex and Discrete Geometry

98 Convex Bodies
Thus
Second, Minkowski’s theorem 6.5 on mixed volumes shows that, in particular,
V (εC + P) = V (P,...,P) + dV(C, P,...,P)ε + ...
(5) V (C, P,...,P) = 1
d lim
ε→+0
+ dV(C,...,C, P)ε d−1 + V (C,...,C)ε d
= V (P) + dV(C, P,...,P)ε +···+V (C)ε d .
V (εC + P) − V (P)
.
ε
Third, in the following we distinguish two cases. In the first case let o ∈ C.
Choose σ>0 such that C ⊆ σ Bd . The convex body εC + P then may be dissected
into P, into cylinders, possibly slanting, with the facets as bases and into the remaining
part of εC + P. The cylinders can be described as follows: for a facet F of P with
exterior unit normal vector u F choose a point p ∈ C ∩ HC(u F). The corresponding
cylinder then is F + ε[o, p]. The remaining part of εC + P is contained in the union
of sets of the form εC + G ⊆ εσ Bd + G, where G ranges over the faces of P with
dim G ≤ d − 2 and thus has volume O(ε2 ) by (4). Hence
(6) V (εC + P) = V (P) + ε �
hC(u F)v(F) + O(ε 2 ) as ε → 0.
F facet of P
Finally, (5) and (6) together yield (3) in case where o ∈ C.
In the second case let o �∈ C. Choose t ∈ E d such that o ∈ C + t. Then
V (C, P,...,P) = V (C + t, P,...,P)
= 1 �
hC+t(u F)v(F)
d
F facet of P
= 1 �
hC(u F)v(F) +
d
1
d
F facet of P
by Proposition 6.5 and the first case. Now note that
�
F facet of P
t · u F v(F) =�t� �
F facet of P
�
F facet of P
t · u F v(F)
t
�t� · u F v(F) = 0
since the latter sum equals the sum of the area of the projection of bd P onto the
hyperplane through o orthogonal to t, taken once with a + sign and once with a −
sign. This concludes the proof of (3) in case where o �∈ C. ⊓⊔
For later reference we state the following consequence of (5), where for the
Minkowski surface area, see Sect. 6.4.
Corollary 6.1. Let P ∈ P. Then d V (B d , P,...,P) equals the (Minkowski) surface
area of P.
The next result is a refinement of Lemma 6.4.