Gruber P. Convex and Discrete Geometry

282 Convex Polytopes
by (2). Hence ϕ = ψ by induction and therefore
Φ(Q +[o, u]) = ϕ(Q) = ψ(Q) = Ψ(Q +[o, u]) for Q ∈ P(H).
Since this holds for any hyperplane in E d , the proof of (3) is complete.
For each hyperplane H the valuation ϕ is simple. Hence any two such valuations
on different hyperplanes are 0 in the intersection of these hyperplanes and thus
coincide for polytopes in the intersection. Thus all these valuations together yield a
valuation φ, say, on the space of all polytopes Q ∈ P with dim Q ≤ d − 1.
The next proposition refines (3).
(4) Φ(Z) = Ψ(Z) = h φ(Q)
for each right cylinder Z ∈ P with base Q and height h.
For cylinders of height 1 this holds by (3) and its proof. For cylinders of height 1/n
this holds by dissection, Proposition (1) and translation invariance. For cylinders of
height l/n this then holds by dissection, Proposition (1) and translation invariance.
For arbitrary real height, it finally follows by monotony.
After these preliminaries we will prove that
(5) Φ(P) = Ψ(P) for each P ∈ P.
If P is improper, then Φ(P) = 0 = Ψ(P) since both Φ and Ψ are simple. Suppose
now that P is proper. P can be dissected into proper simplices. Since Φ and Ψ are
simply additive by (1), it suffices for the proof of (5) to show that Φ(S) = Ψ(S) for
each proper simplex S. By considering the centre c of the inball of S of maximum
radius and for any facet F of S its convex hull with c, we see that S can be dissected
into simplices T = conv({c} ∪F), where the orthogonal projection b of c into the
hyperplane H = aff F is contained in F. It is thus sufficient to prove that Φ(T ) =
Ψ(T ) for such simplices T .Letn ∈ N. Then hyperplanes
Hi = H + i
(c − b), i = 0,...,n − 1,
n
dissect T into n polytopes T1,...,Tn ∈ Pp,say,
(6) T = T1 ˙∪··· ˙∪Tn.
Consider the cylinders Y1,...,Yn, Z1,...,Zn−1 where
�
Yi = Hi−1 ∩ T + o, 1
�
�
(c − b) , Zi = Hi ∩ T + o, −
n 1
�
(c − b) .
n
Then Yi ⊇ Ti ⊇ Zi. This, the monotonicity of Φ, (6) and (1) together yield that
(7) Φ(Z1) +···+Φ(Zn−1) ≤ Φ(T1) +···+Φ(Tn) = Φ(T )
≤ Φ(Y1) +···+Φ(Yn).
Similar inequalities hold for Ψ . Noting (4), it then follows that
Φ(Z1) +···+Φ(Zn−1) ≤ Ψ(T ) ≤ Φ(Y1) +···+Φ(Yn),