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Gruber P. Convex and Discrete Geometry

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100 <strong>Convex</strong> Bodies<br />

The Valuation Property<br />

Mixed volumes have the following weak additivity property.<br />

Theorem 6.10. Let C, D, D2,...,Dd ∈ C such that C ∪ D ∈ C. Then<br />

V (C ∪ D, D2,...,Dd) + V (C ∩ D, D2,...,Dd)<br />

= V (C, D2,...,Dd) + V (D, D2,...,Dd).<br />

Note that here we only require that additivity holds for certain pairs C, D, whereas<br />

in other context, for example in measure theory, it is required that it holds for all<br />

pairs C, D. GivenD2,...,Dd ∈ C, this weak additivity property of the mapping<br />

C → V (C, D2,...,Dd) is expressed by saying that this mapping is a valuation.For<br />

more on valuations, see Sect. 7.<br />

We present two proofs of this result. One is based on Minkowski’s theorem on<br />

mixed volumes, the other one was suggested by Károly Böröczky, Jr. [157] <strong>and</strong><br />

makes use of the continuity of mixed volumes <strong>and</strong> of Lemma 6.5.<br />

Proof (using Minkowski’s theorem on mixed volumes). The following simple proposition<br />

will be needed later:<br />

(12) Let C, D, E ∈ C such that C ∪ D ∈ C. Then<br />

(C ∪ D) + E = (C + E) ∪ (D + E) <strong>and</strong><br />

(C ∩ D) + E = (C + E) ∩ (D + E).<br />

The first assertion is clear. Similarly, the inclusion (C ∩ D)+ E ⊆ (C + E)∩(D + E)<br />

is obvious. To show the reverse inclusion, let x ∈ (C + E) ∩ (D + E). Then x =<br />

c + e = d + f with c ∈ C, d ∈ D, e, f ∈ E. Since C ∪ D is convex, there is a point<br />

p = (1 − λ)c + λd ∈[c, d]∩(C ∩ D). Hence x = (1 − λ)(c + e) + λ(d + f ) =<br />

p + (1 − λ)e + λf ∈ (C ∩ D) + E. Thus (C + E) ∩ (D + E) ⊆ (C ∩ D) + E,<br />

concluding the proof of (12).<br />

To prove the theorem, note that<br />

V � �<br />

λ(C ∪ D) + λ2 D2 +···+λd Dd<br />

d�<br />

� �<br />

d<br />

= λ<br />

j<br />

j=0<br />

j<br />

d�<br />

V (C ∪ D,...,C ∪ D,<br />

Di � �� � j+1<br />

i j+1,...,id=2<br />

j<br />

,...,Did )λi j+1 ···λid ,<br />

V � �<br />

λ(C ∩ D) + λ2 D2 +···+λd Dd<br />

d�<br />

� �<br />

d<br />

= λ<br />

j<br />

j=0<br />

j<br />

d�<br />

V (C ∩ D,...,C ∩ D,<br />

Di � �� � j+1<br />

i j+1,...,id=2<br />

j<br />

,...,Did )λi j+1 ···λid ,<br />

V (λC + λ2 D2 +···+λd Dd)<br />

d�<br />

� �<br />

d<br />

= λ<br />

j<br />

j<br />

d�<br />

V (C,...,C,<br />

Di j+1 ,...,Did )λi j+1 ···λid ,<br />

j=0<br />

i j+1,...,id=2<br />

� �� �<br />

j<br />

V (λD + λ2D2 +···+λd Dd)<br />

d�<br />

� �<br />

d<br />

= λ<br />

j<br />

j=0<br />

j<br />

d�<br />

V (D,...,D,<br />

Di � �� � j+1<br />

i j+1,...,id=2 j<br />

,...,Did )λi j+1 ···λid .

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