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Gruber P. Convex and Discrete Geometry

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192 <strong>Convex</strong> Bodies<br />

halfspace {z : un · z ≤ un · xn}. Now, noting that Cn → C, un → u ∈ F ⊆ Sd−1 ,<br />

<strong>and</strong> xn → x ∈ bd C, the halfspace {z : u · z ≤ u · x} is a support halfspace of C at x.<br />

Thus x ∈ n −1<br />

C (u) ⊆ n−1<br />

C (F), concluding the proof of (10).<br />

It follows from (4), (9) <strong>and</strong> (10) that<br />

� � −1<br />

lim sup µd−1 ϱ nCn n→∞<br />

(F)�� � −1<br />

≤ µd−1 nC (F) � .<br />

Since<br />

� −1<br />

µd−1 nC (S d−1 ) � � � −1<br />

= µd−1 ϱ nCn (Sd−1 ) �� = µd−1(bd C),<br />

� −1<br />

the measures µd−1 ϱ(nCn (· )� are probability measures on Sd−1 up to some constant.<br />

� � −1<br />

Thus (6) implies that the measures µd−1 ϱ nCn (· )�� converge weakly to the measure<br />

� −1<br />

µd−1 nC (·) � . Taking into account (7), this implies that the area measures σCn (· ) =<br />

� −1<br />

µd−1 nCn (· )� � −1<br />

converge weakly to the area measure σC(· ) = µd−1 nC (· ) � . ⊓⊔<br />

Corollary 10.1. Let C, D ∈ Cp. Then<br />

V (C, D,...,D) = 1<br />

�<br />

d<br />

S d−1<br />

hC(u) dσD(u).<br />

Proof. Choose convex polytopes Pn ∈ Pp, n = 1, 2,..., such that Pn → D. By<br />

Proposition 10.2 the area measures σPn<br />

Hence, in particular:<br />

converge weakly to the area measure σD.<br />

�<br />

�<br />

(11) hC(u) dσPn (u) → hC(u) dσD(u).<br />

S d−1<br />

S d−1<br />

Lemma 6.5 <strong>and</strong> the definition of the area measure of polytopes show that<br />

(12) V (C, Pn,...,Pn) = 1<br />

d<br />

�<br />

hC(u F)v(F) = 1<br />

d<br />

�<br />

hC(u) dσPn (u).<br />

F facet of Pn<br />

According to Theorem 6.8 mixed volumes are continuous in their entries. Since<br />

Pn → D, we thus have:<br />

(13) V (C, Pn,...,Pn) → V (C, D,...,D).<br />

The corollary is now an immediate consequence of Propositions (11)–(13). ⊓⊔<br />

Alex<strong>and</strong>rov’s <strong>and</strong> Fenchel–Jessen’s Generalization of Minkowski’s Theorem<br />

Minkowski [736, 739] proved the following result: Let κ : Sd−1 → R + be a continuous<br />

function<br />

�<br />

such that<br />

u<br />

(14)<br />

κ(u) dµd−1(u) = o (componentwise).<br />

S d−1<br />

S d−1

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