Gruber P. Convex and Discrete Geometry

Define δn : B → R, n = 1, 2,...,by
6 Mixed Volumes and Quermassintegrals 87
δn(x) = dist(x, Cn) = min{�x − u� :u ∈ Cn} for x ∈ B.
In order to apply (7) to the functions δn, some properties have to be proved first. To
see that
(9) δn is convex,
let x, y ∈ B and 0 ≤ λ ≤ 1. Choose u,v ∈ Cn such that δn(x) = �x− u�
and δn(y) =�y− v�. Since (1 − λ)u + λv ∈ Cn by the convexity of Cn, it then
follows that
� � � �
(1 − λ)x + λy ≤�(1− λ)x + λy − (1 − λ)u + λv �
δn
≤ (1 − λ)�x − u�+λ�y − v� =(1 − λ)δn(x) + λδn(y),
concluding the proof of (9). For the proof that
(10) |δn(x) − δn(y)| ≤�x − y� for x, y ∈ B,
let x, y ∈ B and choose u,v ∈ Cn such that δn(x) =�x − u� and δn(y) =�y − v�.
Then
δn(x) ≤�x − v� ≤�x − y�+�y − v� =�x − y�+δn(y),
or δn(x) − δn(y) ≤�x − y�. Similarly, δn(y) − δn(x) ≤�x − y� and the inequality
(10) follows. (Alternatively, one may use Lemma 4.1.) Next,
(11) |δn(x)| ≤diam B for x ∈ B.
Note that δn(x) = 0forx∈ Cn and take (10) into account.
Propositions (10), (11) and (7) imply that there is a subsequence δn1 ,δn2 ,...,
converging uniformly to a function δC : B → R. As the uniform limit of a sequence
of non-negative, convex and continuous functions on B, alsoδCis non-negative,
convex and continuous on B. Hence
C = � x ∈ B : δC(x) = 0 � ∈ C or C =∅.
If C =∅, then δC(x) >0 for all x ∈ B. Hence δn j (x) >0 for all sufficiently large
j and all x ∈ B. This contradiction shows that
(12) C ∈ C.
We finally show that
(13) Cn1 , Cn2 , ···→C.
Let ε>0. Since δC is continuous on the compact set B and thus uniformly continuous
and since δC is 0 precisely on C, there is δ>0 such that
� x ∈ B : δC(x) ≤ δ � ⊆ C + εB d .
Since δC(x) ≤ δnk (x) + δ for all sufficiently large k,
(14) Cnk = � x ∈ B : δnk (x) = 0� ⊆ � x ∈ B : δC(x) ≤ δ � ⊆ C + εB d
for all sufficiently large k.