Gruber P. Convex and Discrete Geometry

8 The Brunn–Minkowski Inequality 165
Using the definition of Minkowski surface area, this inequality readily yields the
isoperimetric inequality. Expressed otherwise, (1) says that the volume of the εneighbourhood
Cε = C + εB d of the compact set C is at least as large as the
volume of the ε-neighbourhood of a (solid Euclidean) ball of the same volume as C.
This result may be considered as a sort of isoperimetric inequality, where the notion
of surface area is not needed. This is the starting point for general isoperimetric
inequalities in metric probability spaces.
The Problem on Metric Probability Spaces
Let 〈M,δ,µ〉 be a space M withametricδ and thus a topology and a Borel probability
measure µ.Bytheε-neighbourhood Cε of a compact set C in M we mean the
set
Cε ={x ∈ M : δC(x) ≤ ε}, where δC(x) = dist(x, C) = min{δ(x, y) : y ∈ C}.
Then the following general isoperimetric or Brunn–Minkowski problem arises:
Problem 8.2. Given α, ε > 0, for what compact sets C in M with µ(C) = α has the
ε-neighbourhood Cε of C minimum measure?
The Spherical Case
Let S d−1 be endowed with its chordal metric, i.e. the metric inherited from E d , and
let S be the normalized area measure on S d−1 .Anisoperimetric inequality for S d−1
of Lévy [653], p.269, and Schmidt [891] is as follows: Let α > 0. Then, for all
compact sets C ⊆ S d−1 with S(C) = α, wehave,
S(Cε) ≥ S(Kε) for ε ≥ 0,
where K is a spherical cap with S(K ) = α. The equality case was studied by Dinghas
[269].
This result has the following surprising consequence: Let α = 1 2 , so that C has
the measure of a hemisphere H. Then, for each ε ≥ 0, we have that S(Cε) ≥ S(Hε).
An easy proof shows that the complement of Hε has area measure about e− 1 2 dε2 .
Hence S(Cε) is about 1 − e− 1 2 dε2 . Thus, for any given ε>0 and large d almost all
of Sd−1 lies within distance ε of any given compact set C in Sd−1 of measure 1 2 .In
other words, for large d the area of Sd−1 is concentrated close to any compact set of
measure 1 2 . As a consequence, for large d the area of Sd−1 is concentrated near any
great circle.
The situation just described becomes even more striking if it is interpreted in
terms of a Lipschitz function f : Sd−1 → R with Lipschitz constant 1. There is
a number m, themedian of f , such that there are compact sets C, D ⊆ Sd−1 with
S(C ∩ D) = 0, S(C) = S(D) = 1 2 and C ∪ D = Sd−1 , where f (u) ≤ m for u ∈ C
and f (v) ≥ m for v ∈ D. Then S(Cε) and S(Dε) both have area about 1 − e− 1 2 dε2 .