Gruber P. Convex and Discrete Geometry

196 Convex Bodies
Thus Minkowski’s first inequality, see Theorem 6.11, shows that
V (C) d = V (C, D,...D) d ≥ V (C)V (D) d−1 , or V (C) ≥ V (D).
Similarly, V (D) ≥ V (C). Hence
V (C) = V (D) = V (C, D,...,D).
Thus, in Minkowski’s first inequality, we have equality. This, in turn, implies that C
and D are homothetic, see Theorem 6.11. Since V (C) = V (D), we see that D is a
translate of C.
(ii)⇒(i) Choose convex polytopes Pm ∈ Pp, m = 1, 2,...,such that Pm → C.
By Proposition 10.2, the area measures σPm
σ = σC. Hence, in particular,
converge weakly to the area measure
�
�
udσPm (u) → udσC(u) as m →∞(componentwise).
S d−1
S d−1
By Minkowski’s theorem for polytopes,
�
udσPm (u) = o.
S d−1
Hence �
S d−1
udσC(u) = o.
If σC were concentrated on a great circle, say the equator of Sd−1 ,letBbe the
open northern hemisphere. Then σC(B) = 0. On the other hand, n −1
C (B) consists
of � all points of bd C with exterior unit normal vectors in B. Thus σC(B) =
−1
µd−1 nC (B) � > 0, a contradiction. ⊓⊔
Related Open Problems
The given solution of Minkowski’s problem is rather satisfying, but it does not settle
the following question. If κ : S d−1 → R + is of class C α , to what class does the
corresponding convex body C belong and if it is of class C β with β ≥ 2, is κ then its
ordinary Gauss curvature? There is a large body of pertinent results, see the surveys
in Pogorelov [806], Gluck [381], Su [976] and Schneider [908]. Selected references
are Pogorelov [806], Caffarelli [186, 187] and Jerison [545].
Blaschke Multiplication and Blaschke Addition
Theorem 10.1 led Blaschke [124], p.112, to introduce, besides multiplication with
reals and Minkowski addition, a second type of multiplication with reals and addition