By Georges de Rham, F.R. Smith, S.S. Chern

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**Example text**

Let X be a two-dimensional normed space with unit disc B and area form Ω ∈ Λ2 X ∗ . Of all convex bodies in X with a given perimeter the one that encloses the largest area is, up to translations, a dilate of I := (iΩ B)∗ . VOLUMES ON NORMED AND FINSLER SPACES 33 Proof. Let K ⊂ X be a convex body and let B (∂K) = 2V (K, I) be its perimeter. By Minkowski’s mixed volume inequality, we have B (∂K) 4 2 = V (K, I)2 ≥ λ(K)λ(I) with equality if an only if K and I are homothetic. Thus, the area enclosed by K is maximal for a given perimeter if and only if K is a dilate of I.

VOLUMES ON NORMED AND FINSLER SPACES 19 the k-dimensional normed space “spanned by a”, a := {x ∈ X : a ∧ x = 0} ⊂ X, (provided with the induced norm), and compute µ a (a). Thus, once we have chosen a way to define volume in all finite-dimensional normed spaces, we have a way to associate to each norm on an n-dimensional vector space X a family of k-volume densities, with 1 ≤ k ≤ n. The Busemann k-volume densities. Let X be a normed space of dimension n with unit ball B, and let k be a positive integer less than n.

12 characterizes the Holmes–Thompson definition. However, the following question remains open. Problem [Thompson 1996]. Is the Holmes–Thompson definition of volume characterized by the fact that the area of the unit sphere of a normed space equals the area of the unit sphere of its dual ? The first result extending Gol¸ab’s theorem to higher dimension is the following sharp upper bound for the Busemann area of a unit sphere. 13 [Busemann and Petty 1956]. The Busemann area of the unit sphere of an n-dimensional normed space is at most 2nεn−1 with equality if and only if B is a parallelotope.