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# On the Inequality for Volume and Minkowskian Thickness

Published:2006-06-01
Printed: Jun 2006
Given a centrally symmetric convex body $B$ in $\E^d,$ we denote by $\M^d(B)$ the Minkowski space ({\em i.e.,} finite dimensional Banach space) with unit ball $B.$ Let $K$ be an arbitrary convex body in $\M^d(B).$ The relationship between volume $V(K)$ and the Minkowskian thickness ($=$ minimal width) $\thns_B(K)$ of $K$ can naturally be given by the sharp geometric inequality $V(K) \ge \alpha(B) \cdot \thns_B(K)^d,$ where $\alpha(B)>0.$ As a simple corollary of the Rogers--Shephard inequality we obtain that $\binom{2d}{d}{}^{-1} \le \alpha(B)/V(B) \le 2^{-d}$ with equality on the left attained if and only if $B$ is the difference body of a simplex and on the right if $B$ is a cross-polytope. The main result of this paper is that for $d=2$ the equality on the right implies that $B$ is a parallelogram. The obtained results yield the sharp upper bound for the modified Banach--Mazur distance to the regular hexagon.