with equality if and only if f and g are proportional. For p <0, we assume that f(x), g (x) >0. An (almost) improvement of Minkowski’s inequality, for p Î R\{0}, is obtained in the following Theorem: Theorem 1.2 Let f(x), g(x) ≥ 0 and p >0, or f(x), g(x) >0 and p <0. Let s, t Î R\{0}, and s ≠ t. Then (i) Let p, s, t Î R be different, such that s, t >1 and (s t)/(p t) >1. Then ∫ (f (x)+g(x))pdx ...

We propose a novel method for the description of spatial patterns formed by a coverage of point sets representing galaxy samples. This method is based on a complete family of morphological measures known as Minkowski functionals, which includes the topological Euler characteristic and geometric descriptors to specify the content, shape and connectivity of spatial sets.

We consider a different L-Minkowski combination of compact sets in R than the one introduced by Firey and we prove an L-BrunnMinkowski inequality, p ∈ [0, 1], for a general class of measures called convex measures that includes log-concave measures, under unconditional assumptions. As a consequence, we derive concavity properties of the function t 7→ μ(t 1 pA), p ∈ (0, 1], for unconditional con...

The above norm induces a metric d where d(f, g) = ‖f − g‖p. Note that d(f, g) = 0 if and only if f = g a.e. μ, in which case we identify f with g. The Lp norm, like all worthy norms, satisfies the triangle inequality: ‖f + g‖p ≤ ‖f‖p + ‖g‖p ; this is precisely Minkowski’s inequality. For random variables X, Y defined on the same probability space and having finite p’th moments, Minkowski’s ineq...

Moreover, we introduce a curvature-dimension condition CD(K, N) being more restrictive than the curvature bound Curv(M,d, m) ≥ K. For Riemannian manifolds, CD(K, N) is equivalent to RicM (ξ, ξ) ≥ K · |ξ|2 and dim(M) ≤ N . Condition CD(K,N) implies sharp version of the Brunn-Minkowski inequality, of the Bishop-Gromov volume comparison theorem and of the Bonnet-Myers theorem. Moreover, it allows ...

Suppose two bounded subsets of IR are given. Parametrise the Minkowski combination of these sets by t. The Classical BrunnMinkowski Theorem asserts that the 1/n-th power of the volume of the convex combination is a concave function of t. A Brunn-Minkowski-style theorem is established for another geometric domain functional.

Then dp is a metric. To prove this one must check the axioms. First, since |xk − yk| = |yk − xk| ≥ 0, it is obvious that dp(x, y) = dp(y, x) ≥ 0 for all x and y. Furthermore, since ∑ k |xk−yk| = 0 if and only if all the terms |xk−yk| are zero, we see that dp(x, y) = 0 if and only if x = y. To verify the remaining axiom we use Minkowski’s Inequality. Let x, y, z ∈ C, and define ak = yk − xk and ...