Slicing inequalities for measures of convex bodies
Slicing inequalities for measures of convex bodies
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凸体测量的切片不等式
DOI：
10.1016/j.aim.2015.07.019
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发表时间：
2014
期刊：
影响因子：
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通讯作者：
A. Koldobsky
中科院分区：
文献类型：
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作者：
A. Koldobsky
我们考虑以下问题。是否存在一个绝对常数C，使得对每个n∈ N，每个整数1≤ k< n，Rn中的每个原点对称凸体L，Rn中的每个具有非负偶连续密度的测度μ，（1）μ（L）≤ Ckmax H∈ Gr n− k <$μ（L <$H）|L| k/n，其中Gr n− k是R n的（n− k）维子空间的格拉斯曼，并且|L|代表音量吗这个问题是一个扩展到任意措施（代替体积）和部分的任意余维k的切片问题，一个主要的开放问题凸几何。[18]、[19]证明了（1）对任意原点对称凸体成立，所有k和所有μ，C≤ O（n）。本文对无条件凸体和有界体积比的凸体证明了不等式（1）。证明了对任意λ∈（0，1），存在一个常数C= C（λ），使得不等式（1）对任意n∈ N，Rn中的任意原点对称凸体L，任意连续密度测度μ和截面k≥ λ n的余维数成立.证明是基于广义相交机构的稳定性结果和估计的外体积比距离从任意凸体类的广义相交机构。在最后一节中，我们证明了对于某些测度，极小截面的行为可能与体积的情况非常不同。
We consider the following problem. Does there exist an absolute constant C so that for every n∈ N, every integer 1≤ k< n, every origin-symmetric convex body L in R n, and every measure μ with non-negative even continuous density in R n,(1) μ (L)≤ C k max H∈ Gr n− k μ (L∩ H)| L| k/n, where Gr n− k is the Grassmanian of (n− k)-dimensional subspaces of R n, and| L| stands for volume? This question is an extension to arbitrary measures (in place of volume) and to sections of arbitrary codimension k of the slicing problem, a major open problem in convex geometry. It was proved in [18],[19] that (1) holds for arbitrary origin-symmetric convex bodies, all k and all μ with C≤ O (n). In this article, we prove inequality (1) with an absolute constant C for unconditional convex bodies and for duals of bodies with bounded volume ratio. We also prove that for every λ∈(0, 1) there exists a constant C= C (λ) so that inequality (1) holds for every n∈ N, every origin-symmetric convex body L in R n, every measure μ with continuous density and the codimension of sections k≥ λ n. The proofs are based on a stability result for generalized intersection bodies and on estimates of the outer volume ratio distance from an arbitrary convex body to the classes of generalized intersection bodies. In the last section, we show that for some measures the behavior of minimal sections may be very different from the case of volume.