---
tag: MV58A1F52E
kind: problem
name: Busemann–Petty problem
msc: [52A20, 52A40, 52A38]
uses: [437FSG3419, 2Z2QVWZ262]
about: [437FSG3419]
sources:
  - url: https://en.wikipedia.org/wiki/Busemann–Petty_problem
history:
  posed_by: Herbert Busemann and Clinton Myers Petty
  first_stated: "1956"
---

## Problem

Let $n \ge 2$ and let $K, L \subseteq \R^n$ be compact [[convex bodies|437FSG3419]] with nonempty interior, both symmetric about the origin.
Assume that for every hyperplane $H$ through the origin,
$$\operatorname{vol}_{n-1}(K \cap H) \le \operatorname{vol}_{n-1}(L \cap H),$$
where $\operatorname{vol}_{n-1}$ is the $(n-1)$-dimensional [[Lebesgue measure|2Z2QVWZ262]] on $H$.
Does it follow that $\operatorname{vol}_n(K) \le \operatorname{vol}_n(L)$?

## Remarks

The problem is solved: the answer is yes for $n \le 4$ and no for $n \ge 5$.
For $n = 2$ the answer is yes: a chord through the center in direction $u$ has length $2\rho(u)$, where $\rho$ is the radial function, so the hypothesis gives $K \subseteq L$.
Lutwak showed that the answer in $\R^n$ is yes if and only if every origin-symmetric convex body in $\R^n$ is an [[intersection body|Y4ZE8FVZ8D]].
Counterexamples in high dimensions were found first by Larman and Rogers, and the remaining dimensions were settled through work of Ball, Bourgain, Giannopoulos, Papadimitrakis, Gardner and Zhang, among others.
A unified Fourier-analytic solution in all dimensions was given by Gardner, Koldobsky and Schlumprecht.

## Explanations

### L1 human

Suppose that every slice through the center of one symmetric convex body is smaller than the matching slice of another body.
It seems obvious that the first body must then have smaller volume.
Surprisingly, this is true only up to dimension four: in dimension five and higher there are counterexamples.