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The problem.

Suppose you have a centrally symmetrical (with respect to the origin) convex body in $\mathbb{R}^n$, and you take the sections by intersecting it with hyperplanes in a fixed direction $u$. I want to prove that the section with the biggest volume is exactly the section that passes through the origin (or in general, the center of symmetry).

A more precise formulation would be to consider $K$ the convex symmetrical body, and $H_c=\{x\in\mathbb{R}^n\colon\langle x,u\rangle=c\}$ the hyperplane that is orthogonal to the direction $u$.

If we assume that $|u|=1$ then $c$ is precisely the distance from the hyperplane $H_c$ to the origin. Then, I want prove that $\text{vol}_{n-1}(K\cap H_c)$ is maximum when $c=0$, that is to say, the section that passes through the origin ($\text{vol}_{n-1}$ is the $n-1$-dimensional volume).

Ideas.

I have the feeling that, in fact, that function should be concave, that is, is will be increasing (not necessarily strictly) until $c=0$, and then decreasing.

Based on this, Brunn-Minkowski's inequality pops up, since it essentially says that the volume function, raised to an appropriate power, is concave, in particular $$\text{vol}_{n}(A+B)^{1/n}\geq\text{vol}_n(A)^{1/n}+\text{vol}_{n}(B)^{1/n}.$$

But I can't find a way to apply it properly, or any other way to solve it.

F.Webber
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  • A particular case : if the volume is a cuboid cnetered at the origin, all sections correspond to a maximal volume. – Jean Marie Jan 07 '18 at 01:41
  • Thanks for the comment, I just came up with a solution, I'll post it in a minute after I formalize it, so you guys can see it and maybe check it. – F.Webber Jan 07 '18 at 01:48

1 Answers1

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I finally came up with a proof, hopefully correct.

Let's establish some further notation. Let $c>0$ be fixed, $K_c=K\cap H_c$ the section of $K$ at distance $c$ from the origin, and $v(c)=\text{vol}_{n-1}(K_c)$ its volume. We will prove that $v(c)\leq v(0)$.

To start with, I assert that $v$ is an even function. To see this, it suffices to show that $K_{-c}$ is just the reflection of $K_c$ with respect to the origin, and hence both have the same volume. But this is clear, since if $x\in K_c=K\cap H_c$, then $\langle x,u\rangle=c$, and since $K$ is symmetrical with respect to the origin, $-x\in K$, and also $\langle -x,u\rangle=-c$, hence $-x\in K\cap H_{-c}=K_{-c}$.

Now, let's consider $C=\text{conv}(K_{-c},K_c)$, due to the convexity of $K$ it's obvious that $C\subset K$. Since both prior sections lie on parallel hyperplanes, the sections of $C$ are precisely convex linear combinations of $K_{-c}$ and $K_c$. Concretely, the section given by $C_\lambda=C\cap H_{(2\lambda-1)c}$ with $\lambda\in[0,1]$ verifies $$C_\lambda=(1-\lambda)K_{-c}+\lambda K_c.$$ The numerator merely translates and dilates the index so that $\lambda$ moves between $0$ and $1$. In particular, $C_0=K_{-c}$, $C_1=K_c$ and $$C_{1/2}=\frac{1}{2}K_{-c}+\frac{1}{2}K_c.$$

Since, in general, $C_\lambda\subset H_{(2\lambda-1)c}$, then one has $C_{1/2}\subset H_0$, and since $C\subset K$ also $C_{1/2}\subset K$, then $C_{1/2}\subset K_0$, and thus $\text{vol}_{n-1}(C_{1/2})\leq\text{vol}_{n-1}(K_0)=v(0)$.

On the other hand, applying Brunn-Minkowski's inequality and using that $v(c)=v(-c)$ we have \begin{alignat*}{2} \text{vol}_{n-1}\left(C_{1/2}\right)&=\text{vol}_{n-1}\left(\frac{1}{2}K_{-c}+\frac{1}{2}K_c\right)\geq \left(\left(\text{vol}_{n-1}\left(\frac{1}{2}K_{-c}\right)\right)^{\frac{1}{n-1}}+\left(\text{vol}_{n-1}\left(\frac{1}{2}K_c\right)\right)^{\frac{1}{n-1}}\right)^{n-1}\\ &=\left(\frac{1}{2}v(-c)^{\frac{1}{n-1}}+\frac{1}{2}v(c)^{\frac{1}{n-1}}\right)^{n-1}=v(c). \end{alignat*}

Using both of the proved inequalities we arrive at our result $$v(c)\leq\text{vol}_{n-1}(C_{1/2})\leq\text{vol}_{n-1}(K_0)=v(0).$$

F.Webber
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