Measure of intersection of convex set with hyperplane is concave function

Question

Let $\Omega \subset \mathbb{R}^n$ be convex. We write points of $\mathbb{R}^n$ as $(x_1, x_2, \dots, x_n)$. Set $p(x) = m(\Omega \cap \{x_1 = x\})$, where $m$ is the $n-1$ dimensional Lebesgue measure and $\{x_1 = x\}$ is the hyperplane $\{(x_1, x_2, \dots, x_n) \in \mathbb{R}^n : x_1 = x\}$.

Geometrically, it is not hard to see that this function is concave (i.e. $p(\lambda x + (1-\lambda)y) \ge \lambda p(x) + (1-\lambda) p(y)$, but I can't figure out how to make a formal proof of that fact.

This is stated as a fact in page 197 of this paper: https://www.jstor.org/stable/1193994, but I could not figure out how to prove it or find a reference for it.

Answer 1

It is not true in general that the function $p$ is concave. For instance, let $\Omega$ be the conical hull of the ball of radius $1$ centered at the point $(2,0,\dots,0)\in\mathbb R^n$. Then $p(x)=c_nx^{n-1}$ for some real $c_n>0$ depending only on $n$ and for all real $x\ge0$. So, $p$ is not concave even on the interval $[0,\infty)$ if $n\ge3$.

However, for $n\ge2$, it is true that the function $p^{1/(n-1)}$ is concave on the interval where it is positive. This follows immediately from the Brunn--Minkowski inequality.