Is it necessarily true that the maximal section of a centrally symmetric convex body is always bigger than its minimal projection?

Question

I hope everyone is doing well.

Let $K \subset \mathbb{R}^n$ be a centrally symmetric convex body $(K = -K)$. Denote by $K \mid H$ the orthogonal projection of $K$ onto $H$, where $H$ is an $n - 1$ dimensional linear subspace. It is easy to see that the section $K \cap H$ is always contained in $K \mid H$ and therefore $\lambda(K \cap H) \leq \lambda(K \mid H)$, where $\lambda$ is the Lebesgue measure.

This leads to a natural question that I was thinking about today.

The problem is posed as follows:

Denote by $G(n - 1)$ the set of all $n-1$ dimensional linear subspaces of $\mathbb{R}^n$. Is it true that for every convex body $K \subset \mathbb{R}^n$, one has that $\min_{H \in G(n -1)}\lambda(K \mid H) \leq \max_{H \in G(n -1)}\lambda(K \cap H)?$

What I have done so far:

The case $n = 2$ is not too difficult to work out (True): Let $H_M \in \{H \in G(n-1) : \lambda(K \cap H) = \max_{A \in G(n-1)} \lambda(K \cap A) \}$.

Claim: $\lambda(K \cap H_M) = \lambda(K \mid H_M)$. Indeed, if not then we would have $K \cap H_M \subsetneq K \mid H_M.$ Consider $\mathrm{bd}({K \mid H_M}) = \{a, -a\}$. Let $\pi$ be the orthogonal projection operator from $K$ onto $H_M$ and consider $\pi^{-1}$(a). Choose any element $b \in \pi^{-1}(a)$, then $-b \in \pi^{-1}(-a)$ and the line segment $\ell$ from $b$ and $-b$ lies in $K$ by convexity. Then $\lambda(K \cap H_M) < \lambda(K \cap \ell) \leq \lambda(K \cap \mathrm{span}(\ell)),$ a contradiction to the maximality of $H_M$. Claim shown.

Then $\min_{H \in G(n -1)}\lambda(K \mid H) \leq \lambda(K \mid H_M) = \lambda (K \cap H_M) = \max_{H \in G(n-1)} \lambda(K \cap H).$

Where I am at now:

For higher dimensions, this argument breaks down of course, but I still think the original statement might be true. After working on this problem for a little bit today I thought that maybe this problem is well known. I figured I would ask those who are more knowledgeable in the field just in case so I don't reinvent the wheel.

UPDATE: Thanks to fedja (see below), the question has a negative result in general.

Answer 1

Here is a cheap probabilistic example of a symmetric convex body with the minimal projection greater than the maximal section.

Take the unit ball (say, in $\mathbb R^3$) and remove $m$ pairs of symmetric random caps of radius $r>0$ chosen independently with the centers uniformly distributed over the sphere. Take any plane through the origin. It is not hard to see that the maximal reduction in area one such removed cap can inflict on the corresponding section or projection is of size $r^3$, that the probability that this reduction will occur is about $r$, and that if we have the reduction at all, typically the section is reduced by $cr^3$ more than the projection. Thus, we can say that for each such pair of caps and each fixed plane, the reduction in the section is $r^3 X_i$ and in the projection $r^3 Y_i$ where $X_i$ are i.i.d., bounded by some constant, and having the expectation $\alpha r$ while $Y_i$ have the same properties, but now the expectation is $\beta r$ with $0<\beta<\alpha$.

Now we have the classical Bernstein bound for mean zero i.i.d. bounded random variables $Z_j$ with variance $\delta$ (using $t\in(0,1)$): $$ P(\sum_{j=1}^m Z_j>\gamma m\delta)\le e^{-t\gamma m\delta}E\prod_{j=1}^m e^{tZ_j} \\ =e^{-t\gamma m\delta}[1+E(e^{tZ_1}-1-tZ_1)]^m \\ \le e^{-t\gamma m\delta}(1+Kt^2EZ_j^2)^m\le e^{(Kt^2-\gamma t)m\delta} $$ Applying that to the appropriately shifted $X_i$ and $Y_i$ and $\gamma<(\alpha-\beta)/5$, we see that for any fixed plane, the probability that the reduction in the projection is greater than $(\alpha+\gamma)mr^4$ is at most $e^{-cmr}$.

For sections, we need a bound from below. The Bernstein inequality implies that the sum of the individual reductions is $\ge (\alpha-\gamma)mr^4$ outside of the event of probability $\le e^{-mr}$, but there may be overlaps. We can easily afford about $q=\tau mr$ overlapping pairs of pairs of caps influencing the given plane with sufficiently small $\tau$. What is the probability of such event for a fixed plane? The crude bound is $m^{2q}(Cr^3)^q=(Cm^2r^3)^q$ (the number of possibilities to choose $q$ disjoint pairs of pairs of caps times the probability that all $q$ chosen pairs are overlapping and influencing the given plane). If $Cm^2r^3<1/2$, say, we see that this probability is $\le e^{-cmr}$ as well.

Moral: If $m^2r^3$ is small, for any fixed plane, the probability that either the section reduction is $<(\alpha-2\gamma)mr^4$ or the projection reduction is greater than $(\beta+\gamma)mr^4$ is less than $e^{-cmr}$. Now choose the net of planes so that each plane is, say, $r^{10}$-close to some plane in the net (about $r^{-12}$ planes total), so that the deviation from the net plane changes the numbers by at most $r^{10}\ll r^4$, set $m=r^{-5/4}$, let $r\to 0+$, and enjoy the union bound.

The question that may be not so trivial is whether we have the desired inequality with some constant factor independent of the dimension. I don't know the answer and will be quite interested if something figures it out or finds it in the literature (I would be astonished if my example above hasn't been published before).