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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.

For those interested in the problem, here is my current approach: Let $\rho_{A}$ be the radial function of a convex body $A$ and $\xi^\perp$ be the orthogonal complement of $\xi \in S^{n-1}$. The goal is to construct a continuous tangent vector field on $S^{n-1}$ and conclude things by the Poincaré–Hopf theorem using the function $f(\xi) = \max_{\theta \in \xi^\perp \cap S^{n-1}}(\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta))$ (where $f$ is defined on $S^{n-1}$) as a backbone of the construction. The vector field $F$ would be given by something like $F(\xi) = \theta$ (where $\theta$ and $\xi$ are above), but generally there could be many $\theta$ that maximize $\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta)$ (in fact for centrally symmetric convex bodies there are at least $2$), so I am trying to work around this. If one can work around this then the problem would be solved for $n$ odd.

Thanks for reading! Please let me know if you have any comments, or questions.

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.

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.

For those interested in the problem, here is my current approach: Let $\rho_{A}$ be the radial function of a convex body $A$ and $\xi^\perp$ be the orthogonal complement of $\xi \in S^{n-1}$. The goal is to construct a continuous tangent vector field on $S^{n-1}$ and conclude things by the Poincaré–Hopf theorem using the function $f(\xi) = \max_{\theta \in \xi^\perp \cap S^{n-1}}(\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta))$ (where $f$ is defined on $S^{n-1}$) as a backbone of the construction. The vector field $F$ would be given by something like $F(\xi) = \theta$ (where $\theta$ and $\xi$ are above), but generally there could be many $\theta$ that maximize $\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta)$ (in fact for centrally symmetric convex bodies there are at least $2$), so I am trying to work around this. If one can work around this then the problem would be solved for $n$ odd.

Thanks for reading! Please let me know if you have any comments, or questions.

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.

For those interested in the problem, here is my current approach: Let $\rho_{A}$ be the radial function of a convex body $A$ and $\xi^\perp$ be the orthogonal complement of $\xi \in S^{n-1}$. The goal is to construct a continuous tangent vector field on $S^{n-1}$ and conclude things by the Poincaré–Hopf theorem using the function $f(\xi) = \max_{\theta \in \xi^\perp \cap S^{n-1}}(\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta))$ (where $f$ is defined on $S^{n-1}$) as a backbone of the construction. The vector field $F$ would be given by something like $F(\xi) = \theta$ (where $\theta$ and $\xi$ are above), but generally there could be many $\theta$ that maximize $\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta)$ (in fact for centrally symmetric convex bodies there are at least $2$), so I am trying to work around this. If one can work around this then the problem would be solved for $n$ odd.

Thanks for reading! Please let me know if you have any comments, or questions.

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 $H_M \in \{H \in G(n-1) : \lambda(K \cap H) = \max_{H \in G(n-1)} \lambda(K \cap H) \}$.

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.

For those interested in the problem, here is my current approach: Let $\rho_{A}$ be the radial function of a convex body $A$ and $\xi^\perp$ be the orthogonal complement of $\xi \in S^{n-1}$. The goal is to construct a continuous tangent vector field on $S^{n-1}$ and conclude things by the Poincaré–Hopf theorem using the function $f(\xi) = \max_{\theta \in \xi^\perp \cap S^{n-1}}(\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta))$ (where $f$ is defined on $S^{n-1}$) as a backbone of the construction. The vector field $F$ would be given by something like $F(\xi) = \theta$ (where $\theta$ and $\xi$ are above), but generally there could be many $\theta$ that maximize $\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta)$ (in fact for centrally symmetric convex bodies there are at least $2$), so I am trying to work around this. If one can work around this then the problem would be solved for $n$ odd.

Thanks for reading! Please let me know if you have any comments, or questions.

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.

For those interested in the problem, here is my current approach: Let $\rho_{A}$ be the radial function of a convex body $A$ and $\xi^\perp$ be the orthogonal complement of $\xi \in S^{n-1}$. The goal is to construct a continuous tangent vector field on $S^{n-1}$ and conclude things by the Poincaré–Hopf theorem using the function $f(\xi) = \max_{\theta \in \xi^\perp \cap S^{n-1}}(\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta))$ (where $f$ is defined on $S^{n-1}$) as a backbone of the construction. The vector field $F$ would be given by something like $F(\xi) = \theta$ (where $\theta$ and $\xi$ are above), but generally there could be many $\theta$ that maximize $\rho_{K \mid \xi^\perp}(\theta) - \rho_{K \cap \xi^\perp}(\theta)$ (in fact for centrally symmetric convex bodies there are at least $2$), so I am trying to work around this. If one can work around this then the problem would be solved for $n$ odd.

Thanks for reading! Please let me know if you have any comments, or questions.