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Let $K,L$ be two origin-symmetric, convex bodies in $\mathbb{R}^3$.

Let $\xi\in\mathbb{S}^2$ be a vector on the unit sphere in $\mathbb{R}^3$. Denote $\xi^{\perp}$ the subspace in $\mathbb{R}^3$ orthogonal to $\xi$.

Denote $K\mid_{\xi^{\perp}}$ as the projection of the body $K$ onto the subspace $\xi^{\perp}$. Same for $L\mid_{\xi^{\perp}}$.

Suppose that for any vector $\xi$, there exists a rotation $\phi_{\xi}$ of $K\mid_{\xi^{\perp}}$ such that $\phi_{\xi}(K\mid_{\xi^{\perp}})\subset L\mid_{\xi^{\perp}}$. This rotation is taken in the $2$-dimensional sense on the plane $\xi^{\perp}$.

Does it follow that the volume of $K$ is less than or equal to the volume of $L$?

VShaw
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This doesn't answer the question, but can be useful:

Let $K, L \subset \mathbb{R}^3$ be convex bodies containing the origin as an interior point such that for every $\xi \in S^2$, the projection $K|_{\xi^{\perp}}$ can be rotated about the origin into $L|_{\xi^{\perp}}$ . Then either $K = L$ or $K$ can be obtained by reflecting $L$ about the origin.

see here.

EDIT: if $K$ is centrally symmetric there is an easier proof of the theorem outlined in the book Uniqueness Questions in Reconstruction of Multidimensional Objects from Tomography-Type Projection Data (see theorem 2.1.1.).

The main idea is to express the perimeter of the projection as an integral equation that contains the width function (i.e. the function that, for any given direction gives you the length of the affine diameter) and showing that this implies that the bodies have the same width in any direction.

Maybe it is possible to generalize this concept by using some Gronwall-like result

Moreover it is true a weaker version. If for each $\xi \in S^2$ $K_{ \xi^\perp} \subseteq L_{\xi^\perp}$ than $K \subset L$

A chord of a convex body is called an affine diameter if it is the longest chord of $K$ in a given direction

If a convex body $K$ is symmetric about a point $x$, then any chord through $x$ is an affine diameter of K (see section 4 of this paper)

Now suppose $K \not \subset L$ than there exists a point $x \in K$ such that $x \notin L$. Consider the chord $c:=\ell(x,0) \cap K$. As $K$ is origin symmetric this is the longest chord in a given direction.

Let $P$ any plane that contains such chord and $n$ its normal Than there is a chord of $K |_{n^\perp}$ that has the same length of $c$.

you have $c_L := c \cap L \subset c$ where the inclusion is strict as $x \notin L$, but $c_L$ is a chord passing througth the origin and has the same direction of $c$. So there is a chord of $L |_{n^\perp}$ that has the same length of $c_L$. So $K |_{n^\perp} \not \subset L |_{n^\perp}$ that is absurd by the hypothesys. This implies $K \subset L$

Marco
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    That first result is interesting, and seems like it might be helpful. But, I don't see how you're using it. The wording "can be rotated into" is perhaps a little ambiguous, but from the result, I strongly suspect it refers to when the image of $K|{\xi^\perp}$ under the rotation equals that of $L|{\xi^\perp}$, rather than just set containment. Otherwise, we could use the theorem to show $L$ is equal to a strictly smaller ball contained in itself, for example. So, you'd need to construct a superset of $K$ where each projection rotates to equal the projection of $L$, or a subset of $L$. – Theo Bendit Feb 13 '24 at 18:16
  • Basically, the result does seem promising, I'm not really following how you're applying it. I certainly buy that, if $K|{\xi^\perp} \subseteq L|{\xi^\perp}$ for all $\xi \neq 0$, then $K \subseteq L$, but I can't see the connection with the theorem quoted. – Theo Bendit Feb 13 '24 at 18:18
  • @TheoBendit The idea is to built $K'$ as the convex body such that its projections are $\phi_{\xi}(K|_{\xi^\perp})$ for any $\xi$ By the theorem such body is congruent to $K$. Now that I think about it I don't have proven that such body exists – Marco Feb 13 '24 at 18:24
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    I think constructing such a $K'$ would be difficult. I suspect the best way would be to deep-dive into the paper you provided and modify the result so that it works with set inclusion, rather than just with set equality. I'm going to be bed now, but I should be up before the bounty properly expires. If you can't fix the proof, leave your answer up anyway. I'm happy enough to let you get the half-bounty for bringing that paper to my attention. – Theo Bendit Feb 13 '24 at 18:28