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Given any $d$-dimensional shape $X$, let $V(X)$ be its $d$-dimensional volume, and let $\ell(X)$ be the length of the longest line segment connecting two points of $X$.

Let $\mathcal{S}_C$ be the set of all $d$-dimensional shapes such that their minimum bounding box is a $d$-dimensional cube $C$. I am interested in quantifying the trade-off between $\frac{V(X)}{V(C)}$ and $\frac{\ell(X)}{\ell(C)}$ over $X\in\mathcal{S}_C$ (informally, how much $\frac{V(X)}{V(C)}$ can be large while $\frac{\ell(X)}{\ell(C)}$ is small).

Question: Can we prove that for $d\gg 1$ and for all $X\in\mathcal{S}_C$ there exists a constant $c$ such that the following inequality always holds? $$\left(\frac{V(X)}{V(C)}\right)^{\tfrac1d}\le c\cdot\frac{\ell(X)}{\ell(C)}$$

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    $\begingroup$ A candidate for the extreme is $X=$ the sphere. $\endgroup$
    – Joseph O'Rourke
    Commented Aug 20, 2020 at 12:03
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    $\begingroup$ For $X$ a $d$-ball, $\frac{\ell(X)}{\ell(C)} = d^{-1/2}$ and $\frac{V(X)}{V(C)} = \frac{\pi^{d/2}}{2^d\Gamma(d/2 + 1)}$. For large $d$, this is about $\frac{1}{\sqrt{\pi d}} (\pi e/2d)^{d/2}$ $\endgroup$
    – S. Carnahan ♦
    Commented Aug 20, 2020 at 12:30
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    $\begingroup$ I cannot find a good reference with a proof, but some web searching suggests that if $\frac{\ell(X)}{\ell(C)} = d^{-1/2}$, then $X$ is contained in a body of constant width, and that volumes of such bodies are bounded above by that of the ball. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 20, 2020 at 13:17
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    $\begingroup$ From what I said it follows that you always have $(V(X)/V(C))^{1/d}\le (1+o(1))K(\ell(X)/\ell(C))$ where the best $K$ is $\sqrt{\pi e/2}$ if $\ell\le\sqrt{\frac d{2\log d}}$, when the intersection is essentially the entire ball of diameter $\ell$, and then it gradually drops to $\sqrt 3$ at $\ell=\sqrt{d/3}$, when the intersection becomes essentially the entire unit cube. After that nothing interesting happens but the exact value of $K$ in the range $[\sqrt{\frac d{2\log d}},\sqrt{\frac d 3}]$ is still unclear to me though, I suspect, it is not too important for the question as asked. $\endgroup$
    – fedja
    Commented Aug 21, 2020 at 17:59
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    $\begingroup$ Yes, Penelope, you cannot go worse than if you just consider an unrestricted ball of diameter $\ell$ against the unit cube, so $\sqrt{\pi e/2}$ is a sure upper bound. The point was rather that it usually doesn't get much better than that even with all the extra conditions you imposed. $\endgroup$
    – fedja
    Commented Aug 21, 2020 at 23:32

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This is a bit too long for the comment box, so I'm posting it as an answer.

The worst case scenario is when $X$ is the intersection of a ball of radius $r\ge 1$ with the cube $C=[-1,1]^d$. Indeed, if we take the difference body $\frac{X-X}{2}$ of any body $X$ contained in the cube and of diameter $\ell=2r$, we'll get a body contained in the cube and also in the ball of radius $r$ and the volume will not decrease by Brunn-Minkowski. Also, since any such body contains the unit ball, the standard cube is, indeed, the minimal box for it. Since $\frac{\sqrt n}r X\supset C$, we see that for that body the reverse inequality always holds.

It would be nice to find a decent approximation for the volume of that intersection to see what happens in the regime when $r/\sqrt d$ stays fixed and $d\to\infty$, say.

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    $\begingroup$ I guess when $1\le r\le\sqrt{2}$ then the intersection is the ball with cut $2d$ hats of height $r-1$, so its volume can be easily calculated. When $\sqrt{2}\le r\le \sqrt{d}$ then the intersection seems to be the cube with spherically cut $2^d$ corners. Maybe there is a way to calculate a volume of such corner too, for instance, by induction with respect to $d$. $\endgroup$
    – Alex Ravsky
    Commented Aug 24, 2020 at 8:01
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    $\begingroup$ @AlexRavsky your intuition is correct and is explained here: math.stackexchange.com/questions/1996000/… However, it is not clear how to formally exploit this recursion for the final goal of the problem (for instance by using an induction based on this recursion). $\endgroup$
    – Penelope Benenati
    Commented Aug 24, 2020 at 11:17

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