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Let $B$ be a convex body in $\mathbb{R}^d$, equipped with its standard Euclidean form, and assume that

$$\intop_B x \, dx = 0$$ $$\frac{1}{|B|_d} \intop_B x_i x_j \, dx = \delta_{ij},$$

a normalization that can always be achieved by an appropriate affine transformation. Here $|\cdot|_d$ means $d$-dimensional Lebesgue measure, of course.

Now let's look at the areas of $(d-1)$-dimensional projections of $B$. It's clear that

$$ \inf_{\Vert \xi \Vert = 1} \frac{|\mathrm{pr}_{\xi^\perp} B|_{d-1}}{|B|_d} \le c_d,$$

where $c_d$ is some universal constant, depending only on dimension.

Question: is it true that

$$\lim_{d \to \infty} c_d < \infty$$

As far as I know, the answer is known to be negative if instead of second-moment normalization as above we use a volume normalization (i.e. $|B|_d=1$).

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  • $\begingroup$ Is this false with $\sup$ instead of $\inf$? $\endgroup$ – Will Sawin Jan 26 '14 at 20:54
  • $\begingroup$ @WillSawin: Yes, for the $d$-dimensional cube the largest projection has area of order $\sqrt d$. $\endgroup$ – Alexander Shamov Jan 26 '14 at 20:55
  • $\begingroup$ Are you sure you have the right expression? The homogeneity looks funny. $\endgroup$ – Mark Meckes Jan 29 '14 at 15:00
  • $\begingroup$ @MarkMeckes: Homogeneity is not an issue here exactly because of the second moment normalization, i.e. I'm fixing the "width" of the set in any direction. $\endgroup$ – Alexander Shamov Jan 29 '14 at 19:22
  • $\begingroup$ @MarkMeckes: What I wrote at the bottom line made no sense exactly for the reason that you indicated. Please see the updated version. $\endgroup$ – Alexander Shamov Jan 31 '14 at 9:50

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