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Let $B_p^N$ be the unit ball of $\mathbb{R}^N$ under the $\ell_p^N$ norm.

Question: Let $C_N$ be the infimum of all $C$ for which there is a homeomorphism $f_N$ from $B_\infty^N$ onto $B_2^N$ so that the Lipschitz constant of $f_N$ is $C$. Is $C_N$ bounded as $N\to \infty$?

Notice that the $\ell_\infty^N$ norm is $2$-equivalent to the $\ell_p^N$ norm with $p = \log_2 N$, and the Mazur map from $B_p^N$ to $B_2^N$ has Lipschitz constant about $p$, so $C_N$ grows no faster than $\log N$.

(The Mazur map from $\ell_p$ to $\ell_2$ is defined for unit vectors to be $x\mapsto |x|^{p/2}\rm{sign}\ x$ and extended to be positively homogeneous on $\ell_p$.)

flag	asked Apr 3 2011 at 11:27 Bill Johnson 14.3k●1●33●60

Lovely question! (BTW, what is 2-equivalent?) – Gil Kalai Apr 3 2011 at 12:52

Just to be sure: the Lipschitz constant $C$ refers to the $\ell^N_\infty$ distance on the domain and $\ell^N_2$ distance on the co-domain of $f_N$, right? – Pietro Majer Apr 3 2011 at 13:28

$\|x\|_\infty \le \|x\|_p \le 2 \|x\|_\infty$ for all $x$, Gil. Right, Pietro. – Bill Johnson Apr 3 2011 at 13:53

is it clear that if one of those balls may be covered by balls of radii $r_1,r_2,\dots,r_n$, then the other may be covered by balls of radii $Cr_1,\dots,Cr_n$ for some universal $C$? I.e. that both conditions are equivalent (up to constant) to some $\sum r_i^n=O(1)$? – Fedor Petrov Apr 3 2011 at 20:16

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@Pietro: a standard volume comparison argument shows that in any $N$-dimensional normed space, the ball of radius $R\ge 1$ can be covered by $(3R)^N$ unit balls. This seems to give only a constant lower bound on the $C_N$. – Mark Meckes Apr 4 2011 at 14:19

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