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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 $\ell_p$.)

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  • $\begingroup$ Lovely question! (BTW, what is 2-equivalent?) $\endgroup$
    – Gil Kalai
    Apr 3, 2011 at 12:52
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    $\begingroup$ @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 $C_N$. $\endgroup$
    – Mark Meckes
    Apr 4, 2011 at 14:19
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    $\begingroup$ By the way, is the same question for the homeo with uniformly bounded bi-Lipschitz (not just Lipschitz) constant obvious? I do not see any simple argument (nor hard one, too). $\endgroup$
    – Fedor Petrov
    Apr 5, 2011 at 21:23
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    $\begingroup$ @Fedor Petrov: Differentiate to get a linear isomorphism to see that the bi-Lipschitz equivalence constant is no smaller than the isomorphism constant, which is $n^{1/2}$. $\endgroup$
    – Bill Johnson
    Apr 6, 2011 at 8:31
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    $\begingroup$ ah, sorry, this approach fails, since the unit cube, inscribed in a Euclidean ball, has a volume $C^N$-comparable to that of the unit ball. $\endgroup$
    – Fedor Petrov
    Apr 11, 2011 at 17:02

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