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Pietro Majer
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Best Hölder exponents of surjective maps from the unit square to the unit cube

The Peano's square-filling curve $p:I\to I^2$ turn's out to be Hölder continuous with exponent $1/2$ on the unit interval $I$ (a quick way to see it, is to note that $p$ is a fixed point of a suitable contraction $T:C(I,I^2)\to C(I,I^2)$, and the non-empty, closed subset of curves with modulus of continuity $\omega(t):=ct^{1/2}$ is $T$-invariant, for a suitable choice of $c$, so that $p$ is therein).

For the same reason, the more general analogous $n$-cube-filling curves $I\to I^n$ (e.g. described in the same Peano paper) are Hölder continuous with exponent $1/n$.

On the other hand, for any $1\le k\le n$, by elementary considerations on Hausdorff measures, no $\alpha$-Hölder continuous map $I^k\to I^n$ with exponent $\alpha < k/n $ can be surjective. The natural questions are therefore:

Given $1\le k\le n$, does there exist an $\alpha$-Hölder continuous map $I^k\to I^n$ with exponent $\alpha=k/n$? Otherwise, what is the best exponent $\alpha$ obtainable for such a surjective map? In particular, is there a simple construction for the case $I^2\to I^3$? (Actually, we may focus on this last question, which appears to be the simpler non-trivial case).

Summing up the above remarks, the answer is affirmative if $k=1$ or if $k=n$, and we may also note that if for a pair $(k,n)$ there is such a surjective map $q:I^k\to I^n$, then the map $(x_1,\dots,x_m)\mapsto (q(x_1), q(x_2),\dots ,q(x_m))$ is also a surjective map $I^{mk}\to I^{mn}$ with the same exponent $k/n$ of $q$.

Pietro Majer
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