# 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 $q$. Also, we may consider compositions of maps, so that affirmative answers for $(k,n)$ and $(n,m)$ imply the affirmative answer for $(k,m)$.

Update 08.01.16.   The only answer received so far suggests a nice article, yet not related with this question (the only theorem in that paper that deals with Hölder maps is Thm 2.1, but has nothing or very little to do with the present problem, since it is about $\mathbb{R}$-valued functions, that is $n=1$ (existence of Hölder functions on a metric space which map surjectively onto an interval is a non-trivial problem only for totally disconnected spaces).

• That's a great question! It reminds of the open question whether there is an embedding from $\mathbb{R}^k$ with the snowflaked metric $\lVert\cdot\rVert^{\alpha}$ (where $\alpha <1$) into $\mathbb{R}^n$ whenever $k/\alpha< n$. Your question feels a bit of a dual to that. Apr 27, 2015 at 9:07

There are such surjections with critical Hölder exponent for any pair of dimensions k < n. Stong showed that there is a bijection $$\mathbb Z^k \to \mathbb Z^n$$ that is Hölder continuous with exponent $$k/n$$:

R. Stong, Mapping $$\mathbb Z^r$$ into $$\mathbb Z^s$$ with Maximal Contraction, Discrete Comput Geom 20:131–138 (1998)

A limit construction can then be used to obtain surjections from $$\mathbb R^k$$ to $$\mathbb R^n$$ of the same regularity, which also implies the surjection result for cubes. Some details and further interesting discussions about such maps are contained in section 9.1 of the following notes by Semmes:

S. Semmes, Where the Buffalo Roam: Infinite Processes and Infinite Complexity, arXiv:math/0302308v1 (2003)

(This is not a complete answer, but I cannot comment.)

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Edit: To clarify, I should first observe, as Willie Wong points out in the comments, that there is a typo in the original question. Simple Hausdorff dimension arguments show that there can be no $$\alpha$$-Hölder map from $$I^k$$ onto $$I^n$$ with $$\alpha>k/n$$. (The question says the opposite.) The question is therefore: what is the largest value of $$\alpha$$ for which we can find such a map?

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It follows immediately from the main result of

Tamás Keleti, András Máthé, Ondřej Zindulka: Hausdorff dimension of metric spaces and Lipschitz maps onto cubes, https://arxiv.org/pdf/1203.0686.pdf

that for every $$\alpha there is an $$\alpha$$-Hölder map from $$I^k$$ onto $$I^n$$.

In the general metric space case of that theorem, a map with optimal Hölder exponent need not exist.

In this very specific case, I would bet that it does, but I don't have a specific construction in mind.

• The question has a typo. It is always easier to get maps with smaller Hölder exponent. Apr 27, 2015 at 12:45
• I can believe that it might follow from the main theorem of that paper, but I don't see that it follows immediately! I am a bit rusty - maybe I am missing a well-known fact about maps between cubes in $\mathbb{R}^n$? The theorem in the paper does not talk about surjective maps - although in the case of maps to the interval, surjectivity is obvious, I can imagine that this need not be the case in higher dimensions. Can you provide more details of why that theorem implies your claim?
– jwg
Apr 29, 2015 at 15:59
• Yes, the only theorem in that paper that deals with Hoelder maps is Thm 2.1, but has nothing or very little to see with the present problem. Jan 8, 2016 at 10:25

Arnold posed a problem (1988-5 in "Arnold's problems") if there is a surjective map $$[0,1]^2 \to [0,1]^3$$ with Holder exponent $$2/3$$. E. V. Shchepin proved that one can get arbitrarily close to that (and to $$n/m$$ in generic case $$[0,1]^n\to[0,1]^m$$). See:

Shchepin, E.V. On Hölder maps of cubes. Math Notes 87, 757–767 (2010). https://doi.org/10.1134/S0001434610050135

The problem of construction of a map that attains exact exponent n/m -- remains open, as far as I know.