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There is a measure preserving map from the unit interval onto the unit cube that is Lipschitz of order 1/2, that is $|f(x)-f(y)| \leq A |x-y|^{1/2}$. By considering the image of small intervals, one can see that one could not have a smoother map.

Now consider maps from $[0,1]^2$ onto $[0,1]^3$ that preserve measure. By looking at the image of small balls we see that f can't be smoother than Lipschiz 2/3.

Does there exist a measure preserving map from $[0,1]^2$ onto $[0,1]^3$ that is Lipschitz 2/3?

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  • $\begingroup$ I am not familiar with that stuff but maybe you could approximate 2/3 binary and combine this with that what you know. For instance: $id \times f: R^2 \times R \rightarrow R^2 \times R^2$ brings you $R^3 \rightarrow R^4$. Doubling it with $f$ brings with bring $R^6 \rightarrow R^8$. Concatenating one $f$ brings $R^6 \rightarrow R^9$. Taking third root brings a first approximation $R^2 \rightarrow R^3$. Etc. - Maybe its a phantasy. $\endgroup$
    – Hans
    Commented Mar 4, 2013 at 19:50
  • $\begingroup$ Interesting question... I'm guessing you mean that there's a map onto the unit square that's Lipschitz of order 1/2? I think your argument would say you can't have anything onto the cube of order greater than 1/3. $\endgroup$
    – Anthony Quas
    Commented Mar 5, 2013 at 5:22
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    $\begingroup$ Do you have an example of an onto map $[0,1]^2\to [0,1]^3$ with the Hölder's exponent $2/3$? $\endgroup$
    – ε-δ
    Commented Mar 7, 2013 at 21:58
  • $\begingroup$ No I do not have this either. It may be somewhat easier to drop the measure preserving requirement. $\endgroup$
    – Mike Steele
    Commented Mar 20, 2013 at 19:37

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There seems to be a $\frac23$-Hölder (or $\frac23$-Lipschitz) onto map $[0,1]^2\to[0,1]^3$. It was apparently found in this paper. I don't know if it has been done elsewhere.

The space filling curves due to Hilbert, Peano and Sierpinski preserve measure and the cited article contains generalizations of these, so it seems quite possible that measure is preserved by some of the corresponding surfaces as well.

I have only been able to access the abstract. It would be great if someone with full text access could check the exact results in the paper.

Here is Fig.12b of the cited paper, showing their Hilbert surface, whose fractal dimension approaches $3$. (*Image added by J.O'Rourke*):
      ![Fig12Hilbert][1]
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  • $\begingroup$ Appropriately, the Hilbert surface has H-shaped regions :-) $\endgroup$
    – Qfwfq
    Commented Mar 29, 2019 at 12:00

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