# Measure-preserving maps from the square to the cube

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?

• 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.
– Hans
Mar 4, 2013 at 19:50
• 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. Mar 5, 2013 at 5:22
• Do you have an example of an onto map $[0,1]^2\to [0,1]^3$ with the Hölder's exponent $2/3$?
– ε-δ
Mar 7, 2013 at 21:58
• No I do not have this either. It may be somewhat easier to drop the measure preserving requirement. Mar 20, 2013 at 19:37

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.
Here is Fig.12b of the cited paper, showing their Hilbert surface, whose fractal dimension approaches $$3$$. (*Image added by J.O'Rourke*):