Lipschitz space-filling maps First, some preliminaries and context.
Let $f \colon [0,1]\to[0,1]^2$ be a space-filling curve. If we put on $[0,1]$ and $[0,1]^2$ the standard Euclidean metrics induced by $\mathbb{R}$ and $\mathbb{R}^2$ respectively, it is somewhat well known that $f$ can't be Lipschitz, because Lipschitz maps don't increase Hausdorff dimension, and we have $\dim_H([0,1])=1$ and $\dim_H([0,1]^2)=2$, where $\dim_H$ denotes the Hausdorff dimension.
To hope for such a map to be Lipschitz we can put on $[0,1]$ the metric given by $|\cdot|^{\frac12}$, where $|\cdot|$ denotes the standard metric. In this case, since Hausdorff dimension relies on the metric, we have $\widetilde{\dim}_H([0,1])=\dim_H([0,1]^2)=2$, where $\widetilde{\dim}_H$ denotes the Hausdorff dimension with respect to the metric $|\cdot|^{\frac12}$ on $[0,1]$. Hence, it is a priori possible to find a Lipschitz space-filling curve $f$. The fact that $\widetilde{\dim}_H([0,1])=2$ is an easy calculation using the fact that $\dim_H([0,1])=1$.

I should be able to prove, using some nice properties of dyadic numbers that the Hilbert curve is indeed Lipschitz in this case (with $L=1$). I couldn't extend such proof to other curves which do not have any nice dyadic structure, for instance the Peano curve.  I could attempt to write down the proof, but for the time being this is not helpful to the purpose of my question so I will postpone until someone requires it.

I have two questions:

*

*Does anybody have a proof of the fact that a particular space-filling curve is Lipschitz in the sense described above? I would be happy to see any proof, elementary or not.

*Is it true that any space-filling curve is Lipschitz? If yes, why? If not, is there a counterexample?


I'm interested in this because such a map is a typical example of a Lipschitz map which does not have any biLipschitz piece.
Any comment or partial answer is very welcome.
 A: First, notation/terminology: Talking about Lipschitz maps with respect to that nonstandard metric is a bad idea, simply because there exists a simple and much more standard way to say the same thing. Your function is Lipschitz with respect to that funny metric if $$|f(x)-f(y)|\le c|x-y|^{1/2}.$$
This is exactly the definition of $$f\in \mathrm{Lip}_{1/2}.$$That $\mathrm{Lip}_{1/2}$ thing is known as a Holder class.
Is every space-filling curve $\mathrm{Lip}_{1/2}$? Of course not. Silly counterexample: Say $f:[0,1]\to[0,1]^2$ is surjective. Say $g:[0,2]\to[0,1]^2$ and $g|_{[0,1]}=f$. Then $g$ is space-filling, and $g$ can do whatever you want on $[1,2]$.
Yes, the Hilbert curve $f$ is $\mathrm{Lip}_{1/2}$. This follows from the following: If $I=[j4^{-n},(j+1)4^{-n}]$ then $f(I)=[k2^{-n},(k+1)2^{-n}]\times[l2^{-n},(l+1)2^{-n}]$. Now if $x,y\in[0,1]$ then $x,y\in I_1\cup I_2=[j4^{-n},(j+2)4^{-n}]$ where $|x-y|\sim 4^{-n}$; since $f(I_2)\cap f(I_2)\ne\emptyset$ the diameter of $f(I_1\cup I_2)$ is less than $c2^{-n}$, and there you are.
I suspect the same applies to the Peano curve, with $3$ and $9$ in place of $2$ and $4$. Not sure, not really familiar with the construction.
