# 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:

1. 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.
2. 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.

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.
• Thanks a lot for your answer. The proof for the Hilbert curve you provide is exactly the one I had in mind, and I am also not familiar with the construction for the Peano curve, as I mentioned above. Regarding the Hölder class $H^{\frac12}$, I am well aware of its existence, but in my field (geometric measure theory) we care about Lipschitz maps between metric spaces, hence my perspective on the question. Aug 10, 2015 at 15:51