Does there exist a Lipschitz map from the unit interval onto the unit square?

Question

It is well-known that continuous space-filling curves exist. But can they be Lipschitz?

Specifically, is there a Lipschitz map from [0,1] onto [0,1]x[0,1]?

Answer 1

It's a general fact that if $f$ is a Lipschitz mapping, then the Hausdorff dimension satisfies

$$\dim_{H} f(F) \le \dim_H F$$

for all $F$. If $F$ is the unit interval, then $\dim_H F = 1$, while the Hausdorff dimension of the square is $2$; this means no such Lipschitz map can exist.

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The above inequality can be found, e.g. as Corollary $2.4$ in Falconer's Fractal Geometry.

Answer 2

Thanks for your answers. The answer I came up with doesn't mention Hausdorff dimension, although you might say it's there in disguise:

Suppose there is such a map. Given a positive integer $N$, subdividing $[0,1]$ into $N$ equal intervals and considering the images of these will give a cover of $[0,1]^2$ having area $\leq NC/N^2$ where $C$ is a constant independent of $N$ (this is since the image of a sub-interval lies in a ball of diameter proportional to $1/N$). Letting $N \to \infty$ gives $\big| [0,1]^2 \big|=0$, which cannot be.