Proof that no differentiable space-filling curve exists Could someone provide a reference or a sketch of a proof that no differentiable space-filling curve exists?
Or piecewise differentiable?
Must every continuous space-filling curve be nowhere differentiable?
 A: This is answered for $C^1$ curves here, and then the $C^1$ condition is weakened here.
A: There is a theorem by Michal Morayne saying that there is a space filling function $$f:\mathbb R\to\mathbb R^2;x\mapsto(f_1(x),f_2(x))$$ 
such that for all $x$ at least one of $f_1'(x)$ and $f_2'(x)$ exists if and only if the continuum hypothesis holds. 
This is proved here:
https://www.infona.pl/resource/bwmeta1.element.desklight-90a9a45c-fcc9-4b83-8ebf-cbd61a258fd9/content/partContents/8f463644-c16a-36c1-b87f-0d42dcd1b3c7
Note that the surjection $f:\mathbb R\to\mathbb R^2$ constructed by Morayne assuming CH
is not continuous, though.
However, Morayne's proof also shows that no space filling curve can be differentiable in both components in every point.
A: The image of an interval under a Lipschitz map has finite $1$-dimensional Hausdorff measure.
EDIT: Here's a corrected version of Pablo Shmerkin's construction. Suppose $f: \mathbb R \to \mathbb R^d$ is differentiable.
For positive integers $m,n$ let $A_{m,n} = \{x: |y -x| \le 1/n \implies \|f(y) - f(x)\| \le m |y - x|\}$. 
For $k \in \mathbb Z$ let $A_{m,n,k} = A_{m,n} \cap [(k-1)/n, k/n]$.  Then 
  $\bigcup_{m,n,k} A_{m,n,k} = \mathbb R$, and $f$ is Lipschitz on $A_{m,n,k}$ with Lipschitz constant $m$.  
We conclude that $f(\mathbb R)$ has $\sigma$-finite $1$-dimensional Hausdorff measure, which in particular implies that it has $2$-dimensional Lebesgue measure $0$.
A: Using the Sard's theorem.
Assume that your curves exists. Then Each point in the domain is a critical point and so each point in the image is a critical value.
But this is a contradiction with the sard's theorem since the set of critical values has measure zero.
A: Robert mentions the crucial issue that a Lipschitz map cannot increase Hausdorff dimension.
In the other direction we might ask, how well behaved can a space-filling curve be?  Lebesgue's space filling-curve is smooth on the compliment of Cantor's middle third set.  Thus, a space-filling curve can be differentiable almost everywhere.  The idea is simple.  If $x$ in the Cantor set has base three expansion
$$0.(2d_1)(2d_2)(2d_3)\cdots,$$
where each digit $d_i$ is zero or one, then define $f$ in terms of its binary expansion by
$$f(x) = (0.d_1d_3d_5\ldots,0.d_2d_4d_6\ldots).$$
Then, $f$ maps the Cantor set onto the unit square continuously and easily extends to the interval by connecting the dots.
Hans Sagan's Space-filling curves is, perhaps, the definitive reference on the topic and Lebesgue's curve is covered in chapter 5 of that text.
