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?
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asked Mar 29, 2015 at 21:54
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10$\begingroup$ The answer to your last question must be no, since we could trivially extend a space filling curve by defining it on an extra segment, on which it was very nice, or by inserting a nice smooth function on a segment in the middle, before picking up the space-filling behavior again right at the place we left off. $\endgroup$– Joel David HamkinsCommented Mar 29, 2015 at 21:57
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10$\begingroup$ en.wikipedia.org/wiki/Sard%27s_theorem $\endgroup$– Qiaochu YuanCommented Mar 29, 2015 at 22:02
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7$\begingroup$ @QiaochuYuan isn't Sard's theorem about $C^k$ maps? And the question asks for differentiable, not $C^1$... $\endgroup$– Yemon ChoiCommented Mar 29, 2015 at 22:54
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1$\begingroup$ How research question was it? :-) $\endgroup$– Włodzimierz HolsztyńskiCommented Mar 30, 2015 at 7:24
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2$\begingroup$ @LennartMeier Yes, I know that Sard would apply in this case for $C^1$. My point (although perhaps not one that the OP had in mind) is that $C^1$ is strictly stronger than being everywhere differentiable, and for these kinds of functions I have no good intuition whether that makes a difference to the original question $\endgroup$– Yemon ChoiCommented Mar 30, 2015 at 14:06
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5 Answers
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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 $0$.
answered Mar 30, 2015 at 6:34
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3$\begingroup$ However, a differentiable map (even on a compact interval) is not necessarily Lipschitz. $\hspace{1.27 in}$ $\endgroup$– user5810Commented Mar 30, 2015 at 8:08
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1$\begingroup$ @Ricky - a differentiable map $f$ from $\mathbb{R}$ is not necessarily Lipschitz, but one can partition $\mathbb{R}$ into countably many sets $A_i$ such that the restriction of $f$ to $A_i$ is Lipschitz. $\endgroup$ Commented Mar 30, 2015 at 20:08
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$\begingroup$ @PabloShmerkin : $\:$ How can one show that? $\;\;\;\;$ $\endgroup$– user5810Commented Mar 31, 2015 at 2:17
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$\begingroup$ @Ricky - First partition $\mathbb{R}$ into the sets on which the differential has norm $\le M$ (these are Borel sets). Let $A=A(M)$ be one of these sets. For $x\in A$, there is $\delta(x)>0$ (a Borel function) such that $|f(x)-f(y)|<2M$ when $|y-x|<\delta$. Cover $A$ by $A_n=\{ x\in A: \delta(x)<1/n\}$. Then $f$ is (locally) Lipschitz on $A_n$. $\endgroup$ Commented Mar 31, 2015 at 12:45
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$\begingroup$ That's not quite right, but it can be fixed. I'll edit my answer. $\endgroup$ Commented Mar 31, 2015 at 15:51
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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.
answered Mar 30, 2015 at 7:33
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$\begingroup$ A great theorem! But only a "Peano type function", not really a curve, since neither $f_1$ nor $f_2$ is continuous. $\endgroup$ Commented Mar 30, 2015 at 9:57
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1$\begingroup$ The cited paper: Michal Morayne, "On differentiability of Peano type functions," Colloquium Mathematicum, Volume LIII, 1987. $\endgroup$ Commented Mar 30, 2015 at 10:09
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$\begingroup$ @Goldstern: Thank you for the comment. I edited my answer accordingly. Of course, that doesn't affect the argument that there is no differentiable surjection from $\mathbb R$ to $\mathbb R^2$. $\endgroup$ Commented Mar 30, 2015 at 13:14
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$\begingroup$ Apologies for asking this, but could you point out where in Morayne's arguments one can see that no space filling curve can be differentiable in both components in every point? Thanks. $\endgroup$ Commented Mar 30, 2015 at 23:58
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2$\begingroup$ @Joseph O'Rourke: Unfortunately I am unable to access Morayne's article right now. But in the proof of CH from the existence of a Peano type function that is differentiable in at least one component at every point, he constructs sets $S_0$ and $S_1$ (might be called differently) and argues that these two sets satisfy Sierpinski's condition. I.e., $S_0\cup S_1=M_0\times M_1$ where the $M_i$ are of size $2^{\aleph_0}$ and the sections of $S_i$ in the direction of the $M_i$-axis are countable. If the function were differentiable, these sections would have to be uncountable. $\endgroup$ Commented Mar 31, 2015 at 4:18
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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.
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$\begingroup$ "a space-filling curve can be differentiable almost everywhere": Interesting! $\endgroup$ Commented Mar 30, 2015 at 11:30
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4$\begingroup$ Alternatively, if $f : [0,1] \to [0,1]^2$ is a space-filling curve and $\phi : [0,1] \to [0,1]$ is the Cantor staircase function, then $f \circ \phi$ is a space-filling curve whose derivative exists and vanishes almost everywhere. But this raises an interesting question: is there an easy way to show that a space-filling curve cannot be absolutely continuous? $\endgroup$ Commented Mar 30, 2015 at 15:21
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3$\begingroup$ Oh, yes there is: an absolutely continuous curve is rectifiable (its length is given by $\int_0^1 |f'(t)|\,dt$). $\endgroup$ Commented Mar 30, 2015 at 21:59
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$\begingroup$ Of course a Cantor set can be continuously mapped on any cube of any finite or countable dimension. Then you can extend such a map to a continuous map about any way you want it. There is really nothing interesting at all that the extended part can be differantiable. $\endgroup$ Commented Apr 2, 2015 at 6:23
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$\begingroup$ Must a space-filling curve be non-differentiable uncountably often? $\endgroup$ Commented Jan 25, 2019 at 21:23
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answered Mar 30, 2015 at 0:20
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1$\begingroup$ @GeorgeLowther's proof (your 1st link, in response to an MSE question, "Range of curve on a compact interval is nowhere dense"), is very clean! $\endgroup$ Commented Mar 30, 2015 at 0:33
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1$\begingroup$ I am not completely convinced by the second argument; I made an objection in a comment there, I would be interested in a fix. $\endgroup$ Commented Mar 30, 2015 at 7:13
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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.
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$\begingroup$ Sorry I read now that this solution is already contained in the comments! $\endgroup$– CepuCommented Mar 30, 2015 at 12:15