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How does one find the arc length of the Cantor function? Wikipedia says that the length is $2$. I can "see" that the length is at most $2$ by a simple triangle inequality argument. I am struggling to come up with a partition $P$ such that the arc length is at least 2. I tried a partition of the form $\{ 1/ 3^n : 0 \le k \le n \}$ but I guess I am making some mistake in my calculation so that I get the length as $3/4$ instead of close to $2$.

hbghlyj
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Shibi Vasudevan
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  • You can approximate the Cantor function by a sequence of piecewise linear functions whose arc lengths converge to 2, forming a lower bound on the arc length of the Cantor function. –  Mar 18 '11 at 22:35
  • Thanks for the hint. Could you please elaborate a little bit here. How does one come up with the piecewise functions. – Shibi Vasudevan Mar 18 '11 at 23:15

2 Answers2

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Possibly the easiest way to see this is to note that any partition divides $[0,1]$ into two kinds of intervals: those on which $f$ is constant, and those on which it is not. The total length of the constant intervals can be made arbitrarily close to 1, while the total length of the nonconstant intervals in any partition is at least 1 because they add up to a displacement of 1 on the $y$-axis. The result follows.

  • Yes, that is much easier than my method, and more general. – Jonas Meyer Mar 19 '11 at 01:45
  • Why 'nonconstant intervals must add up to at least 1"? It is not true if the 'slop' of the nonconstant part is great than 1. At least it is not 'clear' – user60933 Jan 06 '14 at 22:43
  • The total length of the constant intervals can be made arbitrary close to $1$ from below, so your argument does not prove that the length is at least $2$. – uniquesolution May 19 '23 at 22:43
  • @uniquesolution If the length is at least $2-\varepsilon$ for every $\varepsilon >0$, then the length is at least $2$. This is a common argument. – Brian Moehring Jun 06 '23 at 19:00
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Partitions of the form $\{\frac{k}{3^n}:0\leq k\leq n\}$ would do the job of showing that the arclength is at least $2$. I don't know what you did exactly, but notice that the partition $\{0,1\}$ shows that the arclength is at least $\sqrt 2$. The partition $\{0,\frac{1}{3},\frac{2}{3},1\}$ shows that the arclength is at least $\frac{1}{3}(\sqrt{13}+1)$. And so on; but we would need some good way to keep track of things to see that the sum goes to $2$ as $n$ goes to infinity.

Edit: The previous version of my answer had a serious error, which has been fixed.

Here is an approach using the symmetry of the function. Let $A$ denote the desired arclength, and for each positive integer $k$, let $A_k$ denote the arclength of the restriction to $\left[0,\frac{1}{3^k}\right]$. I assert without proof something that is geometrically clear from the graph: $A_k=2A_{k+1}+\frac{1}{3^{k+1}}$. This comes from splitting the interval $\left[0,\frac{1}{3^k}\right]$ into thirds, and noticing that the portions of the graph on the outside thirds are congruent, while on the middle third the graph is a horizontal segment. This leads to $A$ being expressed in terms of $A_k$ as $$A=2^kA_k+\frac{1}{3}\sum_{j=0}^{k-1}\left(\frac{2}{3}\right)^j,$$ which is easily proved by induction using the prior equation. Notice that $A_k\geq \frac{1}{2^k}$ for all $k$, and $\displaystyle{\sum_{j=0}^{k-1}\left(\frac{2}{3}\right)^j}$ goes to $3$ as $k$ goes to infinity, so letting $k$ go to infinity shows that the arclength is at least $2$.

Jonas Meyer
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