Elementary ways to calculate the arc length of the Cantor function (and singular function in general)

Cantor's function: http://en.wikipedia.org/wiki/Cantor_function

There is an elementary way to prove that the arc length of the Cantor function is 2?

In this article (http://www.math.helsinki.fi/analysis/seminar/esitelmat/stat0312.pdf) they use the following result:

If $f:[a,b] \rightarrow \mathbb{R}$ is a continuous monotone function, then $f$ is singular if and only if $$L_a^b = |f(a)-f(b)|+|a-b|$$

But, there is a way for calculate the arc length of singular function without using this property? like using the arc length definition

If $X$ is a metric space with metric $d$, then we can define the ''length'' of a curve $\!\,\gamma : [a, b] \rightarrow X$ by $$\text{length} (\gamma)=\sup \left\{ \sum_{i=1}^n d(\gamma(t_i),\gamma(t_{i-1})) : n \in \mathbb{N} \text{ and } a = t_0 < t_1 < \cdots < t_n = b \right\}. $$

where the sup is over all $n$ and all partitions $t_0 < t_1 < \cdots < t_n$ of $[a, b]$.

Solution 1:

Heh. Yes, what's below is so an answer to the question. He didn't say he didn't want to use that theorem, he said he wanted an elementary proof. We give a totally elementary proof of the relevant half of the theorem:

Theorem Suppose $f:[a,b]\to\Bbb R$ is nondecreasing and satisfies $f'=0$ almost everywhere. Then $L(f)=b-a+f(b)-f(a)$.

Note In fact a monotone function must be differentiable almost everywhere, but we're not using that here. To be quite explicit the hypothesis is this: There exists a set $E$ of measure zero such that if $x\in[a,b]\setminus E$ then $f'(x)$ exists and equals zero.

The idea of the proof is simple: We cover most of the set where $f'=0$ by intervals where $f$ is close to constant. The sum of the lengths of those intervals is close to $b-a$ since $f'=0$ almost everywhere. And since $f$ is close to flat on those intervals, the total variation of $f$ on those intervals is small, so the variation of $f$ on the rest of $[a,b]$ is close to $f(b)-f(a)$. This is so far really what's going on in various arguments we've seen for the Cantor function. But since now the set where $f'=0$ need not be open or closed and we don't want to use measure theory there are some technicalities:

Proof: Say $\gamma(x)=(x,f(x))$. If $s<t$ then $||\gamma(s)-\gamma(t)||\le t-s+f(t)-f(s)$ by the triangle inequality; hence $$L(f)\le b-a+f(b)-f(a).$$

For the other inequality, let $\epsilon>0$. Suppose $E$ has measure zero and $f'=0$ on $[a,b]\setminus E$. Choose intervals $I_j=(a_j,b_j)$ with $$E\cup\{a,b\}\subset\bigcup_{j=1}^\infty I_j$$and $$\sum(b_j-a_j)<\epsilon.$$Wew can assume the $I_j$ are disjoint (If necessary, replace $(I_j)$ by the connected components of $\bigcup I_j$.)

Let $K=[a,b]\setminus\bigcup I_j$. Note that $K$ is a compact subset of $(a,b)$. If $x\in K$ then $f'(x)=0$, hence there exists $\delta_x>0$ such that $$f(x+\delta_x)-f(x-\delta_x)<2\epsilon\delta_x.$$Since $K$ is compact we have $K\subset\bigcup_{k=1}^NJ_k$ where $$J_k=(\alpha_k,\beta_k)$$and $$f(\beta_k)-f(\alpha_k)<\epsilon(\beta_k-\alpha_k).$$We may assume the $J_k$ are disjoint, since if two of them should intersect then their union is $(\alpha,\beta)$ with $f(\beta)-f(\alpha)<\epsilon(\beta-\alpha)$.

Since we have finitely many disjoint intervals, we may relabel them from left to right: $$\beta_k\le\alpha_{k+1}.$$

Note that if $s<t$ then $$||\gamma(s)-\gamma(t)||\ge t-s$$and $$||\gamma(s)-\gamma(t)||\ge f(t)-f(s).$$So considering the partition $$ a\le\alpha_1<\beta_1\le\alpha_2\dots<\beta_N\le b$$shows that $$L(f)\ge f(\alpha_1)-f(a)+\sum(\beta_k-\alpha_k)+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N).\quad(1).$$

Now $$f(b)-f(a)= f(\alpha_1)-f(a)+\sum(f(\beta_k)-f(\alpha_k))+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N)$$and $$\sum(f(\beta_k)-f(\alpha_k)) <\epsilon\sum(\beta_k-\alpha_k)\le\epsilon(b-a),$$so $$f(\alpha_1)-f(a)+\sum(f(\alpha_{k+1})-f(\beta_k))+f(b)-f(\beta_N)>f(b)-f(a)-\epsilon(b-a)\quad(2).$$

Since $[a,b]\setminus\bigcup I_j=K\subset\bigcup J_k$ we have $$[a,b]\subset\bigcup I_j\cup\bigcup J_k.$$

Now, it's easy to see by induction on $n$ that if $[a,b]$ is covered by $n$ open intervals then the sum of the lenghts of the intervals must be at least $b-a$. (Say $(\alpha,\beta)$ is one of the $n$ intervals and $b\in(\alpha,\beta)$; then $[a,\alpha]$ is covered by the $n-1$ other intervals...) So, since $I_j$ and $J_k$ are open and $[a,b]$ is compact the previous display implies that $$b-a\le\sum(b_j-a_j)+\sum(\beta_k-\alpha_k).$$Hence$$\sum(\beta_k-\alpha_k)\ge b-a-\epsilon\quad(3).$$ Combining (1), (2), and (3) shows that $$L(f)\ge b-a+f(b)-f(a)-\epsilon(b-a+1);$$hence $L(f)\ge b-a+f(b)-f(a)$.

Solution 2:

Here is an elementary way to derive the arc length of the Cantor Function:

We know that the following sequence of functions converges uniformly to the Cantor Function. Letting $c_0(x) = x$ and $n\ge 0$, we define $$c_{n+1}(x)= \begin{cases} \frac{1}{2}c_n(3x) & 0 \le x \le \frac{1}{3},\\ \frac{1}{2} & \frac{1}{3} \le x \le \frac{2}{3},\\ \frac{1}{2}c_n(3x-2)+\frac{1}{2} & \frac{2}{3}\le x \le 1. \end{cases} $$

We also know that once the sequence of functions goes constant on an interval, it stays constant on that interval. Then from the definition, we see that at $c_n$, $n\ge1$, we add $2^{n-1}$ constant portions of length $3^{-n}$ each. So for the totally horizontal (i.e. constant) portions of $c_n$, we have that they have total length $\displaystyle \sum_{i=0}^n \frac{2^k}{3^{k+1}}$. To find the lengths of the positively sloped portions for a given $c_n$, we note that from the definition, there are $2^n$ sloped portions, which increase the function $2^{-n}$ over a length of $3^{-n}$. Then we have that the total length of the positively sloped portions is $$2^n\sqrt{(\frac{1}{2^n})^2 + \frac{1}{3^n})^2} = \sqrt{\frac{(2^n)^2((2^n)^2 + (3^n)^2)}{(2^n)^2(3^n)^2}} = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}.$$

So then $L(c_n) = \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$, and so $$\lim_{n \to \infty}L(c_n) = \lim_{n \to \infty} \sqrt{\frac{(2^n)^2 + (3^n)^2)}{(3^n)^2}}+ \displaystyle \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k}$$ $$ = \sqrt{\lim_{n\to \infty}\frac{2^{2n}+3^{2n}}{3^{2n}}} + \lim_{n \to \infty} \frac{1}{3}\sum_{i=0}^n \frac{2^k}{3^k} = \sqrt{\lim_{n \to \infty}(\frac{2}{3})^n +1} + \frac{1}{3}\left(\frac{1}{1-\frac{2}{3}}\right)$$ $$= 1 + \frac{1}{3}\left(\frac{1}{\frac{1}{3}}\right) = 2.$$

Letting $c$ be the Cantor function, we have $L(c) \ge 2$, since $L(c) \ge L(c_n)$ for all $n$. To see this, note that on any constant portion of $c_n$, $c$ agrees with $c_n$. But on the diagonal portions of $c_n$, the graph of $c$ is broken up into more and more horizontal and diagonal portions. In essence, if on any $(a,b)$ on which $c_n$ is diagonal, $c$ is less diagonal. Then since the shortest distance between any two points is given by the straight line between them, we have that the length of $c$ is greater than the length of $c_n$ on that portion. Then we have that $L(c) \ge L(c_n)$ for all $n$, so $L(c) \ge 2$. To see that $L(c) \le 2$, note that we have $L(c) = \displaystyle \sup_{P} \sum_{i=1}^n \sqrt{(\triangle x_i)^2 + (\triangle c(x_i))^2} \le \sup_{P}\sum_{i=1}^n |\Delta x_i| + |\Delta c(x_i)| = \sup_{P}\sum_{i=1}^n \Delta x_i + \Delta c(x_i) =2$, since the total length of the horizontal segments is $1$, and since $c$ is monotonically increasing the total length of the vertical growth is $1$. Thus $L(c) =2$.

Solution 3:

It's trivial that $L_{a}^{b}\le (b-a) + |f(b)-f(a)|$ if $f$ is any monotone function on $[a,b]$, because $$||(s,f(s))-(t,f(t))||\le t-s+|f(t)-f(s)|$$if $s<t$. So the length of the Cantor function is less than or equal to $2$.

For the reverse inequality, note first that if $a=a_0<\dots<a_n=b$ then $$L_a^b=\sum_{j=1}^nL_{a_{j-1}}^{a_{j}}.$$So for the Cantor function we have $$L_0^1=\sum_{j=1}^{3^n}L_{(j-1)3^{-n}}^{j3^{-n}}.$$

Split that sum into two parts in the obvious way: Say $I_j=[(j-1)3^{-n},j3^{-n}]$. Write $L_j$ for the term in the sum corresponding to $I_j$.

Now, if $I_j$ is one of the intervals that has been deleted at the $n$th stage of the construction of the Cantor set it's clear that $L_j$ is just the length of $I_j$. The sum of those lengths tends to $1$ as $n\to\infty$, since the complement of the Cantor set has measure $1$.

On the other hand, if $I_j$ is one of the intervals that was not deleted then $$L_j\ge f(j3^{-n})-f((j-1)3^{-n})=2^{-n}.$$Since there are precisely $2^n$ non-deleted intervals $I_j$, the sum of $L_j$ for those values of $j$ is at least $1$.

So the sum above is the sum of the first part, which tends to $1$, plus the second part, which is greater than or equal to $1$; hence the limit of the sum is at least $2$.

Solution 4:

A rather simple proof can be given using the self-similarity of the Cantor function $c$. Of course, those self-similar parts scale differently in $x-$ and $y-$direction, and that scaling doesn't preserve length, so we have to parameterise: let $l(t)$ be the length of the graph of $t\,c(x)$ for $x\in[0,1]$. Because of monotony and the triangle inequality, we have $$\sqrt{1+t^2}\le l(t)\le 1+t\tag1$$ for $t>0$. The graph consists of two self-similar parts and a horizontal line of length $\dfrac13$ between them, so $$l(t)=\frac13\,l\left(\frac32\,t\right)+\frac13+\frac13\,l\left(\frac32\,t\right)=\frac23\,l\left(\frac32\,t\right)+\frac13,$$ implying $$l(t)-1=\frac23\,\left[l\left(\frac32\,t\right)-1\right]=\left(\frac23\right)^n\,\left[l\left(\left(\frac32\right)^n\,t\right)-1\right]\tag2$$ for $n\ge1$. Because of (1), we have $$\lim_{t\to\infty}\frac{l(t)}t=1,$$ so letting $n\to\infty$ in (2), we obtain $l(t)-1=t$.