We want to prove that the following curve is not rectifiable.
$$ f\colon [0,1] \rightarrow \mathbb{R}^2, \quad f(x) = \begin{cases} (0,0) \, &\quad \text{for } x= 0 \\ (x,x^2\cos(\frac{\pi}{x})) &\quad \text{for } x> 0.\end{cases} $$
My idea: I need to show that the Length $L > \sum\limits_{i=1}^n \|f(x_j) -f(x_{j-1})\| $ and the sum diverges for one partition of $[0,1]$. First, I choose $0 < \frac{1}{k} < \cdots <\frac{1}{2} <1$ as my partition. Then $\cos(\pi j) = \pm 1$ and $f(x_j) = (\frac{1}{j},\frac{1}{j^2})$. So $\|f(x_j) – f(x_{j-1}) \| = |\frac{1}{j^2} + \frac{1}{(j+1)^2} |$. But this sum will not diverge. And if I choose $x_j = \frac{1}{\sqrt k}$, then $\cos$ will not be $\pm 1$.
Could you provide me a partition to show the non-rectifiability of $f$. Thanks in advance!
Best Answer
It is easy to see that a curve $(x(t),y(t))$ is rectifiable if and only if $x(t), y(t)$ are of bounded variation. In our case obviously $x(t)$ is of bounded variation, but the same holds for $y(t)=t^2\cos(\pi/t), t>0$ and $y(0)=0$.
Edited
One can easily see that the limit $\lim_{t\to0+}y(t)/t$ is $0$, therefore $y$ is differentiable on $[0,1]$ and $y'(0)=0$. Now for $t>0$ it is $y'(t)=2t\cos(\pi/t)+\pi\sin(\pi/t)$ therefore $|y'(t)|\leq2+\pi$ for all $t\in(0,1]$. Since $0<2+\pi$, we have that $|y'(t)|\leq 2+\pi$ for all $t\in[0,1]$. Now by a simple application of the mean value theorem, you get that the variation of $y(t)$ over $[0,1]$ is less than $2+\pi$, a finite quantity.