We've been given a function and asked to find whether it is a rectifiable path or not. Let $\gamma : [0,1]\to \mathbb{C} $ defined as

$\gamma (t)= t + \iota t \sin(\frac{1}{t})$

and $\gamma (0)=0$

Ofcourse it is a path (as by definition, it is continuous in the domain, so a path)

Now we need to show it is not rectifiable (this is given in the answer key)

I have a definition that a function $f$ is rectifiable if it is a function of bounded variation.

So we need to find a partition P of $[0,1]$ such that

$V(\gamma , P)=\sum_{k=1}^n |\gamma (t_k)-\gamma (t_{k-1})|$

is not bounded

I am not able to find such a partition.

Please help me if there is some other approach. Please use basic techniques as I am new to this course of complex analysis. Thanks!

  • $\begingroup$ What is iota ??? $\endgroup$ – Rene Schipperus Sep 27 '14 at 13:40
  • $\begingroup$ The question doesn't mention. So, i guess $\iota=√-1$ $\endgroup$ – user141561 Sep 27 '14 at 13:42

Take the partition $t_n=\frac{2}{n\pi}$ Then for $n$ odd we have the point, $(\frac{2}{n\pi},\frac{2}{n\pi})$ and for $n$ even we have $(\frac{2}{n\pi},0)$ So

$$|\gamma(t_n)-\gamma(t_{n-1})|=\frac{2}{n\pi}\sqrt{(\frac{1}{n-1})^2+1}$$ and it is easy to see that $$\sum_n\frac{2}{n\pi}\sqrt{(\frac{1}{n-1})^2+1}$$ diverges by a limit comparison with $\frac{1}{n}$.

  • $\begingroup$ Thank you! :) I had tried with $\frac{1}{2n\pi}$ which lead nowhere. $\endgroup$ – user141561 Sep 27 '14 at 21:45

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