3
$\begingroup$

So I thought I had proven that every Lipschitz curve had finite length, but then I read what I think is a counterexample:

Let $\gamma :[0,1] \rightarrow \mathbb{R}^2$ the parametrization of a path such that $\gamma(t)=(t,t \sin(\frac{1}{t}))$ while $t>0$ and $\gamma(0)=(0,0)$.

This curve is continuous and $\mathcal{C}^1$ everywhere in [0,1] , so it is Lipschitz, but it's length is infinite, isn't it? Am i missing something?

$\endgroup$
4
  • $\begingroup$ It is not $C^1$: it has no derivative in $t=0$. It is not Lipschitz, too. $\endgroup$
    – Mizar
    Commented Jun 14, 2014 at 8:58
  • $\begingroup$ Oh yeah, that's right, there is no derivative in 0. I still don't see why it is not Lipschitz, though. $\endgroup$
    – Alubeixu
    Commented Jun 14, 2014 at 9:37
  • $\begingroup$ Well, in $(0,1]$ $\gamma(t)$ is differentiable. You can easily convince yourself that if $\gamma$ were Lipschitz in $(0,1]$ then its derivative would be bounded here. But, if you compute it, you get $\gamma'(t)=(1,\sin(\frac{1}{t})-\frac{1}{t}\sin(\frac{1}{t}))$, which is not bounded. $\endgroup$
    – Mizar
    Commented Jun 14, 2014 at 10:14
  • $\begingroup$ Thanks! I get it now! (Shouldn't there be a cosine in your derivative? :P) $\endgroup$
    – Alubeixu
    Commented Jun 14, 2014 at 10:26

1 Answer 1

1
$\begingroup$

Just so this has an answer: A Lipschitz path has bounded velocity. The velocity of $\gamma$, $$ \gamma'(t) = \left(1, \sin(\tfrac{1}{t}) - \tfrac{1}{t}\cos(\tfrac{1}{t})\right), $$ is unbounded: If $k > 0$ is an integer, then $0 < \frac{1}{2k\pi} < 1$, but $\|\gamma'(\frac{1}{2k\pi})\| = \left\|\left(1, -2k\pi\right)\right\| \geq 2k$ can be made arbitrarily large. That is, $\gamma$ is not Lipschitz on $[0, 1]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .