Suppose $\alpha$ is a curve parametrized by arc length and there is some $s_0$ such that $||\alpha(s)||\le ||\alpha(s_0)||$, $\forall s$ near $s_0$. Show that: $$\kappa(s_0) \ge \frac{1}{||\alpha(s_0)||}.$$

I know the curvature that is parametrized by arc length equals $\kappa = ||\frac{dT}{ds}||$ and since $||\alpha(s)||\le ||\alpha(s_0)||$ then the point $s_0$ is a local maximum of $f(s)=||\alpha(s)||?$

  • $\begingroup$ Do you mean the curvature at $s_0$? $\endgroup$
    – lhf
    Commented Sep 17, 2013 at 2:03

1 Answer 1


You correctly observed that the key fact is that $\Vert \alpha(s) \Vert$ has a local maximum at $s_0$. Thus we have (using $\Vert\alpha(s)\Vert^2$ instead for computational convenience) in particular

$$\frac12 \frac{d^{2}}{ds^{2}}\bigg|_{s=s_{0}}\left\Vert \alpha\left(s\right)\right\Vert ^{2}=\left\Vert \alpha'\left(s_{0}\right)\right\Vert ^{2}+\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le0.$$

The first term here is $1$ because the curve is arclength parametrized; thus we have $\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le-1$ and so from Cauchy-Schwarz we have $\left\Vert \alpha''\left(s_{0}\right)\right\Vert \left\Vert \alpha\left(s_{0}\right)\right\Vert \ge1$. Noting that the curvature of $\alpha$ at $s_{0}$ is just $\left\Vert \alpha''\left(s_{0}\right)\right\Vert $ (once again because $\alpha$ is arclength parametrized), we are done.

  • $\begingroup$ Thank you! I would have never guess to use $$\frac{d^{2}}{ds^{2}}\bigg|_{s=s_{0}}\left\Vert \alpha\left(s\right)\right\Vert ^{2}=\left\Vert \alpha'\left(s_{0}\right)\right\Vert ^{2}+\left\langle \alpha\left(s_{0}\right),\alpha''\left(s_{0}\right)\right\rangle \le0.$$ How did you derive that? $\endgroup$
    – Lays
    Commented Sep 17, 2013 at 8:06
  • $\begingroup$ @Lays: the fact that the second derivative is non-positive is from the local maximum. The formula I found for the second derivative comes from the product rule $\langle u, v \rangle' = \langle u',v \rangle + \langle u, v' \rangle$ applied twice. Also I left a factor of two, I'll correct my answer. $\endgroup$ Commented Sep 17, 2013 at 8:52
  • $\begingroup$ I see that you divided by $1/2$, where did that come from? Because when I took the product rule I got $$2\langle \alpha'(s), \alpha'(s)\rangle + \langle \alpha''(s), \alpha(s)\rangle + \langle \alpha(s), \alpha''(s)\rangle?$$ $\endgroup$
    – Lays
    Commented Sep 17, 2013 at 19:43
  • 1
    $\begingroup$ @Lays: that's correct: the inner product is commutative, so you should be able to factor out a 2 from the whole expression. I just stuck a $1/2$ on the other side because it was the fastest way to fix it up with an edit. $\endgroup$ Commented Sep 18, 2013 at 2:02

You must log in to answer this question.

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