Let $c:I\rightarrow\mathbb R^2$ be an arc length parameterized curve. Assume also that $c$ stays inside a disk of radius $r$ and at sone $t_0$, $|c(t_0)|=r $.

I have shown that $c(t)$ and $c''(t)$ are collinear but I'm having trouble showing that the curvature at $t_0$ verifies $\lvert \rho(t_0)\rvert \geq \dfrac{1}{r}$.

Any ideas on how to get there? I think I need to différentiate but I can't seem to get thé right result.


1 Answer 1


You want to show that the curvature of $c$ at $c(t_0)$ is bigger than that of the circle.

After a translation and rotation, assume that $c(t_0)$ is the origin and the circle is given by $$ x^2 + (y-r)^2 = r^2.$$ (So it touches the origin from above). Assume that $c$ and the circle are locally given by

$$ (x, f_1(x)), (x, f_2(x))$$

respectively, so that $f_i(0) = f'_i(0) = 0$ for $i=1,2$. The assuption that $c$ is inside the circle gives $f_1(x) \ge f_2(x)$. All these imply $f{''}_1(0)\ge f_2^{''}(0)\ge 0$. (See remark)

The curvature of a graph is given by (here)

$$\kappa_i = \frac{|f_i^{''}|}{(1+ f_i')^{\frac 32}}.$$

Thus at the origin we have

$$ \kappa_1 = |f_1^{''}(0)| = f_1^{''}(0)\ge f_2^{''}(0) = \kappa_2 = \frac 1r.$$

Remark: Consider $g = f_1 - f_2$. Then $g(0) = g'(0) =0$ and $g(x)\ge 0$. This imply $g''(0)\ge 0$: If not, then $g''(0) = -\epsilon$ for some $\epsilon >0$. By definition, there is $\delta>0$ so that

$$ \frac{g'(x) - g'(0)}{x} < -\frac{\epsilon}{2} x $$

for all $x$ with $0<x<\delta$. This implies $g'(x) < \frac{\epsilon}{2} x$ and so

$$ g(x) = \int_0^x g'(s) ds < -\frac{\epsilon}{2} \int_0^x sds = -\frac{\epsilon}{4} x^2 <0$$

for all $0<x<\delta$, which is a contradiction. Thus $f_1^{''}(0)\ge f_2^{''}(0)$. That $f_2^{''}(0) \ge 0$ is a directly calculation ($f_2$ represents the circle).

  • $\begingroup$ I don't see why “all these” implies the inequality of the second derivatives. $\endgroup$ Commented Mar 16, 2017 at 15:04
  • $\begingroup$ @MichaelHoppe Please see the edit. $\endgroup$
    – user99914
    Commented Mar 16, 2017 at 18:48

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