# Hint as to how to formalize that the curvature of a curve inside the unit circle is bounded below by the curvature of the circle

I am trying to self teach differential geometry and to that effect I am trying to do the Homework in the MIT open course.

The specific question I am struggling with is:

Let $$c$$ be a regular curve such that $$|c(s)| ≤ 1$$ for all $$s$$. Suppose

that there is a point $$t$$ where $$|c(t)| = 1$$. Prove that the curvature at that

point satisfies $$|κ(t)| ≥ 1$$

This is what I have so far: The curve being regular implies that it can be arc length parametrized, which means that it's curvature is just the norm of the second derivative, or in 2D $$|x'y'' - x''y'|$$.

The curve magnitude being upper bounded by 1 means the curve is fully contained within the unit disk.

The curve being one at $$t$$ means that point is ON the unit circle. Thus my intuition says this curve can only "bend inwards" at a rate equal or higher than the circle, otherwise it will be an epsilon outside of the circle at a point infinitesimally close to $$t$$. In other words it seems to me that a if a curve with a point on the unit circle has a smaller curvature than that of the unit circle (i.e 1) the curve will "pop out" of it for some infinitesimal amount. But I am not sure if A) this is correct B) how to formalize it.

I am looking mostly for a hint and advice, not for the full solution. Thank you lots in advance.

Let $$t_0$$ be the point satisfying $$\| \alpha(t_0)\| = 1$$. Then, at $$t_0$$, the function $$t \mapsto \|\alpha(t) \|^2$$

assumes a maximum. Thus, its first order derivative at $$t_0$$ vanishes, that is:

$$2 \langle \alpha(t_0), \alpha'(t_0)\rangle = 0$$

And by the second derivative test, its second order derivative at $$t_0$$ is at most $$0$$, that is:

$$2 \langle \alpha'(t_0), \alpha'(t_0)\rangle + 2 \langle \alpha''(t_0), \alpha'(t_0) \rangle \leq 0$$

Assuming without loss of generality that $$\alpha$$ has unit speed, we get:

$$1 \leq \langle \alpha''(t_0), \alpha'(t_0) \rangle$$

Taking the norm of both sides and using the Cauchy-Schwarz inequality, we have:

$$1 \leq | \langle \alpha''(t_0), \alpha'(t_0) \rangle | \leq \|\alpha''(t_0)\| \cdot \|\alpha'(t_0)\| = |\kappa(t_0)|$$

as desired.

• Sorry to be nitpicky but the second derivative test states: $f''(x_0) < 0 \implies x_0$ is a maximum, however you are stating the converse, i.e $x_0$ is a maximum $\implies f''(x_0) \leq 0$ Is that claim true? Can you please link a source, that;s the only part I don't fully follow. – Makogan May 29 '20 at 6:51
• You have it backwards. The second derivative test gives necessary conditions for a point to be a local maximum: namely, that the first order derivative is 0 at the point and that the second one is non positive. As for a source, I'm on mobile right now but you can see a recent question I asked on my profile here where I proved the most general case possible. – Matheus Andrade May 29 '20 at 6:54
• I think there's an issue Imma edit your answer a little bit. – Makogan May 30 '20 at 3:35