If $\| \alpha(s) \| \leq \| \alpha(s_0) \| = R$, then the curvature $k(s_0)$ is greater than $1/R$

Question

This question is in Ted Shifrin's A first course in curves and surfaces, page 18, exercise 7:

Suppose $\alpha$ is an arclength-parametrized space curve with the property that $\| \alpha(s) \| \leq \| \alpha(s_0) \| = R$ for all $s$ sufficiently close to $s_0$ . Prove that $k(s_0) \geq 1/R$. (Hint: Consider the function $f(s)=\| \alpha(s)\|^2$. What do you know about $f''(s_0)$?)

Here's what I have tried:

$$f(s) = \| \alpha(s)\|^2 = \langle \alpha(s), \alpha(s) \rangle \implies f'(s) = 2 \langle T(s), \alpha(s) \rangle $$

And using the Cauchy-Schwartz inequality :

$$f'(s_0) = 2 \langle T(s_0), \alpha(s_0) \rangle \leq 2 \|T(s_0)\|\|\alpha(s_0)\|=2R \implies \langle T(s_0), \alpha(s_0) \rangle \leq R$$

Now we find $f''(s)$:

$$f''(s) = 2 \langle k(s)N(s), \alpha(s) \rangle + 2 (\langle T(s), \langle T(s \rangle )= 2 k(s) \langle N(s), \alpha(s) \rangle + 2$$

And because $s_0$ is maximum of $\| \alpha(s)\|$:

$$2 k(s_0) \langle N(s_0), \alpha(s_0) \rangle + 2 \leq 0 \implies -1/k(s_0) \geq \langle N(s_0), \alpha(s_0) \rangle$$

I don't know how to continue from here.

Answer 1

This question is a classic from curve and surface geometry. Good to run into it here, like an old friend one hasn't see in awhile.

These things being said:

We note the hypothesis that

$\Vert \alpha(s) \Vert \le \Vert \alpha(s_0) \Vert \tag 1$

essentially tells us that $s_0$ is a local maximum of the function $\Vert \alpha(s_0) \Vert$; since the real function $w \to w^2$ is strictly monotonically increasing for $w \ge 0$, $s_0$ will also be a local maximum for $\Vert \alpha(s) \Vert^2$; we shall work with this function in lieu of $\Vert \alpha(s) \Vert$.

We set

$f(s) = \Vert \alpha(s) \Vert^2 = \langle \alpha(s), \alpha(s) \rangle; \tag 2$

then

$f'(s) = 2\langle \alpha'(s), \alpha(s) \rangle = 2\langle T(s), \alpha(s) \rangle, \tag 3$

where

$T(s) =\alpha'(s) \tag 4$

is the unit tangent vector field to the curve $\alpha(s)$; we further have

$f''(s) = 2\langle T'(s), \alpha(s) \rangle + 2\langle T(s), T(s) \rangle; \tag{5}$

we may take this equation a step further by recalling that $T(s)$ is a unit vector and

$T'(s) = \kappa(s) N(s), \tag 6$

where $N(s)$ is the unit normal field to, and $\kappa(s) > 0$ is the curvature of, $\alpha(s)$. Thus,

$f''(s) = 2\kappa(s) \langle N(s), \alpha(s) \rangle + 2; \tag 7$

since $s_0$ is a local maximum of $f(s)$, we must have

$f''(s) \le 0; \tag 8$

thus,

$\kappa(s_0) \langle N(s_0), \alpha(s_0) \rangle + 1 \le 0, \tag 9$

or

$\kappa(s_0)\langle N(s_0), \alpha(s_0) \rangle \le -1; \tag{10}$

taking absolute values of both sides, we find

$\kappa(s_0)\vert \langle N(s_0), \alpha(s_0) \rangle \vert \ge 1; \tag{11}$

by Cauchy-Schwarz,

$\vert \langle N(s_0), \alpha(s_0) \rangle \vert \le \Vert N(s_0) \Vert \Vert \alpha(s_0) \Vert = \Vert \alpha(s_0) \Vert \tag{12}$

since

$\Vert N(s_0) \Vert = 1; \tag{13}$

combining (11) and (12), we obtain

$\kappa(s_0) R \ge 1, \tag{14}$

whence

$\kappa(s_0) \ge \dfrac{1}{R}. \tag{15}$

Answer 2

First of all, what is $f'(s_0)$, precisely? No inequalities here. You should be thinking about Cauchy-Schwarz in the analysis of the second derivative. Several warnings: First, $s_0$ is a maximum, not a minimum; but $f''(s_0)\le 0$ (not strict inequality). Second, think about the sign of $\langle N(s_0),\alpha(s_0)\rangle$. Last, don't divide by $k(s_0)$ if you want to solve for $k(s_0)$ :)