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If a closed plane curve $C$ is contained inside a disk of radius $r$, prove that there exists a point $p \in C$ such that the curvature k of C at p satisfies $\lvert k\rvert \ge$ $1/r$.

I understand that the curvature of a circle (or disk) is always 1/r but I don't know how to go about comparing the arbitrary curve to the curvature of the circle. Is there possible a curvature comparison theorem I'm missing because I've been unable to find anything in Stack Exchange or elsewhere?

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1 Answer 1

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The squared distance of the point $C(s)$ on the curve from the origin is

$f(s) = \langle C(s), C(s)\rangle, \tag 1$

where

$C:J \to D(0, r), \tag 2$

$J \subset \Bbb R$ an open interval; we assume as is usual that $C(s)$ is parametrized by its arc-length $s$ and that $C(s)$ is regular; thus differentiable with $C'(s) \ne 0$. Since $C(s)$ is closed, it is compact; hence $f(s)$ attains a maximum at some $s_0$; we have

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

where $T(s) = C'(s)$ is the unit tangent field to $C(s)$; at $s_0$,

$f'(s_0) = 0 \Longrightarrow \langle C(s_0), T(s_0) \rangle = 0; \tag 4$

$s_0$ a maximum yields

$f''(s_0) \le 0; \tag 5$

we have, via the Frenet-Serret equation $T'(s) = \kappa(s)N(s)$,

$f''(s) = 2\langle T(s), T(s) \rangle + 2\langle C(s), T'(s) \rangle = 2 + 2\langle C(s), \kappa(s) N(s) \rangle; \tag 6$

$2 + 2\langle C(s_0), \kappa(s_0) N(s_0) \rangle \le 0; \tag 7$

$1 + \langle C(s_0), \kappa(s_0) N(s_0) \rangle \le 0; \tag 8$

$\langle C(s_0), \kappa(s_0) N(s_0) \rangle \le -1; \tag 9$

(4) implies $C(s_0)$ collinear with $N(s_0)$; since $\kappa(s) > 0$, (9) implies

$C(s_0) = -\vert C(s_0) \vert N(s_0); \tag{10}$

since $\langle N(s), N(s) \rangle = 1$, (9) becomes

$-\vert C(s_0) \vert \kappa(s_0) = -\langle \vert C(s_0) \vert N(s_0), \kappa(s_0) N(s_0) \rangle \le -1, \tag{11}$

whence

$\vert C(s_0) \vert \kappa(s_0) \ge 1; \tag{12}$

finally,

$\kappa(s_0) \ge \dfrac{1}{\vert C(s_0) \vert} \ge \dfrac{1}{r}. \tag{13}$

Note Added in Edit, Tuesday 13 October 2020 5:00 PM PST: We inquire into whether there is an analogous result in $\Bbb R^3$; that is, whether an arc-length parametrized regular curve $C(s)$ contained in a ball $B(0, r)$ of radius $r$ centered at the origin $(0, 0, 0)$,

$C:J \to B(0, r) \tag{14}$

satisfies the curvature condition

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

for some

$s_0 \in J. \tag{16}$

It is relatively easy to see that the argument given above ca. (1)-(9) carries through to the three-dimensional case, but that now (4) implies

$C(s_0) = \langle C(s_0), N(s_0) \rangle N(s_0) + \langle C(s_0), B(s_0) \rangle B(s_0), \tag{17}$

where

$B(s_0) = T(s_0) \times N(s_0) \tag{18}$

is the unit binormal vector to the curve $C(s)$; we substitute (17) into (9) and obtain

$\langle \langle C(s_0), N(s_0) \rangle N(s_0) + \langle C(s_0), B(s_0) \rangle B(s_0), \kappa(s_0) N(s_0) \rangle \le -1; \tag{19}$

in light of (18), from which

$\langle B(s_0), N(s_0) \rangle = \langle T(s_0) \times N(s_0), N(s_0) \rangle = 0, \tag{20}$

(19) becomes

$\langle \langle C(s_0), N(s_0) \rangle N(s_0), \kappa(s_0) N(s_0) \rangle \le -1, \tag{21}$

or

$\langle \langle C(s_0), N(s_0) \rangle \kappa(s_0) \le -1; \tag{22}$

we recall that

$\vert \langle C(s_0), N(s_0) \rangle \vert = \vert C(s_0) \vert \vert N(s_0) \vert \vert \cos \theta \vert, \tag{23}$

where $\theta$ is the angle 'twixt $C(s_0)$ and $N(s_0)$; taking norms in (22) we find

$\vert \langle \langle C(s_0), N(s_0) \rangle \vert \kappa(s_0) \ge 1; \tag{24}$

we combine (23) and (24):

$\vert C(s_0) \vert \vert N(s_0) \vert \vert \cos \theta \vert \kappa(s_0) \ge 1; \tag{25}$

since

$\vert C(s_0) \vert \le r, \tag{26}$

(25) yields

$r \vert \cos \theta \vert \kappa(s_0) \ge 1, \tag{27}$

whence, since this implies

$\vert \cos \theta \vert \ne 0, \tag{28}$

$\kappa(s_0) \ge \dfrac{1}{\vert \cos \theta \vert r} \ge \dfrac{1}{r}, \tag{29}$

$OE\Delta$. End of Note.

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    $\begingroup$ Very neatly explained, thanks :D $\endgroup$
    – Azur
    Aug 25, 2020 at 10:04
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    $\begingroup$ Both results can be explained by finding the point on the curve farthest away from a fixed point inside the ball. It's intuitively clear (and of course you've written a proof) that the curvature must be at least $1/R$, where $R$ is that maximum distance. For the 3D case you can use normal curvature of spheres and mostly avoid Frenet. So there's a bit more challenge for you :P $\endgroup$ Oct 14, 2020 at 4:52

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