Curvature inequality involving a Curve within a disk

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?

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.

• Very neatly explained, thanks :D
– Azur
Aug 25, 2020 at 10:04
• 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 Oct 14, 2020 at 4:52