A good lower bound on the maximum curvature in a loop Suppose $\alpha: \mathbb{R} \rightarrow \mathbb{R}^3$ is a $C^\infty$
curve, parameterized by arc length ($\left\|\alpha'(t)\right\| = 1$),
and with $\alpha(0) = \alpha(\ell)$.  Show that
there exists a $t_0 \in [0,\ell]$ such that
$
\left\| \alpha''(t_0)\right\| \geq \frac{2\pi}{\ell}.
$

If we had that $\alpha'(0) = \alpha'(\ell)$, then the result would
follow from Fenchel's theorem, but we don't. Actually there is a nice
extension to Fenchel's theorem described in Toponogov's book
``Differential Geometry of Curves and Surfaces: A Concise Guide'',
under the name Fenchel-Reshetnyak, which gives lower bounds on the
integral curvature in an open curve. This can be adapted to our
situation to show that $\left\| \alpha''(t_0)\right\| \geq
\frac{\pi}{\ell}$, but that bound is only half as good as it
presumably could be. It makes sense that the integral curvature bound
will be only half as good if we don't constrain $\alpha'(0)$
w.r.t. $\alpha'(\ell)$ (in particular, one may be the negative of the
other), but nonetheless it seems that the original bound on the
maximum curvature in the loop should still stand. How can we
recover it?
 A: This smells like a problem in the Calculus of Variations and can be solved as such.  We therefore make a slight reformulation: among all $C^3$ curves of length $l$ originating and terminating at the same point, find the one(s) for which the maximum curvature is smallest.
Let's begin by developing some intuition.  In general, a minimax problem like this has a solution where the objective (the curvature in this case) is as constant as possible.  A circle is an obvious candidate for a solution and indeed a circle of length $l$ has constant radius $l / (2 \pi)$ and constant curvature of $2 \pi / l$.
(As a tiny simplification, by choosing our linear units of measurement let's henceforth assume $l=1$.)
The technique of the Calculus of Variations is to assume we have a solution and perturb it a little, showing that any perturbation, no matter how small, cannot decrease the value of the objective.  To motivate this in the present case, though, I find it more appealing to consider how one might go about reducing the maximum curvature of any unit-length loop.  A point of maximum curvature is part of a relatively sharp "bump" on the curve.  If we push that bump inwards a little, we should be able to decrease its sharpness and at the same time we slightly decrease the total length of the curve.  To make up for the latter, uniformly dilate the entire curve relative to its point of origin until the dilated curve again has unit length.  This uniformly decreases all curvatures along the loop.  In this fashion the maximum curvature has strictly decreased.  The only way this operation can fail is when there is no "bump": the curvature everywhere is constant.
To make this go through rigorously, use an adapted orthonormal frame for the curve: $T(t) = \alpha'(t)$ is the tangent, $N(t)$ is the inward-pointing unit normal with $T'(t) = \kappa(t)N(t)$, and $B(t)$ is the unit binormal, with $N'(t) = -\kappa(t)T(t) + \tau(t)B(t)$ and $B'(t) = -\tau(t)N(t)$.  ($\tau$ is the torsion.  These Serret-Frenet formulae generalize readily to higher dimensions and in two dimensions just forget about $B$.)  Notice that $\kappa \ge 0$.  Let $\epsilon > 0$ be small and let $\delta(t)$ be a smooth non-negative function in an open neighborhood of $[0,1]$, vanishing at $0$ and $1$ (in order to keep the endpoints of the new curve fixed).  Set the "pushed" curve to be
$$\psi(t) = \alpha(t) + \epsilon \delta(t) N(t).$$
Compute the curvature of $\psi$ to first order in $\epsilon$ using the Serret-Frenet formulae.  I find that its square becomes
$$\kappa^2 + 2 \kappa \left( \delta'' - \delta \tau^2 \right) \epsilon + O(\epsilon^2).$$
Suppose the curvature is not constant.  Then there is a closed interval of maximum curvatures (perhaps reducing to a point) and within some neighborhood of that interval all curvatures are strictly less than the maximum.  We can make $\delta$ zero outside this larger neighborhood, let it increase slowly enough (and choose $\epsilon$ sufficiently small) so that $\delta''$ keeps the new curvature less than the maximum within the outer neighborhood, and make $\delta''$ strictly negative at all points within the interval of maximum curvature.  This procedure strictly decreases the maximum curvature within the outer interval.  You can also check, again to second order in $\epsilon$, that the length of $\psi'$ decreases by $\epsilon \delta \kappa \ge 0$, so we will be able to apply our dilation trick in order to keep the total loop length constant after this perturbation.
A little more formally (and using less work, actually), the preceding equation shows that when $\kappa$ is constant and $\tau = 0$, we would need $\delta'' \lt \delta \tau^2 = 0$ everywhere in order to decrease the curvature, which implies $\delta$ is identically zero in the unit interval.  Otherwise (namely, if $\kappa$ is not constant or $\tau \gt 0$), it will be possible to decrease the curvature subject to maintaining the unit length of the curve.  Curves of constant curvature and zero torsion are circular arcs, and the only circular arc that returns to its starting point is a full circle.  Therefore the maximum curvature in any sufficiently smooth unit-length loop can be no less than $2 \pi$ and is strictly greater than that if the loop is not a circle, QED.
I believe the same technique works in higher dimensions, too, although I have not done the calculations.  (There are additional vectors in the adapted frame, and therefore additional torsion terms, but this doesn't seem to present any obstacle to carrying out the same program.)
