Closed curve with minimum average curvature This problem has been bothering me for a few weeks. Any ideas how to attack? Has this been solved/proved unsolvable/etc? I couldn't really find much online.
Given: Closed curve $C$ in $\mathbb{R}^n$, parametrized as $x = \{x_1(t), x_2(t) \dots x_n(t)\}$. Its curvature then is the derivative of the tangent vector with respect to the arc length element: $k = |\frac{\mathrm{d}{\bf T}}{\mathrm{d}s}|$. $S$ is the length of $C$ (found by $S = \int_C |\mathrm{d}s|$).
Question: Find the curve $C$ that minimizes the integral:
$$
\frac{1}{S}\int_Ck\mathrm{d}s
$$
Speculation: Essentially this means finding the curve with the lowest average curvature (because integrating curvature along length, then dividing by length). To me this sounds like a calculus of variations problem. But there is a norm of a derivative under the integral sign. And $S$ is also an integral.
Okay, let's write it out (denote average curvature $\hat{k}$):
$$
\hat{k} = \frac{\int_C\left|\frac{\mathrm{d}{\bf T}}{\mathrm{d}s}\right|\mathrm{d}s}{\int_C|\mathrm{d}s| }
$$
So I'm trying to minimize $\hat{k}$. But I'm not sure how to attack this fraction of integrals.
I'm not necessarily looking for a solution, just ideas how to approach this problem.
 A: Consider a circle of radius $r$. Its average curvature will clearly be $1/r$ (up to a scalar multiple, which I believe is "1" in your formulation). 
Now instead, consider
$$
c(t) = \frac{1}{k}(\cos k(2\pi t) , \sin k(2 \pi t ) )
$$
which traverses, $k$ times, a circle of radius $1/k$. Its average curvature will be $k$. Since $k$ can be arbitrary, there is no curve with max average curvature. 
You might complain that this curve is selfintersecting (in a big way), but making an arbitrarily small random perturbation will resolve this; as an alternative, stretch it so that it becomes helical, and make the last half-turn be a little wider and join the top of the helix to the bottom. This'll only slightly reduce the average curvature, so the argument still works. 
The question of MINIMUM average curvature is more interesting (to me).  Surely the answer is "a circle" if there's any answer at all. 
A: It suffices to restrict to plane curves, but see the last section for a concrete example.
I'm assuming that you are talking about the signed curvature in this answer.
In fact, for any closed curve $C$, we have
$$ \int_C k \, ds = 2\pi N $$
for some integer $N$, this is called the total curvature or the turning number of $C$. I offer two ways of proving this: one is a simple application of the Gauss–Bonnet theorem,
$$ \int_C k \, ds = \int_{\partial M} k_g \, ds = 2\pi \chi(M)-\int_{M} K \, dA = 2\pi \chi(M), $$
where $M$ is the set of which $C$ is the boundary, and $K$, the curvature of $M$, is obviously zero for a plane. $\chi(M)$, of course, is discrete, taking integer values, and since the integral is continuous for curves with nonsingular curvature (and this means that you can't introduce another loop, for example, as it would have to grow from a cusp), it is constant on a class of such curves.

Now, a second, more constructive proof is to consider the explicit form of the curvature: we have (with $'=d/dt$)
$$ k=\frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}. $$
Therefore
$$ k \, ds = \frac{x'y''-y'x''}{x'^2+y'^2} \, dt. $$
Now, there are two ways to go about noticing something about this: Green's theorem or complex analysis. I'll do it by complex analysis because the theorems are probably more familiar. Notice that (and you can work back to this: I did):
$$ \frac{1}{i} \frac{x''+iy''}{x'+iy'} = \frac{(y''-ix'')(x'-iy')}{(x'+iy')(x'-iy')} = \frac{(x'y''-y'x'')-i(x'x''+y'y'')}{x'^2+y'^2} $$
Therefore we can write
$$ \int_C k \, ds = \Re\left( \frac{1}{i} \int_C\frac{x''+iy''}{x'+iy'} \, dt \right) $$
Ah, but this is the real part of the winding number of $x'+iy'$ about zero: i.e., how many rotations the tangent vector goes through. Well, that has to be a whole number, or the curve isn't smooth. (One criticism that might be levelled here is that we don't know that $x'+iy'$ extends to an analytic function on the interior. Fair enough, but I expect the Green's theorem argument will work anyway. The result is certainly true (as the Gauss–Bonnet argument shows): I'm just trying to make it believable with the minimum framework. See also Total curvature and Turning number on Wikipedia.) 

Right, so now that we know the total curvature $\int_C k \, ds$ is an integer, we can answer the question. The homotopy given by $\gamma:[0,1) \to \{ \text{curves} \}$, $\gamma( \lambda) = (1-\lambda)C$, where $ (1-\lambda)C = [ t \mapsto ((1-\lambda)x(t),(1-\lambda)y(t)) ]$, preserves smoothness of the curve, and is continuous, so it must preserve the total curvature (continuous map to a discrete space is locally constant). But it is intuitively obvious (and easy to verify by calculation) that $S(C)$, the length of $C$, undergoes a linear scaling under the action of $\gamma$:
$$ S((1-\lambda)C) = (1-\lambda)C. $$
(This should not be surprising: think of perimeters of circles.) It follows that
$A(C) := S(C)^{-1} \int_C k \, ds $ has neither maximum nor minimum, even among curves of identical shape (consider $\gamma$, and the similar homotopy that grows $C$ by mapping $\lambda \mapsto (1-\lambda)^{-1}C$).

Now, if you want to talk about the absolute value of the curvature, the result that there is no minimum will still hold, because the circle of radius $r$ has constant curvature $1/r$, so the total curvature is $2\pi$, and the length is $2\pi r$, so $A(C(r)) = 1/r$, which can be made as large or as small as you like.
(Oh, and by the way, you can check that the result about turning number doesn't hold for non-planar curves, just by considering something like $(\cos{t},\sin{t},f(t))$ for some twice-differentiable $f$ with $f(x+2\pi)=f(x)$. This tells you that there's something topologically specific to 2D going on.)
