Suppose we have a non-intersecting closed curve in the plane of fixed length 1 with continuous second derivative. Its mean squared curvature is $$\langle \kappa \rangle = \int_C |\kappa|^2 ds = \int_0^1 \left|\frac{d^2 \textbf{x}(s)}{ds^2}\right|^2 ds,$$ where the curve is parameterized by arclength.

Does the circle with circumference 1 minimize this quantity over all twice continuously differentiable closed curves?


Yes, the circle do minimize the integral of curvature squared.

Recall the total absolute curvature of any closed curve is at least $2\pi$.
For any closed curve of length $1$, we have

$$\int_0^1 |\kappa(s)|ds \ge 2\pi$$ By Cauchy Schwarz, this leads to $$\int_0^1 |\kappa(s)|^2 ds = \left(\int_0^1 |\kappa(s)|^2 ds\right)\left(\int_0^1 ds\right) \ge \left(\int_0^1 |\kappa(s)| ds \right)^2 \ge (2\pi)^2$$ Since a circle of unit circumference has constant curvature $2\pi$, its integral of curvature squared takes the smallest possible value $(2\pi)^2$.


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