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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?

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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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