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Suppose we have two strictly convex closed curves $C_{1}$ and $C_{2}$, $C_{1}$ contains $C_{2}$, then can we conclude $\int_{C_{1}} \kappa_{1}^{p} ds\geq \int_{C_{2}} \kappa_{2}^{p} ds$, $\kappa_{1}$ and $\kappa_{2}$ are corresponding curvatures of $C_{1}$ and $C_{2}$, $p$ is between 0 and 1

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up vote 4 down vote accepted

No. Let $C_2$ be the unit circle and $C_1$ be a big square with corners rounded off by circle arcs of radius $r\ll 1$. The integral for $C_2$ is the same as for the $r$-circle and this is very small.

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Actually, small circles have large curvature so it is not so straight forward. For example for $p=1$ the integrals are equal (to $2\pi$). For $p = 0$ one compares arc-lengths and so we also have the inequality. – alvarezpaiva Mar 4 '12 at 12:32
@alvarezpaiva, the inequality holds only for $p=0$ or $1$; and Sergei's example shows that it does not hold for any $p\in(0,1)$. – Anton Petrunin Mar 4 '12 at 16:59
That's right. I mixed up the scaling factor (of 1-p). – alvarezpaiva Mar 4 '12 at 17:06
I think alvarezpaiya's 1st comments make sense. for Sergei's example, when p=0 the inequality in my question obviously right, and in fact it is a strict inequality. Then notice that the curves are strictly convex, so at least for when p very colose to 0, for sergei's example, the inequality still holds. – user13289 Mar 4 '12 at 18:53
Of course I assumed that $p$ is strictly between 0 and 1. The strict convexity is irrelevant because one can approximate. – Sergei Ivanov Mar 4 '12 at 19:34

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