# a question about Lp norm of curvature on convex curves

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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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.
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
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