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Given two simple, closed, convex, planar curves $C_1$ and $C_2$, let their lengths be $\ell_1$ and $\ell_2$, respectively, and their Fréchet distance be $d_f$. We are trying to bound $|\ell_1 - \ell_2|$, their difference in length. Our current conjecture is that $|\ell_1 - \ell_2| \leq 2\pi d_f$, but we are unable to find anything in the literature about it. Any help regarding a proof would be appreciated.

Note that since both $C_1$ and $C_2$ are convex, the Fréchet distance between them is the same as the Hausdorff distance according to this paper.

As a special case, if $C_2$ is a parallel curve of $C_1$ with distance $d_p$, $|\ell_1 - \ell_2| = 2\pi d_p$ by this paper. Since both curves are simple, closed, and convex, the Fréchet distance $d_f = d_p$ (this is easily provable). Thus, this special case would reach the upper bound of our conjecture.

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  • $\begingroup$ For simple convex closed curves the length is $\pi$ times the mean width, but the widths in every direction can differ by $2d_f$ at most. $\endgroup$
    – fedja
    Commented Jan 19, 2023 at 5:03

2 Answers 2

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Let $C_p$ be the parallel curve of $C_1$ at distance $d_f$ with length $\ell_p$. By this paper, $\ell_p = \ell_1 + 2\pi d_f$.

Lemma: $C_p$ encloses $C_2$.

Proof: we use contradicion. Assume to the contrary there exists a point $P$ on $C_2$ not enclosed by $C_p$. Because $C_1$ is convex, there exists a unique point $P'$ on $C_1$ such that $\overline{PP'}$ is perpendicular to the tangent line $L_{tan}$ of $C_1$ at point $P'$. Since $C_1$ is convex, it lies on one side of $L_{tan}$, so $\overline{PP'}$ gives the shortest distance between point $P$ and curve $C_1$. By the definition of a normal curve, the length of $\overline{PP'}$ must therefore be greater than $d_f$ for $P$ to be exterior to $C_p$. Thus, the Fréchet distance between $C_1$ and $C_2$ is also greater than $d_f$, which contradicts our assumption. Therefore, $C_p$ encloses $C_2$, proving this lemma.

By this paper, a closed convex curve is always enclosed by a curve of greater length. Thus, $\ell_2 \leq \ell_p = \ell_1 + 2\pi d_f$. By a similar argument, we can show $\ell_1 \leq \ell_2 + 2\pi d_f$. Combining these two results, we get $|\ell_1 - \ell_2| \leq 2\pi d_f$.

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Your conjecture is true, it is a special case of the "length theorem" in:

http://pub.ist.ac.at/~edels/Papers/2007-J-02-InequalitiesCurvature.pdf

which addresses the case of non necessarily convex curves.

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