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For a curve $C$, its barycenter is $$\text{Bar}(C) = \frac{1}{\text{length}(C)}\int\limits_C x d \mathcal H^1(x).$$ Does there exist a constant $L$ such that for $C_1,C_2$ convex curves in the plane, $$|\text{Bar}(C_1) - \text{Bar}(C_2)| \leq L d_H(C_1,C_2),$$ where $d_H$ is the Hausdorff distance?

Justthisguy
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  • I do not have time to think about this, but take a look at Lee's solution here: First use it for convex polygonal curves and see if you can get some uniform estimates. – Moishe Kohan Oct 03 '19 at 15:53
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    I don't follow. The sum of the distances between vertices of convex polygonal curves could be much greater than their Hausdorff distance. Take, e.g., the polygonal curves with vertices $(-1,0),(0,\epsilon),(1,0)$ and $(-1,0),(1,\epsilon),(1,0)$. – Justthisguy Oct 03 '19 at 18:24
  • This question has been posted to MathOverflow here, where it has been answered. – RavenclawPrefect Feb 23 '21 at 02:42

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