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Crossposted from Math Stack Exchange

For a convex curve $C$, define its barycenter to be $$b(C) = \frac{1}{\mathcal H^1(C)} \int\limits_C x d \mathcal H^1(x)$$ Is there a constant $L$ such that for $C_1,C_2$ convex curves, $|b(C_1) - b(C_2)| \leq L d_H(C_1,C_2)$? Note that the corresponding claim fails if "convex curve" is replaced by "convex body" or "smooth curve", and is true if the barycenter of the curve is replaced by the barycenter of the measure on the curve weighted by the curvature, in which case we recover the Steiner curvature centroid.

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The answe is "yes".

Assume $d_H(C_1,C_2)<\varepsilon$. Let $\lambda_1$ and $\lambda_2$ be the length-measures of $C_1$ and $C_2$.

We can assume $C_1$ is surrounded by $C_2$; the general case can be reduced to this one.

Consider the closest-point projection of $C_2$ to $C_1$. Let $\mu$ be the push-forward of $\lambda_2$ for this projection. Clearly $\lambda_1\le \mu$. Therefore $$|b(\lambda_1)-b(\mu)|<\frac{|\mu-\lambda_1|}{|\mu|}\cdot \mathrm{daim}\, C_1.$$

It remains to apply the following observations:

  • $|b(\lambda_2)-b(\mu)|\le \varepsilon$;
  • $|\mu-\lambda_1|<2\cdot\pi\cdot\varepsilon$;
  • $2\cdot \mathrm{daim}\, C_1\le |\lambda_1|\le \pi\cdot\mathrm{daim}\, C_1$.
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    $\begingroup$ P.S. I would expect that $L=1$. The above argument seems to give $\pi$ instead. $\endgroup$ – Anton Petrunin Oct 4 '19 at 19:18

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