# Continuity of barycentre in Hausdorff metric

Let $K_1$, $K_2$ be two convex compact sets in $\mathbb{R}^d$, and $p_1,p_2$ be their barycenters. Is it true that the distance between $p_1$ and $p_2$ does not exceed a Hausdorff distance between $K_1$ and $K_2$? If not, maybe there is some weaker estimate, say, uniform continuity of the map (convex body)$\rightarrow$ (its barycenter) for bodies inside, say, unit ball?

UPDATE. As Anton pointed out, the answer is obviously no, just take rectangle $\varepsilon\times 1$ and divide it by diagonal onto two triangles. Let me ask another question: is the barycentre of a closed $\varepsilon$-neighborhood close to the barycentre of initial body uniformly for all bodies inside unit ball?

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Yes, the barycenter depends continuously on the Hausdorff metric. You should be able to show this directly from the definition of the barycenter as the expected value of a point in $R^n$ with respect to the uniform probability density on $K$. –  Deane Yang May 7 '12 at 19:25
@Deane: You're right, but I think you're implicitly using that the Hausdorff distance and the distance defined as the volume of the symmetric difference of two convex bodies define the same topology on the space of convex compact sets. It may be worthwhile to point this out. –  alvarezpaiva May 7 '12 at 19:40
@alvarezpaiva: sorry, what volumes are you saying about? Bodies may even have different dimensions. –  Fedor Petrov May 7 '12 at 19:47
@Deane: you mean something like "if $K_1$ and $K_2$ are on Hausdorff distance $c$, then the probability [for a random point in $K_1$ have first coordinate less then $x$] does not exceed the probability [for a random point in $K_2$ have first coordinate less then $x+c$]"? I probably should be able to prove such things, but alas, just now I feel myself disabled. –  Fedor Petrov May 7 '12 at 19:50
@Fedor: sorry, I was thinking of something else. –  alvarezpaiva May 7 '12 at 20:25
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No, the distance can be as big as you want. Take two isosceles triangles with small bases, which "look" in the opposite directions.

For the updated question, the answer is still "NO". Again take long thin isosceles triangle with small base $\ll\varepsilon$. The distance between the barycentre of triangle and its $\varepsilon$-neigborhood is about $$\tfrac16{\cdot}\mathop{\rm diam}.$$

This is the upper bound for $\mathbb R^2$; for $\mathbb R^n$ you should get $$(\tfrac12-\tfrac1{n+1}){\cdot}\mathop{\rm diam}.$$

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Thanks, indeed! So I change the question a bit, though probably the answer is again negative. –  Fedor Petrov May 7 '12 at 20:15