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?
 A: *

*This question has been first discussed in the paper [ABB] below. They show that, in the plane, the barycenter of the boundary has the desired property: It is Lipschitz-continuous with respect to the Hausdorff distance. (However, the Lipschitz constant is larger than 1.)

*Other such "reference points" which are Lipschitz-continuous w.r.t. the Hausdorff-distance, or w.r.t. some other metric (like the area of the symmetric difference) have been investigated. The objective is to get a fast heuristic or initial solution for matching two shapes under translation. A good starting point into the literature might be Oliver Klein's Ph.D. thesis "Shape Matching With Reference Points" from 2008.

*The best Lipschitz constant (for the Hausdorff distance) is achieved by the so-called Steiner point (or Steiner center). In functional analysis, such "reference points" are known under the name "continuous selectors". It has been proved (by non-constructive methods, however) that the Steiner point has the smallest possible Lipschitz constant w.r.t. the Hausdorff distance.

*One can show by an elementary example that a reference point (a mapping from compact convex sets to points that is equivariant under, say, isometries) cannot have Lipschitz-constant 1 in dimension 2 or larger. (Thus the answer to the first, strong, part of the original question is NO, even if you think of other points than the barycenter.)

[ABB] Helmut Alt, Bernd Behrends, Johannes Blömer. Approximate matching of polygonal shapes.
Ann. Math. Artif. Intell., Volume 13, Pages 251-266, 1995,
doi:10.1007/BF01530830. The conference version may be easier to access:
Proceedings of the seventh annual symposium on computational geometry, pp.186-193, June 10-12, 1991, North Conway, New Hampshire. ACM Press. doi:10.1145/109648.109669.
A: 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 $$(\tfrac12-\tfrac1{n+1}){\cdot}\mathop{\rm diam}.$$
