Timeline for Continuity of barycentre in Hausdorff metric
Current License: CC BY-SA 4.0
10 events
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| Oct 18, 2022 at 16:08 | comment | added | Fabian Wirth | Boundedness of the map $F$ is assumed. This at least I quoted correctly. So letting the triangles become arbitrarily large is not covered by the quoted Theorem 1 and thus does not constitute a counterexample. | |
| Oct 18, 2022 at 12:19 | comment | added | Günter Rote | Well, if you scale the triangles to be very big, adding the (relatively small) unit disk changes neither the Hausdorff distance nor the barycenters very much. So the counterexample remains valid to show that this map is not Lipschitz-continuous. Unless you add more assumptions, like that the sets have to be uniformly bounded. (It must be interesting to see how their proof goes.) | |
| Oct 16, 2022 at 18:53 | comment | added | Fabian Wirth | You are correct. I have misquoted Aubin and Cellina. Apologies to you and to them as well. I read superficially, ts, ts. Indeed, for their selection they do not take the barycenter of F, but the barycenter of "F+unit ball". This then gives the Lipschitz selection. The barycenter itself is not even continuous, as the example above shows. | |
| Oct 16, 2022 at 11:04 | comment | added | Günter Rote | In the example of the triangles in the other answer, the Hausdorff distance is as small as you want (on the order of ϵ), while the distance between the barycenters of the two triangles can remain constant (1/3 of the horizontal extension (altitude) of the triangles if the triangles are vertically aligned). Thus, there can be no Lipschitz constant at all for this map. | |
| Oct 16, 2022 at 10:56 | history | edited | Günter Rote | CC BY-SA 4.0 |
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| Oct 14, 2022 at 21:09 | comment | added | Fabian Wirth | It just shows that the Lipschitz constant is not $1$, doesn't it. | |
| Oct 13, 2022 at 20:32 | history | edited | Günter Rote | CC BY-SA 4.0 |
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| Oct 13, 2022 at 20:25 | comment | added | Günter Rote | @FabianWirth. This theorem has more general assumptions, but one can apply it to the current problem: E.g. take the metric space to be the set of compact nonempty convex subsets of $R^n$ itself (with Hausdorff metric), and let $F$ be the identity map. Nevertheless I am surprised by the claimed proof: The example of the triangles in the other answer shows clearly that the barycenter does not work. | |
| Oct 12, 2022 at 14:32 | comment | added | Fabian Wirth | It is of course always hard to say which paper is the "first" reference on a given subject. The merits of [ABB] notwithstanding, they are hardly the first. For instance, Aubin & Cellina: "Differential inclusions", Springer, 1984 has this theorem as Theorem 1 in Chapter 1, Section 9: Let F be a bounded, Lipschitz map from a metric space to the compact convex subsets of $R^n$ (endowed with Hausdorff metric). Then $F$ has a Lipschitz selection $f$, i.e. $f(x) \in F(x)$ for all $x$. This is proved by showing that $f(x) = barycenter(F(x))$ is the desired selection. | |
| Jan 29, 2013 at 18:21 | history | answered | Günter Rote | CC BY-SA 3.0 |