Assuming standard euclidean metric, the geodesic diameter, of simply connected polygons in the plane is realized by a shortest connection between two vertex points. This result is e.g. referenced in "The Geodesic Diameter of Polygonal Domains" ( https://arxiv.org/pdf/1001.0695.pdf ) where also counter examples for multiply connected polygons are given.

I feel like this should be true in more general than for polygons in the plane, e.g. for simply connected domains in general. Not necessarily that the diameter is realized by vertex points, but by points on the boundary; I suppose, in locally convex regions of the boundary. Does there exists a more general (classical?) statement (preferably in citable form)?

E.g.: "For every simply connected domain, subset of $\Bbb R^n$ with boundary there exist two points $a, b$ on the boundary such that $\operatorname{dist}(a, b)$ is the diameter of the domain?"

Thinking of surfaces with boundary I see that it might not be so self-evident - e.g. on a sphere with a small area removed, forming a boundary, the diameter would still be given by the largest geodesic loop rather than having a relation to boundary points.

Still I think this should be true for "many" cases. Is there some theorem characterizing the domains for which the diameter is realized by boundary points (possibly more general than just requiring convex domains)?

The paper named above gives counter examples for multiply connected domains. It appears to me that for an $n-$dimensional domain, at least $(n+1)-$connectedness is required to provoke an interior point to participate in realizing the diameter (because $(n+1)$ is the number of vertices of the n-dimensional simplex). Is there some result in this direction?

Thanks in advance!

  • $\begingroup$ To rule out the counter example with the spheres, a constraint on the curvature of the metric space might be required. $\endgroup$ – stewori May 5 '18 at 1:07
  • $\begingroup$ mathoverflow.net/questions/163333/… may be loosely related $\endgroup$ – stewori May 10 '18 at 17:29
  • $\begingroup$ To relate the curvature to the question, I suspect that some constraint on the radius of injectivity might be appropriate. $\endgroup$ – John Hughes Jan 11 at 17:07

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