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Is there a standard term for a metric space in which $\rho(p,r) < \rho(p,q) + \rho(q,r)$ for any distinct $p$, $q$, $r$? Sort of the opposite of metric convexity.

For instance, a subset of euclidean space with the inherited distance function has this property if and only if no three points are colinear. Another example: $\rho^\alpha$ has this property for any metric $\rho$ and any $\alpha \in (0,1)$.

(Context: I'm working on the second edition of my book Lipschitz Algebras, and I realized that a property I called "concavity" in the first edition is equivalent to this simple condition.)

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  • $\begingroup$ Welp, "concave" it is, then. $\endgroup$
    – Nik Weaver
    Feb 28, 2016 at 22:21
  • $\begingroup$ I had to look for this recently: the only reference I found was this paper arxiv.org/pdf/1003.5087.pdf, where the property is gracefully called "has no co linear triples of pairwise distinct points". $\endgroup$ Feb 29, 2016 at 9:57
  • $\begingroup$ @JamesKilbane: thanks! It sounds like there's no standard term, then. $\endgroup$
    – Nik Weaver
    Feb 29, 2016 at 16:36
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    $\begingroup$ what about "discollinear"? $\endgroup$
    – YCor
    Mar 3, 2019 at 13:23
  • $\begingroup$ @YCor: I'd be sorely tempted to use that, but my book is already in print ... $\endgroup$
    – Nik Weaver
    Mar 3, 2019 at 13:39

2 Answers 2

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The only thing which I recall in this connection is: Blumenthal in his "Theory and Applications of Distance Geometry" (page 56) calls three points a linear triple if the triple is congruent with a triple in $\mathbb{R}^1$. See also page 242 in Blumenthal-Menger, "Studies in Geometry". (Possibly one can find in these books something more relevant, but I do not recall such things now.) So one can call such spaces without linear triples.

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    $\begingroup$ Or "linear tripleless"? $\endgroup$
    – Nik Weaver
    Mar 1, 2016 at 23:31
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Strictly convex seems to be the name, at least according to A Short Course on Banach Space Theory:

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    $\begingroup$ No, this is different. The definition you quote refers to normed spaces and adds a non-colinearity requirement. $\endgroup$
    – Nik Weaver
    Feb 28, 2016 at 16:32
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    $\begingroup$ A quick Google search shows that the term "strict convexity" is usually used in the general metric context to mean that for any $p,r$ and any $0 < t < 1$ there is a unique $q$ such that $\rho(p,q) = t\rho(p,r)$ and $\rho(q,r) = (1-t)\rho(p,r)$. $\endgroup$
    – Nik Weaver
    Feb 28, 2016 at 16:34
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    $\begingroup$ Or it could mean that if $\rho(p,z) = \rho(q,z) > 0$ then $\rho(r,z) < \rho(p,z)$ where $r$ is the midpoint of $p$ and $q$ (which would not exist if the triangle inequality were strict). Maybe this is the more common usage. Anyway, very different from what I'm asking. $\endgroup$
    – Nik Weaver
    Feb 28, 2016 at 16:40
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    $\begingroup$ I guess the connection is that a normed linear space is strictly convex iff its unit sphere has the strict triangle inequality property. $\endgroup$
    – Nik Weaver
    Feb 28, 2016 at 18:18

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