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Let $(X,d)$ be a metric space. For any $0<\epsilon<1$, we call the metric space $(X,d^{\epsilon})$; where $d^{\epsilon}(x,y)\triangleq (d(x,y))^{\epsilon}$ the $\epsilon$-snowflake of $(X,d)$.

My question is, given $(X,d)$ and some $\epsilon\in(0,1)$, under what conditions can we deduce that there exists some other metric space $(Z,\rho)$ such that $(X,d)$ is the $\epsilon$-snowflake of $(Z,\rho)$; i.e.: $$ (X,d)=(Z,\rho^{\epsilon})? $$

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    $\begingroup$ I suppose you don't consider “there exists $e>1$ such that $d^e$ satisfies the triangle inequality” a satisfactory answer to your question. 😁 $\endgroup$
    – Gro-Tsen
    Commented Mar 15, 2021 at 12:14
  • $\begingroup$ Haha, nope also note that I require that $0<\epsilon<1$ ;). $\endgroup$
    – Bernard_Karkanidis
    Commented Mar 15, 2021 at 12:21
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    $\begingroup$ @Gro-Tsen In the question, $\epsilon$ is fixed, so instead of "there exists $e>1$", you want $e=1/\epsilon$. $\endgroup$
    – Andreas Blass
    Commented Mar 15, 2021 at 18:27
  • $\begingroup$ @Bernard_Karkanidis Why not? I can hardly imagine a simpler condition than the one Gro-Tsen suggests. If there is a triple satisfying $d^{1/\epsilon}(x,z) > d^{1/\epsilon}(x,y) + d^{1/\epsilon}(y,z)$ then the answer is no, otherwise it is yes. $\endgroup$
    – Nik Weaver
    Commented Mar 17, 2021 at 12:59

1 Answer 1

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You may be interested in the following paper:

Jeremy T. Tyson and Jang-Mei Wu, Characterizations of Snowflake Metric Spaces. Annales Academiae Scientiarum Fennicae Mathematica. Volume 30, 2005, 313-336. https://www.emis.de/journals/AASF/Vol30/tyson.pdf

The paper contains a number of equivalent characterizations of the property that a metric space $X$ is bi-Lipschitz equivalent to a snowflake. Some of the characterizations in the paper require that the space is embedded in a certain Banach space, and some do not.

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