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Let $X,Y$ be two compact metric spaces. Suppose there is a sequence of bi-Lipschitz homeomorphisms $f_n: X\to Y$, and $c\in(0,1]$, satisfying $$c\cdot d(x_1,x_2)\le d(f(x_1),f(x_2))\le \frac{1}{c}\cdot d(x_1,x_2),$$ for any $x_1,x_2\in X$.

Is is true that one can choose a sub-sequence $f_{n_i}$ that uniformly converges to a bi-Lipschitz homeomorphism $f$ (with the same constant $c$)?

Such a statement would imply that in case the Lipschitz distance between two compact metric spaces is bounded, it is actually realised by a bi-Lipschitz homeomorphism (which is better than taking an inf over all bi-Lipschitz homeomorphisms).

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    $\begingroup$ This should follow from an appropriate version of the Arzela-Ascoli theorem for maps between metric spaces. Even if you haven't seen such a version, it would be a good exercise to adapt the proof for real-valued functions. $\endgroup$
    – Nate Eldredge
    Commented Jan 4, 2020 at 18:53
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    $\begingroup$ Yes: all you need is a subsequence that converges uniformly (or even just pointwise). The fact that the limit function must be bi-Lipschitz is easy. $\endgroup$
    – Nik Weaver
    Commented Jan 4, 2020 at 19:48
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    $\begingroup$ Dear Nate and Nik, I guess you are right. I wonder though why the proof of Theorem 7.2.4 in Burago, Burago, Ivanov citeseerx.ist.psu.edu/viewdoc/… doesn't use this fact. It is a tiny bit more involved than that. This theorem states, in particular, that two compact metric spaces on Lipschitz distance $0$ from each other are isometric. $\endgroup$
    – aglearner
    Commented Jan 4, 2020 at 19:48
  • $\begingroup$ @aglearner That's a striking application ! Did not see that at first. $\endgroup$
    – M. Dus
    Commented Jan 5, 2020 at 10:27

1 Answer 1

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Consider a subsequence $\,g_n\,$ of $\,f_n\,$ which is uniformly convergent. Then consider a subsequence $\,h_n\,$ of $\,g_n\,$ such that sequence $\,h^{-1}_n\,$ is uniformly convergent. Then the limit function $\,h\,$ of $\,h_n\,$ is bilipschitz with the bi-constant equal the same $\,c.$   Great

REMARK This proves what @NikWeaver has already said in his comment under the OP's Question. (Yes, I agree with the comment by @NatEldredge).

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  • $\begingroup$ Note that the result also holds not assuming that the $f_n$ are homeomorphisms and in this case, you cannot use $h_n^{-1}$. The $f_n$ certainly are one-to-one but necessarily onto in general. However, you don't need to play this trick to obtain a bi-Lipschitz limit. Since $g_n$ converges pointwise (note that this is enough) to a limit $g$, fixing $x_1$ and $x_2$, you get $1/c d(x_1,x_2)\leq d(g_n(x_1),g_n(x_2))\leq c d(x_1,x_2)$ for every $n$, so that the same holds with $g$. $\endgroup$
    – M. Dus
    Commented Jan 5, 2020 at 9:05
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    $\begingroup$ @aglearner Actually the answer proves that and I think that's the point of the answer (although not explicitely stated). Denote by $g$ the limit of $g_n$ and by $h$ the limite of $h_n^{-1}$. Then, $h_n$ also converges to $g$. Also, $h_n \circ h_n^{-1}$ converges pointwise to $Id$ and converges pointwise to $g\circ h$, so that $g\circ h=Id$. Similarly, $h\circ g=Id$. Since both $g$ and $h$ are Lipschitz, they are in particular continuous and so they are inverse homeomorphism. $\endgroup$
    – M. Dus
    Commented Jan 5, 2020 at 9:56
  • $\begingroup$ You are right M. Dus, thanks! $\endgroup$
    – aglearner
    Commented Jan 5, 2020 at 10:16
  • $\begingroup$ @Dirk, did I say "bilinear"? :) === seriously, thank you Dirk for the correction. $\endgroup$
    – Wlod AA
    Commented Jan 5, 2020 at 11:52

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