Timeline for Choosing a convergent sub-sequence from a sequence of bi-Lipschitz homeomorphisms

8 events

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Jan 5, 2020 at 11:52	comment	added	Wlod AA		@Dirk, did I say "bilinear"? :) === seriously, thank you Dirk for the correction.
Jan 5, 2020 at 10:22	vote	accept	aglearner
Jan 5, 2020 at 10:16	vote	accept	aglearner
Jan 5, 2020 at 10:18
Jan 5, 2020 at 10:16	comment	added	aglearner		You are right M. Dus, thanks!
Jan 5, 2020 at 9:56	comment	added	M. Dus		@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.
Jan 5, 2020 at 9:05	comment	added	M. Dus		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 $g$.
Jan 5, 2020 at 8:40	history	edited	Dirk	CC BY-SA 4.0	added 3 characters in body
Jan 5, 2020 at 6:01	history	answered	Wlod AA	CC BY-SA 4.0