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Apr 15, 2019 at 6:33 history edited Pietro Majer CC BY-SA 4.0
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Apr 13, 2019 at 16:41 comment added Pietro Majer (I edited and added details and rectified the unclear sentence about infinite variation. Thank you!)
Apr 13, 2019 at 16:35 history edited Pietro Majer CC BY-SA 4.0
details added
Apr 13, 2019 at 12:49 history edited Pietro Majer CC BY-SA 4.0
m
Apr 13, 2019 at 11:43 comment added Mizar Ah yes, now it's clear! So at each iteration you are reparametrizing by arclength, right? If you don't do that, having infinite variation does not give $|\phi(x)-\phi(y)|=o(|x-y|)$ (this fails if you don't reparametrize by arclength, as I was saying in my previous comment).
Jun 8, 2018 at 16:24 comment added Mizar $\kappa$ having infinite variation is not sufficient to get the estimate on $\phi$: build "one third" of the Koch snowflake starting from the trivial map $\gamma_0:[0,1]\to\mathbb R^2$ (with $\gamma_0(t)=(t,0)$), subdividing $[0,1]$ into three thirds and replacing the straight segment with a tent on the middle interval (thus obtaining $\gamma_1$), and so on. The limiting map $f_\infty:[0,1]\to\mathbb R^2$ will have $f_\infty(0)=(0,0)$ and $f_\infty(3^{-k})=(3^{-k},0)$! Maybe you are reparametrizing by arclength before taking the limit? (I don't know how to deal with $f_\infty$ in that case...)
Mar 16, 2018 at 20:50 comment added Piotr Hajlasz For generalizations of this example see mathscinet.ams.org/mathscinet-getitem?mr=1991757
Dec 27, 2016 at 23:13 comment added BigM Very interesting. I actually had not heard about Whitney's extension theorem (Tras. AMS 1934) before. It is a very neat analogue to Tietze-Urysohn extension theorem.
Dec 27, 2016 at 22:30 history edited Johannes Hahn CC BY-SA 3.0
grammar++;
Dec 27, 2016 at 20:02 history answered Pietro Majer CC BY-SA 3.0