Let $f\colon[0,1]\to \mathbb{R^2}$ be continuous such that $f(0)=f(1)$.

If want to find a 1-Lipschitz function $g : [0,a]\to f([0,1])$ such that $g(0)=g(a)$ and $g$ is surjective ($a>0$).

I had the following idea using the total variation of $f$: Denote $V_T(f) = \sup \left\lbrace\sum_{i=1}^n \lVert f(x_i) - f(x_{i-1})\rVert_2 \ \biggm| \ n\in\mathbb{N}, \ 0=x_0<x_1<\dots<x_n=1\right\rbrace$. Suppose morevover that $V_T(f)<+\infty$.

Define $g:[0,V_T(f)]\to f([0,1])$ such that $g(x) = f\left(\frac{x}{V_T(f)}\right)$.

Then $g(0)=g(V_T(f))$ and $g$ is surjective. But I cannot prove that $\Vert g(x_1)-g(x_2)\Vert_2\leq |x_1-x_2|$. Maybe it is not true, in that case is there another definition of $g$ that would be suitable?

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    $\begingroup$ Try normalizing so that the total variation of $f$ on $[0,1]$ is $1$, letting $V_x$ be the total variation of $f$ on the interval $[0,x]$, and setting $g(x) = f(V_x)$. $\endgroup$ – Nik Weaver Oct 25 '16 at 11:59
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    $\begingroup$ I think that you may have a problem if $f$ is a periodic Peano curve. The image of $f$ has Hausdorff dimension $2$ while a the image of an interval via a Lipschitz map has Hausdorff dimension $\leq 1$. $\endgroup$ – Liviu Nicolaescu Oct 25 '16 at 12:12
  • $\begingroup$ @LiviuNicolaescu: surely a Peano curve has infinite length? He's assuming the total variation is finite. $\endgroup$ – Nik Weaver Oct 25 '16 at 14:12
  • $\begingroup$ I think that the graph of the Cantor function might produce a counterexample. You need to look in some books of geometric measure theory. $\endgroup$ – Liviu Nicolaescu Oct 25 '16 at 15:16
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    $\begingroup$ This is a classical construction, btw, and holds in any metric space in place of $\mathbb{R}^2$. Up to reparametrizations, curves $[0,1]\to X$ of length $L$ are $L+\epsilon$ Lipschitz. It is a key observation to prove the "Hopf-Rinow theorem" for compact metric spaces. $\endgroup$ – Pietro Majer Oct 25 '16 at 18:45

You want to take the inverse function of the increasing function $x\mapsto V_x$ (let's call it $W$) and then let $g(t)=f(W(t))$. This is obviously Lipschitz with constant $1$.

There is the small technical issue that $V_x$ could be constant on certain intervals and thus fail to be invertible. If that happens, just remove these (open) intervals $I_j$ to obtain a smaller domain $A=[0,1]\setminus \bigcup I_j$, and work with $f_1(x): [0,|A|]\to\mathbb R^2$, $f_1(x)=f(y)$, $|[0,y]\cap A|=x$, instead of $f$.

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    $\begingroup$ The usual trick to bypass the technical issue is to consider (for some $\epsilon>0$) the inverse of the strictly increasing function $x\mapsto V_x +\epsilon x$, which is a homeo $\sigma:[0,V_T+\epsilon]\to[0,1]$ such that $g:=f\circ\sigma$ is 1-Lipschitz. $\endgroup$ – Pietro Majer Oct 25 '16 at 18:38
  • $\begingroup$ @PietroMajer: Yes, or (as I now realize) I just could have said $g(t)=f(x)$, where $x$ is any point with $V_x=t$. $\endgroup$ – Christian Remling Oct 25 '16 at 19:34

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