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Suppose $(X,d)$ is a metric space and $f:[0,1] \rightarrow X$ is a path in $X$ with non-zero finite length $L$. Then, does there always exist a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and satisfies $\operatorname{Length}(g\rvert_{[0,t]})=tL$ for all $t \in [0,1]$?

Any views would be really appreciated.

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    $\begingroup$ But if $g(t)$ is constant for $t$ in some interval $J\subset [0,1]$, then the length of $g$ on $[0,t]$ is also constant for $t$ in the same interval $J$. $\endgroup$
    – Pietro Majer
    Commented Mar 17, 2018 at 16:23
  • $\begingroup$ @PietroMajer Would you like to check a link. It contains a proof but i think it's incorrect. math.stackexchange.com/questions/2696001/… $\endgroup$
    – MathMan
    Commented Mar 17, 2018 at 16:30
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    $\begingroup$ If $f:[0,1]\to (X,d)$ is continuous and has finite total variation $L=L([0,1],f)$, then the length $L([0,t],f)$ is continuous wrto $t$. If $f$ is constant on no open non-empty interval, then $L([0,t],f)$ is also strictly increasing, thus an increasing homeo $[0,1]\to[0,L]$. Its inverse gives an arc-length parametrization $[0,L]\ni t\mapsto f(g(t)$ with $L([0,s],f\circ g)=s$ for all $0\le s\le L$. Check e.g. the nice book Cours d'Analyse (tome II - Topologie) by Gustave Choquet. $\endgroup$
    – Pietro Majer
    Commented Mar 17, 2018 at 16:35

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This is a standard result that every rectifiable curve in a metric space admits an arc length parametrization. The proof can be found in many sources. For examples Theorem 3.2 in Hajłasz - Sobolev spaces on metric-measure spaces. Then the arc-length parametrization is defined on $[0,L]$ and a linear change of variables leads us to $g$ defined on $[0,1]$.

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