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Suppose $(X,d)$ is a metric space with the nearest point property and $a,b \in X$ with $a \ne b$. Suppose there is a path of finite length in $X$ from $a$ to $b$ and let $m$ be the infimum of the lengths of all paths from $a$ to $b$.Then, by Lipschitz reparametrization, there exists a path $g:[0,1] \rightarrow X$ from $a$ to $b$ that satisfies $lth_t(g) = tm ~\forall~t \in [0,1].$ and $g$ is lipschitz with length $m. lth_t(g)$ represents the length of the function $g$ upto a point $t$.

Then is the function $g$ always injective?

I am kind of stuck here and can't think of any counterexamples as well. Any inputs will be appreciated, thanks!

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    $\begingroup$ The minimizng path $g$ may not exists. If it exists, then it must be injective (if it would not be then you could shorten it). (You do not need "nearest point property" no matter what is its meaning.) $\endgroup$
    – Anton Petrunin
    Commented Apr 1, 2018 at 5:25
  • $\begingroup$ @AntonPetrunin thanks. Could you help me prove injectivity $\endgroup$
    – MathMan
    Commented Apr 1, 2018 at 5:28
  • $\begingroup$ Is it your homework? $\endgroup$
    – Anton Petrunin
    Commented Apr 1, 2018 at 5:29
  • $\begingroup$ @AntonPetrunin nope. It is not. I saw this as an open question posed by searcoid in a book $\endgroup$
    – MathMan
    Commented Apr 1, 2018 at 5:31

2 Answers 2

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It is injective because if $g(s)=g(t)$, $s<t$, then you can remove the interval $[s,t]$ from the domain of definition of $g$ and make the curve shorter. Then you rescale the domain to be $[0,1]$. Rescaling does not change the length of the curve. See also Lemma 3.10 in my notes linked to the answer to another related question: does there always exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and ..?

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  • $\begingroup$ Thanks but the problem posed mentions $[0,1]$ as the domain. Aren't we altering the conditions of the problem by changing the domain? $\endgroup$
    – MathMan
    Commented Apr 1, 2018 at 5:40
  • $\begingroup$ @MathMan I edited my answer to address your concern. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 1, 2018 at 5:49
  • $\begingroup$ okay. which means the answer to the problem posed is : $g$ may not be always injective but there exists one $g$ which is injective obtained by shortening the curve? $\endgroup$
    – MathMan
    Commented Apr 1, 2018 at 5:52
  • $\begingroup$ It is injective since it is the shortest one. My argument was by contradiction. $\endgroup$
    – Piotr Hajlasz
    Commented Apr 1, 2018 at 5:54
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Yes, but not because of the reason in the accepted answer. If you remove the portions where $g$ is constant, the length stays the same, you're not making it shorter. Instead, if there was such an interval, then the condition that ${\rm lth}_t(g) = tL$ would be violated since the LHS is constant on the interval in case, while the RHS isn't.

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