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I have a question related to the Skorokhod distance.

Let $\Omega:=D([0,1],R)$ be the space of cadlag functions $x$ defined on $[0,1]$. Let $\Lambda$ be the collection of non-decreasing continuous onto functions $\lambda: [0,1]\to [0,1]$. Then define

$$\rho(x, x'):=\inf_{\lambda\in\Lambda}\Big\{\max\Big(||x\circ \lambda-x'||, ||\lambda-I||\Big)\Big\},~ \forall x, x'\in\Omega$$

Here $\circ$ denotes the composition of functions. Compared to the Skorokhod distance, $\rho$ is not a distance since it is no longer symetric, i.e.

$$\rho(x,x')\neq \rho(x',x)$$

Now it is easy to show the triangle inequality. I would like to know whether we have

$$\rho(x,x')=0\Longrightarrow x=x'$$

Thx a lot for the reply!

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  • $\begingroup$ You write that $\Lambda$ is the set of all non-increasing onto functions $\lambda:[0,1]\to[0,1]$, but I suppose you mean non-decreasing because otherwise we would need $\lambda(0) = 1$ which implies $||\lambda-I|| = 1$ implying $\rho(x,x') \geq 1$ even if $x=x'$. $\endgroup$
    – Dominic van der Zypen
    Commented Nov 13, 2014 at 9:48
  • $\begingroup$ Sure, thx for pointing out this mistake. $\endgroup$
    – CodeGolf
    Commented Nov 13, 2014 at 9:49
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    $\begingroup$ Doesn't this take care of your question? math.stackexchange.com/questions/163678/… . $\endgroup$
    – Dominic van der Zypen
    Commented Nov 13, 2014 at 10:00
  • $\begingroup$ Exactly what i am looking for, thx for this comment!math.stackexchange.com/questions/163678/… $\endgroup$
    – CodeGolf
    Commented Nov 13, 2014 at 10:05
  • $\begingroup$ Cool - pls accept my answer so that this question is not shown as unanswered any more. Thanks! $\endgroup$
    – Dominic van der Zypen
    Commented Nov 13, 2014 at 10:16

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In the answer to the following comment, it is shown why $\rho(x,x') = 0$ implies $x = x'$: https://math.stackexchange.com/questions/163678/a-proof-that-skorohod-metric-is-a-metric

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