Let $\Omega:=D([0,1],R)$ be the space of cadlag functions $x$ defined on $[0,1]$. Let $\rho$ be the Skorokhod metric on $\Omega$, see e.g.


Now define two family of functions $\{f_{\varepsilon}\}_{\varepsilon}$ and $\{b_{\varepsilon}\}_{\varepsilon}$, where

\begin{eqnarray} f_{\varepsilon}(t)&:=&\frac{1}{1-\varepsilon}\big(t-\varepsilon\big)^+,~ t\in [0,1], \\ b_{\varepsilon}(t)&:=&1-\Big(1-\frac{1}{1-\varepsilon}t\Big)^+,~ t\in [0,1]. \end{eqnarray}

For any $x\in\Omega$, I would like to estimate the Skorokhod distance $\rho(x, x\circ f_{\varepsilon})$ and $\rho(x, x\circ b_{\varepsilon})$. Is it possible that

\begin{eqnarray} \lim_{\varepsilon\to 0}|| x\circ f_{\varepsilon}-x||+|| x\circ b_{\varepsilon}-x||=0,~ \forall x\in\Omega. \end{eqnarray}

Thx for the reply!

  • $\begingroup$ What's the definition of $(\ldots)^+$? $\endgroup$ – Dominic van der Zypen Nov 13 '14 at 15:37
  • $\begingroup$ $x^+$ means the positive part of $x$, i.e. $x^+=\max(x,0)$ $\endgroup$ – CodeGolf Nov 13 '14 at 16:48

The answer to the question is No. Let $x:[0,1] \to [0,1]$ be defined by $x(t) = 0$ for $t\in [0,1/2[$ and $x(t) = 1$ for $t\in[1/2, 1]$. Clearly, $x$ is cadlag and therefore a member of $D([0,1], R)$.

Note that for all $\varepsilon > 0$ we have $f_\varepsilon(t) < t$ for $t\in]0,1[$.

So for all $\epsilon >0$ we have $(x\circ f_\epsilon)(1/2) = 0$ and $x(1/2) = 1$ therefore $1 \leq \lim_{\varepsilon\to 0} ||x\circ f_\varepsilon -x|| \leq \lim_{\varepsilon\to 0} ||x\circ f_\varepsilon -x|| + ||x\circ b_\varepsilon -x ||$.

  • $\begingroup$ Is my answer what you were looking for? If yes, could you check the "accept answer" button? Thanks! $\endgroup$ – Dominic van der Zypen Nov 18 '14 at 9:12

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