# Equivalent definition for Skorokhod metric

I have a question about the Skorokod distance on the space $$\mathcal{D}([0,1],\mathbb{R})$$: $$d(X,Y):= \inf_{\lambda \in \Lambda}\left( \sup_{t\in [0,1]}|t-\lambda(t)|\vee \sup_{t\in [0,1]}|X(t)-Y(\lambda(t))|\right)$$ where $$\Lambda$$ is the set of continuous and strictly increasing functions $$[0,1]\to[0,1]$$ such that $$\lambda(0)=0$$ $$\lambda(1)=1$$. If instead of cadlag functions $$\mathcal{D}([0,1],\mathbb{R})$$ we consider the space of continuous functions $$\mathcal{C}([0,1],\mathbb{R})$$, I am wondering how necessary the condition that $$\lambda$$ be continuous and strictly increasing is.

My reasoning is that if we had an increasing and discontinuous function $$\gamma:[0,1]\to[0,1]$$ such that $$\gamma(0)=0$$ and $$\gamma(1)=1$$ such that: $$\sup_{t\in [0,1]}|t-\gamma(t)|\vee \sup_{t\in [0,1]}|X(t)-Y(\gamma(t))|<\epsilon$$ then we could just mollify the function to get some $$\gamma_\delta^\star$$ where $$\gamma_\delta^\star$$ is continuous and strictly increasing and $$\gamma_\delta^\star \to \gamma$$ uniformly on $$[0,1]$$ as $$\delta \downarrow 0$$. Since $$X,Y,\gamma$$ are all continuous on $$[0,1]$$ and hence uniformly continuous on $$[0,1]$$ (since $$[0,1]$$ is compact), then we could surely choose $$\delta$$ to be small enough to make: $$\sup_{t\in [0,1]}|t-\gamma_\delta^\star(t)|\vee \sup_{t\in [0,1]}|X(t)-Y(\gamma_\delta^\star(t))|<\epsilon$$ and hence the metrics would be equivalent on $$\mathcal{C}([0,1],\mathbb{R})$$.

Is this something that has been used before? Can anyone point me towards any references - I couldn't find any myself? Is my reasoning okay?

• There should be only one question in one post. Your post contains at least two questions. Please fix this. Anyhow, I don't think this was used anywhere, because it does not seem to make sense to use Skorokhod's metric for $\mathcal C$. Commented Jan 19 at 15:09

$$\newcommand\ep\varepsilon\newcommand\la\lambda\newcommand\La\Lambda\newcommand\tla{\tilde\lambda}\newcommand\tLa{\tilde\Lambda}\newcommand\ga\gamma\newcommand\Ga\Gamma\newcommand\de\delta$$Your conjecture is true, but your reasoning is incorrect.

Let us first show that your conjecture is true. Let $$\La$$ denote the set of all nondecreasing functions $$\la\colon I\to I$$ such that $$\la(0)=0$$ and $$\la(1)=1$$, where $$I:=[0,1]$$. Let $$\tLa$$ denote the set of all continuous strictly increasing functions $$\la\in\La$$.

Take any $$\la\in\La$$. The set $$D_\la$$ of the points of discontinuity of $$\la$$ is at most countable, and the sum $$\sum_{t\in D_\la}J_\la(t)$$ of the jumps $$J_\la(t):=\la(t+)-\la(t-)$$ of the function $$\la$$ is $$\le1$$. So, $$\la$$ can be approximated uniformly by a function in $$\La$$ with only finitely many points of discontinuity -- see details on this at the end of this answer. So, without loss of generality (wlog) the set $$D_\la$$ is finite.

Next, the function $$\la$$ (with only finitely many points of discontinuity) can be approximated uniformly (on $$I$$) by a strictly increasing function in $$\La$$ with only finitely many points of discontinuity -- say by the function $$\la_h$$ for small real $$h>0$$ such that $$\la_h(t)=(\la(t)+ht)/(1+h)$$ for $$t\in I$$.

So, wlog $$\la$$ is strictly increasing and the set $$D_\la$$ is finite.

Take some real $$\de>0$$. For each point $$t_*\in D_\la$$, consider its $$\de$$-neighborhood $$N_\de(t_*):=I\cap[t_*-\de,t_*+\de]=:[a_{t_*},b_{t_*}]$$. Choose $$\de$$ so that these neighborhoods are pairwise disjoint. Let the function $$\la_\de\in\tLa$$ be defined by the following conditions:

• on each interval $$[a_{t_*},b_{t_*}]$$, $$\la_\de$$ equals the linear interpolation of $$\la$$ over this interval;

$$$$\text{ \la_\de=\la on I_\de:=I\setminus\bigcup_{t_*\in D_\la}(a_{t_*},b_{t_*}).} \tag{1}\label{1}$$$$

Take any real $$\ep>0$$. If $$\de$$ is small enough, then for each $$t_*\in D_\la$$ and all $$t$$ and $$s$$ in $$[a_{t_*},b_{t_*}]$$ we have $$|X(t)-X(s)|\le\ep$$, and hence
\begin{aligned} &\sup\{|X(t)-Y(\la_\de(t))|\colon t\in[a_{t_*},b_{t_*}]\} \\ &\le\sup\{|X(t)-Y(l)|\colon t\in[a_{t_*},b_{t_*}], l\in[\la_\de(a_{t_*}),\la_\de(b_{t_*})]\} \\ &=\sup\{|X(t)-Y(l)|\colon t\in[a_{t_*},b_{t_*}], l\in[\la(a_{t_*}),\la(b_{t_*})]\} \\ &=\sup\{|X(t)-Y(\la(s))|\colon t\in[a_{t_*},b_{t_*}], s\in[a_{t_*},b_{t_*}]\} \\ &\le\ep+\sup\{|X(s)-Y(\la(s))|\colon s\in[a_{t_*},b_{t_*}]\}. \end{aligned} Similarly (or in particular), \begin{aligned} &\sup\{|t-\la_\de(t)|\colon t\in[a_{t_*},b_{t_*}]\} \\ &\le\ep+\sup\{|s-\la(s)|\colon s\in[a_{t_*},b_{t_*}]\}. \end{aligned} Recalling now \eqref{1} and that $$\ep>0$$ was arbitrary, and letting $$\rho_\la(X,Y):=\sup_{t\in [0,1]}|t-\la(t)|\vee \sup_{t\in [0,1]}|X(t)-Y(\la(t))|,$$ we see that $$\inf\{\rho_\la(X,Y)\colon\la\in\tLa\}\le\inf\{\rho_\la(X,Y)\colon\la\in\La\}.$$ The reverse inequality, $$\inf\{\rho_\la(X,Y)\colon\la\in\La\}\le\inf\{\rho_\la(X,Y)\colon\la\in\tLa\}$$, holds because $$\La\supseteq\tLa$$. Thus, your conjecture is proved.

Your reasoning was incorrect, because it is not true that "we could just mollify the function to get some $$\gamma_\delta^\star$$ where $$\gamma_\delta^\star$$ is continuous and strictly increasing and $$\gamma_\delta^\star \to \gamma$$ uniformly on $$[0,1]$$ as $$\delta \downarrow 0$$" if $$\ga$$ is discontinuous: then such an approximation cannot be uniform, because the uniform limit of continuous functions is continuous.

Detail: Take any nondecreasing function $$f$$ on $$[0,1]$$ with $$f(0)=0$$. Let $$f_0:=f$$. Take then any $$t_1$$ in the set $$D_f$$ of points of discontinuity of $$f$$, with the jump $$J_f(t_1)$$ of $$f$$ at $$t_1$$. Let $$f_1:=f_{t_1}$$ be the function obtained from $$f_0$$ by the removal of the jump of $$f_0=f$$ at $$t_1$$; that is, let $$f_1(t)=f_0(t)$$ for $$t\in[0,t_1)$$ and $$f_1(t)=f_0(t)-J_{f_0}(t_1)$$ for $$t\in[t_1,1]$$. Then $$f_1$$ is nondecreasing, $$f_1(0)=0$$, $$f_1(1)=f_0(1)-J_f(t_1), $$f_0-J_f(t_1)\le f_1\le f_0$$, $$D_{f_1}=D_{f_0}\setminus\{t_1\}$$, and $$J_{f_1}(t)=J_{f_0}(t)=J_f(t)$$ for all $$t\in D_{f_1}$$.

Continuing thus, we can remove the jumps of $$f$$ at all points in any finite subset, say $$T$$, of $$D_f$$, to get a nondecreasing function $$f_T$$ on $$[0,1]$$ such that $$f_T(0)=0$$, $$f_T(1)=1-\sum_{t\in T}J_f(t)$$, $$f-\sum_{t\in T}J_f(t)\le f_T\le f$$, $$D_{f_T}=D_{f}\setminus T$$, and $$J_{f_T}(t)=J_f(t)$$ for all $$t\in D_{f_T}$$. In particular, we have $$0=f_T(0)\le f_T(1)=1-\sum_{t\in T}J_f(t)$$, so that $$\sum_{t\in T}J_f(t)\le1$$ for any finite $$T\subseteq D_f$$. So, $$\sum_{t\in D_f}J_f(t)\le1<\infty$$ and hence $$D_f$$ is at most countable, so that we can write $$$$D_f=\{t_1,t_2,\dots\}$$$$ for some distinct $$t_j$$'s in $$[0,1]$$. Take now any natural $$n$$. Let $$f_n,f_{n+1},\dots$$ be obtained from $$f$$ by the successive removals of the jumps of $$f$$ at $$t_n,t_{n+1},\dots$$. Then $$f_n\ge f_{n+1}\ge\cdots$$ and hence there is the pointwise limit $$g_n:=\lim_{k\to\infty}f_{n+k}$$. Then $$g_n$$ is nondecreasing, $$g_n(0)=0$$, $$g_n(1)=1-\sum_{k\ge0}J_f(t_{n+k})$$, $$f-\sum_{k\ge0}J_f(t_{n+k})\le g_n\le f$$, $$D_{g_n}=D_{f}\setminus\{t_n,t_{n+1},\dots\}=\{t_1,\dots,t_{n-1}\}$$, and $$J_{g_n}(t)=J_f(t)$$ for all $$t\in D_{g_n}$$.

Now take any $$\la\in\La$$ in place of $$f$$ above. Take any real $$h>0$$. Take any natural $$n$$ such that $$\sum_{k\ge0}J_\la(t_{n+k}). Then $$\la-h\le g_n\le\la$$, so that $$g_n$$ uniformly approximates $$\la$$ if $$h$$ is small enough. In particular, $$g_n(1)\ge f(1)-h=1-h>0$$ if $$h\in(0,1)$$. So, for small enough $$h>0$$, the function $$\la_h:=g_n/g_n(1)$$ uniformly approximates $$\la$$. Moreover, $$\la_h$$ is in $$\La$$ and the set $$D_{\la_h}=\{t_1,\dots,t_{n-1}\}$$ is finite.

Thus, indeed, any $$\la\in\La$$ can be approximated uniformly by a function in $$\La$$ with only finitely many points of discontinuity.

• Thanks for you comprehensive answer Iosif. Can you just clarify your two claims about approximations: "so, without loss of generality (wlog) λ is strictly increasing" and "So, wlog λ is strictly increasing and the set Dλ is finite". I can see them intuitively, but are these standard results with standard references? Commented Jan 22 at 12:22
• @user1598 : I have added the requested details. I have not been able to find references covering these details. This is one of the cases when it is much easier for me to prove something than to find it in the literature. Commented Jan 22 at 15:22
• Thanks Iosif - this is very helpful. Answer is accepted. Commented Jan 23 at 13:42