Let $D(0,1)$ be the Skorohod space with the Skorohod topology, i.e. the space of real-valued càdlàg-functions on $[0,1]$ with topology induced by the metric $$d(f,g) = \inf_{\varphi \in \Lambda} \left\{ \lVert \varphi - \operatorname{Id} \rVert_{\infty} \lor \lVert f - g \circ \varphi \rVert_{\infty} \right\},$$ where the infimum is taken over all strictly increasing continuous functions $\varphi$ mapping $[0,1]$ onto itself.

In an article I'm reading, it is stated without proof that $D(0,1)$ is continuously embedded in $L^2(0,1)$. I tried to prove it but did not succeed. Does anybody know a proof or a reference?

Thanks a lot!

  • Is there a version of the closed graph theorem that can be used here? Both modes of convergence imply a.e. convergence (of a subsequence), right? – Nate Eldredge Dec 5 at 20:08
  • Is the "usual topology" the weak $*$ topology on the measures induced by these functions? – Christian Remling Dec 5 at 20:33
  • @ChristianRemling I have added a description of the topology for clarity – r_faszanatas Dec 5 at 20:44
  • @Nate Eldredge yes, the implication should be true. I will look into your idea to use the closed graph theorem. In any case, it does not seem to be as straightforward as the article made me believe. – r_faszanatas Dec 5 at 20:46
  • One question is whether Skorokhod space is a Frechet space. If it is, then the closed graph theorem applies. I feel like this should be well known, but I don't know myself and couldn't find the answer with 30 seconds of Google. – Nate Eldredge Dec 5 at 22:39
up vote 2 down vote accepted

I think the following works, but please check me.

Note first that cadlag functions are measurable and bounded, so $D(0,1) \subset L^2(0,1)$.

Suppose $f_n \to f$ in the Skorokhod metric. Note that $f$ has at most countably many discontinuities, and $f_n(t) \to f(t)$ for each continuity point $t$ of $f$ ((*), see below). In particular, $f_n \to f$ almost everywhere. If we can show the sequence $f_n$ is uniformly bounded, then the dominated convergence theorem implies $f_n \to f$ in $L^2$.

Convergent sequences in a metric space are bounded, so there is some $R$ such that $d(f_n, 0) < R$ for all $n$. Thus for each $n$ there is a $\varphi$ such that, in particular, $\|f_n - 0 \circ \varphi\|_\infty < R$. Since $0 \circ \varphi = 0$, we have $\|f_n\|_\infty < R$. So the dominated convergence theorem applies and we are done.

To see why (*) is true (which I think is a fairly well-known fact), fix $\epsilon > 0$ and $t \in [0,1]$. If $f$ is continuous at $t$, there is $\delta > 0$ such that $|f(t) - f(s)| < \epsilon/2$ whenever $|s-t| < \delta$. Now if $f_n \to f$ in Skorokhod metric, then for any sufficiently large $n$, we have $d(f_n, f) < \min(\epsilon/2, \delta)$. This means there exists $\varphi$ (depending on $n$) such that $\|\varphi - Id\|_\infty < \delta$ and $\|f_n - f \circ \varphi\|_\infty < \epsilon/2$. In particular, $|\varphi(t) -t|<\delta$, and so $|f(\varphi(t)) - f(t)| < \epsilon/2$. Then $$\begin{align*} |f_n(t) - f(t)| &\le |f_n(t) - f(\varphi(t))| + |f(\varphi(t)) - f(t)| \\ &\le \epsilon/2 + \epsilon/2. \end{align*}$$

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    Sorry, I don't see why "$f_n(t) \to f(t)$ for each continuity point $t$ of $f$". This is much stronger than the convergence in $L^2$ in question, since the $f_n$'s are uniformly bounded. – Iosif Pinelis Dec 6 at 0:48
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    @IosifPinelis: I think it's kind of a standard property of the Skorokhod space, but I added a proof. – Nate Eldredge Dec 6 at 1:05
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    Thank you. Yes, this is indeed this simple. – Iosif Pinelis Dec 6 at 1:23
  • @NateEldredge Thank you very much for sharing this elegant proof, the situation is now very clear to me. – r_faszanatas Dec 6 at 21:44

$\newcommand{\de}{\delta} \newcommand{\vp}{\varepsilon}$

Take any $x,y$ in $D:=D[0,1]$ with $d(x,y)\le\vp$ for some $\vp\in(0,1)$, so that for some strictly increasing continuous function $\ell$ mapping $[0,1]$ onto itself we have $\|y\circ\ell-x\|\le\vp$ and $\|\ell-\text{id}\|\le\vp$, where $\|\cdot\|:=\|\cdot\|_\infty$.

Since $x\in D$, it is easy to see that $\|x\|<\infty$. So, $\|y\|=\|y\circ\ell\|\le\|x\|+\vp<\infty$.

For $t\in[0,1]$, let \begin{equation} y^+(t):=\sup_{0\le s\le t}y(s) \end{equation} and $y^-:=y^+-y$. Then $y^\pm$ is a nondecreasing function in $D$. So, there exists a unique nonnegative (Lebesgue--Stieltjes) measure $\mu^\pm$ such that $\mu^\pm([0,t])=y^\pm(t)-y(0)$ for all $t\in[0,1]$. Moreover, $\|\mu^+\|_{tv}=y^+(1)-y(0)\le2\|y\|$ and similarly $\|\mu^-\|_{tv}\le2\|y\|$, where $\|\cdot\|_{tv}$ denotes the total variation norm. Letting now \begin{equation} \mu:=\mu^0+\mu^+-\mu^-, \end{equation} where $\mu^0:=y(0)\de_0$ and $\de_0$ is the Dirac measure supported at $0$, we have $\|\mu\|_{tv}\le5\|y\|$ and \begin{equation} y(t)=\int_{[0,t]}d\mu=\int I_{[0,t]}d\mu \end{equation} for $t\in[0,1]$, where $\int:=\int_{[0,1]}$ and $I$ denotes the indicator.

Hence, for the $L^2$ norm $\|\cdot\|_2$ we have \begin{align} \|y\circ\ell-y\|_2^2&=\int dt\,(y(\ell(t))-y(t))^2 \\ &=\int dt\Big(\int\mu(du)\,\big(I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big)\Big)^2 \\ &=\iint \mu(du)\mu(dv)\,\int dt\, J(u,v,t), \end{align} where \begin{equation} J(u,v,t):=(I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big)\big(I_{[0,\ell(t)]}(v)-I_{[0,t]}(v)\big), \end{equation} so that \begin{align} \big|J(u,v,t)\big| &\le\big|I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big| \\ &=I\{u\text{ btw }\ell(t)\text{ and }t\} \\ &= I\{t\text{ btw }\ell^{-1}(u)\text{ and }u\} \\ \end{align} where btw means "is between". So, \begin{equation} \int dt\, |J(u,v,t)|\le\max_u|\ell^{-1}(u)-u|=\max_t|\ell(t)-t|\le\vp \end{equation} and hence \begin{equation} \|y\circ\ell-y\|_2^2\le\|\mu\|_{tv}^2\,\vp\le(5\|y\|)^2\vp \le25(\|x\|+\vp)^2\vp. \end{equation} Also, \begin{equation} \|y\circ\ell-x\|_2\le\|y\circ\ell-x\|\le\vp. \end{equation} Thus, \begin{equation} \|y-x\|_2\le \|y\circ\ell-x\|_2+\|y\circ\ell-y\|_2\le\vp+5(\|x\|+\vp)\sqrt{\vp} \le6(\vp+\|x\|\sqrt{\vp}\,), \end{equation} which proves the desired continuity.

One may also note that the latter bound is optimal up to a universal constant factor. Indeed, let $\vp\in(0,\frac12)$, $x=cI_{[\frac12,1]}$, and $y=\vp+cI_{[\frac12+\vp,1]}$, where $c$ is any real number. Then $x$ and $y$ are in $D$, and $d(x,y)\le\vp$ (consider the function $\ell$ that is affine on each of the intervals $[0,\frac12]$ and $[\frac12,1]$ and maps $0,\frac12,1$ to $0,\frac12+\vp,1$, respectively). On the other hand, it is easy to see that here $\|y-x\|_2\asymp\vp+|c|\sqrt{\vp}\asymp\vp+\|x\|\sqrt{\vp}$ -- which proves the mentioned optimality.

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    Nate Eldredge has now given a much simpler answer, using a very different approach. However, I have decided to retain this answer as well, since it provides an explicit and optimal bound on the rate of convergence. – Iosif Pinelis Dec 6 at 5:03
  • Indeed the answer from Nate Eldredge is simpler but I also learned a lot from your answer (and the same is true for all the other questions I had in the past). Thank you very much, I really appreciate it! – r_faszanatas Dec 6 at 21:43

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