# Continuous embedding of the Skorohod space D(0,1) into L^2(0,1)

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

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*}

• 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
• @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
• 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 $$$$y^+(t):=\sup_{0\le s\le t}y(s)$$$$ and $$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}$$ denotes the total variation norm. Letting now $$$$\mu:=\mu^0+\mu^+-\mu^-,$$$$ where $$\mu^0:=y(0)\de_0$$ and $$\de_0$$ is the Dirac measure supported at $$0$$, we have $$\|\mu\|_{tv}\le5\|y\|$$ and $$$$y(t)=\int_{[0,t]}d\mu=\int I_{[0,t]}d\mu$$$$ for $$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 $$$$J(u,v,t):=(I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big)\big(I_{[0,\ell(t)]}(v)-I_{[0,t]}(v)\big),$$$$ so that \begin{align} \big|J(u,v,t)\big| &\le\big|I_{[0,\ell(t)]}(u)-I_{[0,t]}(u)\big| \\ &=I\{u\text{ btw }\ell(t)\text{ and }t\} \\ &= I\{t\text{ btw }\ell^{-1}(u)\text{ and }u\} \\ \end{align} where btw means "is between". So, $$$$\int dt\, |J(u,v,t)|\le\max_u|\ell^{-1}(u)-u|=\max_t|\ell(t)-t|\le\vp$$$$ and hence $$$$\|y\circ\ell-y\|_2^2\le\|\mu\|_{tv}^2\,\vp\le(5\|y\|)^2\vp \le25(\|x\|+\vp)^2\vp.$$$$ Also, $$$$\|y\circ\ell-x\|_2\le\|y\circ\ell-x\|\le\vp.$$$$ Thus, $$$$\|y-x\|_2\le \|y\circ\ell-x\|_2+\|y\circ\ell-y\|_2\le\vp+5(\|x\|+\vp)\sqrt{\vp} \le6(\vp+\|x\|\sqrt{\vp}\,),$$$$ 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.

• 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