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!
 A: 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
$$\begin{align*}
|f_n(t) - f(t)|  &\le |f_n(t) - f(\varphi(t))| + |f(\varphi(t)) - f(t)| \\ &\le \epsilon/2 + \epsilon/2.
\end{align*}$$
A: $\newcommand{\de}{\delta}
\newcommand{\vp}{\varepsilon}$
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.
