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!



Continuous embedding of the Skorohod space D(0,1) into L^2(0,1)
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
$$\begin{align\*}
|f\_n(t) - f(t)| &\le |f\_n(t) - f(\varphi(t))| + |f(\varphi(t)) - f(t)| \\ &\le \epsilon/2 + \epsilon/2.
\end{align\*}$$


