Question about the relation between the Skorohod norm and the uniform norm Is it true that convergence of random elements in $D[0,1]$, equipped with the Skorohod norm, to a limit with continuous paths (i.e., a limit in $C[0,1]$) implies that the same convergence holds with respect to the uniform norm on $D[0,1]$? 
I vaguely remember that I read this statement somewhere, but I could not find a reference. I could not find such a theorem in Billingsley, for instance (but I might have overlooked it of course). 
Any help would be much appreciated ! 
Thanks! 
 A: Let $(f_n)_{n \in \mathbb{N}}$ in $D[0,1]$ and $f \in C[0,1]$ such that $f_n \to f$ with respect to the Skorohod topology. This is equivalent to
$$\begin{align*} \lim_{n \to \infty} \|f_n-f \circ \lambda_n\|_{\infty} = \lim_{n \to \infty} \|\lambda_n-\text{id}\|_{\infty}=0 \end{align*}$$
for some sequence $(\lambda_n)_n$ of strictly increasing continuous mappings of $[0,1]$ onto itself where $\circ$ denotes the composition. Let $\varepsilon>0$. Since $f$ is uniformly continuous, we may choose $\delta>0$ such that for any $s,t \in [0,1]$ $|s-t| < \delta$, we have
$$|f(t)-f(s)| < \varepsilon$$
Now choose $n_0 \in \mathbb{N}$ such that
$$\begin{align*} \|f_n - f \circ \lambda_n\|_{\infty} &< \varepsilon & \|\lambda_n-\text{id}\|_{\infty} < \delta \end{align*}$$
for all $n \geq n_0$. Then,
$$\begin{align*} |f_n(t)-f(t)| &\leq |f_n(t)- f \circ \lambda_n(t)| + |f(\lambda_n(t))-f(t)| \leq 2 \varepsilon \end{align*}$$
for $n \geq n_0$. Since $n_0$ does not depend on $t$, this shows that $f_n \to f$ with respect to the uniform norm on $D[0,1]$.
(See e.g. Billingsley, Section 14, right after the definition of the Skorohod topology.)
