Are coordinate projections in the Skorokhod space continuous? I was wondering whether coordinate projections in the Skorokhod space $D[0,1]$ are actually continuous (and, if so, how can this be proven)?
many thanks for any comments/ideas.
cheers!
 A: If $\pi_x: D[0,1]\to \Bbb R$ is the projection map sending $f$ to $f(x)$, then it is continuous everywhere if $x=0$ or $x=1$.  If $0<x<1$, it is continuous at $f$ if $f$ is continuous at $x$ but discontinuous at $f$ if $f$ is discontinuous at $x$.
Proof: The Skorokhod topology can be defined by the metric
$$
d(f,g):=\inf_\lambda \max(||\lambda-I||_\infty, ||f-g\circ \lambda||_\infty)
$$
where $I$ is the identity on $[0,1]$ and the infimum is over all strictly increasing bijective functions $\lambda$ from $[0,1]$ to itself.  All such functions $\lambda$ send $0$ to $0$ and $1$ to $1$.  Therefore,
$$
d(f,g)\ge \max(|f(0)-g(0)|,|f(1)-g(1)|),
$$
so $\pi_0$ and $\pi_1$ are continuous.
If $0<x<1$ and $f$ is continuous at $x$, then continuity of $f$ implies that, for all $\epsilon>0$, we can find $\delta>0$ such that $|f(y)-f(x)|<\epsilon$ whenever $|y-x|<\delta$.  Then, if $d(g,f)<\min(\delta,\epsilon)$, there must be some
$\lambda$ such that $$|\lambda(x)-x|<\delta,\ \ |g(x)-f(\lambda(x))|<\epsilon.$$
Setting $y:=\lambda(x)$, it follows that $|g(x)-f(x)|<2\epsilon$ whenever $d(g,f)<\min(\delta,\epsilon)$, so $\pi_x$ is continuous at $f$.
If $0<x<1$ and $f$ is discontinuous at $x$, then since $f$ is càdlàg, it must have a jump discontinuity at $x$: $$\lim_{t\to x^{-}} f(t)\ne f(x).$$
Then, let $f_\epsilon$ be the shifted version of $f$ defined by
$$f_\epsilon(t):=\left\{\begin{array}{ll}
f(0), & t\le \epsilon,\\
f(t-\epsilon), & \rm otherwise.\end{array}\right.$$
Using the fact that $f$ is right-continuous at $0$, you can prove that
$$d(f_\epsilon,f)\le \epsilon,$$
so, in the Skorokhod topology, $$\lim_{\epsilon\to 0^+} f_\epsilon=f,$$ but
$$\lim_{\epsilon\to 0^+} \pi_x(f_\epsilon)=
\lim_{\epsilon\to 0^+} f_\epsilon(x)=\lim_{t\to x^-} f(t)\ne \pi_x(f)=f(x).
$$
Therefore, $\pi_x$ is discontinuous at $f$.
