Are coordinate projections in the Skorokhod space continuous?

Question

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!

Answer 1

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 $f$.