Why is the canonical projection Lipschitz?

Question

Let $A=\{w:[0,\infty)->\mathbb{R}^d, \ w\text{ is continuous and } w(0)=0\}$, and let's use this metric of locally uniform convergence:

$\rho(w,v)=\sum_{n\geq 1} (1\wedge \sup_{i\in [0,n]}|w(i)-v(i)|)2^{-n} $.

Let also $\pi_t(w):=w(t)$(canonical projection).

Is the $\pi_t$ Lipschitz? Why?

I'm reading, Brownian Motion by Schlling and Partzsch, and they state that it is, without any proof. I've tried, but couldn't, and in fact I'm convinced it's not. Maybe they were thinking of Locally Lipschitz?

Answer 1

It's certainly not globally Lipschitz, since $\rho(w,v)\le 1$ always, but $\pi_t$ is unbounded. Yes, I suspect they mean locally Lipschitz.