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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?

An old man in the sea.
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1 Answers1

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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.

Robert Israel
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  • https://math.stackexchange.com/questions/3085502/canonical-projection-on-the-set-of-continuous-functions-that-equal-0-at-x-0/3096293#3096293 Robert, I wrote an answer to a question equal to this. In my answer I try to prove that's it's locally lipschitz. Could you please check if it's ok? I hope I'm not being bothersome. ;) – An old man in the sea. Feb 01 '19 at 14:41
  • Also, Thanks for your answer. =D ;) – An old man in the sea. Feb 01 '19 at 14:41