TITLE: Proof of Lipschitz continuity of coordinate map in $C[0,\infty)$
QUESTION [3 upvotes]: consider continuous function space $C[0,\infty)$ with metric
$$d(\omega_1,\omega_2)=\sum_{n=1}^\infty\frac{1}{2^n}\left[(\sup_{t\in[0,n]}|\omega_1(t)-\omega_2(t)|)\wedge 1\right]$$
where $\omega_1,\omega_2\in C[0,\infty)$
I want to show that for any $t$, the coordinate map $\pi_t(\omega):=\omega(t)$ is Lipschitz continuous.

I'm only able to show that $\pi_t$ is continuous without using $\epsilon-\delta$:
if $d(\omega_n,\omega)\to 0$, then we can have they converge uniformly in each bounded interval. If not, there exist $n_0$ and $\epsilon\in(0,1)$ such that $\sup_{t\in[0,n_0]}|\omega_n(t)-\omega(t)|\ge \epsilon$,so  $(\sup_{t\in[0,n]}|\omega_n(t)-\omega(t)|)\wedge 1\ge \epsilon$ is true for any $n\ge n_0$.
thus we have 
$$d(\omega_n,\omega)\ge\sum_{n=n_0}^\infty\frac{1}{2^n}\left[(\sup_{t\in[0,n]}|\omega_n(t)-\omega(t)|)\wedge 1\right]\ge \sum_{n=n_0}^\infty\frac{1}{2^n}\epsilon\ge \frac{\epsilon}{2^{n_0-1}}$$
is a contradiction.
then $\omega_n\to\omega$ uniformly in $[0,t]$, $\pi_t(\omega_n)\to\pi_t(\omega)$ in particular.

but I can't prove the continuity above using $\epsilon-\delta$ which I think is more rigorous.
(if $d(\omega_n,\omega)<\epsilon$, what can we say about $\sup_{t\in[0,N]}|\omega_n(t)-\omega(t)|$? )
moreover , it is actually lipschitz continuous, but I don't know how to get the lipschitz constant.
Thanks a lot!

REPLY [2 votes]: The function $\pi_t$ is actually not Lipschitz continuous: Note that by definition of $d$, $d(\omega_1, \omega_2) \le 1$ for any $\omega_1, \omega_2 \in C[0,\infty)$. Thus we let $\omega_0 ,\omega_1, \cdots, \omega_n \in C[0,\infty)$ so that $\omega_i(t) = i$, then 
$$|\omega_n (t) - \omega_0(t)| = n \ge n d(\omega_n, \omega_1)$$
for all $n$. Thus there isn't a $M>0$ so that 
$$|\pi_t (\omega_1) - \pi_t(\omega _2)| \le M d(\omega_1, \omega_2)$$
for all $\omega_1, \omega_2$.
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