Canonical projection on the set of continuous functions that equal $0$ at $x=0$ to $\mathbb R ^d $ is Lipschitz continuous?

Question

I would like to know if the following is true.

Let $\mathcal C_{(0) } $ denote the set of continuous functions from $[0,\infty)$ to $\mathbb R^d $ that sends $0 $ in $[0,\infty)$ to $0 $ in $\mathbb R^d $ and equip $\mathcal C_{(0) } $ with the metric $$\rho(f,g):= \sum _{n=1 } ^{\infty } 2^{-n }\text{min}\{1, \sup_{0 \le t \le n } |f(t)-g(t)|\} $$ then the canonical projections $\pi_t \ :\mathcal C_{(0) } \to \mathbb R^d \ ,\pi_t(f)=f(t) $ is Lipschitz continuous (with respect to the metric $\rho $)?

I would think it isn't since the distance between the constant zero function and the sequence of functions $f_n(t)=(mt,...,mt) $ increases without bound at the point $t $ in $\mathbb R^d $, but the book I'm reading, Brownian Motion by R. Schlling and L Partzsch, claims it is:

then later

Have I misunderstood something?

For what it's worth, I agree with you. I think they mean locally Lipschitz, which would then imply continuity. ;) — An old man in the sea., Feb 01 '19 at 12:00
https://math.stackexchange.com/questions/1480971/proof-of-lipschitz-continuity-of-coordinate-map-in-c0-infty?rq=1 — An old man in the sea., Feb 01 '19 at 14:39
https://math.stackexchange.com/questions/3096130/why-is-the-canonical-projection-lipschitz/3096196#3096196 — An old man in the sea., Feb 01 '19 at 14:39
https://math.stackexchange.com/questions/530605/locally-lipschitz-implies-continuity — An old man in the sea., Feb 01 '19 at 15:11

An old man in the sea. · Answer 1 · 2019-02-01T16:11:31.390

I'm reading the same book. I think it's a typo, since the projection is not Lipschitz...However, it's locally lipschitz.

Here's my take on it.

We have that $|w(t)-v(t)|<\infty$, for all $t \in [0,\infty[$, since $w,v \in (\mathbb{R}^d)^I$.

$\sup_{i \in [0,n]} |w(i)-v(i)|$ is non-decreasing with $n$, and $1 \wedge \sup_{i \in [0,n]} |w(i)-v(i)|\leq 1$. So, $ \sup_{i \in [0,1]} |w(i)-v(i)|\leq \rho(w,v)\leq\sum_{i\geq 1} 2^{-i}=1$

The Locally Lipschitz definition applied to our case is: for all $z \in C_{(0)}$, there exists $\delta_t$ such that $\displaystyle \exists_K \forall_{w,v\in B(z,\delta_t)}|\pi_t(w)-\pi_t(v)|<K\rho(w,v)$

Let's fix $t$, with $[t]$ being the smallest integer greater or equal than $t$.

If $\sup_{i \in [0,[t]]} |w(i)-v(i)|\geq 1$, then $\displaystyle \rho(w,v)\geq \frac{1}{2^{[t]-1}}$.

So, let's choose $\displaystyle \delta_t\in \left]0,\frac{1}{2^{[t]-2}}\right[$, and we're sure that $\displaystyle \forall_{w,v\in B(z,\delta_t)}$ we have

$$\rho(w,v)\leq \rho(w,z)+\rho(v,z)<2\cdot \frac{1}{2^{[t]-2}}=\frac{1}{2^{[t]-1}}$$

Now, we know that $|w(t)-v(t)|\leq \sup_{i \in [0,[t]]} |w(i)-v(i)|= \frac{2^{[t]}}{2^{[t]}} \sup_{i \in [0,[t]]} |w(i)-v(i)|< 2^{[t]} \rho(w,s)$

Then we just need to choose $K=2^{[t]}$

Canonical projection on the set of continuous functions that equal $0$ at $x=0$ to $\mathbb R ^d $ is Lipschitz continuous?

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