In the book on Brownian motion by Schilling and Praetzsch there is following statement: Let $\mathcal{C}_{(0)}:=\{f\in\mathcal{C}[0,\infty):\ f(0)=0\}$ be the space of all continuous functions starting in zero. This space is equipped with the metric
$d(f,g)=\sum_{n=0}^\infty(1\wedge\sup_{0\leq s\leq n}|f(s)-g(s)|)\frac{1}{2^n}$
which is called metric of uniform local convergence.
Then the projection on time $t$, i.e. $\pi_t:\mathcal{C}_{(0)}\to\mathbb{R},\ \pi_t(f)=f(t)$ is Lipschitz continuous with respect to the metric $d$.
I cannot show that and I am really confused because of the minimum $\wedge$ in the metric. Let $f,g$ are such that after both starting in zero, they differ pointwise by at least 1 after some time $T$ onwards, i.e. $|f(t)-g(t)|>1$ for all $t\geq T$ large enough. Then $d(f,g)\leq \sum_{n=1}^\infty \frac{1}{2^n}=1$ and
$|f(t)-g(t)|\geq d(f,g)$ for $t>T$
and I dont see how a constant could uniformly correct that.
Thank you for your help!
The authors probably meant that $\pi_t$ is locally Lipschitz continuous. Indeed, let $n = \lceil t \rceil$. Then for any $f,g \in \mathcal{C}_{(0)}$ such that $d(f,g) < 2^{-n}$, we have $\sup_{0\leq s\leq n}|f(s)-g(s)| < 1$, which implies that \begin{align} |f(t) - g(t)| &\le \sup_{0\leq s\leq n}|f(s)-g(s)| \\ &\le 2^n d(f,g) \end{align}