I'm having trouble seeing the justification for what I assume should be a very simple step in a paper's calculations.
Let n be a natural number and define $\phi^n_s:=(ns)\wedge1\vee0=\max(\min(ns,1),0)$. Show that for any two arbitary real numbers $t_1$ and $t_2$ greater than or equal to zero, $$|\phi^n_{t_1-s}-\phi^n_{t_2-s}|\leq n|(t_1-s)-(t_2-s)|=n|t_1-t_2|.$$
Let $\phi(n,t):= \phi_t^n. $The variable $s$ is irrelevant. Consider the following cases:
$$nt<0 \implies \phi(n,t) = 0$$ $$0\le nt \le 1 \implies \phi(n,t) = n t$$ $$nt > 1 \implies \phi(n,t) = 1$$
Now, you can see that for any combination $nt_1, nt_2$ that you pick from above, the inequality holds. Take for example $0\le nt_1, nt_2 \le1$. Then:
$$|\phi(n, t_1) - \phi(n, t_2)| = |nt_1 - nt_2| = n|t_1 - t_2|$$
Now convince yourself that for any other combination, we get: $$|\phi(n, t_1) - \phi(n, t_2)| \le n|t_1 - t_2|$$
Now pick any $s \in \mathbb{R^n}$, then: $$|\phi(n, t_1 -s) - \phi(n, t_2-s)| \le n|(t_1-s) - (t_2-s)| = n |t_1 - t_2|$$