Proof from Ethier and Kurtz' Markov Processes on showing uniform convergence on bounded intervals using convergence of a sequence.

Question

This is Corollary 6.6 from Chapter 1 of Ethier and Kurtz' Markov Processes.

In this proof, why is the first sentence sufficient to prove uniform convergence on bounded intervals? I can't figure this claim out. I would greatly appreciate any help.

Answer 1

Suppose that $V(t/n)^n f$ does not converge to $T_t f$ uniformly on compact sets for some $f \in L$. Then there exist $T>0$, $\epsilon>0$ and a sequence $(n_k)_{k \in \mathbb{N}} \subseteq \mathbb{N}$, $n_k \to \infty$, such that

$$\Delta_k(s) := |V(s/n_k)^{n_k} f-T_s f|$$

satisfies

$$\sup_{s \in [0,T]} \Delta_k(s) > \epsilon, \qquad k \in \mathbb{N}.$$

By the very definition of supremum, we can choose $t_k \in [0,T]$ such that

$$\Delta_k(t_k)> \frac{\epsilon}{2}.\tag{1}$$

As $(t_k)_{k \in \mathbb{N}}$ is contained in the compact set $[0,T]$, there exists a convergent subsequence, say $t_{k_j} \to t \in [0,T]$. Because of $(1)$ and the strong continuity of $(T_t)_{t \geq 0}$, we have

$$|V(t_{k_j}/n_{k_j})^{n_{k_j}} f-T_t f| > \frac{\epsilon}{4}$$

for all $j \gg 1$ in contradiction to the statement which was shown in the proof.