Supremum of cadlag functions

Question

Let $f_n,f$, $n\in\mathbb{N}$, be (real-valued) cadlag functions on $[0,1]$ such that $$\sup_{0\le t\le 1}|f_n(t)-f(t)|\to 0\text{ as }n\to\infty.$$ Does someone have an idea how to prove that $$\sup_{0\le t\le 1}|f_n(t)|\to\sup_{0\le t\le 1}|f(t)|\text{ as }n\to\infty?$$

Answer 1

If $g$ and $h$ are real functions defined on a domain $D$, then $$\sup_{x \in D} (g(x) + h(x)) \le \sup_{x \in D} g(x) + \sup_{x \in D} h(x).$$ This is pretty simple to verify using the definition. Thus $$\sup_{0 \le t \le 1} |f_n(t)| \le \sup_{0 \le t \le 1} |f_n(t)| + \sup_{0 \le t \le 1} |f_n(t) - f(t)|$$ and $$\sup_{0 \le t \le 1} |f(t)| \le \sup_{0 \le t \le 1} |f_n(t)| + \sup_{0 \le t \le 1} |f_n(t) - f(t)|$$ for all $n$. In particular, $$ \left| \sup_{0 \le t \le 1} |f_n(t)| - \sup_{0 \le t \le 1} |f(t)| \right| \le \sup_{0 \le t \le 1} |f_n(t) - f(t)|$$ for all $n$. The conclusion follows from the squeeze theorem. There is nothing special about cadlag functions here.