How to prove a sequence of functions converges uniformly to a function $f$ if and only if $\lim_{n\to\infty}\sup\{|f(x)-f_n(x)|:x\in S\}=0$?

Question

How to prove a sequence of functions on a set $S$ of real numbers converges uniformly to a function $f$ on $S$ if and only if $\lim_{n\to\infty}\sup\{|f(x)-f_n(x)|:x\in S\}=0$?

My thought:

$\rightarrow$: Suppose uniform convergence, then $\forall \epsilon>0,\exists N, \forall x\in S,\forall n>N,|f_n(x)-f(x)|<\epsilon$. Let $\epsilon $ be given, fix $N$, since it's true for each $n>N$, we know $\sup\{|f_n(x)-f(x)|\}<\epsilon$, but then it's the exact definition of $\lim\sup$.

$\leftarrow$ I feel like it's the reverse of the above process.

Could someone give any idea?

Answer 1

Here are some detailed hints:

For the only-if direction: Suppose $\{ f_n \}_{n \in \mathbb{N}}$ converges uniformly to $f$ on $S$. We must then show that $\lim_{n \rightarrow \infty} \sup_{x \in S} |f_n(x) - f(x)| = 0$. So take an $\epsilon > 0$. We must then find an $n_0 \in \mathbb{N}$ such that $\sup_{x \in S} |f_n(x) - f(x)| < \epsilon$ whenever $n > n_0$. Now use the assumption about uniform convergence and the definition of least upper bounds to find this $n_0$.

For the if-direction: Suppose $\lim_{n \rightarrow \infty} \sup_{x \in S} |f_n(x) - f(x)| = 0$. Take an $\epsilon > 0$. We must then find an $n_0 \in \mathbb{N}$ such that $|f_n(x) - f(x)| < \epsilon$ for all $x \in S$. Now use the assumption about the convergence (and the definition of convergence of a sequence) to find this $n_0$.