My book says
Convergence in sup norm $||f_n-f||\to 0$ is equivalent to uniform convergence and this follows immediately from definitions.
but I just want to check:
$\Rightarrow$ If lim$_{n\to\infty}||f_n-f||=0$, then sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0\Rightarrow |f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow (f_n)\to f$ uniformly.
And then running in reverse:
$\Leftarrow$ If $f_n\to f$ uniformly, then $|f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow$sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0\Rightarrow ||f_n-f||\Rightarrow 0$.
My question is, why $|f_n(x)-f(x)|\to 0 \forall x\in[a,b]\Rightarrow$sup$\{|f_n(x)-f(x)|:x\in[a,b]\}\to 0$. I think it's because sup is really max here because functions are continuous and $[a,b]$ is compact. Is this right? Does it hold in general if the functions aren't continuous?