TITLE: On the distance function $d(f,g) = \sup_{x\in[0,2\pi]}\{|f(x)−g(x)|\}$
QUESTION [5 upvotes]: Consider the distance function 
$$
d(f,g) = \sup_{x\in[0,2\pi]}\{|f(x)−g(x)|\}.
$$
Let $f_n(x) = \sin(2^nx)$ where $n\in \Bbb N$: show $d(f_n,f_m)≥1$ when $n \neq m$.
I know for every $(n,m)$, the function image does show their distance is greater than 1. And when $n=0$, I can prove it.

REPLY [2 votes]: Let me define $d_r(f,g) = sup\{|f(x)-g(x)|:x\in[0,2 r \pi]\}$. I am going to use two properties of this distance: 

$d_r(f,g) = d(f(rx), g(rx))$.
For $r>1$, $d_r(f,g) > d(f,g)$,    (you are taking sup over larger interval).

Since you already know how to prove for $n = 0$, without loss of generality, assume $m>n$. Then $d(f_n,f_m) = d_{2^{n}}(f_0,f_{m-n}) > d(f_0, f_{m-n}) > 1$.