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.
If $n>m\ge 0$ then by letting $x=\pi/2^{m+1}\in [0,2\pi]$, we find $$\begin{align*} \sup_{x\in[0,2\pi]}\{|\sin(2^nx)−\sin(2^m x)|\}&\geq |\sin(2^n (\pi/2^{m+1})))−\sin(2^m (\pi/2^{m+1}))|\\ &=|\sin(2^{n-m-1} \pi)−\sin(\pi/2)|=|0-1|=1. \end{align*}$$
Hint: Use that $$\sin(x)-\sin(y)=2\sin\left(\frac{x-y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$
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:
Since you already know how to prove for $n = 0$, without loss of generality, assume $m>n$. Then $d(f_n,f_m) = d_{2^{n}}(f_0,f_{m-n}) > d(f_0, f_{m-n}) > 1$.
