On the distance function $d(f,g) = \sup_{x\in[0,2\pi]}\{|f(x)−g(x)|\}$

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:

• $d_r(f,g) = d(f(rx), g(rx))$.
• 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 $d(f_n,f_m) = d_{2^{n}}(f_0,f_{m-n}) > d(f_0, f_{m-n}) > 1$.