Let $f:\mathbb{R}\to\mathbb{R}$ be a $2\pi$ periodic function where $$ f(x)=|x|= \begin{cases} -x&x\in[-\pi,0[\\ x&x\in[0,\pi] \end{cases} $$ Show that $\sum_{n=0}^{\infty} f(2^nx)/2^n$ converges uniformly on $\mathbb{R}$.
My problem is that I can't see why $f(2^ nx)/2^n\neq |x|$. Any help on getting started is appreciated.
You seem to have forgotten that $f$ is periodic of period $2\,\pi$. This implies for instance that $f(2^3)=f(8-2\,\pi)=8-2\,\pi$. With this in mind, it is clear that $|f(x)|\le\pi$ for all $x\in\Bbb R$ and the series converges uniformly by Weirstrass's $M$-test.