Given $0<\alpha\leq1$. Show that the function $$f(x)=\sum_{j=1}^\infty 2^{-j\alpha}\sin(2^jx)$$ is nowhere differentiable.
I have solved the case $x=0$. Taking $t_l=2^{-l-1}\pi$, then $f(t_l)-f(0)=f(t_l)$ can be broken into three terms: the sum from $1$ to $l-1$, the single summand $j=l$, and the sum from $l+1$ to $\infty$.
The third term is $0$. The middle term is $2^{-l\alpha}$, which dominates the first term as $l$ goes to $\infty$. This makes the Newton quotient blow up and shows nondifferentiability at $x=0.$
However, when $x$ is not a multiple of $\pi$ (or, rather, not $\frac{k\pi}{2^n}$) the (single-summand) middle term is very difficult to control for me (The third term is still $0$).
Please help. This is not homework. Thank you very much.