Define $f(x) = \sum_{i=1}^\infty 2^{-i}\sin(4^i x)$. I am trying to show that $f$ is nowhere differentiable. The exercise contains the following hints:
Show that for $x_n = \pi 4^{-n}$ we have $f(x_n)\geq \frac{1}{\sqrt{2}}2^{-n}$;
Conclude that $f$ does not have a limit at $0$;
Use steps 1. and 2. to show that $f$ is nowhere differentiable.
Out of these steps I have only managed to do 2.), so 1.) and 3.) are missing.
I have managed to show that $$f(x_n) > 2^{-n}(\sqrt{2} - 1).$$ This is because if we define $$f_i(x) = 2^{-i}\sin(4^i x)$$ then
i. for $i > n$ we simply have $$f_i(x_n) = 2^{-i}\sin(4^{i-n}\pi) = 0$$ as the argument of $\sin$ is an integer multiple of $2\pi$.
ii. for $i = n$ we have $$f_i(x_n) = 2^{-n}\sin(\pi) = -2^{-n}$$
iii. for $1\leq i < n$ we have $$f_i(x_n) = 2^{-i}\sin\left(\frac{\pi}{4^{n-i}}\right)$$
Hence,
$$f(x_n) = \sum_{i=1}^\infty f_i(x_n) = -2^{-n} + \sum_{i=1}^{n-1}f_i(x_n) > -2^{-n} + f_{n-1}(x_n) = -2^{-n} + 2^{-n + 1}\sin\left(\frac{\pi}{4}\right) = -2^{n} + 2^{-n}\sqrt{2} = 2^{-n}(\sqrt{2} - 1)$$
I am choosing only the last term of the series since I do not quite see how to get a meaningful bound with the $\sin$ terms being potentially close to $0$, at least at the start, for large values of $n$. While the lower bound $2^{-n}(\sqrt{2} - 1)$ is not what the step requires, it does still show that
$$\frac{f(x_n) - f(0)}{x_n - 0} = \frac{f(x_n)}{x_n} > \frac{4^n}{\pi}2^{-n}(\sqrt{2} - 1) = \frac{2^n(\sqrt{2} - 1)}{\pi} \to \infty, n\to\infty$$
since clearly $f(0) = 0$. Thus $f$ is not differentiable at zero.
However I do not quite know how I could use these steps to then conclude that $f$ it not differentiable at any point $x\in\mathbb{R}$. Is the missing lower bound the hint suggested somehow related to that?
