I got stuck on this exercise from Prof. Tao's real analysis notes. Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be the function $$f:= \sum_{n=1}^\infty 4^{-n} \sin(8^n\pi x)$$ Show that for every 8-dyadic interval $[\frac{j}{8^k},\frac{j+1}{8^k}]$ with $k\geq 1$, one has $|f(\frac{j+1}{8^k})-f(\frac{j}{8^k})| \geq C4^{-k}$ for some absolute constant $C > 0$.
I can see that all terms $n\geq k$ in the summation are $0$; the problem can be deduced to $$\left|\sum_{n=1}^{k-1} 4^{-n} \sin(8^{n-k}(j+1)\pi ) - \sum_{n=1}^{k-1} 4^{-n} \sin(8^{n-k}(j)\pi )\right|\geq C4^{-k}.$$ I also tried to use sum of angles formula since we know that $0 < 8^{n-k}\pi\leq \frac{\pi}{8}$.
This is similar to a step in an exercise in his textbook Analysis II, but with the function $$f(x) = \sum_{n=1}^{\infty}4^{-n}\cos(32^n \pi x),$$ which I think makes it a bit easier to prove. In this case you are trying to prove that
$$|f(\frac{j+1}{32^k})-f(\frac{j}{32^k})| =\\ \big|\sum_{n=1}^\infty 4^{-n}[\cos(32^{n-k}(j+1)\pi) -\cos(32^{n-k}j\pi)]\big| \ge C4^{-k}$$
when $j,k$ are positive integers. A middle term of the sum, when $n=k$ is $$2\cdot4^{-k}\cdot(-1)^j$$ the behavior when $n>k$ is the same as in your function - all the terms disappear - and for the terms where $n < k$ you use the inequality $$|cos(x) - \cos(y)|\le |x-y|.$$
This ends up giving that $$|\sum_{n=1}^{k-1}4^{-n}[\cos(32^{n-k}(j+1)\pi) -\cos(32^{n-k}j\pi)]| \le \frac 1 7 4^{-k}$$
which completes this step.