It's again from Tao's book.
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 $C > 0$.For $n>k, |f(\frac{j+1}{8^k})-f(\frac{j}{8^k})|=0$. But how to get the inequality when $n<k$?
Disclaimer
This is a proof for the version of $f$ with $\sin$ replaced by $\cos$. The estimates get more awkward if sines are used, without any added insight.
Let $f_n(x) = 4^{-n} \cos (8^n \pi x)$. Observe the following:
$f_n((j+1)/8^k) = f_n(j/8^k)$ for $n > k$.
Since $\cos \pi m = (-1)^m$, it follows that
$$|f_n((j+1)/8^k) - f_n(j/8^k)|=2\cdot 4^{-k},\quad n=k$$
Cosine function has Lipschitz constant $1$, hence $$|f_n((j+1)/8^k) - f_n(j/8^k)|\le 4^{-n}8^{n-k}\pi ,\quad n<k$$
Finally, use the reverse triangle inequality: $$ \begin{split} &|f ((j+1)/8^k) - f (j/8^k)| \\ & \ge |f_k ((j+1)/8^k) - f_k (j/8^k)| - \sum_{n=1}^{k-1} |f_n ((j+1)/8^k) - f_n (j/8^k)| \\ &\ge 2\cdot 4^{-k} - \pi 4^{-k}\sum_{n=1}^{k-1} 4^{n-k} \ge 4^{-k}(2 - \pi/3) \end{split}$$ which is the desired estimate.