Suppose $f(x) = |x|$ for $x \in [-1,1]$ with a periodic extension $f(x+2) = f(x)$ for $x \in \mathbb{R}$. We are given the property that for any x, either: $|f(x+h) - f(x)| = h$ or $|f(x-h) - f(x)| = h$ for $0\leq h \leq \frac{1}{2}$. For $h_k = \frac{1}{2}4^{-k}$, this gives us the property \begin{align*} |f(4^n (x+h)) - f(4^n x) |= \begin{cases} 0 \quad n>k\\ 4^n |h| \quad n\leq k \end{cases}. \end{align*}
Suppose we have the series $g(x) = \sum \limits_{n=0}^{\infty}\left(\frac{3}{4}\right)^n f(4^n x)$. We are asked to prove the lower bound for $\left|\frac{g(x+h)-g(x)}{h}\right| \geq \frac{3^k + 1}{2}$ using $h=h_k$ or $h=-h_k$.
Attempt: Since the tail terms of $g(x+h) - g(x)$ go to $0$ then we have \begin{align*} \left|\frac{g(x+h)-g(x)}{h}\right| = \left|\sum \limits_{n=0}^{k}\left(\frac{3}{4}\right)^n \left(f(4^n (x+h)) - f(4^n x)\right) \right|. \end{align*}
I am thinking to apply the result $|x-y|\geq |x| - |y|$ for some choice of term in the summation but this is where I'm stuck.
Note that $$\tag{1} \left(\frac34\right)^n\left\lvert f(4^n(x+h))-f(4^nx)\right\rvert\leq\left(\frac34\right)^n4^n\lvert h\rvert=3^n\lvert h\rvert $$ and equality is achieved for at least one choice of sign of $h$, $\lvert h\rvert\leq h_n$.
So we have \begin{align*} \left\lvert\frac{g(x+h)-g(x)}{h}\right\rvert &=\left\lvert\frac1h\sum_{n=0}^k\left(\frac34\right)^n\Big(f(4^n(x+h))-f(4^nx)\Big)\right\rvert\\ &\geq 3^k-\sum_{n=0}^{k-1}3^n\\ &=3^k-\frac{3^k-1}{3-1}=\frac{3^k+1}{2}. \end{align*} where $h=h_k$ or $h=-h_k$ is chosen so we have equality in (1) with $n=k$.