$f(x)=\sum_{n=1}^{\infty} {4^{1-n}} {h(4^{n-1}x)}$ is continuous and nowhere monotonic,$h(x)=\vert x \vert$ with period 1,$\vert x \vert \leq 1/2$ Given $h: \mathbb{R} \to \mathbb{R}$ be a periodic function with period 1 defined for $\vert x \vert \leq 1/2$ by $h(x)=\vert x \vert$ .
To show $f: \mathbb{R} \to \mathbb{R}$ be such that
$f(x)=\sum_{n=1}^{\infty} \frac{h(4^{n-1}x)}{4^{n-1}}$, for all $x \in \mathbb{R}$ is continuous and nowhere monotonic.
We have, $h(x+1)=h(x)$ for  $\vert x \vert \leq 1/2$.
Consider $f_n(x)=\frac{h(4^{n-1}x)}{4^{n-1}}$ and let $x \leq y$. Then $f_n(x)-f_n(y)=\frac{h(4^{n-1}x)}{4^{n-1}}-\frac{h(4^{m-1}x)}{4^{m-1}}= \vert x \vert -\vert y \vert$.
I am not sure how to use the periodicity to prove continuity and nowhere monotonic.
 A: Here is a fairly detailed proof.
Proof that f is continuous:
This is the easier half of the proof. First note that $ h(x)\leq1$ for all x. Thus, we have $$ \Big\lvert\frac{h(4^{n-1}x)}{4^{n-1}}\Big\rvert\ \leq \frac{1}{4^{n-1}}$$.
Since $\sum_{n=1}^{\infty}\frac{1}{4^{n-1}}<\infty$, the Weirstrass M-test implies $\sum_{n=1}^{N}\frac{h(4^{n-1}x)}{4^{n-1}}$ converges uniformly to f as N goes to $\infty.$ Thus, f is the uniform limit of continuous functions, so f is continuous.
Proof that f is not monotone on any interval:
This is the more delicate part. First, we fix some notations. Let
$$f_{1}(x)=h(x)$$ and $$f_{n}(x)=\frac{f_{1}(4^{n-1}x)}{4^{n-1}},$$ so that
$$ f(x) = \sum_{n=1}^{\infty}f_{n}(x)=\sum_{n=1}^{\infty}\frac{f_{1}(4^{n-1}x)}{4^{n-1}}.$$ Note that $f_{n}$ is bounded above by $\frac{1}{2}4^{-n+1}$ and is periodic with period $4^{-n+1}$.
Now let $$A= \left\{x\in\mathbb{R}:x=k 4^{-m}, k\in\mathbb{Z},m=0,1,2,.. \right\}$$ and note that $A$ is dense in $\mathbb{R}$ by the Archimedean property of $\mathbb{R}$ for instance. Now note that for $a\in A$, say $a=k 4^{-m}$, we have $$f_{n}(a)= 0$$ for all $n>m$, since $f_{n}$ vanishes at integers. Thus for such $a$, we have $$f(a)= \sum_{n=1}^{m}f_{n}(a).$$ Now let $m$ be a positive integer and let $$x_{m}=4^{-2m-1}.$$ Then for similar reasons as above, we have $$f_{n}(a+x_{m})=0$$ for all $n>2m+1.$ It then follows that $$f(a+x_{m})-f(a)=\sum_{n=1}^{m}f_{n}(a+x_{m})-f_{n}(a) + \sum_{n=m+1}^{2m+1}f_{n}(a+x_{m}) \geq -mx_{m}+(m+1)x_{m} = x_{m}>0.$$
Similarly, $$ f(a-x_{m})-f(a) \geq -mx_{m}+(m+1)x_{m}=x_{m}>0.$$ Thus by the density of $A$, it follows $f$ is not monotone on any interval.
