Consider the function $f(x)=\sum_{n=1}^{\infty} 2^{-n}g(2^{2^{n}}x)$ where \begin{equation} g(x)=\begin{cases} 1+x &-2 \le x \le 0 \\ 1-x &0 \le x \le 2 \end{cases} \end{equation} where $g(x)$ has period $4$. Is $f$ continuous on $\mathbb{R}$? Is it differentiable at some $x$?
My Take
If one insert any value $x \in \mathbb{R} \backslash \{0\}$ into $f$ one gets a divergent series.
As such, I came to considered the partial sum $f^{*}(x)=\sum_{n=1}^{N} 2^{-n}g(2^{2^{n}}x)$. Here it is possible to add the two series together to form a single new entity, namely:
\begin{align} f^{*}(x+h)-f^{*}(x)&=\sum_{n=1}^{N}2^{-n}\left(1-2^{2^{n}}(x+h) \right)-\sum_{n=1}^{N}2^{-n}\left(1-2^{2^{n}}x \right) \\ &=h\sum_{n=1}^{N}\left(-2^{2^{n}-n}\right) \tag{*} \end{align} Now, as $h \to 0$ we have that $|f^{*}(x+h)-f^{*}(x)| \to 0$ as each term in $(*)$ will tend to zero. Thus, if we let $N \to \infty$ then $(*)$ will be zero, which is less than any $\epsilon>0$, so $f$ is continuous.
If this argument holds, then I claim the only point at which $f$ will be differentiable is $x=0$, as $f'(0)=0$. Otherwise, we end up with a divergent series.
Note $|g(x)|\leq 1$ on $\mathbb{R}$, so $|2^{-n}g(2^{2^{n}}x)|\leq\frac{1}{2^n}$ and $\sum \frac{1}{2^n} $ converges, hence by M-test, the series uniformly converges. Since each term is continuous, $f(x)$ is continuous.
$x\in[0,1],h=\pm 2^{-2^k}$, where we choose the sign to make $x$ and $x + h$ be on the same linear segment of $g(2^{2^k}x)$. Then we have
$\text{d}\, g(2^{2^n}x) = 0$ for $n > k$, since the period of $g(2^{2^n}x)$is $4\cdot 2^{-2^n}.$
$|\text{d}\, g(2^{2^k}x)| =1.$
$|\text{d}\,\sum_{n=1}^{k-1}g(2^{2^n}x)| \leq (k-1) \text{max}_{1\leq n\leq k-1} |\text{d} g(2^{2^n}x)| \leq (k - 1)2^{2^{k-1}}2^{-2^k} < 2^k2^{-2^{k-1}}.$
Hence $|\frac{\text{d} f}{h}| \geq \frac{|\text{d}\, f(2^{2^k}x)|-|\text{d}\,\sum_{n=1}^{k-1}f(2^{2^n}x)|}{h}\geq 2^{-k}2^{2^k} - 2^k2^{2^{k-1}}\to \infty$
So it's nowhere differentiable.