I recently ran into this interesting exercise:
Define$h(x)=|x|$on the interval $[-1,1]$ and extend the definition of $h$ on all of $\mathbb{R}$ by requiring that $h(x+2)=h(x)$. The result is a periodic "saw tooth" function.
Now, define$g(x)=\sum_{n=0}^{\infty}\frac{1}{2^n}h(2^nx).$Consider the sequence $x_m=1/2^m$, where $m\in\mathbb{N}\cup\{0\}$. Show that$\frac{g(x_m)-g(0)}{x_m-0}=m+1$and use this to prove that g'(0) does not exist.
I solved this the following way:$g(x_m)=\sum_{n=0}^{\infty}\frac{1}{2^n}h\left(\frac{2^n}{2^m}\right)=\frac{1}{2^m}(m+1),$$\frac{g(x_m)-g(0)}{x_m-0}=\frac{1/2^m(m+1)}{1/2^m}=m+1,$$g'(0)=\lim_{m\to\infty}\frac{g(x_m)-g(0)}{x_m-0}=\lim_{m\to\infty}m+1=\infty.$Therefore, the equality holds, and g'(0) does not exist.
I also extended the above 'proof' to how that neither g'(1) nor g'(1/2) exist. However, I now want to show that if $x=p/2^k$, where $p\in\mathbb{Z}$ and $k\in\mathbb{N}\cup\{0\}$, then g'(x) does not exist, but I am running into the problem where I am having to consider lots of cases, such as when $k
Edit 1: I have not been able to progress any further than showing that the summation boils down to$g(x)=\sum_{n=0}^{k}\frac{1}{2^n}h\left(\frac{2^n}{2^k}p\right).$