Let there be a function f(x) so that:
$$f(x)=\int_0^x{\textrm{sgn}\left(\cos\left(\frac{\pi}{2}t\right)\right)\textrm dt}$$
Graphing it in Desmos, you will see it's like a sin function - the difference being it's 'made' of $x$ and $-x$ lines, and has a cycle of $4$, instead of $2π$.
Suppose we have the infinite nested function:
$$k(x)=\left|\cdots\left|\left|\left|\left| f(x) \right| +\frac{1}{2}\right|+\frac{1}{4}\right|+\frac{1}{8}\right|\cdots\right|$$
Basically, add $1/2^n$ and put an absolute value.
If you were to write a couple of terms of that and graph them, you would see the triangles get smaller and smaller. Each time, the absolute value turns the negative part positive with a sharp break, and the $1/2^n$ centers it again.
Let's say we have another function:
$$j(x)=\lim_{a\to\infty} \frac{1}{a}f(ax+1)$$
graphing with a small $a$ value, then gradually increasing it, you can see the entire graph gets smaller and smaller. Just like $k(x)$ gradually gets smaller.
I have yet to learn how to prove if a function is differentiable nowhere. And yet, intuitively, it seems to me like $k(x)$ will be differentiable nowhere, while $j(x)$ will always have sections that can be differentiated - as if the 'line' sections of $j(x)$ will never disappear, but will only get smaller.
Am I correct? Which one of them, if any, will be differentiable nowhere?
Thanks.