Set $$\Phi_0=\begin{cases}x\quad\text{if}\:0\le x\le\frac 12\\ 1-x\quad\text{if}\:\frac12\le x\le 1\end{cases}$$ And extend this function to the entire real line as to have period $1$.
We denote such a function by $\varphi_0$
Further, we let $\varphi_n(x)=\frac 1{4^n}\varphi_0(4^n x)$.
Where the function $\varphi_n$ has period $4^{-n}$ and a derivative equal to $+1$ or $-1$ everywhere except at points $x=\frac{k}{2^{2n+1}},k\in\mathbb{Z}.$ Let $$f(x)=\sum_{n=1}^{\infty}\varphi_n(x)$$
Show that the function $f$ is defined and continuous on $\mathbb{R}$, but does not have a derivative at any point of $\mathbb{R}$.
My noticings:
Since derivatives can be taken additively, we can claim that $f'(x)=\sum_{i=1}^{\infty}a_i$, where $a_i\in \{-1,1\}\forall i$, since the randomness (but I don't think it suffices to prove anyhow) we can sense that the derivative might be undefined, since the sum is divergent.
Maybe we can use the fact that $0\le\varphi_0(x)\le \frac 12\forall x\in\mathbb{R}$ to construct inequalities to show it continuity
The incomplete answer in the question linked below said Weierstrauss M-test might be utilized to solve the second part of the problem
My attempt
We notice that $|\varphi_n(x)|\le \frac{1}{2^{2n+1}}$
And since sums of absolute value is greater or equal to absolute value of sum, by cauchy series convergence criteria, and the arithmetics of sum of geometric series we conclude that each $f_n(x)=\sum_{i=1}^n f_n(x)$ is uniformly convergent to $f$, since that its difference with $f$ would be smaller and smaller. Therefore as $f_i$ is continuous for all $i$, we claim that $f$ is continuous as well.
I have no idea how to prove the second part though, any advice, hint, answer of solution would be greatly appreciated.
A similar question may be found here, with no explicit answer been given.
Proving Van der Waerden’s Example of a Continuous Nowhere Differentiable Function
Edit:
A proof of a generalized result can be found here https://educ.jmu.edu/~querteks/seniorthesis.pdf
