This is a continuation of this exercise.
Let the zigzag function defined as
$$zz(x):=\left|\lfloor x+1/2\rfloor-x\right|,\quad x\in\Bbb R$$
Then for $k\in\Bbb Z$ we have that $zz(k)=0$, and $zz(\Bbb R)=[0,1/2]$, and $zz$ is increasing in any interval of the kind $[k,k+1/2]$, and decreasing in $[k+1/2,k+1]$ for $k\in\Bbb Z$.
Then we define the function
$$F(x):=\sum_{n=0}^\infty \frac{zz(4^nx)}{4^n},\quad x\in\Bbb R$$
Now I must prove that $F$ is continuous only using the fact that
$$|F(x)-F(y)|\le \frac1{4^m3}+\sum_{n=0}^m |f_n(x)-f_n(y)|\tag{1}$$
where $f_n(x)=4^{-n}zz(4^nx)$.
The book have not stated any theorem about uniform convergence so it cannot be used here, at least not directly.
Proof
I will show that $zz$ is Lipschitz continuous. It is enough to consider $x,y\in[0,1]$ because the function $zz$ is cyclic with period $1$.
First observe that if $x\in [0,1/2)$ then $zz(x)=x$, then for $x,y\in[0,1/2)$ the function $zz$ is Lipschitz continuous with $K=1$. If $x\in[1/2,1]$ then $zz(x)=1-x$, so if $x,y\in[1/2,1]$, so $zz$ is again Lipschitz continuous with $K=1$.
Then we need to confirm that if $x\in[0,1/2)$ and $y\in[1/2,1]$ then $zz$ is Lipschitz. Then it is enough to observe that
$$|zz(x)-zz(y)|=|x+y-1|\le|x-y|=\begin{cases}x+y-1\le y-x\implies x\le 1/2\\1-x-y\le y-x\implies y\ge 1/2\end{cases}$$
because both cases are true then $zz$ is Lipschitz, with $K=1$. Then choosing enough big $m$ such that $\frac1{4^m3}<\frac{\epsilon}2$ and choosing $\delta=\frac{\epsilon}{2(m+1)}$ we have that
$$|f_n(x)-f_n(y)|=4^{-n}|zz(4^nx)-zz(4^ny)|\le 4^{-n}|4^nx-4^ny|<\frac{\epsilon}{2(m+1)}$$ we finally have that
$$|x-y|<\delta\implies |F(x)-F(y)|\le \frac1{4^m3}+\sum_{n=0}^m|f_n(x)-f_n(y)|<(m+1)\frac{\epsilon}{2(m+1)}+\frac{\epsilon}2=\epsilon$$
so $F$ is continuous.$\Box$
Questions:
It is the proof correct? (It seems that it is correct but a confirmation will be welcome).
There is a easier way to prove this in the context of this exercise (i.e. not using theorems about uniform convergence)?