Show that $\phi$ is nowhere differentiable in $\mathbb R$ if $\alpha \beta >1$.

Question

Let $f(x)=|x|$ on $[-1,1]$ and $F$ be the extension of $f$ on the entire real line defined inductively as $:$

$$F(x)=f(x+2).$$

Let $\phi_n(x)={\alpha}^nF({\beta}^n x)$ for some $0 < \alpha <1$. Prove that $\sum \phi_n$ is uniformly convergent in $\mathbb R$ to a continuous function $\phi$. Further, if $\alpha \beta >1$, then show that $\phi$ is nowhere differentiable in $\mathbb R$.

Would anybody please make it clear to me that how $F$ is defined here in an inductive way? I dont clearly understand the defination of $F$. If it is well understood I think I can proceed. Please help me in understanding it first. Then let me try it for a while. If any problem will occur then I will ask for further clarification.

Thank you in advance.

Answer 1

It is inductive because evaluating $F$ of some number reduces to finding it for another value. Repeating this will end in finitely many steps, thus defining it for all real numbers.

Perhaps this is best seen through an example. How will you evaluate $F(-10.5)$? Well, $$F(-10.5)=F(-8.5)=F(-6.5)=F(-4.5)=F(-2.5)=F(-0.5)=\lvert -0.5\rvert=0.5$$