In This question I asked
Define $g(x)= |x|$ for $|x|\in [-1,1]$ , $g(x+2)=g(x)$$$f(x)= \sum_{n \ge 1} \frac{3^n g\left(4^n x\right) }{4^n}$$ what is the $\sup \{\alpha\}$ such that$$\lim\limits_{h \to 0 }\frac{f(x+h) -f(x)}{h^\alpha}= 0$$ for all $x$?
Does the absolute value of the $\alpha-$derivative exist at that value of $\alpha$ ?
I expected that at this value of $a$ the absolute value of the function might have $\alpha-$derivative at all points but it turns out I was wrong as this answers shows (the supremum is $\frac{\ln\left(\frac 4 3\right)}{\ln 4}$) .
My question is: Is there a continuous nowhere differentiable function with the $\alpha-$derivative exist at that value of $\sup \{\alpha\}$ where is the sup is taken on all reals such that $\lim\limits_{h \to 0 }\frac{f(x+h) -f(x)}{h^\alpha}= 0$ for all $x$?
If it is impossible to have a function like that how to prove it ?