I asked this question on MSE [here][1].


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>Define $g(x)= |x|$ for $|x|\in [-1,1]$ , $g(x+2)=g(x)$ 
$$f(x)= \sum_{n \ge 1} \frac{3^n g\left(4^n x\right) }{4^n}$$

This function is a famous example of a  continuous nowhere differentiable function on $\mathbb{R}$. There is a proof that the absolute value secant line's slope goes to infinity as it approaches the point so this made me wonder: will the absolute value of the $\alpha-$derivative exist if $\alpha<1$?

One can pick a very large number like $\operatorname{TREE}(3)$ and say that this function has $\alpha-$derivative $=0$ when $\alpha=1/\operatorname{TREE}(3)$ and he will probably be right so let me phrase the question in other way:<br>
what is the $\sup \{\alpha\}$ such that 
$$
\lim\limits_{h \to 0 }\frac{f(x+h)  -f(x)}{h^\alpha}= 0
$$ for all $x$?<br>
Does the absolute value of the $\alpha-$derivative exist at that value of $\alpha$ ?
 


  [1]: https://math.stackexchange.com/questions/4900413/what-is-sup-a-such-that-for-continuous-non-differentiable-function-f-l