I asked this question on MSE here.

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 $TREE(3)$ and say that this function has $\alpha-$derivative $=0$ when $\alpha=1/TREE(3)$ and he will probably be right so let me phrase the question in other way:

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 $a-$derivative exist at that value of $\alpha$ ?

I asked this question on MSE here.

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:
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 asked this question on MSE here.

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 $TREE(3)$ and say that this function has $\alpha-$derivative $=0$ when $\alpha=1/TREE(3)$ and he will probably be right so let me phrase the question in other way:

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 $a-$derivative exist at that value of $\alpha$ ?

I asked this question on MSE here.

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 $TREE(3)$ and say that this function has $\alpha-$derivative $=0$ when $\alpha=1/TREE(3)$ and he will probably be right so let me phrase the question in other way:

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 $a-$derivative exist at that value of $\alpha$ ?