Now asked on MO here
The definition of the derivative of a function $f$ at a point $x$ is : $\lim\limits_{h \to 0 }\frac{f(x+h) -f(x)}{h}$ but what if we change that $h$ to be any continuous function $g(h)$ that goes to $0$ as $h$ goes to $0$? Lets focus our attention at powers of $h$ so $g(h)=h^a$ for some $a$. If $f$ is a differentiable function at $x$ any function other than $h$ the absolute value of the derivative will go to either $0$ or $\infty$. I will refer to $\lim\limits_{h \to 0 }\frac{f(x+h) -f(x)}{g(h)}$ as $a-$derivative.
A question that came to my mind Given any continuous function nowhere differentiable on $\mathbb{R}$ is there a $g$ such that the absolute value of the $a-$derivative" exist almost everywhere and how to determine such functions ?
The case where $g(x)$ is big enough to make this $a-$derivative $0$ is annoying because one can say a very big number like $a=1/TREE(3) $ and say, for example, at this number the $a-$derivative exist and it is $0$ for the Weierstrass function,
so let me clear that : Given any continuous function nowhere differentiable what is the $\sup g(x)$ such that $\lim\limits_{h \to 0 }\frac{f(x+h) -f(x)}{g(h)}= 0$ for all $x$ where the $\sup $ is taken of all real $a$ does the absolute value of the $a-$derivative exist at that value ?
Lets take L. van der Waerden's function for example
Define $g(x)= |x|$ for $|x|\in [-1,1]$ , $g(x+2)=g(x)$ $$f(x)= \sum_{n \ge 1} \frac{3^n g\left(4^n x\right) }{4^n}$$
Is an example of continuous function nowhere differentiable, 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 $a-$derivative exist if we decrease the power of $h$?
