For this continuous non differentiable function $f$ How to determine $\sup\{a\}$ s.t $\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h^\alpha}=0$ for all $x$?

Question

I asked this question on MSE here.

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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:
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 $\alpha$ ?

Answer 1

The supremum you are looking for is $-\frac{\ln(3/4)}{\ln(4)}$. I leave my proof below, although as user479223 mentions, this is the same as showing Hölder continuity of the Weierstrass function, which is well known (see this MSE answer).

Note that for all $x,y\in\mathbb{R}$, $|g(x)-g(y)|\leq\min(|x-y|,1)$. Thus, for all $x,h\in\mathbb{R}$, $$ \begin{split} |f(x+h)-f(x)|&=\sum_n\frac{3^ng(4^n(x+h))}{4^n}-\frac{3^ng(4^nx)}{4^n} \\ &\leq \sum_n\frac{3^n}{4^n}\min(|4^nh|,1)\\ &=\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right). \end{split} $$ Note that when $h$ is close to $0$, the sequence $\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)$ first increases exponentially and then decreases exponentially as follows:

$$h,3h,9h,27h,\dots,3^{k_1}h,\left(\frac{3}{4}\right)^{k_2},\left(\frac{3}{4}\right)^{k_2+1},\left(\frac{3}{4}\right)^{k_2+2},\dots$$ where both $3^{k_1}h$ and $\left(\frac{3}{4}\right)^{k_2}$ are, within a factor of ten, equal to $h^{\frac{-\ln(3/4)}{\ln(4)}}$. Thus,

$$|f(x+h)-f(x)|\leq\sum_n\min\left(|3^nh|,\left(\frac{3}{4}\right)^n\right)\leq100h^{\frac{-\ln(3/4)}{\ln(4)}}.$$

This means that any constant $\alpha<\frac{-\ln(3/4)}{\ln(4)}$ should make the limit of the question be $0$ for all $x$. However, if $\alpha=\frac{-\ln(3/4)}{\ln(4)}$ and my computations below are correct, then the limit $\lim\limits_{h\to0}\frac{f(h)}{h^\alpha}$ need not exist. E.g. taking $h_m=1/4^m$; then we have

$$f(h_m)=\sum_{n=1}^m\frac{3^n}{4^n}g(4^{n-m}) =\sum_{n=1}^m\frac{3^n}{4^n}4^{n-m}=\sum_{n=1}^m\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(h_m)}{h_m^\alpha}=\lim_m\frac{\left(\frac{3^m\cdot\frac{3}{2}}{4^m}\right)}{(3/4)^m}=\frac{3}{2}. $$

However, $$f(2h_m)=\sum_{n=1}^{m-1}\frac{3^n}{4^n}g(2\cdot4^{n-m})=2\sum_{n=1}^{m-1}\frac{3^n}{4^m}.$$ Thus, $$ \lim_m\frac{f(2h_m)}{(2h_m)^\alpha}=\lim_m\frac{\left(\frac{3^{m-1}\cdot\frac{3}{2}}{4^m}\right)} {2^\alpha(3/4)^m}=\frac{1}{2\cdot2^\alpha}=\frac{\sqrt{3}}{4}. $$

Answer 2

This is just local Holder continuity.

Let $f$ be $\gamma$ Holder at $x$. If $\alpha<\gamma$ then

$$\lim\limits_{h\to0}\frac{|f(x+h)-f(x)|}{h^\alpha}\leq\lim\limits_{h\to0} \frac{C h^\gamma}{h^\alpha}=0.$$

Conversely, suppose that

$$\lim\limits_{h\to0}\frac{|f(x+h)-f(x)|}{h^\alpha}=0.$$

By choosing $|h|<\delta$ for some $\delta>0$ and the definition of limit we have that

$$\frac{|f(x+h)-f(x)|}{h^\alpha}<1,$$

or $$|f(x+h)-f(x)|<h^\alpha.$$

See this Q&A for the relationship with fractional differentiation.