I have recently defined a function on the unit interval $[0,1]$ recursively. The definition is based on repeated subdivision of the interval in two equally-sized parts. The definition is as follows:
The initial function is just $f_0(x)=x$ and in the following step, one divides the domain in halves and multiplies the slope of the function on the first half by 1/2 and on the second half by 2-1/2=3/2 (and adjusting the height of the second interval such that the function is continuous). This step is then repeated in each new subinterval.
Therefore the function $f_n$ in the nth step is always continuous and piecewise linear between all multiples $k\cdot2^{-n}$ for $k\in\left\lbrace0,1,\dots,2^n\right\rbrace$. The function $f_{25}$ (black) is shown in the following picture along with some intermediate functions (gray scales). Example sequence of functions
I basically just want to know how to show the differentiability of the function at non-dyadic points, i.e. points with infinite binary representation. It is easy to show that for dyadic points the right-side slope is 0 and the left-side slope approaches $\infty$, such that the function is not differentiable at these points.
The idea was to contruct a function that is differentiable in a point but in no neighbourhood of that point. If one extends the domain to $[-1,1]$ and extends the function either symmetrically or anti-symmetrically, then the new function is differentiable in 0. I was up to now unable to prove the non-differentiability (or differentiability) of the function at non-dyadic points.
There are several facts about the function that make it interesting. For example, the parts of the graph between each two kinks of the function are a rescaled version of the whole function itself. So the function is self-similar up to rescaling. On the other hand, the recursion rule implies that the limiting function is continuous and monotonuously increasing (but not strictly). The function is constructed such that the values of any dyadic point are defined by only finitely many iterations by $$f(x)=\frac{\sum\limits_{l=0}^{k-1} 3^{a_l}}{2^{2n}},\, x=k\cdot 2^{-n},\; k=0,1,\dots,2^n,\; n\in\mathbf{N}$$ where $a_l$ is the Hamming weight or number of ones in the binary representation of l. (the formula possibly contains minor errors)
To make the prove possibly easier, I also found the formula for the slope left of any dyadic point x in the nth iteration $f_n$ to be $$\lim\limits_{h\downarrow 0}\left(\frac{f_n(x+h)-f_n(x)}{h}\right)=\left(\frac{3}{2}\right)^{a_x}\left(\frac{1}{2}\right)^{n-a_x}$$ where again $a_x$ is the Hamming weight of x. Since every dyadic number can be represented in exactly two ways in binary (namely the one with infinitely many zeroes and the one with infinitely many ones), one can conclude the non-differentiability of these points. Maybe the fact that all other points have only one representation in binary is crucial for their differentiability. But as I said, I don't know if the function is differentiable at non-dyadic points.
The exact method, I used to prove the differentiability of the function in 0 was as follows: for any $h<2^{-n}$, the function $f:=\lim\limits_{n\to\infty}f_n$ will always be smaller than $g(h)=2^{-n}h$ on the open interval $(0,2^{-n})$ and therefore the left-side derivative of f in 0 (as well as the right-side derivative when appriately extending f) is: $$f'(0)=\lim\limits_{h\downarrow 0}\left(\frac{f(h)-f(0)}{h}\right)<\lim\limits_{h\downarrow 0}\left(\frac{g(h)-g(0)}{h}\right)=2^{-n}\to 0\; \text{as} \; n\to\infty.$$ Therefore, by the sandwich theorem, we can conclude that $f'(0)=0$.
P.S.: I appreciate any help and maybe some more interesting facts about the function. I was unable to find the fractal dimension of the limiting curve for example.