Hello I have an exercise to do but I’m stuck on a few questions. The statement is as follows :
Consider the Cantor function f: [0,1] —> [0,1] With $f(x) = \sum_{j=1}^{N(x)} \frac{1}{2^j} 1(x_j \ge 1) $.
a) Calculate the derivative f’(x) of the function f for each point x $\in$ (0,1) \ K which is a Cantor set.
b) For x = 2/3 compute both of the one sided limits $Lim_{h->0^{+}} \frac{f(2/3 + h)-f(2/3)}{h}$ and $Lim_{h->0^{-}} \frac{f(2/3 + h)-f(2/3)}{h}$.
Define the symmetric derivative of f at x by SD(f)(x) = $Lim_{h->0} \frac{f(x + h)-f(x-h)}{2h}$.
c) For each x $\in$ (0,1) \ K calculate SD(f)(x) (if it exists).
For the question a) I have write that f’(x) = $0$ For all x $\in [0,1]\K $ but for the rest I cn’t do .... Someone can help me please ?
$f'(x)$ does not exist when $x$ belongs to the Cantor set $K.$ Let $x=\sum_{n=1}^{\infty} 2x_n3^{-n}$ where each $x_n\in\{0,1\}.$ Then $f(x)=\sum_{n=1}^{\infty}x_n2^{-n}.$ For any $\epsilon >0$ we have $$\sup\left\{\left|\frac {f(x)-f(y)}{x-y}\right|: y\in [0,1]\cap (x-\epsilon,x+\epsilon)\right\}=\infty.$$ Proof: Given any $r>0,$ take $n_1\in \Bbb N$ large enough that $2\cdot 3^{-n_1}<\epsilon$ and $\frac {1}{2}\cdot (3/2)^{n_1}>r.$ Let $y_n=x_n$ when $n\ne n_1$ and let $y_{n_1}=1-x_{n_1}.$ Let $y=\sum_{n=1}^{\infty}2y_n3^{-n}.$
Then $y\in K$ and $|y-x|=2\cdot 3^{-n_1}<\epsilon,$ while $|f(x)-f(y)=2^{-n_1}.$ So $$\left|\frac {f(x)-f(y)}{x-y}\right|=\frac {1}{2}(3/2)^{n_1}>r.$$