I know that the Cantor function is differentiable a.e. but I want to prove it without using the theorem about monotonic functions. I have already proved that $f'(x) = 0$ for all $x \in [0,1] \backslash \mathbb{C}$ where $\mathbb{C}$ is the Cantor set.
But I'm not sure how to go about proving that if $x \in \mathbb{C}$ then $f$ is not differentiable at $x$.
Actually, upon reflection, I think I have already proved differentiability a.e. but I would still like to know how to finish this part.
Also, the definition I am using for the function:
$$f:[0,1] \to [0,1]$$
Let $x \in [0,1]$ with ternary expansion $0.a_1a_2…$ Let $N$ be the first $n \in \mathbb{N}$ such that $a_n = 1$. If for all $n \in \mathbb{N}$, $a_n \in \{0,2\}$, let $N = \infty$.
Now define $b_n = \frac{a_n}{2}$ for all $n < N$ and $b_N = 1$.
Then $$f(x) = \sum_{i=1}^{N} \frac{b_n}{2^n}.$$
Best Answer
Consider a right-hand endpoint of one of the intervals removed to form the Cantor set. It has a ternary representation
$$x = 0.(2a_1)(2a_2)\ldots(2a_n)2000\ldots$$
where the $a$'s are all $0$ or $1$,
and the binary representation of $f(x)$ is
$$f(x) = 0.(a_1)(a_2)\ldots(a_n)1000\ldots.$$
Pick $m > n$ and $h>0$ with $3^{-(m+1)} < h < 3^{-m}.$
Then as $m \rightarrow \infty$ and $h \rightarrow 0+$
$$\frac{f(x+h)-f(x)}{h}>\frac{3^m}{2^{m+1}}\rightarrow \infty$$
and the right-hand derivative $f'_+(x) = \infty$.
You can make a similar argument for a left-hand endpoint of a removed interval.