Differentiability of the Cantor Function

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.

Here is a proof that the Cantor function $f$ is not differentiable at non-endpoints of the Cantor set.

Let $C_0=[0,1]$, and let $C_n$ be constructed from $C_{n-1}$ by removing an open interval from each closed interval in $C_{n-1}$, in particular the middle third. The Cantor set $C$ is the intersection of the $C_n$.

Let $x$ be a point of $C$, but not an endpoint. Then for each $n$, $x$ is contained in some interval of $C_n$ (of maximal size), i.e. $x\in [a_n,b_n]\subset C_n$.

We then have the following properties:

P1: $b_n-a_n = \frac{1}{3^n}$

P2: $f(b_n)-f(a_n)=\frac{1}{2^n}$

P3: $x-a_n<\frac{1}{3^n}$

P4: $x$ must be in the left third of $(a_n,b_n)$ infinitely many times. Similarly for the right third.

P5: if $x$ is in the right third of an interval $(a_n,b_n)$, then $x-a_n>\frac{2}{3^{n+1}}$

P6: if $x$ is in the right third of an interval $(a_n,b_n)$, then $f(b_n)-f(x)<\frac{1}{2^{n+1}}$

Let us consider a subsequence $a_{n_k}$ where $x$ is in the right third of the interval $(a_{n_k},b_{n_k})$

\begin{align*} \frac{f(x)-f(a_{n_k})}{x-a_{n_k}} &= \frac{f(b_{n_k})-f(a_{n_k})}{x-a_{n_k}} - \frac{f(b_{n_k})-f(x)}{x-a_{n_k}} \end{align*}

Both fractions are positive due to the monotonicity of $f$. Using the properties above

\begin{align*} \frac{f(x)-f(a_{n_k})}{x-a_{n_k}} &\geq \frac{\frac{1}{2^{n_k}}}{\frac{1}{3^{n_k}}} - \frac{\frac{1}{2^{n_k+1}}}{\frac{2}{3^{n_k+1}}}\\ &= \left(\frac{3}{2}\right)^{n_k} - \frac{3}{4}\left(\frac{3}{2}\right)^{n_k} \\ &= \frac{1}{4}\left(\frac{3}{2}\right)^{n_k} \end{align*}