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 $$f(x) = \sum_{i=1}^{N} \frac{b_n}{2^n}.$$
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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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*}
$\endgroup$ $\begingroup$You stated it correctly:
However you proved it (using the fact that it is a monotonic function or whatever), if you already proved that the function is differentiable on $[0,1]-\mathcal{C}$ then there is no need to continue (who cares what happens on $\mathcal{C}$ it is a set of measure zero) and when we say that a property $\mathcal{P}$ occurs almost everywhere what we mean is that $\mathcal{P}$ is true everywhere except $perhaps$ on a set of measure zero.
Suppose $(x_n)\to x$ and $(y_n)\to x$ such that $x_n < y_n$ for all $n$. Then by way of contradiction assume the derivative exists $$\frac{f(y_n)-f(x_n)}{y_n-x_n} \to f'(x)$$
So if $x\in \mathcal{C}$ such that $x\in I_n:=[x_n,y_n]$ for all $n$. Then we immediately see it cannot be differentiable because $y_n-x_n=\frac{1}{3^n}$ but $f(y_n)-f(x_n)=\frac{1}{2^n}$.
$\endgroup$ 3 $\begingroup$On the other hand, suppose that one uses a fat Cantor set to produce a similar Cantor function. So now the Cantor set has a positive Lebesgue measure. I would conjecture that now, except for the endpoints of the complementary intervals, the derivative would exist and be finite. Then it would be finite. It would likely be infinite at the endpoints. Remember that, for a function of bounded variation, the derivative exists and is finite almost everywhere. Certainly something of this sort must happen. The devil's staircase
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