What is the derivative of Cantor function on Cantor set? Although on Cantor set $C$, the Cantor function $f|_C$ is non-differentiable, but I wonder for $x,y\in C$, the limit:
$$\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=$$
$$\begin{cases}
=+\infty\\\mathrm{or}\\\text{does not exist?}
\end{cases}$$
I have tried to use the property that $f=\lim\limits_{n\to \infty} F_n$
and $\lim\limits_{n\to\infty}\frac{F_n(y)-F_n(x)}{y-x}=+\infty$,but since $\phi_n(y)= \frac{F_n(y)-F_n(x)}{y-x}$ is not uniformly convergent, we cannot say:$$\lim\limits_{y\to x} \lim\limits_{n\to \infty }\phi_n(y)=\lim\limits_{n\to \infty}\lim\limits_{y\to x}\phi_n(y)$$
so I do not know – is there any ways to research the property of the derivative of $f|_C$.
 A: If $x\in C$, then $x$ has a base 3 expansion consisting of only $0$s and $2$s (and we will always use this particular base 3 expansion, in case $x$ has two different base 3 expansion).  The behavior of $f'(x)$ then depends on whether $x$ has sufficiently long streaks of identical digits arbitrarily far out in its its base 3 expansion (where "sufficiently long" means roughly $\frac{\log(3/2)}{\log(2)}$ times how far out in the expansion the streak is).
To be precise, say a streak in $x$ is a sequence of  consecutive digits in its base 3 expansion that are all the same (either all $0$s or all $2$s).  If a streak in $x$ starts at the $(n+1)$st digit and has length $d$, we say this streak has weight $$d\log 2-n\log(3/2).$$ So, longer streaks have higher weight, but streaks that start later in the expansion have lower weight.  The origin of this mysterious formula will become clear in the proof below.
Theorem: Let $x\in C$.  Then $f'(x)=\infty$ iff for all $r\in\mathbb{R}$, $x$ has only finitely many streaks of weight greater than or equal to $r$.  In the case where this does not happen, $f'(x)$ does not exist (even as an infinite value).
(As long as $x$ is not one of the "endpoints" of $C$, you can think of its base 3 expansion as consisting of alternating streaks of $0$s and $2$s.  This theorem then says that $f'(x)=\infty$ if the weights of these alternating streaks approach $-\infty$, and otherwise $f'(x)$ does not exist.  Also, $f'(x)$ does not exist in the case where $x$ is one of the "endpoints" so its base 3 expansion is eventually constant.)
Proof of Theorem: First, let me note that the lim sup of the difference quotient for $f'(x)$ is always $\infty$.  Indeed, suppose the base $3$ expansion of $x$ has $0$ as the $n$th digit.  Then $f(x+2/3^n)=f(x)+1/2^n$ (since $x+2/3^n$ is just $x$ with its $n$th base $3$ digit changed to $2$) and so the difference quotient for $y=x+2/3^n$ is $\frac{1/2^n}{2/3^n}=\frac{3^n}{2^{n+1}}$.  This goes to $\infty$ as $n\to\infty$, so as long as $x$ has infinitely many digits that are $0$ this shows the lim sup of the difference quotient is $\infty$.  But if $x$ does not have infinitely many digits that are $0$ then it has infinitely many digits that are $2$, and then similarly by considering $y=x-2/3^n$ we again find the lim sup of the difference quotient is $\infty$.
Now suppose that there exists $r\in\mathbb{R}$ such that $x$ has infinitely many streaks of weight at least $r$; let us assume $x$ has such streaks of $2$s (the case of streaks of $0$s is similar).  If $x$ has only finitely many $0$s in its base 3 expansion, then $f(x+y)=f(x)$ for all sufficiently small $y$ so the nonexistence of $f'(x)$ is immediate.  Thus we may assume that $x$ has infinitely many $0$s in its base 3 expansion.  Now let $\epsilon>0$ and choose $N$ such that the $N$th digit of $x$ is $0$ and $2/3^N<\epsilon$.  By hypothesis, we can find a streak of weight at least $r$ in $x$ which starts after the $N$th digit of $x$.  Extending this streak backwards as far as possible (which can only increase its weight), we may assume the digit right before the streak starts is $0$.  So, for some $n\geq N$, the $n$th digit of $x$ is $0$ and then the next $d$ digits are all $2$ for some $d$ such that $$d\log 2-n\log(3/2)\geq r.$$  Let $a$ be the number whose first $n$ digits are all the same as $x$'s and then all the rest of the digits are $0$.  Because $x$ has a streak of $d$ $2$s after its $n$th digit, we have $$f(x)-f(a)\geq\sum_{i=n+1}^{n+d}1/2^i=1/2^n-1/2^{n+d}.$$ Also, since the first $n$ digits of $x$ are the same as those of $a$, $$x\leq a+1/3^n.$$ Now let $y=a+2/3^n$.  We then have $$f(y)-f(x)=f(a)+1/2^n-f(x)\leq 1/2^{n+d}$$ and $$y-x=a+2/3^n-x\geq 1/3^n$$ so $$\frac{f(y)-f(x)}{y-x}\leq \frac{3^n}{2^{n+d}}.$$  The logarithm of this difference quotient is precisely the negative of the weight of the streak we chose, so this difference quotient is at most $\exp(-r)$.  Also, we have $|y-x|<\epsilon$ since $n\geq N$.  So, there are values of $y$ arbitrarily close to $x$ which make the difference quotient at most $\exp(-r)$, so the lim inf of the difference quotient is at most $\exp(-r)$.  Since the lim sup is $\infty$, this means $f'(x)$ does not exist.
Now suppose that for all $r\in\mathbb{R}$, $x$ has only finitely many streaks of weight at least $r$.  Fix $r$ and choose $N$ such that every streak in $x$ starting after the $N$th digit has weight less than $r$.  There is then some $\epsilon>0$ such that if $|y-x|<\epsilon$ then $y$ has the same first $N$ digits as $x$ (here we use the fact that $x$ must have both infinitely many $0$s and infinitely many $2$s in its expansion).  Suppose $|y-x|<\epsilon$ and let us assume $y>x$ (the case $y<x$ is similar).  Say that $y$ first differs from $x$ at its $n$th digit, so the $n$th digit of $x$ is $0$ and the $n$th digit of $y$ is $1$ or $2$.  Take the maximal streak of $2$s in $x$ starting with the $(n+1)$st digit, so for some $d$ we have the $(n+1)$st through $(n+d)$th digits of $x$ are all $2$ and then the $(n+d+1)$st digit is $0$. (This includes the possibility that the $(n+1)$st digit is $0$, in which case $d=0$.)  By hypothesis, this streak has weight less than $r$. Let $a$ be the number which has the same first $n$ digits as $x$ and then the rest of the digits are all $0$.  Since the $(n+d+1)$st digit of $x$ is $0$, we have $$f(x)-f(a)\leq \sum_{i=n+1}^{n+d+1}1/2^i=1/2^n-1/2^{n+d+1}.$$  On the other hand, $y\geq a+1/3^n$, so $$f(y)-f(a)\geq 1/2^n.$$  Thus $$f(y)-f(x)\geq 1/2^{n+d+1}.$$  Also, since $y\leq a+2/3^n$, we have $$y-x\leq y-a=2/3^n.$$  So our difference quotient satisfies $$\frac{f(y)-f(x)}{y-x}\geq\frac{1/2^{n+d+1}}{2/3^n}=\frac{1}{4}\frac{3^n}{2^{n+d}}.$$  Since our streak has weight less than $r$, this is at least $\frac{1}{4}\exp(-r)$.
That is, for any $r$ there exists $\epsilon>0$ such that all difference quotients for $y$ within $\epsilon$ of $x$ are at least $\frac{1}{4}\exp(-r)$.  Taking $r\to-\infty$, this proves that $$\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=\infty.$$
