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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 $f|_C$.

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    $\begingroup$ If $(a,b)$ is one of the intervals removed in the construction of $C$ then the right hand derivative at $a$ is $0$. $\endgroup$ Jun 26 at 9:30
  • $\begingroup$ Thanks. What about other points in Cantor set? Not all points in Cantor set are points like 'a' as the interval's endpoint. $\endgroup$
    – LEY
    Jun 26 at 14:09
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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 $$\lim_{y\to x}\frac{f(y)-f(x)}{y-x}=\infty.$$

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  • $\begingroup$ Hi, is the $f$ in the ANS the Cantor function or it's restriction on Cantor set? $\endgroup$
    – LEY
    Sep 1 at 10:08
  • $\begingroup$ It seems that the answer poster has been careful in using points of the Cantor set to evaluate the different quotient. So basically yes, he is using the restriction of $f$ to the Cantor set. Only in the last paragraph (the proof of the "if" part of the theorem) he takes an arbitrary point $y$ that could also be outside the Cantor set, but that is perfectly fine because in that case we want to prove that the limit of the different quotient goes to infinity (so it still works asking in addition for $y$ to be in the Cantor set). $\endgroup$ Sep 1 at 10:58
  • $\begingroup$ That's not quite correct. In the case where $x$ is one of the endpoints of the Cantor set, I am using the limit from the side that is outside the Cantor set to show that the derivative does not exist at $x$. If you restrict the difference quotient limit to only consider $y\in C$, then the limit should be $\infty$ instead by calculations similar to those in the final part of the answer. $\endgroup$ Sep 1 at 15:21

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