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Define $f:[0,1]\to [0,1]$ by

$$f(x)=\begin{cases}0, &x=0,\\ \\ \sum\limits_{r_n<x } 2^{-n}, & 0 \lt x \le 1, \end{cases} $$

where $\{r_n \}_{n\in \mathbb N} =\mathbb Q \cap (0,1) $.

How to show that the derivative $f'(x)=0$ a.e.?

I can show this function is increasing and discontinuous at every rational, and how to word on?

Srivatsan
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Leitingok
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  • I'm not convinced that $f(x)$ is continuous anywhere, except perhaps at 0. –  Nov 18 '11 at 17:31
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    The same function is used in an exercise on p.13 of Carothers' Real analysis. The reader is asked in this exercise to show that "$f$ is everywhere discontinuous on $[0,1]$ but that $f$ is everywhere continuous when considered as a function on only $[0,1]\setminus\mathbb Q$." – Martin Sleziak Nov 18 '11 at 18:45
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    And the same function is used in Understanding analysis by Stephen Abbott in an exercise on p.169, where the author claims that it is continuous at every irrational point. – Martin Sleziak Nov 18 '11 at 18:50
  • But ignoring the question of continuity raised in the above comments, $f$ is obviously monotone and there is a result by Lebesgue that monotone function on interval is differentiable almost everywhere. – Martin Sleziak Nov 18 '11 at 19:12
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    @Martin: It’s not hard to show that $f$ is continuous at every irrational. However, it need not be differentiable at every irrational, since if $a$ is any irrational in $(0,1)$, you can use a Hilbert’s hotel trick to construct the enumeration of $\mathbb{Q}\cap(0,1)$ in such a way that $f$ is not differentiable at $a$. Thus, the naive approach can’t work. – Brian M. Scott Nov 18 '11 at 19:35
  • @Brian : I think it's just enough to show $\lim\limits_{n\to\infty}\cfrac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}=0 $,for every every differentiable point. – Leitingok Nov 19 '11 at 03:20
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    It seems this has been asked and answered at mathoverflow http://mathoverflow.net/questions/81411/showing-the-derivative-of-this-function-is-equal-to-0-a-e-closed – David Mitra Nov 20 '11 at 21:15
  • @David: I copied that answer over here. Also, can I convince you to undelete the link that you posted? I think an obvious modification to example 3 in that paper shows that also that the function is differentiable almost everywhere. – Willie Wong Nov 21 '11 at 17:04
  • On the other hand that paper also illustrates well @BrianM.Scott 's comment: in fact there is a dense set $C$ such that $C\subset (\mathbb{R}\setminus \mathbb{Q})$ and that $f$ is not differentiable on $C$. – Willie Wong Nov 21 '11 at 17:05
  • @willie wong Sure... – David Mitra Nov 21 '11 at 17:11
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    @MartinSleziak: So it appears the statement in Carothers is incorrect, since the function is continuous at the irrationals. – Nate Eldredge Nov 21 '11 at 19:27

2 Answers2

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The following is the elementary answer that Pietro Majer gave to this question on MO. I copy the answer here so we can consider this question answered.

Consider the nested family of open nbd's of $(0,1)\cap\mathbb{Q}\ :$ $$A_\epsilon:=\cup_{n\in\mathbb{Z} _ + } (r_n- \epsilon 2^{-n/3},r_n+ \epsilon 2^{-n/3})\ , \qquad \epsilon > 0\ . $$ So $|A _\epsilon|=O(\epsilon)$ and $A:=\cap _ {\epsilon > 0} A _ \epsilon$ has measure zero. Let $x \in (0,1) \setminus A$: There exists $\epsilon > 0$ such that for any $n\in\mathbb{Z}_+$ there holds $ \epsilon 2^{-n/3}\le |x-r_n|$. Thus, for any $y\in (0,1)$ $$|f(x)-f(y)|\le \sum_{|x- r _ n|\le|x- y| } 2^{-n}= \frac{1}{\epsilon^2}\sum_{|x- r _ n|\le|x- y| } 2^{-n/3}(\epsilon 2^{-n/3})^2\le $$ $$\le \frac{1}{\epsilon^2}\bigg(\sum_{n=1}^\infty 2^{-n/3}\bigg)|x-y|^2= \frac{|x-y|^2}{\epsilon^2(2^{1/3}-1))}\ ,$$ showing that $f'(x)=0\ .$

Community
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Willie Wong
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You won't get anywhere if you try to prove that $f$ is differentiable (with 0 derivative) at every irrational point. See here, whose result implies that there is a subset of the irrational numbers, dense on the interval, over which $f$ is not differentiable. (This question, however, neatly illustrates the difference between small in the sense of Baire category and small in the sense of measure.)

Willie Wong
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David Mitra
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  • Note, though, that the example in part $3$ of that paper shows that a function can be discontinuous on a dense set and differentiable almost everywhere even though the set of points of differentiability must then be of first category. – Brian M. Scott Nov 19 '11 at 18:38
  • Ack; so, I'm completely off here. – David Mitra Nov 19 '11 at 18:54