I am trying to construct a function $f(x)$ which

  1. has derivative = $0$ almost everywhere, and
  2. is strictly increasing.

I realize one can do this by playing with the Cantor function, but I would like to do it directly. However, I am having difficulty seeing if the beautiful monster has a derivative at all!

Let $(a_k)$ be a sequence of numbers, such that $\sum |a_k| < \infty$, and $a_k > 0$. Define an ordering on $q_k \in \mathbb{Q}$, which one can do because $\mathbb{Q}$ is countable and totally ordered.

Define $f: \mathbb{R} \to \mathbb{R}$, $$f(x) = \sum\limits_{\{k \text{ : } q_k < x\}} a_k$$

We see that $f_k$ is strictly increasing (thus is both increasing and non-constant on any open interval), for $\mathbb{Q}$ is dense in $\mathbb{R}$, and any open interval in $\mathbb{R}$ contains an element of $\mathbb{Q}$.

I would think this function would have derivative zero on all but $\mathbb{Q}$, which are totally disconnected and countable, and a disjoint union of countably many points has Lesbegue measure 0.

The issue is that I don't know how to rigorously show that $|f'| = 0$ a.e., indeed, I don't know how to check that $f$ is differentiable. I don't know how to see that this $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{ x - x_0}$$

has limit $0$ on $S \subset \mathbb{R}$ such that the Lesbegue measure of $\{ \mathbb{R} - S \}$ is $0$.

Here is my question: How do I take the derivative of this function? Does this function satisfy the desired criterion?

  • $\begingroup$ You will need $\sum_k |a_k|<\infty$ to ensure that $f$ is defined, and $a_k>0$ to guarantee that $f$ is strictly increasing. I don't see how $x\in Q$ has any bearing on whether $f'(x)$ exists. $\endgroup$ – DanielWainfleet Oct 23 '15 at 2:31
  • $\begingroup$ now i see that f is discontinuous on Q $\endgroup$ – DanielWainfleet Oct 23 '15 at 2:45
  • $\begingroup$ Thank you, I made the appropriate edits to reflect the restrictions on $a_k$. $\endgroup$ – Catherine Ray Oct 23 '15 at 3:00
  • 1
    $\begingroup$ This is a great question! As a result of the stuff I've been working on, I've become quite fond of functions like this. I won't post an answer until I get a chance to think about it some more, but here are some off-the-cuff remarks. $\endgroup$ – Vectornaut Oct 23 '15 at 3:04
  • $\begingroup$ First, like @user254665 said, you need $\sum_k a_k$ to be finite to have a chance. How does it help you? My intuition is that the bound on $\sum_k a_k$ means "there's only so much $a_k$ to go around," so if you look at smaller and smaller intervals, you'll find less and less "total $a_k$" inside them. That means $f(a) - f(b)$ at least goes to zero as $(a, b)$ shrinks. I'd start by proving that, as a warm-up. $\endgroup$ – Vectornaut Oct 23 '15 at 3:09


As noted below in the original version, the fact that $f'$ exists almost everywhere is immediate from standard results, for example in Folland. Those results rely on other results - putting it all together into a proof requires a substantial fraction of the results in that chapter. Here's an entirely self-contained ad hoc proof.

If $I$ is an interval we let $3I$ denote the interval with the same center but three times the length: If $I=(a-r,a+r)$ then $3I=(a-3r,a+3r)$.

Lemma 1 Suppose that $K\subset\Bbb R$ is compact and $C$ is a collection of open intervals covering $K$. Then there exist finitely many pairwise disjoint $I_1,\dots,I_n\in C$ such that $3I_1,\dots,3I_n$ cover $K$.

Proof: We can suppose $C$ is finite. Let $I_1$ be an element of $C$ of maximal length. "Discard" any element of $C$ that intersects $I_1$. Note that if $I$ was discarded just now then $I\subset 3I_1$, since $I$ intersects $I_1$ and $I$ is no longer than $I_1$.

Now let $I_2$ be one of the "remaining" intervals in $C$ of maximal length. Discard any remaining interval that intersects $I_2$. Note that any interval discarded at this stage is contained in $3I_2$. Also note that $I_1$ and $I_2$ are disjoint, since $I_2$ was not discarded at the first stage.

Etc. QED.

Now for $f:[0,1]\to\Bbb R$ define $$Mf(x)=\sup_{y\ne x}\left|\frac{f(x)-f(y)}{x-y}\right|$$and $$\omega f(x)=\limsup_{y\to x}\left|\frac{f(x)-f(y)}{x-y}\right|.$$Note that $$0\le\omega f(x)\le Mf(x)$$and that $f'(x)=0$ if and only if $\omega f(x)=0$.

Lemma 2 If $f:[0,1]\to\Bbb R$ is nondecreasing then $$m\left(\{x\in[0,1]\,:\,Mf(x)>\lambda\}\right)\le \frac c\lambda(f(1)-f(0)).$$

Proof: Since $m$ is inner regular it is enough to show that $m(K)$ satisfies the same inequality, where $K$ is a compact set with $Mf(x)>\lambda$ for every $x\in K$.

For each $x\in K$ there exists $y\ne x$ with $|(f(x)-f(y))/(x-y)|>\lambda$. Let $J_x=[x,y]$ or $J_x=[y,x]$, whichever makes sense. Let $I_x$ be an open interval containing $J_x$, with $|I_x|\le 2|J_x|$.

Now the $I_x$ for $x\in K$ form an open cover of $K$. Hence, writing $I_j$ in place of $I_{x_j}$, there exist finitely many disjoint $I_1,\dots I_n$ such that $3I_1,\dots,3I_n$ cover $K$.

Aargh, we need more notation because we didn't know whether $x<y$ or $y<x$ at the start of this. Each $I_j$ is an open interval containing $J_j$, where $|I_j|\le 2|J_j|$. Write $J_j=[a_j,b_j]$. Now

$$m(K)\le\sum m(3I_j)\le 6\sum m(J_j)=6\sum_{j=1}^n(b_j-a_j).$$But $(f(b_j)-f(a_j)/(b_j-a_j)>\lambda$, so that $b_j-a_j\le(f(b_j)-f(a_j))/\lambda$. So we have $$m(K)\le\frac6\lambda\sum_{j=1}^n(f(b_j)-f(a_j)).$$But since $f$ is nondecreasing and the $[a_j,b_j]$ are disjoint, $$\sum(f(b_j)-f(a_j))\le f(1)-f(0).$$QED.

Note of course both lemmas are analogous to results in that chapter in Folland, adapted to the present context.

And now we can show that your $f$ has $f'=0$ almost everywhere. Write $f=\sum f_n$ in the obvious way. Write $$f=s_N+r_N,$$where $$s_N=\sum_{n=1}^Nf_n.$$

For every $N$ we certainly have $s_N'=0$ almost everywhere, so $\omega s_N=0$ almost everywhere. Hence $$\omega f\le\omega s_N+\omega r_N=\omega r_N\le Mr_N$$almost everywhere. Let $\epsilon>0$. Choose $N$ so that $$\frac c\epsilon(r_N(1)-r_N(0))<\epsilon.$$Then the set where $Mr_N>\epsilon$ has measure less than $\epsilon$. So the previous inequality shows that the set where $\omega f>\epsilon$ has measure less than $\epsilon$. So $\omega f=0$ almost everywhere. QED.


Yes, $f'=0$ almost everywhere. (Assuming $a_k>0$ and $\sum a_k<\infty$.)

This is not quite trivial. I don't know how much real analysis you know; the fact that $f'=0$ almost everywhere is immediate from basic/standard results in reals.

See for example Proposition 3.30 in Folland Real Analysis, as well as related results in that section. Your function $f$ is non-decreasing, hence it has bounded variation. There are fussy details regarding whether or not $f$ lies in what Folland calls NBV, but Theorem 3.23 shows that this doesn't matter; $f$ nondecreasing implies that $f=g$ almost everywhere, where $g$ is in NBV, and also $f'=g'$ almost everywhere. Now in the notation of Proposition 3.30, your $\mu_f$ is a singular measure, being $\sum a_k\delta_{q_k}$ (where $\delta_q$ is a point mass at $q$). So 3.30 says $f'=0$ almost everywhere.

It's not clear to me how difficult a more elementary proof would be (for example a proof accessible to someone taking the course that's called "Advanced Caclulus" here...)

  • $\begingroup$ Normalized Bounded Variation. Look again - it's in that section, honest. $\endgroup$ – David C. Ullrich Oct 23 '15 at 18:48
  • $\begingroup$ @CatherineRay A self-contained proof has just appeared.... $\endgroup$ – David C. Ullrich Oct 29 '15 at 15:29

Your construction might work for some choices of the enumeration $(q_k)$ of $ Q$ and some choices of $(a_k)$ but if $a_k=1/k^2$ and if $q_{2k}\in (\pi, \pi+2^{-k})$ for each $k$, then $f(\pi+2^{-k})-f(\pi)\geq \sum_{j\geq k}1/(4j^2)= 1/(4k) + o(1/k)$ as $k\to \infty$, so $f'(\pi)$ does not exist.

  • 1
    $\begingroup$ As long as this only happens for a disjoint union for a countable set of points, there is hope. $\endgroup$ – Catherine Ray Oct 23 '15 at 3:35
  • $\begingroup$ Maybe you could put this on MathOverflow and see what the pros say. $\endgroup$ – DanielWainfleet Oct 23 '15 at 5:09
  • $\begingroup$ ??? The question was whether $f'=0$ almost everywhere. It's clear that there exist points where $f'$ does not exist: the function is not even continuous at $q_k$. $\endgroup$ – David C. Ullrich Oct 23 '15 at 21:28
  • $\begingroup$ Yes,but the O.P.conjectured that f' might exist and vanish except on Q. $\endgroup$ – DanielWainfleet Oct 24 '15 at 4:23
  • $\begingroup$ @user254665 Oh. I missed that, sorry. $\endgroup$ – David C. Ullrich Oct 29 '15 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.