I was doing some fairly simple research a few hours ago and I almost asked a similar question with the word continuous instead of differentiable in the title, but then I found this question asked by Gro-Tsen where there is an affirmative answer to that question.

Apparently, that is the result of Blumberg, that for every $f: \mathbb R \to \mathbb R$ there exists a dense subset $D$ of $\mathbb R$ such that $f|_D$ is continuous.

Blumberg´s paper can be found here and I have slightly did a research of his arguments, however, I am not sure can they be adapted to show that $f$ is differentiable at at least one point when restricted to some everywhere dense subset of $\mathbb R$.

Honestly, I expect that there are some $f$´s which have the property that when restricted to every possible everywhere dense subset of $\mathbb R$ are non-differentiable everywhere on all such sets

However, I am not sure, and that´s why I ask it here, since I think that´s known, because Blumberg´s result is relatively long time ago established (1922).

Here is the question:

  • Is it true that for every function $f: \mathbb R \to \mathbb R$ there exists at least one everywhere dense set $D \subseteq \mathbb R$ such that $f|_D$ is differentiable at at least one point?
  • 2
    $\begingroup$ What if you demand that $f|_D$ is differentiable everywhere (on $D$): do you know this to be false, are you hedging your question by asking something more modest, or is it simply the particular fact that interests you? $\endgroup$
    – Gro-Tsen
    Commented Apr 2, 2020 at 8:02
  • $\begingroup$ @Gro-Tsen I asked because I would like to see constructions of some $f$´s that are non-differentiable at every point of every everywhere dense subset of $\mathbb R$ on which they are restricted, if they exist at all? The idea of constructing them is what I am mostly interested in, again, if they exist at all? $\endgroup$
    – user153451
    Commented Apr 2, 2020 at 8:12
  • $\begingroup$ One candidate worth looking at are the Brownian path which are nowhere differentiable on an interval with probability $1$. $\endgroup$ Commented Apr 2, 2020 at 10:34
  • 1
    $\begingroup$ I think that if a continuous function is nowhere differentiable, then so should be its restriction to any dense set. The idea is that if the slopes of secant lines behave badly near $x$, then continuity forces them to behave just as badly on a dense set near $x$. In particular, @LiviuNicolaescu's suggestion should work. $\endgroup$
    – Will Brian
    Commented Apr 2, 2020 at 12:31
  • 2
    $\begingroup$ Relevant for all sorts of related issues is Jack Brown's 1995 survey paper Restriction theorems in real analysis (preprint version here). $\endgroup$ Commented Apr 2, 2020 at 17:47

1 Answer 1


The answer is no. This is because, if $f: \mathbb R \rightarrow \mathbb R$ is a continuous, nowhere differentiable function, then $f \!\restriction\! Q$ is nowhere differentiable for any dense $Q \subseteq \mathbb R$.

To see this, fix $x \in \mathbb R$ and, aiming for a contradiction, let us suppose $f \!\restriction\! Q$ is differentiable at $x$, say with derivative $c \in \mathbb R$.

Let $\varepsilon > 0$. Because $f \!\restriction\! Q$ is differentiable at $x$, there is some $\delta > 0$ such that for all $y \in Q \setminus \{x\}$ with $|x-y| < \delta$, we have $|\frac{f(y) - f(x)}{y-x} - c| < \varepsilon$.

Because $f$ is not differentiable at $x$, and in particular does not have derivative equal to $c$ at $x$, there is some $z_0 \in \mathbb R \setminus \{x\}$ with $|x-z_0| < \delta$ such that $|\frac{f(z_0) - f(x)}{z_0-x} - c| > 2\varepsilon$.

Because $f$ is continuous on $\mathbb R$, the function $z \mapsto |\frac{f(z) - f(x)}{z-x} - c|$ is continuous on $\mathbb R \setminus \{x\}$. This means that $\lim_{z \rightarrow z_0} |\frac{f(z) - f(x)}{z-x} - c| = |\frac{f(z_0) - f(x)}{z_0-x} - c| > 2\varepsilon$. Therefore there is some $\eta < \delta-|x-z_0|$ such that if $|z_0-z| < \eta$ then $|\frac{f(z) - f(x)}{z-x} - c| > \varepsilon$.

Let $z \in Q$ with $|z-z_0| < \eta$. Then we have $|x-z| < \delta$ while $|\frac{f(z) - f(x)}{z-x} - c| > \varepsilon$. This contradicts our choice of $\delta$.

  • $\begingroup$ Why is $\lim_{z \rightarrow z_0} |\frac{f(z) - f(x)}{z-x} - c| = 2\varepsilon$? $\endgroup$
    – user153451
    Commented Apr 2, 2020 at 15:23
  • $\begingroup$ Sorry, it should say the limit is $> 2\varepsilon$. (Because it's equal to whatever the function evaluates to at $z_0$.) $\endgroup$
    – Will Brian
    Commented Apr 2, 2020 at 15:51
  • $\begingroup$ Can there exist some $z_1 \neq x$ such that $|x-z_1|< \delta$ and $|\frac{f(z_1) - f(x)}{z_1-x} - c| < 2\varepsilon$? $\endgroup$
    – user153451
    Commented Apr 2, 2020 at 15:58
  • $\begingroup$ Sure -- as long as $|z_1-z_0| > \eta$, there's nothing to prevent this. $\endgroup$
    – Will Brian
    Commented Apr 2, 2020 at 16:01
  • $\begingroup$ +1, I have read carefully the proof and as of now I think that the proof is fine $\endgroup$
    – user153451
    Commented Apr 2, 2020 at 16:33

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