# Is every function $f: \mathbb R \to \mathbb R$ differentiable at at least one point when restricted to some everywhere dense subset of $\mathbb R$?

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?
• 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? Apr 2, 2020 at 8:02
• @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?
– user153451
Apr 2, 2020 at 8:12
• One candidate worth looking at are the Brownian path which are nowhere differentiable on an interval with probability $1$. Apr 2, 2020 at 10:34
• 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. Apr 2, 2020 at 12:31
• Relevant for all sorts of related issues is Jack Brown's 1995 survey paper Restriction theorems in real analysis (preprint version here). Apr 2, 2020 at 17:47

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 $$\delta$$.

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