# 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? Commented 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
Commented Apr 2, 2020 at 8:12
• One candidate worth looking at are the Brownian path which are nowhere differentiable on an interval with probability $1$. Commented 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. Commented 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). Commented 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
Commented 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 $z_0$.) Commented 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
Commented Apr 2, 2020 at 15:58
• Sure -- as long as $|z_1-z_0| > \eta$, there's nothing to prevent this. Commented 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
Commented Apr 2, 2020 at 16:33