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

Question

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?

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 $\delta$.