Does there exist a nowhere smooth function, that has arbitrary many derivatives?

Question

I'm sorry if my title sounds misleading, I don't know a better way to word my question briefly. But I have the following question about functions.

First, as long as $A$ is a dense subset of $\mathbb{R}$, note that we can still make sense of a derivative of a function $f : A \rightarrow \mathbb{R}$. It is also possible for the derivative to exist on removable discontinuities of $f$ that are outside of $A$, if the limit of the differential quotient exists.

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With this in mind, for any function $f : \mathbb{R} \rightarrow \mathbb{R}$, we can define the following sets inductively:

• $D_0 := \mathbb{R}$
• $D_{n+1} := \{x \in \mathbb{R} : f^{(n)} : D_n \rightarrow \mathbb{R} \text{ is differentiable at } x \}$

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My question is whether it is possible for $D_n$ to be a dense subset of $\mathbb{R}$ for all $n \in \mathbb{N}$, but any particular point stops being in these derivative sets eventually. More formally, is it possible that:

1. $D_n$ is dense in $\mathbb{R}$ for all $n \in \mathbb{N}$
2. $\bigcap_{n=0}^\infty \bigcup_{k=n}^\infty D_k = \emptyset$

Answer 1

Note: This is a completely new answer compared to what I wrote previously, but this is a much stronger result, that better answers the question. The original version can be found in this answer's edit history.

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First, consider a shifted version of the so called "Riemann function" $R : \mathbb{R} \rightarrow \mathbb{R}$ given via:

$$R(x) := \frac{1}{2}x + \sum_{n=1}^\infty \frac{\sin(n^2x)}{n^2}$$

This function has the following properties:

1. $R$ is a continuous function.
2. $R$ is differentiable at $x \iff x = \frac{2p+1}{2q+1} \pi$ for $p,q \in \mathbb{Z}$.
3. $R'(\frac{2p+1}{2q+1} \pi) = 0$ for all $p,q \in \mathbb{Z}$.

All of these claimed properties can be found here and also on the Wikipedia page for the Weierstrass function.

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Next, let $a_1, a_2, \dots$ be an enumeration of the set $\{\frac{2p+1}{2q+1} \pi : p,q \in \mathbb{Z}\}$.

Define $R_n : \mathbb{R} \rightarrow \mathbb{R}$ via:

$$R_n(x) := R(x) + \sum_{k=1}^n |x - a_k|$$

Then define $S_n : \mathbb{R} \rightarrow \mathbb{R}$ such that:

1. $S_n^{(n)} = R_n$
2. $S_n(0) = 0, S_n'(0) = 0, \dots , S_n^{(n-1)}(0) = 0$

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Finally the function $f := \sum_{n=0}^\infty S_n$ has the property that:

$$D_{n+1} = \{\frac{2p+1}{2q+1} \pi : p,q \in \mathbb{Z}\} \setminus \{a_1, \dots , a_n \}$$

which gives us the desired result.