The limit of a function with derivative at least $1_\mathbb{Q}$

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable, such that $f'(x) \ge 1_{\mathbb{Q}}(x)$. Is it true that $\lim_{x\to\infty}f(x) = \infty$?

Answer 1

The Pompeiu function $g:[0,1]\to \mathbb{R}$ is strictly increasing, everywhere differentiable, with derivative vanishing on a dense $G_\delta$ set. With minor modifications in the construction, one can make it instead a homeo $G:[0,1) \to [0,+\infty)$. (Alternatively, such modification $G$ can be made starting from the original $g$, see below).

In this case, the function $h(x):=x+G(x)$ is therefore a differentiable, strictly increasing homeo $[0,1)\to [0,+\infty)$ everywhere differentiable with $h'(x)=1$ in a dense $G_\delta$ set. Hence the inverse map $f$ is an increasing homeo $[0,+\infty)\to[0,1)$ with derivative $1$ on a dense $G_\delta$ set. (It may be a bit technical to make this set contain the rationals as required).

How to make an unbounded Pompeiu function: start from the original Pompeiu function $g:[0,1]\to[0,1]$, where we can assume $g(0)=g'(0)=0$, so that we can extend it on the left to a differentiable function vanishing identically for $x\le0$. Define, for any $0\le x<1$, and with coefficients $c_n>0$ large enough to make the sum unbounded $$G(x):=\sum_{n=1}^\infty c_n g\big(x-1+{1\over n}\big) $$ This sum is locally finite for all $x\in[0,1)$, therefore defines a differentiable function on $[0,1)$, which has a derivative vanishig on a dense set, due to the funny properties of the Pompeiu derivatives: Since the derivative of each term of the sum vanishes on a dense $G_\delta$ set, their sum vanishes on the intersection, which is a dense $G_\delta$ set by Baire's theorem.

Answer 2

The answer is "no", and here is how one can adapt Pietro Majer's construction from the other answer to work with the set of rational numbers.

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Step 1. We begin with the original construction of Pompeiu function. Let $q_n$ be an enumeration of rational numbers in $(0, 1)$, and define, as in the original construction, $$ u(x) = a + b \sum_{n = 1}^\infty \frac{\sqrt[3]{x - q_n}}{2^n} \, , $$ where $a$ and $b > 0$ are chosen in such a way that $u(0) = 0$ and $u(1) = 1$. Then $u$ is increasing and differentiable in an extended sense: $u'(x) \in (0, \infty]$ exists for every $x \in [0, 1]$, and $u'(x) = \infty$ for every rational $x \in (0, 1)$.

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Step 2. We now define $$ v(x) = u\biggl(\frac{1}{2} + \frac{x}{1 + 2 |x|}\biggr) . $$ Observe that $x \mapsto 1/2 + x / (1 + 2 |x|)$ is differtentiable with everywhere positive derivative, and it maps rational numbers to rational numbers. Thus, $v : \mathbb{R} \to (0, 1)$ is an increasing homeomorphism, $v$ is everywhere differentiable in the extended sense, $v'(x) \in (0, \infty]$ for every $x$, and $v'(x) = \infty$ whenever $x$ is rational.

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Step 3. We now find an increasing homeomorphism $\phi : (0, 1) \to (0, 1)$ which is differentiable with everywhere positive derivative, and which has the following property: for every $s \in \mathbb{Q}$ there is $x \in \mathbb{Q}$ such that $$x + \phi(v(x)) = x.$$ Such a function $\phi$ can be constructed recursively, as follows.

We begin with $\phi_0(x) = s$. Let $r_n$ be an enumeration of all rational numbers. In step $n$ we suppose that we have already constructed a function $\phi_{n-1}$ with the following properties: $$|\phi_{n-1}' - 1| < 3^{-1} + \ldots + 3^{-(n-1)},$$ and there are rational numbers $x_1, \ldots, x_{n-1}$ such that $$x_j + \phi_{n-1}(v(x_j)) = r_j$$ for $j = 1, \ldots, n-1$. We now define $\phi_n$ to be an appropriate "tiny" modification of $\phi_{n-1}$.

Since $\phi_{n-1}' > 0$, there is a unique number $\tilde{x}_n$ such that $$\tilde{x}_n + \phi_{n-1}(v(\tilde{x}_n)) = r_n.$$ We choose a non-negative, compactly supported, smooth function $\psi_n$ on $(0, 1)$ such that $\psi_n(v(x_j)) = 0$ for $j = 1, \ldots, n - 1$, $\psi_n(v(\tilde{x}_n)) > 0$, and $\|\psi_n'\|_\infty < 3^{-n}$. For every $\epsilon \in [0, 1)$ we have $$\phi_{n-1}' + \epsilon \psi_n' \ge 1 - (3^{-1} + \ldots + 3^{-(n-1)} + \epsilon 3^{-n}) > 0 ,$$ and hence there is a unique solution $x$ of $$x + \phi_{n-1}(v(x)) + \epsilon \psi_n(v(x)) = r_n .$$ This solution depends continuously on $\epsilon$, and since $\psi_n(v(\tilde{x}_n)) \ne 0$, it indeed changes with $\epsilon$. By the intermediate value property, there is an $\epsilon$ such that the corresponding solution $x$ is a rational number. For this $\epsilon$ and $x$, we set $$ \phi_n = \phi_{n-1} + \epsilon \psi_n $$ and $x_n = x$. This way, we have constructed $\phi_n$ with all the desired properties.

By construction, $\psi_n$ converges in $C^1$ to an increasing homeomorphism $\psi : (0, 1) \to (0, 1)$ with the desired properties; namely, $|\psi' - 1| < 1/2$, and $x_n + \psi(v(x_n)) = r_n$.

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Step 4. We set $w(x) = \psi(v(x))$. Then $w : \mathbb{R} \to (0, 1)$ is differentiable everywhere in the extended sense, $w'(x) = \psi'(v(x)) v'(x) \in (0, \infty]$, $w'(x) = \infty$ for every rational $x$, and $$x_n + w(x_n) = r_n.$$ Now we follow Pietro Majer's argument: we set $g : (0, 1) \to \mathbb{R}$ to be the inverse function of $w$, $h(x) = g(x) + x$, and $f : \mathbb{R} \to (0, 1)$ to be the inverse function of $h$.

Clearly, $g$, $h$ and $f$ are everywhere differentiable, with $g' \geqslant 0$, $h' \geqslant 1$ and $0 < f' \leqslant 1$, respectively.

Observe that $h(w(x_n)) = g(w(x_n)) + w(x_n) = x_n + w(x_n) = r_n$. Since $w'(x_n) = \infty$ for every $n$, we have $g'(w(x_n)) = 0$, and hence $h'(w(x_n)) = 1$. Thus, $f'(r_n) = f'(h(w(x_n)) = 1 / h'(w(x_n)) = 1$. Since $r_n$ exhaust all rational numbers, $f'(x) = 1$ for every rational $x$.