Pompeiu derivative

Example

Let $h$ be a function that is differentiable at every point. Its derivative $h'$ is a Pompeiu derivative if $h'$ vanishes on a dense set but is not identically zero. Here is a construction. Let $q_1, q_2, \dots$ list the rational numbers in $[0, 1]$, and write $u^{1/3}$ for the real cube root. For $x \in [0, 1]$ set $$g(x) = \sum_{n=1}^\infty 2^{-n} (x - q_n)^{1/3}.$$ The series converges uniformly, and each term is increasing, so $g$ is continuous and strictly increasing.

Each difference quotient of $u \mapsto u^{1/3}$ at a point $u \ne 0$ is positive and at most $\frac{4}{3} \abs{u}^{-2/3}$, since $a^2 + ab + b^2 \ge \frac{3}{4} a^2$ for real $a, b$. Use this bound with dominated convergence where the series below is finite, and Fatou's lemma where it is infinite. It follows that $g$ has at each $x$ a derivative in $(0, \infty]$, namely $$g'(x) = \sum_{n=1}^\infty \frac{2^{-n}}{3} \abs{x - q_n}^{-2/3},$$ with $g'(q_n) = \infty$ for every $n$ (one-sided at the end points). Since $\abs{x - q_n} \le 1$, every term is at least $2^{-n}/3$, so $g'(x) \ge 1/3$.

Let $h = g^{-1}$ on the interval $[g(0), g(1)]$. By the rule for the derivative of an inverse function, $h$ is differentiable everywhere, with $h'(y) = 1/g'(h(y))$, read as $0$ when $g'(h(y)) = \infty$. So $0 \le h' \le 3$. The derivative $h'$ vanishes at every point $g(q_n)$, and these points are dense in $[g(0), g(1)]$ because $g$ is a homeomorphism. But $h'$ is not identically zero, since $h$ is strictly increasing and a function with zero derivative is constant. So $h'$ is a Pompeiu derivative.