# How quickly can the derivative of an everywhere differentiable function change sign?

Let $f : [a,b] \to \Bbb R$ be everywhere differentiable with $f'(a) = 1$ and $f'(b) =-1$.

By Darboux theorem, we know that $f'([a,b])$ is an interval containing $[-1,1]$. In particular, the set $\{x \in [a,b]: |f'(x)| < 1\}$ is uncountable. But how small can it be? Or to be more formal:

Can $\{x \in [a,b]: |f'(x)| < 1\}$ have measure zero?

I guess not, because I have never heard of such a counterexample. But I don't see how to prove it.

• If $|f'(x)|\le 1$ for all $x\in [a,b]$ then the answer is no. For if the answer were yes, then $f'(x)=1$ a.e. and in particular $f'$ is Lebesgue-integrable. Thus (e.g., Theorem 7.21 of Rudin's Real and Complex Analysis) the fundamental theorem of calculus holds, so $f(x) - f(a) = \int_a^x f'(t)\,dt = x - a$, i.e., $f(x)$ is actually the linear function $x-a+f(a)$ and in particular $f'(b) = 1 \ne -1$. So I expect the answer to be no in general because I don't think it can help you to allow $f'(x)$ to exceed 1. Apr 4, 2017 at 21:00

Fact 1 (Goldowsky-Tonelli): Let $F:(a, b) \to \mathbb{R}$ be continuous and have finite derivative everywhere. Suppose $F' \geq 0$ almost everywhere. Then $F$ is monotonically increasing.

For a proof of this, see Saks, Theory of the integral, Chapter 6, page 206.

Suppose $X = \{x \in [a, b]: -1 < f'(x) < 1\}$ has zero measure. Let $Y = \{x \in [a, b]: f'(x) \leq -1\}$ and $Z = \{x \in [a, b]: f'(x) \geq 1\}$

Claim 1: Every point of $X$ is a limit point of $Y$ and a limit point of $Z$.

Proof: Suppose for example $x \in X$ is not a limit point of $Y$ - the other case is similar. Let $I$ be an open interval around $x$ disjoint with $Y$. Then at almost every $y \in I$, $f'(y) \geq 1$. Using Fact 1, it follows that the function $y \mapsto f(y) - y$ is monotonically increasing on $I$ and hence for every $y \in I$, $f'(y) \geq 1$ which is impossible as $x \in Y \cap X$.

Claim 2: The set of points of continuity of $f' \upharpoonright \overline{X}$ is dense in $\overline{X}$ (the closure of $X$).

Proof: Well known (using Baire category theorem).

Now let $I$ be any open interval around $x \in X$. Then the supremum of $f' \upharpoonright I$ is at least $1$ and its infimum is at most $-1$ by Claim 1. By Darboux theorem, it follows that $f' \upharpoonright I$ and hence also $f' \upharpoonright (I \cap X)$ takes every value in $(-1, 1)$. So $f'$ is everywhere discontinuous on $\overline{X}$ which contradicts Claim 2.

• For lazy readers, this post is showing that the answer to the question (can the set where the absolute value of the derivative is strictly less than $1$ have measure $0$?) is 'no'. Apr 4, 2017 at 21:26
• Great! It seems the result used here is actually older than Goldowsky and Tonelli papers. See de La Vallée Poussin's Cours d'analyse infinitésimale (1914) which is cited by Goldowsky. Apr 7, 2017 at 8:48