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Note: This is an attempt to narrow down conditions under which the conjecture stated in this previous post is true. As stated, it is false as shown by the counterexample provided in the answers by the user mathworker21.

Let $f: \mathbb R \to \mathbb R$ be a $C^1$ function.

We say a point $c \in \mathbb R$ is a mean value point of $f$ if there exists an open interval $(a,b)$ containing $c$ such that $f’(c) = \frac{f(b) - f(a)}{b-a}$.

Question: Suppose $f’^{-1} (x)$ has Lebesgue measure $0$ for every $x \in \mathbb R$. Is it true that almost every point in $\mathbb R$ is a mean value point of $f$?

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    $\begingroup$ Nice diabolical question $\endgroup$
    – Pietro Majer
    Commented Oct 14 at 20:43
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    $\begingroup$ @PietroMajer One of many! Please enjoy. $\endgroup$
    – Nate River
    Commented Oct 14 at 21:34
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    $\begingroup$ If I’m not wrong there are counterexamples even with countable $f’^{-1}(p)$ for all $p$. $\endgroup$
    – Pietro Majer
    Commented Oct 15 at 5:25
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    $\begingroup$ I think one can do counterexamples with $f\in C^{1,\alpha}$ for all $\alpha<1$ but probably not $C^{1,1}$ (I think I have a class now) $\endgroup$
    – Pietro Majer
    Commented Oct 15 at 6:47
  • $\begingroup$ Thanks for the high effort answer as usual! Will read soon. @PietroMajer $\endgroup$
    – Nate River
    Commented Oct 15 at 6:48

2 Answers 2

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A counterexample $f\in C^1(\mathbb R,\mathbb R)$ with denumerable $f’^{-1}(p)$ for all $p$ relies on the following:

There is a non-negative continuous function $u:\mathbb R\to\mathbb R$ with zero-set $C:= \{u=0\}$ of positive measure, which is concave on each component interval of $\mathbb R\setminus C$, and such that $$\mu:= \inf_{a<b}\frac1{(b-a)^2}\int_a^b u(t)dt>0.$$

Pick $0<\nu<\mu$ and define $$f(x):=\int_0^xu(t)dt + \nu x^2.$$ For all $a<c<b$ with $c\in C$ $$\frac{f(b)-f(a)}{b-a}-f'(c)=$$$$=\frac1{ b-a }\int_a^b u(t)dt+\nu(a+b-2c) >(\mu-\nu)(b-a)>0,$$ proving that no point of $C$ is a mean value point.

For all $p\in\mathbb R$ the equation $f'(x)=p$, that is

$$ u(x)=p-2\nu x$$ has at most one solution $x\in C$, namely $x=\frac p{2\nu}$, and it has at most $2$ solutions on each connected component of $\mathbb R\setminus C$, because $u$ is strictly concave on it. Since there are countably many such components, $f^{-1}(p)$ is at most countable.

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  • $\begingroup$ (to be completed with the construction of $u$ ) $\endgroup$
    – Pietro Majer
    Commented Oct 15 at 6:49
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    $\begingroup$ I think this works: Let $k_0$ be a sufficiently large integer and let $C:=R\setminus \bigcup_{k=k_0}^\infty \bigcup_{\ell \in \mathbb{Z}} B(\ell/2^k, (\ell/2^k)^{4/3})$ where $B(x, r)$ for $x\in \mathbb{R}$ and $r>0$ denotes the ball of radius $r$ around the point $x$ in $\mathbb{R}$. A straight-forward calculation shows that $C$ has positive measure for sufficiently large $k_0$. Let $u(x):=\sqrt{dist(x, C)}$. Any interval $[a, b]$ contains a ball $B(\ell/2^k, (\ell/2^k)^{4/3})$ of radius $r=\Omega( (b-a)^{4/3} )$, thus the integral of $u$ is at least $\Omega(r^{1.5})=\Omega( (b-a)^2)$. $\endgroup$
    – Anders Martinsson
    Commented Oct 15 at 16:40
  • $\begingroup$ Excellent. And in fact I think for every $\alpha<1$ one can make the function $u$ Hoelder of exponent $\alpha$. $\endgroup$
    – Pietro Majer
    Commented Oct 15 at 17:40
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    $\begingroup$ Oh, I made a mistake. The radius should be $(1/2^k)^{4/3}$. And yes agreed, one can get any Hölder exponent $<1$ by tweaking the constants. $\endgroup$
    – Anders Martinsson
    Commented Oct 15 at 18:57
  • $\begingroup$ That is one nasty function $u$ @AndersMartinsson. As usual distance functions prove to be really versatile objects. $\endgroup$
    – Nate River
    Commented Oct 15 at 21:56
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Here is a counterexample.

Let $C$ be a Cantor set with positive Lebesgue measure. The function $x \mapsto d(x,C)$ is continuous on $\mathbb{R}$, null on $C$ and (strictly) positive on $\mathbb{R} \setminus C$.

Let $f$ be an antiderivative. Since $\mathbb{R} \setminus C$ is dense in $\mathbb{R}$, the function $f$ is (strictly) increasing. Hence points of $C$ (where $f'$ vanishes) cannot be mean value points of $f$.

[Old answer below, actually to another question]

The function $g : (a,b) \mapsto \frac{f(b)-f(a)}{b-a}$ is continuous on the connected set $D := \{(a,b) \in \mathbb{R}^2\ : a < b\}$, so $g(D)$ is a connected subset of $\mathbb{R}$, hence an interval.

By definition of $f'$, the set $f'(\mathbb{R})$ is contained in the closure of the interval $g(D)$. Hence $f'(\mathbb{R}) \setminus g(D)$ contains at most two points.

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    $\begingroup$ you probably missed $c\in (a,b)$ condition $\endgroup$
    – Fedor Petrov
    Commented Jan 24 at 14:50
  • $\begingroup$ Ah, yes. I will think more about it. $\endgroup$
    – Christophe Leuridan
    Commented Jan 24 at 15:25
  • $\begingroup$ @Nate River I think that I now answer the right question. $\endgroup$
    – Christophe Leuridan
    Commented Feb 5 at 9:17
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    $\begingroup$ But you missed the condition that $\{t\in \mathbb R: f'(t)=x\}$ should have measure $0$ for all $x\in \mathbb R$. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 5 at 12:03

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