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> **Conjecture:** Let $f:\mathbb{R}→\mathbb{R}$ be an **everywhere** differentiable function and assume that $f(x)+f′(x)∈ \{-1,1\}$ **almost everywhere** and $f'(0)=0$. Then is $f$ necessarily a constant function?
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Can you give me a counter-example? I have already asked the question [here on MathSE](https://math.stackexchange.com/q/2859868).



Differential equation changing sign almost everywhere
Your conjecture is **true** and there is no counterexample.




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Suppose, contrary to the above claim, that your conjecture is false. Define $$g(x) = f(x) + \int\_0^x f(y) dy,$$ so that $g'(x) = f'(x) + f(x)$. Thus, $g$ is everywhere differentiable, $g'(x) \in \{-1,1\}$ almost everywhere, and $g'$ is not a constant. (For if $g'$ was constant $\pm 1$, we would have $f(x) = \pm 1 + c e^x$, which, together with $f'(0) = 0$, would imply that $f$ is constant).


[This beautiful result](https://projecteuclid.org/euclid.bams/1183510405) of J.A. Clarkson from 1947 asserts that if $\alpha < \beta$, then $$E(\alpha, \beta) := \{x : g'(x) \in (\alpha, \beta)\}$$ is either empty or it has positive Lebesgue measure.


Since $g'$ is not constant, it takes at least two values, and by [Darboux's theorem](https://en.wikipedia.org/wiki/Darboux's_theorem_(analysis)#Darboux's_theorem), in fact it takes all values in some interval $(\alpha, \beta)$. Changing this interval to a smaller one if necessary, we may assume that $(\alpha, \beta) \cap \{-1, 1\} = \varnothing$. Since $E(\alpha, \beta)$ is non-empty, it has positive Lebesgue measure, and therefore $g'(x)$ cannot belong to $\{-1, 1\}$ almost everywhere — a contradiction.


