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In another thread, I remarked that if $$\int_0^1 \vert f'(x)\vert\mathrm dx=0,$$ then $f'(x)=0$ for all $x\in(0,1)$ can only be concluded if $f'$ is continuous. This would certainly be correct if we were talking about a general integrable function $(0,1)\to\mathbb R$. Take the characteristic function of a singleton set as a counterexample where the integral vanishes, but the function doesn't. But such a function is not a valid derivative of any function, since it doesn't have the mean value property. And I couldn't find any functions where the integral of the absolute value vanishes and which have the mean value property. That's why I'm wondering:

Is there a differentiable function $f:(0,1)\to\mathbb R$ such that $f'$ is not identically $0$, but $$\int_0^1\vert f'(x)\vert\mathrm dx=0,$$ or equivalently, $f'$ is $0$ almost everywhere, but not everywhere.

mathlander
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Vercassivelaunos
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    [Cantor Function](https://en.wikipedia.org/wiki/Cantor_function) has this property. – Surb Nov 14 '20 at 11:27
  • I was looking up the Wikipedia link for the Cantor ternary function, but by the time I returned I found that @Surb had beat me to it. Well, for more than you'd probably want to know about this topic, see the references and discussions to the Stack Exchange "question" [Bibliography for Singular Functions](https://math.stackexchange.com/q/677927/13130). – Dave L. Renfro Nov 14 '20 at 11:29
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    @Surb: If I understood correcly, the OP is looking for a differentiable function, while the Cantor function is only a.e. differentiable –  Nov 14 '20 at 11:30
  • And also (if I'm not mistaken), a function that is not absolutely continuous (otherwise $f(x)=\int_0^x\boldsymbol 1_{\mathbb Q}(t)\,\mathrm d t$ would obviously work. @Caffeine – Surb Nov 14 '20 at 11:35
  • Maybe this: https://en.wikipedia.org/wiki/Pompeiu_derivative and this: https://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be are relevant? Not really sure though – Adam Rubinson Nov 14 '20 at 11:35
  • @Surb: As mentioned, I'm looking for an everywhere differentiable function. Your other example also doesn't seem to work, since the derivative of $\int_0^x1_{\mathbb Q}(t)\mathrm dt$ isn't $1_{\mathbb Q}$, but just the constant $0$ function. – Vercassivelaunos Nov 14 '20 at 11:54
  • wrong, the derivative is $\boldsymbol 1_{\mathbb Q}(x)$ a.e. (which is $0$ a.e., but not $0$). You don't work with function but with class of function. @Vercassivelaunos – Surb Nov 14 '20 at 11:58
  • @Surb: I'm working in $\mathcal L^1((0,1))$, not $L^1((0,1))$ (where asking for something that is almost everywhere but not everywhere wouldn't even make sense). I'm aware that going from $\mathcal L$ to $L$ removes a lot of issues, but right now I'm specifically interested in how to resolve the issue if we stay in the less nice function space. – Vercassivelaunos Nov 14 '20 at 12:03
  • The problem is that the derivative of absolute continuous function is defined a.e. only, so in $\mathcal L^1$ is not well defined.@Vercassivelaunos – Surb Nov 14 '20 at 12:21

1 Answers1

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If $f$ is differentiable at every point of $(0,1)$ and $\int_0^{1} |f'| <\infty$ then $f$ is absolutely continuous. This theorem is proved in Rudin's RCA. So if we also assume that $f'=0$ a.e. then $f$ is a constant and $f'(x)=0$ for every $x$.

Kavi Rama Murthy
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    More precisely, if $f$ is everywhere differentiable and $f'=0$ a.e., then $f$ has a signed-infinite derivative at continuum many points (Theorem $1',$ p. 222 [here](https://doi.org/10.1112/jlms/s1-23.3.222)) and $f'$ does not exist finitely or infinitely at continuum many points (corollary on p. 21 [here](https://www.ams.org/journals/tran/1950-069-00/S0002-9947-1950-0037338-9/)). For even more, see *On singular functions* by K. M. Garg, pp. 1441-1452 in **Revue Roumaine de Mathématiques Pures et Appliquées** 14 #10 (1969)], and papers/books that cite Garg's paper (google "Garg" & paper title). – Dave L. Renfro Nov 14 '20 at 11:55
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    Thanks! Though I don't have the book at hand. Would you mind showing a short sketch of the proof (or just the proof idea would also be fine)? – Vercassivelaunos Nov 14 '20 at 12:11
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    Sorry, this is a non-trivial theorem. @Vercassivelaunos – Kavi Rama Murthy Nov 14 '20 at 12:14