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I have the following question that I want to prove or disprove.

Suppose $F:[0,1]\rightarrow [0,1]$ be a non-constant absolutely continuous function. Then, there must exist an open interval in which $F$ is strictly monotone.

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  • $\begingroup$ Good question! What are your thoughts about this problem? Do you know the characterization of absolutely continuous functions as integral functions of $L^1$ functions? $\endgroup$ – Crostul Mar 12 '17 at 22:16
  • $\begingroup$ The derivative of an AC function belongs to $L^1$. If $f \in L^1([0,1])$ does there necessarily exist an open interval on which $f$ is almost everywhere positive? $\endgroup$ – Umberto P. Mar 12 '17 at 22:17
  • $\begingroup$ I know that the derivative (denoted by $f$) exists and in $L^1$. I believe the statement is correct. Hence, I tried a proof by contradiction technique. For simplicity let us assume that $F$ is nondecreasing. Suppose that the statement is incorrect. Let, $A:=\{x\in [0,1]:f(x)>0\}$. Then $A$ does not contain any interval but $m(A)>0$. Is there a way to construct intervals $(a_n,b_n)$ such that $\sum_n b_n-a_n <\delta$ but $\sum_n F(b_n)-F(a_n)>\epsilon$ so that $F$ is not absolutely continuos? $\endgroup$ – Valentino B Mar 12 '17 at 22:32
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This is false. There are examples of differentiable functions with bounded derivative who are nowhere monotone. These functions would be Lipschitz and so absolutely continuous. The reference is a paper of Y. Katznelson and K. Stromberg. "Everywhere differentiable, nowhere monotone, functions, Am. Math. Monthly, 81 (4), (1974), 349-353. See math overflow.

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  • $\begingroup$ Thank you very much for the answer and the reference. $\endgroup$ – Valentino B Apr 27 '17 at 16:32

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