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I think the answer is negative but I can't find a counterexample. I really apreciatte your help.

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- Thread starter Levi Franco
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In summary, the conversation discusses the question of whether a function is absolutely continuous in a given interval, and the conclusion is that this may not always be true and that there exist continuous, non-decreasing functions that are not absolutely continuous. The conversation also mentions a theorem or lemma from a book by G. Ye. Shilov that may be relevant to this topic.

- #1

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I think the answer is negative but I can't find a counterexample. I really apreciatte your help.

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This is not true.Svein said:

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G. Ye. Shilov: Mathematical Analysis (A Special Course). Pergamon Press 1965, page 306.Krylov said:This is not true.

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The "vice versa" is not true, because there exist continuous, non-decreasing functions that are not absolutely continuous.Svein said:G. Ye. Shilov: Mathematical Analysis (A Special Course). Pergamon Press 1965, page 306.

Let ##f : [0,1] \to \mathbb{R}## be the Cantor function. Then ##f## is continuous, non-decreasing, but not absolutely continuous. The zero function is continuous and non-decreasing. The difference ##f - 0## is not absolutely continuous.

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That dang Cantor. If it wasn't for him, math would be much simpler. ;>)

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i believe Svein is thinking of (continuous) functions of bounded variation, a larger class than absolutely continuous ones.

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I don't have the book Svein suggests to check it myself but perhaps the theorem/lemma is about non-decreasing AND non-identically zero functions.Krylov said:The "vice versa" is not true, because there exist continuous, non-decreasing functions that are not absolutely continuous.

Let ##f : [0,1] \to \mathbb{R}## be the Cantor function. Then ##f## is continuous, non-decreasing, but not absolutely continuous. The zero function is continuous and non-decreasing. The difference ##f - 0## is not absolutely continuous.

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Delta² said:I don't have the book Svein suggests to check it myself but perhaps the theorem/lemma is about non-decreasing AND non-identically zero functions.

I don't think so.

Just replace the ##f## from before by the sum ##g + h##, where ##g## is the Cantor function and ##h## is any absolutely continuous, non-decreasing function that is not identically zero. Then ##g + h## is continuous, non-decreasing but not absolutely continuous. The difference ##(g + h) - h## is not absolutely continuous.

The forward implication in post #4 is true, because absolutely continuous functions are of bounded variation. It is the "vice versa" there that is not true.

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