Timeline for Does a weaker condition than vanishing derivative imply a function being constant?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Dec 13, 2012 at 6:51 | comment | added | George Lowther | You can show the following. If $g$ exists everywhere, then $f$ is absolutely continuous outside of a closed nowhere dense set. This is the best you can do in the general case, as given any closed nowehere dense set $S$ there exists a differentiable function which is absolutely continuous outside of $S$ but has infinite variation on any neighbourhood of each point of $S$. However, if $g$ exists everywhere and is locally integrable, then $f$ is absolutely continuous. You can prove this using the Baire category theorem. I don't have time to post a full answer right now, so just leaving a comment. | |
| Dec 11, 2012 at 7:02 | history | edited | qianzhang | CC BY-SA 3.0 |
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| Dec 10, 2012 at 16:45 | vote | accept | qianzhang | ||
| Dec 10, 2012 at 16:44 | history | edited | qianzhang | CC BY-SA 3.0 |
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| Dec 10, 2012 at 11:46 | comment | added | Daniel Spector | OK. Thanks for the differentiable correction :). So my thinking is the requirement that $f$ is $BV$ is enough to get rid of these kind of pathologies and regain the fundamental theorem, if $g$ is locally integrable (since the fundamental theorem implies absolute continuity). | |
| Dec 10, 2012 at 9:48 | comment | added | Loïc Teyssier | @Daniel Spector : take for $f$ any even function, so that $g(0)=0$ is well-defined. Yet it happens that some even functions are not differentiable at 0. You can elaborate easily on this example to show that there exists continuous functions $f$ nowhere differentiable for which $g$ exists everywhere. You can also extend the construction of Alexander Eremenko given below: if you take for $f$ the indicator function of the rational numbers then $g$ is identically zero on the rational numbers, since both $x+\eps$ and $x-\eps$ are irrational or rationa at the smae time. | |
| Dec 10, 2012 at 9:36 | comment | added | qianzhang | I cannot see why they are close enough, because when $f$ is differentiable everywhere and of bounded variation, I know $f$ is absolutely continuous. | |
| Dec 10, 2012 at 8:55 | comment | added | Daniel Spector | The existence of $g$ for every $x \in [0,1]$ is saying that $f$ is differentiable everywhere, right? I mean, this is not exactly the definition, but close enough, maybe? | |
| Dec 10, 2012 at 1:47 | answer | added | Jack Huizenga | timeline score: 10 | |
| Dec 9, 2012 at 21:06 | answer | added | Alexandre Eremenko | timeline score: 6 | |
| Dec 9, 2012 at 19:16 | answer | added | Bazin | timeline score: 2 | |
| Dec 9, 2012 at 19:00 | history | asked | qianzhang | CC BY-SA 3.0 |